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M.C.Q (1 Marks)

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MCQ 11 Mark
The correct way(s) of representing 2-bromobutane is/ are:
  • A
    CH3CHBrCH2CH3
  • B


  • C

  • D
    All of the above
Answer
  1. All of the above
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MCQ 21 Mark
HCHO and HCOOH are detected by:
  • A
    NaHCO3​
  • B
    CuSO4​/NaOH
  • C
    AgNO3​/NH4​OH
  • D
    All of these
Answer
  1. NaHCO3

Explanation:

Both HCHO and HCOOH react with AgNO3​/NH4​OH

CuSO4​/NaOH solution. But HCOOH gives effervescences of CO2​ when reacts with NaHCO3​, whereas HCHO does not give any effervescence of CO2​ when reacts with NaHCO3​.

HCOOH + NaHCO$\rightarrow$ HCOONa + H2​O + CO$\uparrow$

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MCQ 31 Mark
$\sigma_{\text{C}-\text{C}} :4 ;\ \sigma_{\text{C}-\text{H}}:6;\ \pi_{\text{C}=\text{C}}:1; \ \pi_{\text{C}\equiv\text{C}}:2$  These number of $\sigma$ and $\pi$-bonds are present in which of the following molecule?
  • A
    $\text{CH}\equiv\text{C} -\text{CH}=\text{CH}-\text{CH}_{3}$
  • B
    $\text{CH}_{2}=\text{C}=\text{CH}-\text{CH}_{3}$
  • C
    $\text{CH}_{2}=\text{CH}-\text{CH}=\text{CH}_{3}$
  • D
    None of the above.
Answer
  1. $\text{CH}\equiv\text{C} -\text{CH}=\text{CH}-\text{CH}_{3}$
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MCQ 41 Mark
In homlogous series, successive members differ by:
  • A
    CH3​−
  • B
    CH3​−CH2​−
  • C
    −CH2​
  • D
    −H
Answer
  1. −CH2​−

Explanation:

A homologous series is a series of compounds in which each member differs from the next/previous by −CH2​− or 14 mass units.

Physical properties change in members of homologous series but chemical properties remain almost the same because the functional group remains the same.

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MCQ 51 Mark
Which of the following statements is incorrect regarding the naming of complex compounds?
  • A
    NH2−, will be named as 'amido'
  • B
    'en' will be named as 'ethylenediamine'
  • C
    ​ $\text{C}_2\text{O}_2\text{S}_2^{2-}$will be named as 'dithio oxalato'
  • D
    CH3​− will be named as 'methyl'
Answer
  1. NH2−, will be named as 'amido'

Explanation:

$\ \ \ \ \ \ \ \ \ \ \ \ \text{O}\\\ \ \ \ \ \ \ \ \ \ \ \ \ ||\\ \text{NH}_2-\text{C}-\text{named as amido}$ 

NH2− will be named as amido is incorrect regarding the naming of complex.

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Question 61 Mark
On complete combustion, 0.246g of an organic compound gave 0.198g of carbon dioxide and 0.1014g of water. The percentage composition of carbon and hydrogen in the compound respectively are:
 
%C
(a)
21.95
4.58
(b)
21.95
5.58
(c)
11.95
5.58
(d)
11.95
4.58
Answer
 
%C
(a)
21.95
4.58

Explanation:

Given mass of CO2 = 0.198g and mass of H2O = 0.1014g

As we know % of $\text{C} = \frac{12}{44} \times\frac{\text{m}_{\text{CO}_{2}}}{\text{w}}\times 100$

Percentage of carbon $= \frac{12 \times 0.198\times 100}{44 \times 0.246} = 21.95\%$

Also, $\% \text{ of H} = \frac{2}{18} \times \frac{\text{m}_{\text{H}_{2}\text{O}}}{\text{w}}\times 100$

Percentage of hydrogen $= \frac{2\times 0.1014 \times100}{18 \times 0.246}= 4.58\%$

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MCQ 71 Mark
Amongest the following, the acceptable name as per IUPAC nomenclature is:
  • A
    pent-4-en-1-yne
  • B
    pent-1-en-4-yne
  • C
    2, 2-dimethyl-4-ethylpentane
  • D
    3-hydroxybutan-1-ol
Answer
  1. pent-1-en-4-yne

Explanation:

In naming hydrocarbons, alkenes gets position first priority as compared to alkynes so the correct name is pent-1-en-4-yne. But in numbering carbons, alkynes gets higher priority than alkenes.

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MCQ 81 Mark
Which techniques is based on the difference in the solubilities of the compound and the impurities in a suitable solvent?
  • A
    Sublimation.
  • B
    Crystallisation.
  • C
    Distillation.
  • D
    None of these.
Answer
  1. Crystallisation.

Explanation:

Crystallisation process is based on the difference in the solubilities of the compound and the impurities in a suitable solvent. The impure compound is dissolved in a solvent in which it is sparingly soluble at room temperature but appreciably soluble at higher temperature.

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MCQ 91 Mark
Ethanol is also known as:
  • A
    Ethyl alcohol
  • B
    Ethane
  • C
    Acetaldehyde
  • D
    Formic acid
Answer
  1. Ethyl alcohol

Explanation:

The common name of ethanol is ethyl alcohol.

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MCQ 101 Mark
Ionic species are stabilised by the dispersal of charge. Which of the following carboxylate ion is the most stable?
  • A
    $\ \ \ \ \ \ \ \ \ \ \ \ \text{O}\\\ \ \ \ \ \ \ \ \ \ \ \ ||\\\text{CH}_3-\text{C}-\text{O}^-$
  • B
    $\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{O}\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ||\\\text{CI}-\text{CH}_2-\text{C}-\text{O}^-$
  • C
    $\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{O}\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ||\\\text{F}-\text{CH}_2-\text{C}-\text{O}^-$
  • D
     

Answer
  1.  

Explanation:

The stabilsation of carboxylate ion depends on dispersal of negative charge. The negative charge is dispersed by two factors, i.e., + R effect of the carboxylate ion and Inductive effect of the halogens. In all the above structures, R effect is common but halogen atoms are different. Therefore, dispersal of negative charge depends upon halogen atoms. F is most electronegative, in structure (iv) two Fatoms are present and more dispersal of negative charge is there.

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MCQ 111 Mark
Group showing strongest +M effect:
  • A
    −NH2​
  • B
    −N(CH3​)2​
  • C
    −NHCH3
  • D
    −O-
Answer
  1. −O-

Explanation:

(-) charge is a stronger donor than lone pairs hence the strongest +M effect is shown by −O-.

Also, the other species does not contain any negative charge thus option D is correct.

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MCQ 121 Mark
The colour of bromine solution is:
  • A
    Red
  • B
    Purple
  • C
    Green
  • D
    Blue
Answer
  1. Red

Explanation:

Bromine is a chemical element with symbol Br and atomic number 35.

It is the third-lightest halogen, and is a fuming red-brown liquid at room temperature that evaporates readily to form a similarly coloured gas.

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MCQ 131 Mark
The trends in physical properties of compounds within a homologous series are due to:
  • A
    Change in size
  • B
    Change in weight
  • C
    Functional group
  • D
    Both a and b
Answer
  1. Both a and b

Explanation:

The trends in physical properties of compounds within a homologous series are primarily due to the progression of sizes and therefore, weights of the molecules that form the homologous series.

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MCQ 141 Mark
'I.U.P.A.C.', stands for?
  • A
    Iodine Under Packing.
  • B
    International Union of Pure and Applied Chemistry.
  • C
    International Units of Protein and Carbohydrates.
  • D
    International Understanding on Physical Aspects of Chemistry.
Answer
  1. International Union of Pure and Applied Chemistry.

Explanation:

A chemical nomenclature is a set of rules to generate systematic names for chemical compounds.

The nomenclature used most frequently worldwide is the one created and developed by the International Union of Pure and Applied Chemistry.

The full form of IUPAC is International Union for Pure and Applied Chemistry.

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MCQ 151 Mark
The IUPAC name of the compound is:

  • A
    2 - Amino - 3 - chloro - 2 - methylpent - 2 - enoic acid
  • B
    3 - Amino - 4 - chloro - 2 - methylpent - 2 - enoic acid
  • C
    4- Amino -3 - chloro - 2 - methylpent - 2 - enoic acid
  • D
    All of the above
Answer
  1. 3 - Amino - 4 - chloro - 2 - methylpent - 2 - enoic acid

Explanation:

3-amino-4-chloro-2-methylpent-2-enoic acid.

In IUPAC nomenclature carboxylic acid gets first priority, then -ene, -amino group, -chloro group and -methyl group.

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MCQ 161 Mark
Choose the correct bond-line formula for

  • A
     

    1. B
       

      1. C
         

        1. D
           

        Answer
        1.  

        Explanation:

        The correct bond-line formula for

         

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        MCQ 171 Mark
        Nitrene is a/an:
        • A
          Nucleophile
        • B
          Electrophile
        • C
          Charged species
        • D
          Free atom
        Answer
        1. Nucleophile

        Explanation:

        Nucleophiles are the electron rich species.

        Nitrene is a nucleophile due to the lone pair electrons of Nitrogen.

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        MCQ 181 Mark
        The carbocation stability is:
        $\stackrel{{+}}{\hbox{ CH }_3}<\text{CH}_{3}\stackrel{{+}}{\hbox{ CH}_2}< (\text{CH}_{3})_{2}\stackrel{{+}}{\hbox{ CH }}<(\text{CH}_{3})_{3}\stackrel{{+}}{\hbox{C}}$
        and alkyl radical stability is:
        • A
          $\stackrel{{\bullet}}{\hbox{ CH}}(\text{CH}_{3})_{2} < \stackrel{{\bullet}}{\hbox{ CH}}_{3}<\stackrel{{\bullet}}{\hbox{ CH}}_{2}\text{CH}_{3}<\stackrel{{\bullet}}{\hbox{C}}(\text{CH}_{3})_3$
        • B
          $\stackrel{{\bullet}}{\hbox{ CH}}(\text{CH}_{3})_{3} < \stackrel{{\bullet}}{\hbox{ CH}}_{}(\text{CH}_{3})_{2}<\stackrel{{\bullet}}{\hbox{ CH}}_{3}<\stackrel{{\bullet}}{\hbox{CH}}_{2}\text{CH}_{3}$
        • C
          $\stackrel{{\bullet}}{\hbox{ CH}}(\text{CH}_{3})_{3} < \stackrel{{\bullet}}{\hbox{ CH}}_{}(\text{CH}_{3})_{2}<\stackrel{{\bullet}}{\hbox{ CH}}_{2}\text{CH}_{3}<\stackrel{{\bullet}}{\hbox{CH}}_{3}$
        • D
          $\stackrel{{\bullet}}{\hbox{ CH}}_{3}<\stackrel{{\bullet}}{\hbox{ CH}}_{2}\text{CH}_{3}<\stackrel{{\bullet}}{\hbox{ CH}} (\text{CH}_{3})_{2}<\stackrel{{\bullet}}{\hbox{C}}(\text{CH}_{3})_{3}$
        Answer
        1. $\stackrel{{\bullet}}{\hbox{ CH}}_{3}<\stackrel{{\bullet}}{\hbox{ CH}}_{2}\text{CH}_{3}<\stackrel{{\bullet}}{\hbox{ CH}} (\text{CH}_{3})_{2}<\stackrel{{\bullet}}{\hbox{C}}(\text{CH}_{3})_{3}$
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        MCQ 191 Mark
        Which is not the characteristic of $\pi$−bond?
        • A
          $\pi$ bond is formed when a sigma bond already exists.
        • B
          $\pi$ bonds are formed from hybrid orbitals.
        • C
          $\pi$ bond may be formed by the overlapping of p orbitals.
        • D
          $\pi$ bond results from lateral overlap of atomic orbitals.
        Answer
        1. $\pi$ bonds are formed from hybrid orbitals.

        Explanation:

        $\pi$ bond is formed between two atoms if a sigma bond already exists between them.

        $\pi$ bonds are formed only by pure orbitals (p−orbitals), not from hybrid orbitals.

        They formed by lateral or sideways overlapping.

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        MCQ 201 Mark
        The technique in which extraction of compound takes place on the basis of more solubility of one compound in another solvent, is:
        • A
          Differential extraction.
        • B
          Chromatography.
        • C
          Sublimation.
        • D
          Crystallisation.
        Answer
        1. Differential extraction.
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        MCQ 211 Mark
        In Lassaigne's test, a blue colour is obtained if the organic compound contains nitrogen. The blue colour is due to the formation of:
        • A
          K4[Fe(CN)6]
        • B
          Fe4[Fe(CN)6]3
        • C
          Na3[Fe(CN)6]
        • D
          Cu2[Fe(CN)6]
        Answer
        1. Fe4[Fe(CN)6]3

        Explanation:

        If nitrogen is present in the organic compound then sodium extract contains NaCN.

        $\text{Na}+\text{C}+\text{N}\xrightarrow{\text{Fuse}}\text{NaCN}$

        $\text{FeSO}_4+6\text{NaCN}\rightarrow\text{Na}_4[\text{Fe(CN)}_6]+\text{Na}_2\text{SO}_4$

        It changes to prussian blue Fe4[Fe(CN)6]3 on reaction with FeCl3.

        $4\text{FeCl}_3+3\text{Na}_4[\text{Fe(CN)}_6]\rightarrow\text{Fe}_4[\text{Fe(CN)}_6]_3+12\text{NaCl}$

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        MCQ 221 Mark
        Which one of the following is least stable?
        • A
          $\stackrel{{\ominus}}{\hbox{C}}\text{H}_3$
        • B
          $\text{HC}\equiv\stackrel{{\ominus}}{\hbox{C}}$
        • C
          $(\text{C}_6\text{H}_5)_3\stackrel{{\ominus}}{\hbox{C}}$
        • D
          $(\text{CH}_3)_3\stackrel{{\ominus}}{\hbox{C}}$
        Answer
        1. $(\text{CH}_3)_3\stackrel{{\ominus}}{\hbox{C}}$

        Explanation:

        It is least stable because $\text{C}^\ominus$ is attached to three electron releasing CH3 groups.  

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        MCQ 231 Mark
        The order of priority in IUPAC system:
        • A
          $-\text{CONH}_2,-\text{CHO},-\text{SO}_3\text{H},-\text{COOH}$
        • B
          $-\text{COOH},-\text{SO}_3\text{H},-\text{CONH}_2,-\text{CHO}$
        • C
          $-\text{SO}_3\text{H},-\text{COON},-\text{CONH}_2,-\text{CHO}$
        • D
          $-\text{CHO},-\text{COOH},-\text{SO}_3\text{H},-\text{CONH}_2$
        Answer
        1. $-\text{SO}_3\text{H},-\text{COON},-\text{CONH}_2,-\text{CHO}$
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        MCQ 241 Mark
        Which of the following method is used to purify liquids having very high boiling points and those which decompose at or below their boiling points?
        • A
          Simple distillation.
        • B
          Fractional distillation.
        • C
          Distillation under reduced pressure.
        • D
          None of the above.
        Answer
        1. Distillation under reduced pressure.

        Explanation:

        Distillation under reduced pressure method is used to purify liquids having very high boiling points and those, which decompose at or below their boiling points. Such liquids are made to boil at a temperature lower than their normal boiling points by reducing the pressure on their surface. A liquid boils at a temperature at which its vapour pressure is equal to the external pressure. The pressure is reduced with the help of a water pump or vacuum pump.

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        MCQ 251 Mark
        The indicator which is used to find the strength of caustic soda solution with the help of oxalic acid is:
        • A
          Methyl orange
        • B
          Phenolphthalein
        • C
          Potassium permanganate
        • D
          None of the above
        Answer
        1. Phenolphthalein

        Explanation:

        An indicator is a chemical substance that undergoes a colour change at the endpoint. The endpoint of an acid-base titration can be determined using acid-base indicators.

        Acid Base indicators are either weak organic acids or weak organic bases. The colour change of an indicator depends on the pH of the medium.

        The un-ionized form of an indicator has one colour, but its ionized form has a different colour.

        Phenolphthalein is used to find the strength of caustic soda solution with the help of oxalic acid.

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        MCQ 261 Mark
        Passing H2​S gas into a mixture of Mn2+,Ni2+,Cu2+,Hg2+ ions in acidified aqueous solution precipitates:
        • A
          CuS and HgS
        • B
          MnS and CuS
        • C
          MnS and NiS
        • D
          NiS and HgS
        Answer
        1. CuS and HgS

        Explanation:

        Cu2+ and Hg2+ lie in IInd group of qualitative analysis H2​S in acid medium is reagent for it.
        HCl $\rightarrow$ H+ Cl
        H2​S $\leftrightharpoons$ 2H+ S2−

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        MCQ 271 Mark
        Which of the following does not belong to homologous series of alkanes?
        • A
          C2​H6​
        • B
          C3​H4
        • C
          C4​H10​
        • D
          C5​H12​
        Answer
        1. C3​H4

        Explanation:

        Alkanes have the general formula Cn​H2n+2​

        Thus, C3​H4​ does not belong to homologous series of alkanes.

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        MCQ 281 Mark
        identify the rate of reaction of given compounds in E2​ reaction:

        • A
          a > b > c > d
        • B
          a > c > b > d
        • C
          b > a > c > d
        • D
          b > d > a > c
        Answer
        1. a > b > c > d

        Explanation:

        The rate of reaction depends on the leaving ability of halide ions.

        Leaving Ability: I− > Br− > Cl− > F (Acidic nature).

        Strong acids act as a strong leaving groups.

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        MCQ 291 Mark
        Which of the following is the most reactive towards electrophilic reagent?
        • A
           

          1. B
             

            1. C
               

              1. D
                 

              Answer
              1.  

              Explanation:

              +R-effect of OH > -CH3

              Hence, 2-methyl phenol is more reattive towards electrophilic reagent.

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              MCQ 301 Mark
              The IUPAC name of the molecule:
              $\ \ \ \ \ \ \ \ \ \ \ \ \ \text{O}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{O}\\\ \ \ \ \ \ \ \ \ \ \ \ \ ||\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ||\\\text{CH}_3-\text{C}-\text{C}=\text{C}-\text{C}-\text{OH}\ \text{is }\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ |\ \ \ \ \ \ \ |\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{CH}_3 \ \text{CH}_3$
              • A
                4-oxo-2,3-dimethyl pent-2-en-1-oic acid.
              • B
                2-carboxy-3-methyl pent-2-en -3-one.
              • C
                4-carboxy-3-methyl pent-3-en-2-one.
              • D
                2,3-Dimethyl-4-oxo-pent-2-en-1-oic acid.
              Answer
              1. 2,3-Dimethyl-4-oxo-pent-2-en-1-oic acid.

              Explanation:

              -COOH is given more preference over keto, prefix for ketone group is oxo.

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              MCQ 311 Mark
              Which of the following is a correct representation of condensed formula for$\text{HOCH}_2\text{CH}_2\text{CH}_2\text{CH(CH}_3)\text{CH(CH}_3)\text{CH(CH}_3)\text{CH}_3$?
              Answer
              1. $\text{HO(CH}_2)_2\text{CH(CH}_3)\text{CH(CH}_3)_2$

              Explanation:

              The correct condensed formula for$\text{HOCH}_2\text{CH}_2\text{CH}_2\text{CH(CH}_3)\text{CH(CH}_3)\text{CH(CH}_3)\text{CH}_3$

              is $\text{HO(CH}_2)_3\text{CHCH}_3\text{CH(CH}_3)_2$

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              MCQ 321 Mark
              In which of the following inductive effect is possible:
              • A
                Butane
              • B
                Benzene
              • C
                Cyclohexane
              • D
                Butanal
              Answer
              1. Butanal

              Explanation:

              To show inductive effect compound must contain electronegative element/eletropositive element attatched to C-atom.

              So that movement of electrons away from chain/towards chain would occur. This is possible only in case of Butanal among given options.

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              MCQ 331 Mark
              Sodium nitroprusside reacts with sodium sulphide formed in lassaigne's test to detect presence of sulphur gives violet colour due to:
              • A
                $\text{Na}_2\big[\text{Fe}(\text{CN})_5\text{NO}\big]$
              • B
                $\text{Na}_4\big[\text{Fe}(\text{CN})_5\text{NOS}\big]$
              • C
                $\text{Na}_2\big[\text{Fe}(\text{CN})_5\text{COS}\big]$
              • D
                None of these.
              Answer
              1. $\text{Na}_4\big[\text{Fe}(\text{CN})_5\text{NOS}\big]$

              Explanation:

              $\text{Na}_2\text{S}+\text{Na}_2\big[\text{Fe}(\text{CN})_5\text{NO}\big]\xrightarrow{\ \ \ \ }\text{Na}_4\big[\text{Fe}(\text{CN})_5\text{NOS}\big]\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{Violet coloour}$

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              MCQ 341 Mark
              Consider the following compounds,

              Which of the following statements is/ are true regarding I and II?
              • A
                I shows +R-effect, whereas II shows -R-effect.
              • B
                I shows -R-effect, whereas II shows +R-effect.
              • C
                Both I and II show +R-effect.
              • D
                Both I and II show -R-effect.
              Answer
              1. I shows +R-effect, whereas II shows -R-effect.
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              MCQ 351 Mark
              The principle involved in paper chromatography is:
              • A
                Adsorption
              • B
                Partition
              • C
                Solubility
              • D
                Volatility
              Answer
              1. Partition

              Explanation:

              In paper chromatography, separation of the components of a mixture depends upon their partitioning between water held in the stationary phase (i.e. adsorbent paper) and the liquid present in the mobile phase.

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              MCQ 361 Mark
              Successive members of homologous series differ from one another in their mass by:
              • A
                10 units
              • B
                15 units
              • C
                20 units
              • D
                14 units
              Answer
              1. 14 units

              Explanation:

              A series of organic compounds with the same general formula but differ from adjacent members by "−CH2​−" group are referred to as homologous series of compounds. Successive members of homologous series differ from one another in their mass by 14 units.

              For example: CH4​ and C2​H6​ are homologous series. Its molar mass is 16 and 30gm/mole respectively.

              Hence, it is differed by unit 14.

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              MCQ 371 Mark
              Electronegativity of carbon atoms depends upon their state of hybridisation. In which of the following compounds, the carbon marked with asterisk is most electronegative?
              • A
                CH3 – CH2 – *CH2 –CH3
              • B
                CH3 – *CH = CH – CH3
              • C
                CH3 – CH2 – C ≡ *CH
              • D
                CH3 – CH2 – CH = *CH2
              Answer
              1. CH3 – CH2 – C ≡ *CH

              Explanation:

              Electronegativity increases as the state of hybridization changes from sp3 to spand sp2 to sp. Thus, sp hybridized carbon has the highest electronegativity.

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              MCQ 381 Mark
              Which of the following pairs are members of a homologous series?
              • A
                CH3​OCH3​; CH3​CH2​OH
              • B
                CH3​CHO; CH3​CH2​CHO
              • C
                CH3​CH2​COOH; CH3​COOCH3​
              • D
                (CH3​)2​CHOH; CH3​CH2​OH
              Answer
              1. CH3​CHO; CH3​CH2​CHO

              Explanation:

              Acetaldehyde i.e CH3​CHO and propionaldehyde i.e CH3​CH2​CHO belong from the aldehyde family hence they are homologous series.

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              MCQ 391 Mark
              The bond that consist of an upper and a lower sharing of electron orbitals is called as:
              • A
                A pi bond
              • B
                A sigma bond
              • C
                A hydrogen bond
              • D
                An ionic bond
              Answer
              1. A pi bond

              Explanation:

              The bond that consist of an upper and a lower sharing of electron orbitals is called as a pi bond.

              A pi bond is formed by lateral (side ways) overlap.

              For example, two 2pz​ orbitals of two C atoms laterally overlap to for pi bond in ethene CH2 ​= CH2​

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              MCQ 401 Mark
              In Kjeldahl's method for estimation of nitrogen, CuSO4 acts as:
              • A
                Oxidising agent.
              • B
                Reducing agent.
              • C
                Catalytic agent.
              • D
                Hydrolysis agent.
              Answer
              1. Hydrolysis agent.

              Explanation:

              In Kjeldahl's method (used for the estimation of nitrogen) the organic compound is heated with conc. H2SO4 in the presence of K2SO4 (used to elevate boiling point of H2SO4) and CusO4 (used as catalyst) to convert all the nitrogen into (NH4)2SO4

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              MCQ 411 Mark
              In the detection test of carbon and hydrogen with cupric oxide, carbon is oxidized to:
              • A
                Carbon monoxide
              • B
                Carbon dioxide
              • C
                Carbonic acid
              • D
                Copper sulphate
              Answer
              1. Carbon dioxide

              Explanation:

              The given organic compound is mixed with dry copper oxide (CuO) and heated in a hard glass tube. The products of the reaction are passed over (white) anhydrous copper sulphate and then bubbled through lime water.

              If copper sulphate turns blue due to the formation of CuSO4​.5H2​O (by water vapor) then the compound contains hydrogen.

              If lime water is turned milky by CO2​, then the compound contains carbon.So, the carbon is oxidized to CO2​.

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              MCQ 421 Mark
              Which term is the best match for the color of a chemical substance students make note of:
              • A
                Qualitative data
              • B
                Inference
              • C
                Observation
              • D
                Hypothesis
              Answer
              1. Qualitative data

              Explanation:

              Color of the substance is the qualitative property. Qualitative properties are the properties which can be observed without making any measurements or calculations.

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              MCQ 431 Mark
              Yellow ammonium sulphide can separate:
              • A
                CdS and As2​S3​
              • B
                Bi2​S3​ and CuS
              • C
                PbS and Bi2S3​
              • D
                HgS and PbS
              Answer
              1. CdS and As2​S3​

              Explanation:

              Since CdS belongs to IIA group and As2​S3​ belongs to IIB group, they can be separated with yellow ammonium sulphide as only IIB group sulphides dissolve in yellow ammoium sulphide.

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              MCQ 441 Mark
              In Dumas method, 0.25g of organic compound gave 40ml of N, at 300k and 725mm pressure. If aqueous tension at 300k is 25mm, the percentage of nitrogen in compound is:
              • A
                16.76%
              • B
                15.76%
              • C
                17.36%
              • D
                18.20%
              Answer
              1. 16.76%

              Explanation:

              $\frac{\text{P}_1\text{V}_1}{\text{T}_1}=\frac{\text{P}_2\text{V}_2}{\text{T}_2}\Rightarrow \frac{40\times760}{300}=\frac{\text{V}_2\times760}{273}$

              $\Rightarrow\text{V}_2=\frac{28\times273}{76\times3}$ $[\text{P}=725-25=700\text{mm}]$​​​​​​​

              $\text{V}_2=\frac{7644}{228}=33.53\text{mL}$

              $\%\text{ of }\text{N}=\frac{\text{V}_2}{8\text{w}}=\frac{33.53}{8\times0.25}=16.76\%$

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              MCQ 451 Mark
              Name the two types of chromatography techniques based on the principle of differential adsorption.
              • A
                Column chromatography and thick layer chromatography.
              • B
                Non-column chromatography and thin layer chromatography.
              • C
                Column chromatography and thin layer chromatography.
              • D
                Paper chromatography and thick layer chromatography.
              Answer
              1. Column chromatography and thin layer chromatography.

              Explanation:

              Two types of adsorption chromatography techniques are column chromatography and thin layer chromatography.

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              MCQ 461 Mark
              Which of the following compounds contain all the carbon atoms in the same hybridisation state?
              • A
                H—C ≡ C—C ≡ C—H
              • B
                CH3—C ≡ C—CH3
              • C
                CH2 = C = CH2
              • D
                CH2 = CH—CH = CH2
              Answer
              1. H—C ≡ C—C ≡ C—H
              1.  CH2 = CH—CH = CH2

              Explanation:

              1. $\ \ \ \ \ \ \ \ \text{sp}\ \ \ \ \text{sp}\ \ \ \ \text{sp}\ \ \ \ \text{sp}\\\text{H}-\text{C}\equiv\text{C}-\text{C}\equiv\text{C}-\text{H}$

              2. $\text{sp}^3\ \ \ \ \ \ \ \text{sp}\ \ \ \ \text{sp}\ \ \ \ \text{sp}^3\\\text{CH}_3-\text{C}\equiv\text{C}-\text{CH}_3$

              3. $\text{sp}^2\ \ \ \ \ \ \ \text{sp}\ \ \ \ \ \text{sp}^2\\\text{CH}_2=\text{C}=\text{CH}_2$

              4. $\text{sp}^2\ \ \ \ \ \ \ \text{sp}^2\ \ \ \ \ \ \text{sp}^2\ \ \ \ \ \text{sp}^2\\\text{CH}_2=\text{CH}-\text{CH}=\text{CH}_2$

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              MCQ 471 Mark
              Homologous series is____________.
              • A
                A series of compounds in which the same functional group substitutes for hydrogen in a carbon chain.
              • B
                A series of compounds in which different functional group substitutes for hydrogen in a carbon chain.
              • C
                A series of compounds with same molecular formula and different structural formulae.
              • D
                A series of compounds with same molecular formula but different functional groups.
              Answer
              1. A series of compounds in which the same functional group substitutes for hydrogen in a carbon chain.

              Explanation:

              Homologous series is a series of compounds in which the same functional group substitutes for hydrogen in a carbon chain. The chemical properties of the compounds are very similar.

              Example of a homologous series: CH3​OH,C2​H5​OH,C3​H7​OH,C4​H9​OH.

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              MCQ 481 Mark
              Pi bond is formed:
              • A
                By the overlapping of atomic orbitals on the axis of nuclei.
              • B
                By mutual sharing of p - electrons.
              • C
                By sideways overlapping of half filled p - orbitals.
              • D
                By overlapping of s-orbitals with p - orbital.
              Answer
              1. By sideways overlapping of half filled p - orbitals.

              Explanation:

              If a bond between two atoms is broken when one atom is rotated around the bond axis, that bond is called a pi bond.

              Pi bonds are formed by the sideways overlap of parallel half filled p−orbitals on adjacent atoms as shown in the figure.

              They are not formed from hybrid orbitals.

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              MCQ 491 Mark
              Which general formula represents the homologous series of hydrocarbons that includes the compound 1-heptyne?
              • A
                Cn​H2n−6​
              • B
                Cn​H2n
              • C
                Cn​H2n−2
              • D
                Cn​H2n+2​
              Answer
              1. Cn​H2n−2

              Explanation:

              1-heptyne is alkyne. The general formula for alkyne is Cn​H2n−2.

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              MCQ 501 Mark
              Sodium extract gives blood red colour when treated with FeCl3​. Formation of blood red colour confirms the presence of:
              • A
                Only nitrogen
              • B
                Only sulphur
              • C
                Only halogens
              • D
                Both nitrogen and sulphur
              Answer
              1. Both nitrogen and sulphur

              Explanation:

              Na + N + C + S → NaSCN (Sodium thiocyanate)

              SCN− + Fe3→ [Fe(SCN)]2+ (Ferric thiocyanate) (Blood-red colour)

              It confirms the presence of both N and S.

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              MCQ 511 Mark
              In each of the following, the carbonyl group is present except in:
              • A
                HCHO
              • B
                CH3​COCH3
              • C
                HCH2​OH
              • D
                $\ \ \ \ \ \ \ \ \ \ \ \ \text{O}\\\ \ \ \ \ \ \ \ \ \ \ \ ||\\\text{CH}_3​−\text{C}​−\text{H}$
              Answer
              1. HCH2OH

              Explanation:

              Carbonyl group is −C = O group, which present in aldehydes and ketones but not present in alcohols.

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              MCQ 521 Mark
              Hyperconjugation involves delocalisation of ___________.
              • A
                Electrons of carbon-hydrogen σ bond of an alkyl group directly attached to an atom of unsaturated system.
              • B
                Electrons of carbon-hydrogen σ bond of alkyl group directly attached to the positively charged carbon atom.
              • C
                $\pi$-Electrons of carbon-carbon bond.
              • D
                Lone pair of electrons.
              Answer
              1. Electrons of carbon-hydrogen σ bond of an alkyl group directly attached to an atom of unsaturated system.
              2. Electrons of carbon-hydrogen σ bond of alkyl group directly attached to the positively charged carbon atom.

              Explanation:

              Hyperconjugation, also known as sigma-pi conjugation, is the delocalization of sigma electrons. Presence of -H with respect to double bond, triple bond or carbon containing positive charge (in carbonium ion) or unpaired electron (in free radical) is a condition required for hyperconjugation.

              $\ \ \ \ \ \ \ \ \text{H}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{H}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{H}\\ \ \ \ \ \ \ \ \ \ | \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{H}^\oplus \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ |\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ |\\\text{H}-\text{C}-\text{CH}=\text{CH}_2\leftrightarrow\text{H}-\text{C}=\text{CH}-\stackrel{{\ominus}}{\hbox{C}\text{}}\text{H}_2\leftrightarrow\text{H}^\oplus\text{C}=\text{CH}-\stackrel{{\ominus}}{\hbox{C}}\text{H}_2\leftrightarrow\text{H}-\text{C}-\text{CH}-\stackrel{{\ominus}}{\hbox{C}}\text{H}_2\\\ \ \ \ \ \ \ \ \ |\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ |\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ |\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{H}^\oplus\\\ \ \ \ \ \ \ \ \text{H}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{H}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{H}$

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              MCQ 531 Mark
              The number of structural isomers possible for the molecular formula = C3H9N is:
              • A
                4
              • B
                5
              • C
                2
              • D
                3
              Answer
              1. 4

              Explanation:

              Structural isomers of C3H9N are

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              MCQ 541 Mark
              Which substance has the lowest boiling point?
              • A
                CH3​CH2​CH2​CH2​OCH3​
              • B
                CH3​CH2​OCH2​CH3
              • C
                CH3​CH2​CH2​CH3
              • D
                CH3​CH2​C=OCH3
              Answer
              1. CH3​CH2​CH2​CH3

              Explanation:

              CH3​CH2​CH2​CH3 (n−butane) has the lowest boiling point as it has lowest molecular weight and is non polar. It has weakest intramolecular forces of attraction.

              All others are polar compounds and so has dipole-dipole inbteraction.

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              MCQ 551 Mark
              Which of the following compounds will not give Lassaigne's test for nitrogen?
              • A
                $\text{NH}_2-\text{NH}_2$
              • B
                $\text{C}_6\text{H}_5\text{NHNH}_2$
              • C
                $\text{C}_6\text{H}_5-\text{N}=\text{N}-\text{C}_6\text{H}_5$
              • D
                $\text{NH}_2\text{CONH}_2$
              Answer
              1. $\text{NH}_2-\text{NH}_2$

              Explanation:

              Because it does not have carbon.

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              MCQ 561 Mark
              Arrange the above in the decreasing order of reactivity towards HBr:

              • A
                a > b > c
              • B
                b > a > c
              • C
                b > c > a
              • D
                a > c > b
              Answer
              1. b > a > c

              Explanation:

              Among a, b and c electron donating ability order is −OCH3​ > −CH​> −CH2​OCH3​ more the electron donating ability of substituent more is the stability of carbocation.

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              MCQ 571 Mark
              The compound of the molecular formula, C6​H14​ has two tertiary carbons. Its IUPAC name is:
              • A
                n-hexane
              • B
                2−methylpentane
              • C
                2, 3−dimethylbutane
              • D
                 
              Answer
              1. 2, 3−dimethylbutane

              Explanation:

              Molecular formula of the compound is C6​H14​. The compound has 2 tertiary carbon.

              Then the probable structure of the compound will be CH3​−CH(Me)−CH(Me)−CH3​.

              Name of this compound is 2, 3-dimethylbutane.

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              MCQ 581 Mark
              The fruity smell is of a/an:
              • A
                Aldehyde
              • B
                Ketone
              • C
                Alcohol
              • D
                Ester
              Answer
              1. Ester

              Explanation:

              Esters have a fruity smell due to their volatile nature caused by their chemical composition and conformations.

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              MCQ 591 Mark
              What is the name of tube in which a known mass of an organic compound is heated for the quantitative analysis of sulphur?
              • A
                Borosil tube.
              • B
                Carius tube.
              • C
                Kjeldahl tube.
              • D
                None of these.  
              Answer
              1. Carius tube.

              Explanation:

              Tube in which a known mass of an organic compound is heated for the quantitative analysis of S is known as Carius tube.

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              MCQ 601 Mark
              The molecular formulae for phosgene and tear gas are .......... and ......... respectively. 
              • A
                COCl2​ and CCl3​NO2​
              • B
                SOCl2​ and CCl3​NO2​
              • C
                COCl2​ and CCI3​NO2​
              • D
                SOCl2​ and CCI3​NO2​
              Answer
              1. COCl2​ and CCl3​NO2​
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              MCQ 611 Mark
              Given two compounds, one more polar than the other, which one will move up the silica stationery phase higher than the other in TLC?
              • A
                Polar compound
              • B
                Non polar compound
              • C
                Both will move equally
              • D
                None will move
              Answer
              1. Non polar compound

              Explanation:

              Polar compounds will have more interaction with silica gel and hence polar compounds won't move much in the stationery phase.

              Non polar compounds will have lesser interaction with the silica or the stationery phase and will move higher than the polar compound.

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              MCQ 621 Mark
              Which one is a nucleophilic substitution reaction among the following?
              • A
                $\text{CH}_3\text{CHO}+\text{HCN}\rightarrow\text{CH}_3\text{CH(OH)CN}$
              • B
                $\text{CH}_3-\text{CH}=\text{CH}_2+\text{H}_2\text{O}\xrightarrow{\text{H}^+}\text{CH}_3-\text{CH}-\text{CH}_3\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ |\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{OH}$
              • C
                $\text{RCHO}+\text{R}'\text{MgX}\rightarrow\text{R}-\text{CH}-\text{R}'\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ |\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{OH}$
              • D
                $ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{CH}_3\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ |\\\text{CH}_3-\text{CH}_2-\text{CH}-\text{CH}_2\text{Br}+\text{NH}_3\rightarrow\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{CH}_3\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ |\\\text{CH}_3-\text{CH}_2-\text{CH}-\text{CH}_2\text{NH}_2$
              Answer
              1. $ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{CH}_3\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ |\\\text{CH}_3-\text{CH}_2-\text{CH}-\text{CH}_2\text{Br}+\text{NH}_3\rightarrow\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{CH}_3\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ |\\\text{CH}_3-\text{CH}_2-\text{CH}-\text{CH}_2\text{NH}_2$
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              MCQ 631 Mark
              The main source of organic compounds is:
              • A
                Coaltar
              • B
                Petroleum
              • C
                Both (a) and (b)
              • D
                None of the above
              Answer
              1. Both (a) and (b)

              Explanation:

              Petroleum and coaltar are the main sources of organic compounds.

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              MCQ 641 Mark
              $\text{CH}_3​−\text{CH}=\text{CH}−\text{C}\equiv\text{CH}$ has lUPAC name:
              • A
                pent - 2 - en- 4 - yne
              • B
                pent - 4 - yn - 2 - ene
              • C
                pent - 1 - yn - 3 - ene
              • D
                pent - 3 - en - 1 - yne
              Answer
              1. pent - 3 - en - 1 - yne
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              MCQ 651 Mark
              Which of the statements below best describes a functional group?
              • A
                It is the non-polar section of an organic molecule.
              • B
                It is a highly reactive group of elements with similar outer shell electron configurations.
              • C
                It is an atom or a group of atoms which determines the chemical properties of a compound.
              • D
                It is the part of a chelate ligand attracted to the central ion.
              Answer
              1. It is an atom or a group of atoms which determines the chemical properties of a compound.

              Explanation:

              Functional groups are specific groups or recognizable group of atoms or bonds within molecules that are responsible for the characteristic chemical reactions of those molecules.

              These functional group give characteristic properties to the carbon chain to which it is attached, e.g., aldehydic group or carboxylic group.

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              MCQ 661 Mark
              Which one of the following is not an alcohol?
              • A
                (CH3​)2​CHCH2​CH2​OH
              • B
                (CH3​)3​C−CH2​OH
              • C
                (CH3)2​−CH−OH
              • D
                CH3​CH2CH2​CH2COOH
              Answer
              1. CH3​CH2CH2​CH2COOH

              Explanation:

              (d) is the correct answer as the group associated with it is carboxylic acid group.

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              MCQ 671 Mark
              Which of the following are members of same homologous series?
              • A
                HCOOCH3​ and CH3COOCH3
              • B
                HCOOH and HCOOCH3
              • C
                CH3COCH3​ and CH3​CH2​CHO
              • D
                C2H5​OH and CH3OCH3
              Answer
              1. HCOOCH3​ and CH3COOCH3

              Explanation:

              A homologous series is a family of hydrocarbons with similar chemical properties that share the same general formula.

              Both given compound HCOOCH3​ and CH3​COOCH3​ are homologous series of a simple ester group RCOOH, where a successive methyl group −CH3​ added to previous one.

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              MCQ 681 Mark
              Identify the correct decreasing order of priority of the functional groups from the following:
              • A
                −CONH2​, −C=O, −HC=O, −CN
              • B
                −CONH2​​, −CN, −HC=O, −C=O
              • C
                −CN, −CONH2​​, −C=O, −HC=O
              • D
                −CN, −CONH2​​, −HC=O, −C=O
              Answer
              1. −CONH2​, −CN, −HC=O, −C=O

              Explanation:

              Amides > Nitriles > Aldehydes > ketones

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              MCQ 691 Mark
              Which of the following is correct about the number of bonds formed by atoms of different elements in acetaldehyde?(Report your answer in the ratio of H:O:C)
              • A
                4:2:2
              • B
                4:1:2
              • C
                4:1:3
              • D
                3:3:3
              Answer
              1. 4:1:2

              Explanation:

              CH3​CHO − Acetaldehyde

              No of atoms of H:O:C = 4:1:2

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              MCQ 701 Mark
              In presence of HCl(aq.);H2​S gas results the precipitation of group-2 cations but not of group-4 cations during qualitative analysis. It is due to:
              • A
                Lower concentration of S2-
              • B
                Higher concentration of S2-
              • C
                Lower concentration of H+
              • D
                Higher concentration of H+
              Answer
              1. Higher concentration of S2-

              Explanation:

              Due to common ion effect, the ionisation of H2​S is suppressed and at low concentration of S2- ions, the second group sulphides are precipitated.

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              MCQ 711 Mark
              Which of the following is a homocyclic alicyclic compound?
              • A
                 

                1. B
                   

                  1. C
                     

                    ​​​​​

                    1. D
                       

                       

                    Answer
                    1.  

                    ExPlanation:

                    Alicyclic (aliphatic cyclic) compounds contain carbon atoms joined in the form of a ring. Homocyclic means it contains only C-atoms in the ring, e.g.

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                    MCQ 721 Mark
                    Pick out the wrong statement.
                    • A
                       
                      1. B
                         
                        1. C
                          Hyperconjugation is no bond resonance.
                        2. D
                          Dopole moment of $\text{CH}_2=\text{CH}-\text{Cl}<\text{CH}_3\text{Cl}$
                        Answer
                        1.  

                        Explanation:

                        It is aromatic $\because$ it has $(4\text{n}+2)\pi,$ i.e., $6\pi$ electrons which are delocalised.

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                        MCQ 731 Mark
                        Which type of bond is formed between carbon atom and nitrogen atom?
                        • A
                          Covalent bond.
                        • B
                          Ionic bond.
                        • C
                          Dative bond.
                        • D
                          None of these.
                        Answer
                        1. Covalent bond.

                        Explanation:

                        A covalent bond is formed between carbon and nitrogen atom. If duplet (2) or octet (8) is completed by sharing of electrons between two electronegative elements, the bond formed is called covalent bond.

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                        MCQ 741 Mark
                        Electrophilic addition reactions proceed in two steps. The first step involves the addition of an electrophile. Name the type of intermediate formed in the first step of the following addition reaction.
                        $\text{H}_3\text{C}-\text{H}\text{C}=\text{CH}_2+\text{H}^+\rightarrow?$
                        • A
                          2° Carbanion
                        • B
                          1° Carbocation
                        • C
                          2° Carbocation
                        • D
                          1° Carbanion
                        Answer
                        1. 2° Carbocation

                        Explanation:

                        When the electrophile attacks CH3 – CH = CH2, delocalisation of electrons can take place in two possible ways

                        As 2° carbocation is more stable than 1° carbocation, the first addition is more feasible.

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                        MCQ 751 Mark
                        The atom or the group of atoms by which the characteristic reactions of organic compounds are determined, that atom or group of atoms is called ____________.
                        • A
                          Hydrocarbon
                        • B
                          Functional group
                        • C
                          Atom
                        • D
                          Compound
                        Answer
                        1. Functional group

                        Explanation:

                        Functional groups are specific substituents or moieties within molecules that are responsible for the characteristic chemical reactions of those molecules.

                        Examples: −OH, −COOH, −COOR etc.

                        Hence, the atom or the group of atoms by which the characteristic reactions of organic compounds are determined, that atom or group of atoms is called functional group.

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                        MCQ 761 Mark
                        Which functional group is present in a molecule of CH3​OCH2​CH3​?
                        • A
                          Ether
                        • B
                          Carboxyl
                        • C
                          Aldehyde
                        • D
                          Ester
                        Answer
                        1. Ether

                        Explanation:

                        The functional group present in CH3​−O−CH2​CH3​ is an ether group −O−.
                        Note: In general, ethers can be represented as R−O−R′.

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                        MCQ 771 Mark
                        The IUPAC name of dimethyl ether is:
                        • A
                          Ethoxymethane
                        • B
                          Methoxymethane
                        • C
                          Methoxyethane
                        • D
                          Ethoxyethane
                        Answer
                        1. Methoxymethane

                        Explanation:

                        Dimethyl ether is CH3​OCH3​.

                        It's IUPAC name is methoxymethane.

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                        MCQ 781 Mark
                        Homolyte fission leads to formation of:
                        • A
                          Nucleophile.
                        • B
                          Carbanion.
                        • C
                          Free radical.
                        • D
                          Arbocation.
                        Answer
                        1. Free radical.

                        Explanation:

                        $\text{cl}-\text{cl}\xrightarrow{\ \ \text{hv}\ \ \ }2\text{c}\dot{\text{l}}$

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                        MCQ 791 Mark
                        How many pi bonds are present in phosgene?
                        • A
                          1
                        • B
                          2
                        • C
                          3
                        • D
                          4
                        Answer
                        1. 1

                        Explanation:

                        From the image we can see that there is only 1 pi-bond present in phosgene (cabonyl chrolide).

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                        MCQ 801 Mark
                        Look at the figure below and find the hybridisation of carbon atom.
                        Image
                        • A
                          sp3
                        • B
                          sp2
                        • C
                          sp
                        • D
                          None of these.
                        Answer
                        1. sp2
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                        MCQ 811 Mark
                        Covalent bond can undergo fission in two different ways. The correct representation involving a heterolytic fission of CH3—Br is:
                        • A

                        • B

                        • C

                        • D


                        Answer
                        1.  

                        Explanation:

                        Arrow denotes the direction of movement of electrons

                        Since, Br is more electronegative than carbon, hence heterolytic fission occurs in such a.

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                        MCQ 821 Mark
                        What does A and B refer to, in the figure of differential extraction?

                        Choose the correct option.
                        Answer
                        1. A → Organic compound in aqueous layer.

                        B → Organic compound in solvent layer.

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                        MCQ 831 Mark
                        The IUPAC name of the compound

                        • A
                          5-Formyl hex-2-en-3-one.
                        • B
                          5-methyl-4-oxohex-2-en-5-al.
                        • C
                          3-keto-2 methyl hex-5-enal.
                        • D
                          3-keto-2-methyl hex-4-enal.
                        Answer
                        1. 5-Formyl hex-2-en-3-one.

                        Explanation:

                        Aldehyde group is preferred over keto group.

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                        MCQ 841 Mark
                        The product(s) is / are:

                        • A
                          (A) is an alcohol; (B) is an aldehyde
                        • B
                          (A) is an aldehyde; (B) is an alcohol
                        • C
                          (A) and (B) both are alcohols
                        • D
                          (A) and (B) both are aldehydes
                        Answer
                        1. (A) and (B) both are alcohols

                        Explanation:

                        A reagent used for both reactions is reducing, it will reduce the carbonyl-chloride to the alcohol group.

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                        MCQ 851 Mark
                        IUPAC name of the compound:
                        $\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{CH}_3\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ |\\\text{CH}_3-(\text{CH}_2)_4-\text{CH}-\text{C}-\text{CH}_3\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ |\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{C}_2\text{H}_5(\text{CH}_2)_2-\text{CH}_3$
                        • A
                          5 − ethyl −4, 4−dimethyl decane.
                        • B
                          4 − ethyl 4, 5 dimethyl decane.
                        • C
                          3, 4− dimethyl − 3 − Hexane.
                        • D
                          6, 7− dimethyl decane.
                        Answer
                        1. 5 − ethyl − 4, 4 − dimethyl decane.

                        Explanation:

                        Numbering should be started from right side. 2 methyl and one ethyl groups are attached C4​ and C5​ respectively.

                        So the IUPAC name of this compound is 5 − ethyl − 4, 4 − dimethyldecane.

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                        MCQ 861 Mark
                        In which of the following ways the structure(s) of organic compounds is/ are represented?
                        • A
                          Lewis or dot structure.
                        • B
                          Dash structure.
                        • C
                          Condensed and bond-line structure.
                        • D
                          All of the above.
                        Answer
                        1. All of the above.
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                        MCQ 871 Mark
                        The IUPAC name for,

                        • A
                          1-Chloro-2-nitro-4-methylbenzene.
                        • B
                          1-Chloro-4-methyl-2-nitrobenzene.
                        • C
                          2-Chloro-1-nitro-5-methylbenzene.
                        • D
                          M-nitro-p-chlorotoluene.
                        Answer
                        1. 1-Chloro-4-methyl-2-nitrobenzene.

                        Explanation:

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                        MCQ 881 Mark
                        How do we differentiate between Fe3+ and Cr3+ in group III?
                        • A
                          By taking excess of NH4​OH
                        • B
                          By increasing $\text{NH}_4^+$​ ion cocentration
                        • C
                          By decreasing OH ion concentration
                        • D
                          Both (b) and (c)
                        Answer
                        1. Both (b) and (c)

                        Explanation:

                        $\text{NH}_4^+$ ions are increased to suppress release of OHions, hence solubility product of Fe(OH)3 is attained.

                        Colour of precipitate is different.

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                        MCQ 891 Mark
                         Which of the following is not in accordance to IUPAC system?
                        Answer
                        1. $\text{Br}-\text{CH}_{2}- \text{CH}=\text{CH}_{2}\\ \ \ \ \ \ \ \ \ \text{1-bromoprop 2-ene}$
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                        MCQ 901 Mark
                        The structure of iso-butyl group in an organic compound is:
                        Answer
                        1.  

                        Explanation:

                        'Iso' means one Me group is present in side chain. Hence, the structure of iso-butyl group is

                        $ \ \ \ \ \ \ \ \ \ \ \ \ \text{CH}_3 \\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ |\\\ \text{CH}_3-\text{C}-\text{CH}_2 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ |\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{C}$

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                        MCQ 911 Mark
                        Which one of the following has the most nucleophilic nitrogen?
                        Answer
                        1.  

                        Explanation:

                        $\because$ Lone pair of electrons are easily available. In (b) lone pair is delocalised, in (c) -COCH3 is electron withdrawing, in (d) C6H5- is electron withdrawing.

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                        MCQ 921 Mark
                        The fragrance of flowers is due to the presence of some steam volatile organic compounds called essential oils. These are generally insoluble in water at room temperature but are miscible with water vapour in vapour phase. A suitable method for the extraction of these oils from the flowers is:
                        • A
                          Distillation.
                        • B
                          Crystallisation.
                        • C
                          Distillation under reduced pressure.
                        • D
                          Steam distillation.
                        Answer
                        1. Steam distillation.

                        Explanation:

                        Essential oils are insoluble in water, soluble in steam and have high vapour pressure. Therefore, they can be separated by steam distillation.

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                        MCQ 931 Mark
                        In which of the following, functional group isomerism is not possible?
                        • A
                          Alcohols
                        • B
                          Aldehydes
                        • C
                          Alkyl halides
                        • D
                          Cyanides
                        Answer
                        1. Alkyl halides

                        Explanation:

                        Alkyl halides do not show functional isomerism. Alcohols and ethers; aldehydes and ketones; cyanides and isocyanides are functional isomers.

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                        MCQ 941 Mark
                        The functional group present in organic acid is:
                        • A
                          −OH
                        • B
                          −CHO
                        • C
                          −COOH
                        • D
                          > C = O
                        Answer
                        1. −COOH

                        Explanation:

                        The functional group present every organic acid is −COOH as it dissociate in water as −COO and H+.

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                        MCQ 951 Mark
                        Glycerol can be separated from spent-lye in soap industry by:
                        • A
                          Chromatography.
                        • B
                          Sublimation.
                        • C
                          Fractional distillation.
                        • D
                          Distillaltion under reduced pressure.
                        Answer
                        1. Distillaltion under reduced pressure.

                        Explanation:

                        Glycerol decomposes at its boiling point is separated from spent-lye by using distillation under reduced pressure.

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                        MCQ 961 Mark
                        The difference in energy between the actual structure and lowest energy canonical form is called:
                        • A
                          Resonance energy.
                        • B
                          Localisation energy.
                        • C
                          Both (a) and (b).
                        • D
                          All of these.
                        Answer
                        1. Resonance energy.
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                        MCQ 971 Mark
                        Transition state 2 is structurally most likely as:
                        Transition state 2 is structurally most likely as :
                        • A
                          Intermediate 1
                        • B
                          Transition state 3
                        • C
                          Intermediate 2
                        • D
                          Product
                        Answer
                        1. Intermediate 2

                        Explanation:

                        Generally transition state tries to bring stability with less energy difference. From the figure it is evident that energy difference between TSand intermediate 2 is least.

                        So it is most likely that TSand intermediate 2 have similar structures.

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                        MCQ 981 Mark
                        Give the functional group of the following.-OH
                        • A
                          Aldehydic group
                        • B
                          Alcoholic group
                        • C
                          Ketonic group
                        • D
                          Carboxylic group
                        Answer
                        1. Alcoholic group

                        Explanation:

                        −OH represents an alcoholic  group. It represents alcohols such as ethanol CH3, ​−CH2, ​−OH. The general formula of alcohols is R−OH.

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                        MCQ 991 Mark
                        Why does ethane have a higher boiling point than methane?
                        • A
                          Difference in size
                        • B
                          Functional group
                        • C
                          Difference in shape
                        • D
                          None of above
                        Answer
                        1. Difference in size

                        Explanation:

                        Ethane have a higher boiling point than methane because molecules of ethane (C2​H6​) have more Van der Waals forces (intermolecular forces) with neighboring molecules than methane (CH4​) due to the greater number of atoms present in the molecule of ethane compared to methane.

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                        MCQ 1001 Mark
                        Which of the following should be added in the compound for the detection of carbon and hydrogen?
                        • A
                          Cu(I) oxide.
                        • B
                          Cu(II) oxide.
                        • C
                          Zn dust.
                        • D
                          Activated charcoal.
                        Answer
                        1. Cu(II) oxide.

                        Explanation:

                        Carbon and hydrogen are detected by heating the compound with Cu(II) oxide. Carbon present in the compound is oxidised to carbon dioxide (tested with lime-water, which develop turbidity) and hydrogen to water (tested with anhydrous copper sulphate, which turns blue)

                        $\text{C}+ 2\text{CuO} \xrightarrow[]{\Delta}2\text{Cu + CO}_{2}\uparrow$

                        $\text{2H} + \text{CuO}\xrightarrow[]{\Delta} \text{Cu} + \text{H}_{2}\text{O}$

                        $\text{CO}_{2} + \text{Ca}(\text{OH})_{2} \xrightarrow{\ \ \ \ \ }\text{CaCO}_{3} \downarrow + \text{H}_{2}\text{O}\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{White turbidity}$

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                        MCQ 1011 Mark
                        Carbon and hydrogen are detected by heating the organic compound with ________.
                        • A
                          Magnesium oxide
                        • B
                          Cupric oxide
                        • C
                          Zinc oxide
                        • D
                          Sulphur dioxide
                        Answer
                        1. Cupric oxide

                        Explanation:

                        The detection of various elements present in an organic compound is called qualitative analysis. Carbon and hydrogen are present in almost all the organic compounds.

                        Carbon and hydrogen are detected by heating the organic compound with cupric oxide (CuO) strongly, where carbon is oxidized to carbon dioxide and hydrogen to water.

                        Carbon dioxide is tested by lime water test, whereas water is tested by anhydrous copper sulphate test.

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                        MCQ 1021 Mark
                        Study the structures given below carefully and choose the type of isomerism they represent.
                        $ \ \ \ \ \ \ \ \ \ \ \ \ \text{CH}_3 \\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ |\\\ \text{CH}_3-\text{C}-\text{CH}_3 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ |\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{CH}_3$
                        $ \ \ \ \ \ \ \ \ \ \ \ \ \text{CH}_3 \\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ |\\\ \text{CH}_3-\text{CH}- \text{CH}_2-\text{CH}_3$
                        • A
                          Chain isomerism.
                        • B
                          Position isomerism.
                        • C
                          Functional isomerism.
                        • D
                          Metamerism.
                        Answer
                        1. Chain isomerism.

                        Explanation:

                        $ \ \ \ \ \ \ \ \ \ \ \ \ \text{CH}_3 \\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ |\\\ \text{CH}_3-\text{C}-\text{CH}_3 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ |\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{CH}_3$

                        $ \ \ \ \ \ \ \ \ \ \ \ \ \text{CH}_3 \\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ |\\\ \text{CH}_3-\text{CH}- \text{CH}_2-\text{CH}_3$

                        chain isomers as they have different carbon chains.

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                        MCQ 1031 Mark
                        In the Kjeldahl's method for estimation of nitrogen present in a soil sample, ammonia evolved from 0.75g of sample neutralised 10mL of 1M H2SO. The percentage of nitrogen in the soil is:
                        • A
                          37.33
                        • B
                          45.33
                        • C
                          35.33
                        • D
                          45.33
                        Answer
                        1. 37.33
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                        MCQ 1041 Mark
                        Generally pi bond is formed from the lateral overlap of:
                        • A
                          s−s orbitals
                        • B
                          p−p orbitals
                        • C
                          d−d orbitals
                        • D
                          Both B and C
                        Answer
                        1. Both B and C

                        Explanation:

                        $\pi$ bond is formed by lateral (side ways) overlap of either p−p, d−d or d−p orbitals.

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                        MCQ 1051 Mark
                        The formula of methane is CH4​ and its succeeding member ethane is expressed in C2​H6​. The common difference of succession between them is:
                        • A
                          CH2​
                        • B
                          C2​H2
                        • C
                          CH4
                        • D
                          None of the above
                        Answer
                        1. CH2

                        Explanation:

                        Methane and ethane are alkanes. These are saturated carbon compounds. They have the general formula Cn​H2n+2​.

                        All the alkanes differ by CH2​ group successively in the series.

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                        MCQ 1061 Mark
                        Preceeding and succeeding homologous of C10​H22​ respectively:
                        • A
                          C9​H20​, C11​H22​
                        • B
                          C9​H20​, C11​H24​
                        • C
                          C11​H24​, C9​H20​
                        • D
                          C8​H18​, C9​H20​
                        Answer
                        1. C9​H20​, C11​H24​

                        Explanation:

                        Preceeeding homologue is C9​H20​.

                        Succeeding homologue is C11​H24​​.

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                        MCQ 1071 Mark
                        Homologous have the same:
                        • A
                          Empirical formulae
                        • B
                          Molecular formulae
                        • C
                          Chemical properties
                        • D
                          Physical properties
                        Answer
                        1. Chemical properties

                        Explanation:

                        Members of Homologous series have same chemical properties.

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                        MCQ 1081 Mark
                        In cyclic compounds, the bond-line formula for chlorocyclohexane is represented by which of the following representations?
                        Answer
                        1.  

                        Explanation:

                        Structure of chlorocyclohexane is:

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                        MCQ 1091 Mark
                        The presence of carbon in an organic compound is detected by heating it with:
                        • A
                          Sodium metal to convert it to NaCN.
                        • B
                          CaO to convert it into CO which burns with a blue flame.
                        • C
                          CuO to convert it into CO2​ which turns lime water milky.
                        • D
                          Cu wire to give a bluish-green flame.
                        Answer
                        1. CuO to convert it into CO2​ which turns lime water milky.

                        Explanatioin:

                        The given organic compound is mixed with dry copper oxide (CuO) and heated in a hard glass tube. The products of the reaction are passed over (white) anhydrous copper sulphate and then bubbled through lime water.

                        If copper sulphate turns blue due to the formation of CuSO4​.5H2​O (by water vapor) then the compound contains hydrogen. If lime water is turned milky by CO2​, then the compound contains carbon.

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                        MCQ 1101 Mark
                        The order of -I effect of orbitals is:
                        • A
                          sp3>sp2>sp
                        • B
                          sp>sp2>sp3
                        • C
                          sp2>sp3>sp
                        • D
                          sp2>sp>sp3
                        Answer
                        1. sp>sp2>sp3

                        Explanation:

                        As %s character increases, electronegativity of orbitals increases which means its tendency to pull electron towards itself also increases hence -I effect increases.

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                        MCQ 1111 Mark
                        Linear combination of two hybridised orbitals belonging to two atoms and each having one electron leads to the formation of:
                        • A
                          Sigma bond
                        • B
                          Double bond
                        • C
                          Co-ordinate covalent bond
                        • D
                          pi bond
                        Answer
                        1. Sigma bond

                        Explanation:

                        A sigma bond is a single bond formed by linear combination of hybridised orbitals.

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                        MCQ 1131 Mark
                        Correct IUPAC name for $\text{H}_3\text{C}-\text{CH}-\text{CH}-\text{CH}_3\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ | \ \ \ \ \ \ \ \ \ |\\\ \ \ \ \ \ \ \ \ \ \ \ \ \text{C}_2\text{H}_5\ \text{C}_2\text{H}_5$ ___________.
                        • A
                          2- Ethyl-3-methylpentane.
                        • B
                          3, 4- Dimethylhexane.
                        • C
                          2-Sec-butylbutane.
                        • D
                          2, 3-Dimethylbutane.
                        Answer
                        1. 3, 4- Dimethylhexane.

                        Explanation:

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                        MCQ 1141 Mark
                        Among the following which has greatest -I effect?
                        • A
                          −OH
                        • B
                          −NH2​
                        • C
                          −NO2​
                        • D
                          −F
                        Answer
                        1. −NO2

                        Explanation:

                        −NO2​ is the strongest -I group among all the neutral group because of the presence of + charge on N atom as shown in structure.

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                        MCQ 1151 Mark
                        Marsh gas is:
                        • A
                          C6​H6
                        • B
                          CH4​
                        • C
                          C2​H2​
                        • D
                          CO
                        Answer
                        1. CH4

                        Explanation:

                        Marsh gas is methane, CH4​.

                        It is called so because it is often obtained from marshy places.

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                        MCQ 1161 Mark
                        The first three members of a homologous series are CH4​,C2​H6​,C3​H8​. The fifth member of this series will be: 
                        • A
                          C5​H10​
                        • B
                          C5​H14​
                        • C
                          C5​H12​
                        • D
                          C5​H8
                        Answer
                        1. C5​H12

                        Explanation:

                        General formula of homologous series is Cn​H2n+2​.

                        First member is methane i.e. CH4

                        Second member is ethane i.e. C2​H6

                        Third member is propane i.e. C3​H8

                        Forth member is butane i.e. C4​H10

                        Similarly, fifth member is pentane i.e

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                        MCQ 1171 Mark
                        What is a homologation reaction?
                        • A
                          Conversion of one member of a homologous series to the other compounds.
                        • B
                          Breakdown of compounds in a homologous series.
                        • C
                          Conversion of one member of a homologous series to the next member of homologous series.
                        • D
                          None of above
                        Answer
                        1. Conversion of one member of a homologous series to the next member of homologous series.

                        Explanation:

                        A homologation reaction is a chemical process that converts one member of a homologous series to the next member of the homologous series.

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                        MCQ 1181 Mark
                        Compounds belonging to a homologous series have the same:
                        • A
                          Number of carbon atoms
                        • B
                          Functional group
                        • C
                          Number of hydrogen atoms
                        • D
                          None of above
                        Answer
                        1. Functional group

                        Explanation:

                        Compounds within a homologous series typically have a fixed set of functional groups that gives them similar chemical and physical properties.

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                        MCQ 1191 Mark
                        Write the correct IUPAC name of the following.

                        • A
                          1-chloro-2, 4-dinitrobenzene.
                        • B
                          6-chloro-1, 3-dinitrobenzene.
                        • C
                          1-chloro-4, 6-dinitrobenzene.
                        • D
                          2-chloro-1, 5-dinitrobenzene.
                        Answer
                        1. 1-chloro-2, 4-dinitrobenzene.
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                        MCQ 1201 Mark
                        What are the values of N and M?

                        • A
                          6, 6
                        • B
                          6, 4
                        • C
                          4, 4
                        • D
                          3, 3
                        Answer
                        1. 4, 4

                        Explanation:

                        Since A has four different types of H atoms, so it on chlorinatives gives 4 isomeric products.

                        Again as four products are chemically different, on fractional distillation they give 4 isomeric products. So N = 4 and M = 4.

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                        MCQ 1211 Mark
                        During hearing of a court case, the judge suspected that some changes in the documents had been carried out. He asked the forensic department to check the ink used at two different places. According to you which technique can give the best results?
                        • A
                          Column chromatography.
                        • B
                          Solvent extraction.
                        • C
                          Distillation.
                        • D
                          Thin layer chromatography.
                        Answer
                        1. Thin layer chromatography.

                        Explanation:

                        Thin layer Chomatography (TLC): itis a method for analyzing mixtures by separating the components of the mixture. TLC can be used to determune the number of components in the mixture, the identity of compounds and the purity of compounds by observing the apperance of a product or the disppearance of areactant. it can also be used to monitor the progress of a reaction. TLC is a sensitive technique - microgram (0.000001)g quantities can be analyzwd by TLC,

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                        MCQ 1221 Mark
                        Amongst the following elements present in an organic compound, the one which has no direct test for detection is:
                        • A
                          Cl
                        • B
                          N
                        • C
                          S
                        • D
                          O
                        Answer
                        1. O

                        Explanation:

                        There is no test that can directly indicate the presence of oxygen in an organic compound. The presence of oxygen is indicated indirectly by the presence of groups containing oxygen.

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                        MCQ 1231 Mark
                        The correct order of electronegativity of carbon in ethane, ethene and ethyne is:
                        • A
                          Ethane ethene < ethyne.
                        • B
                          Ethyne < ethene < ethane.
                        • C
                          Ethene < ethyne < ethane.
                        • D
                          Ethene <ethane < ethyne.
                        Answer
                        1. Ethane ethene < ethyne.

                        Explanation:

                        Hydrocarbons
                        Molecular formula
                        % S-Character
                        Ethane
                        CH3-CH3
                        25%
                        Ethene
                        CH2 = CH2
                        33%
                        Ethyne
                        $\text{CH}\equiv\text{CH}$
                        50%

                        As we know that, electronegativity $\propto$ % s-character $\propto$acidic character Hence, correct order is, ethane < ethene < ethyne.

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                        MCQ 1241 Mark
                        Which type of compound is shown by the following structure:

                        • A
                          Alicyclic compound.
                        • B
                          Benzenoid aromatic compound.
                        • C
                          Non-benzenoid aromatic compound.
                        • D
                          Acyclic compound.
                        Answer
                        1. Benzenoid aromatic compound.

                        Explanation:

                        Since, the given compounds contain benzene nucleus so, naphthalene is a benzenoid aromatic compound.

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                        MCQ 1251 Mark
                        Which of the following ion is most resonance stabilized?
                        • A
                          $\text{C}_2\text{H}_5\text{O}^\ominus$
                        • B
                          $\text{C}_6\text{H}_5\text{O}^\ominus$
                        • C
                          $(\text{CH}_3)_3\text{CO}^\ominus$
                        • D
                          $(\text{CH}_3)_2\text{CHO}^\ominus$
                        Answer
                        1. $\text{C}_6\text{H}_5\text{O}^\ominus$

                        Explanation:

                        Phenoxide is stabilized by resonance.

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                        MCQ 1261 Mark
                        Formation of which of the following complex indicates presence of sulphur in the organic compound when sodium nitroprusside is added to sodium extract of the compound?
                        • A
                          $\text{Fe}_{4}[\text{Fe}(\text{CN})_{6}]_{3}$
                        • B
                          $\text{Na}_{2} [\text{Fe}(\text{NO})(\text{CN})_{5}]$
                        • C
                          $\text{Fe}(\text{CNS)}$
                        • D
                          $\text{Na}_{4} [\text{Fe}(\text{CN})_{5}\text{NOS}]$
                        Answer
                        1. $\text{Na}_{4} [\text{Fe}(\text{CN})_{5}\text{NOS}]$

                        Explanation:

                        Formation of sodium thionitroprusside (purple) shows the presence of sulphur in organic compound.

                        $\text{Na}_{2}\text{S} \ \ \ \ + \ \ \ \ \ \text{Na}_{2}[\text{Fe}(\text{NO})(\text{CN})_{5}] \xrightarrow{\ \ \ \ \ \ }\text{Na}_{4}[\text{Fe}(\text{CN})_{5}\text{NOS}]\\\text{Sodium extract}$

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                        MCQ 1271 Mark
                        Which of the following is the correct IUPAC name?
                        • A
                          3-Ethyl-4, 4-dimethylheptane.
                        • B
                          4, 4-Dimethyl-3-ethylheptane.
                        • C
                          5-Ethyl-4, 4-dimethylheptane.
                        • D
                          4, 4-Bis (methyl)-3-ethylheptane.
                        Answer
                        1. 3-Ethyl-4, 4-dimethylheptane.

                        Explelation:

                        While writing IUPAC name, the alkyl groups are written in alphabetical order. Thus lower locant 3 is assigned to ethyl. Prefix, di, tri, and tetra are not included in alphabetical order.

                        The structure is $\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{CH}_3\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ |\\\text{CH}_3-\text{CH}_2-\text{CH}_2-\text{C}-\text{CH}-\text{CH}_2-\text{CH_3}\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ |\ \ \ \ \ \ \ |\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{CH}\ \ \text{C}_2\text{H}_5$

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                        MCQ 1281 Mark
                        Which one of the following can show +I effect?
                        • A
                          CBr3​
                        • B
                          CH3​
                        • C
                          CCl3​
                        • D
                          CHO
                        Answer
                        1. CH3

                        Explanation:

                        Only CH3​ shows +I effect rest of them shows -I effect, since other groups contain electronegative atom which will pull electrons towards it causing -I effect.

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                        MCQ 1291 Mark
                        In Carius method of estimation of halogens, 250mg of an organic compound gave 141mg of AgBr. The percentage of bromine in the compound is (atomic mass Ag = 108 and Br = 80)
                        • A
                          24
                        • B
                          36
                        • C
                          48
                        • D
                          60
                        Answer
                        1. 24
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                        MCQ 1301 Mark
                        All members of a homologous series have the same:
                        • A
                          Chemical formula.
                        • B
                          Empirical formula.
                        • C
                          Boiling point.
                        • D
                          Melting point.
                        Answer
                        1. Empirical formula.

                        Explanation:

                        Homologous series differ by a CH2​ group. It has a mass of 14 units. They can be represented by a common general formula for alkanes, alkenes, alkynes, and other functional groups. They can thus, be represented by a common empirical formula.

                        For example alkanes has empirical formula Cn​H2n+2​ and for alkenes it is Cn​H2n​ etc.

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                        MCQ 1311 Mark
                        In Lassaigne's test, the organic compound is fused with a piece of sodium metal in order to:
                        • A
                          Increase the ionisation of the compound.
                        • B
                          Decrease the melting point of the compound.
                        • C
                          Increase the reactivity of the compound.
                        • D
                          Convert the covalent compound into a mixture of ionic compounds.
                        Answer
                        1. Convert the covalent compound into a mixture of ionic compounds.

                        Explanation:

                        Lassaigne's test is a general test for the detection of halogens, nitrogen and sulphur in an organic compound. These elements are covalently bonded to the organic compounds.

                        In order to detect them, these have to be converted into their ionic forms. This is done by fusing the organic compound with sodium metal.

                        The ionic compounds formed during the fusion are extracted in aqueous solution and can be detected by simple chemical tests. The extract is called sodium fusion extract or Lassaigne's extract.

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                        MCQ 1321 Mark
                        Which of the following elements in an organic compound cannot be detected by Lassaigne's test?
                        • A
                          Nitrogen
                        • B
                          Sulphur
                        • C
                          Chlorine
                        • D
                          Hydrogen
                        Answer
                        1. Hydrogen

                        Explanation:

                        Lassaigne's test is the most reliable test for the detection of nitrogen, halogens and sulphur in an organic compound.

                        The compound is heated with sodium metal to convert the elements present in the organic compound into the water-soluble salts of sodium.

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                        MCQ 1331 Mark
                        The number of sigma and pi bonds in a molecule of cyanogen are:
                        • A
                          4, 3
                        • B
                          3, 4
                        • C
                          5, 2
                        • D
                          3, 5
                        Answer
                        1. 3, 4

                        Explanation:

                        Cyanogen, $\text{N}\equiv\text{C}-\text{C}\equiv\text{N}$has three sigma and four 'pi' bonds.

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                        MCQ 1341 Mark
                        In the given potential energy diagram, the point X represents:

                        • A
                          A transition state
                        • B
                          A reaction intermediate
                        • C
                          A resonance hybrid
                        • D
                          A reactant
                        Answer
                        1. A transition state

                        Explanation:

                        In the given potential energy diagram, the point XX represents the transition state.

                        It is the maximum energy state and below which reaction will not occur.

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                        MCQ 1351 Mark
                        What is the technological applications of fractional distillation?
                        • A
                          To separate different fractions of crude oil in petroleum industry.
                        • B
                          To separate different fractions of volatile and non-volatile solvents.
                        • C
                          To separate mixture of amino acids.
                        • D
                          No technological application of fractional distillation.
                        Answer
                        1. To separate different fractions of crude oil in petroleum industry.
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                        MCQ 1361 Mark
                        Which of the following represents ketones?
                        • A
                          > C = O
                        • B
                          - OH
                        • C
                          - CHO
                        • D
                          - COOH
                        Answer
                        1. > C = O

                        Explanation:

                        COOH = carboxylic acid, −CHO = aldehydes > C = O = ketones, OH = alcohol.

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                        MCQ 1371 Mark
                        The IUPAC name is:

                        • A
                          1-phenyl-1-acetyloxyethane
                        • B
                          1-acetoxyeihylbenzene
                        • C
                          1-phenylethyl ethanoate
                        • D
                          1-methylbenzyl acetate
                        Answer
                        1. 1-phenylethyl ethanoate

                        Explanation:

                        In the outside ring functional group present is an ester of ethyl ethonate at benzene ring position 1 so by taking out one hydrogen it becomes 1-Phenyl ethylethanoate.

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                        MCQ 1381 Mark
                        Acidified sodium fusion extract on addition of ferric chloride solution gives blood red colouration. This confirms the presence of which of the following element(s)?
                        • A
                          Both S and cl.
                        • B
                          Both N and S.
                        • C
                          Only N.
                        • D
                          Only S.
                        Answer
                        1. Both N and S

                        Explanation:

                        Acidified sodium fusion extract on addition of ferric chloride solution gives blood red colouration, which confirms the presence of N and S both.

                        $\text{Na} + \text{C} + \text{N} + \text{S} \xrightarrow{\ \ \ \ \ \ } \text{NaSCN}$

                        $\text{Fe}^{+3} + \text{SCN}^{-} \xrightarrow{\ \ \ \ \ \ } [\text{Fe}(\text{SCN})]^{2+}\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{Blood red}$

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                        MCQ 1391 Mark
                        In the detection of carbon, lime water turns milky due to the formation of:
                        • A
                          CaO
                        • B
                          Ca(OH)2​
                        • C
                          CaCO3
                        • D
                          Ca(HCO3​)2​
                        Answer
                        1. CaCO3

                        Explanation:

                        Carbon dioxide passes through limewater, giving a milky solution.

                        This is due to the insoluble suspension of calcium carbonate formed:

                        Ca(OH)2(aq)+CO2(g)→CaCO3(s)+H2O(l)

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                        MCQ 1401 Mark
                        The order of relative stability of the contributing structures is:

                        • A
                          II > I > III
                        • B
                          I > II > III
                        • C
                          III > II > I
                        • D
                          I = II = III
                        Answer
                        1. I > II > III
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                        MCQ 1411 Mark
                        Which of the following does not belong to the same homologous series?
                        • A
                          CH4
                        • B
                          C2​H6
                        • C
                          C3​H8
                        • D
                          C4​H6​
                        Answer
                        1. ​​​​​C4​H6​

                        Explanation:

                        Methane, ethane, and propane-CH4​, C2​H6​, C3​H8​−Cn​H2n+2​ belong to alkane family with same general formula except C4​H6​

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                        MCQ 1421 Mark
                        The following compounds show:

                        • A
                          Configuration isomerism.
                        • B
                          Conformationat isomerism.
                        • C
                          Structural isomerism.
                        • D
                          Stereo isomerism.
                        Answer
                        1. Structural isomerism.
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                        MCQ 1431 Mark
                        The most common characteristics of homologous series is increase in the number of:
                        • A
                          Propyl units
                        • B
                          Methyl units
                        • C
                          Methylene units
                        • D
                          Butyl units
                        Answer
                        1. Methylene units

                        Explanation:

                        The most common characteristic of homologous series is increase in the number of methylene (−CH2​−) units in saturated chain within the molecule.

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                        MCQ 1441 Mark
                        In which of the following compounds the carbon marked with asterisk is expected to have greatest positive charge?
                        • A
                          *CH3—CH2 —Cl
                        • B
                          *CH3—CH2—Mg+Cl
                        • C
                          *CH3—CH2 —Br
                        • D
                          *CH3—CH2—CH3
                        Answer
                        1. *CH3—CH2 —Cl

                        Explanation:

                        In CH3—CH2—Cl, the carbon marked with asterisk has the greatest positive charge due to high electronegativity of Cl.

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                        MCQ 1451 Mark
                        The chief source of aromatic compounds is:
                        • A
                          Fruits
                        • B
                          Oil of winter green
                        • C
                          Petroleum
                        • D
                          Ants
                        Answer
                        1. Petroleum

                        Explanation:

                        The hydrocarbons in crude oil are mostly alkanes, cycloalkanes and various aromatic hydrocarbons. Fruits contain fructose and glucose derivatives.

                        Oil of wintergreen is an organic ester with fragrance. Ants have formic acid.

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                        MCQ 1461 Mark
                        Define homologous series.
                        • A
                          Compound with same melting point.
                        • B
                          Compound with same molecular mass.
                        • C
                          Compounds with different general formula.
                        • D
                          Compounds with same functional group substitutes for hydrogen in a carbon chain.
                        Answer
                        1. Compounds with same functional group substitutes for hydrogen in a carbon chain.

                        Explanation:

                        A homologous series is a series of compounds with the same general formula.

                        But they have different melting points (physical properties) and different molecular masses. However, they have the same functional group substitutes for hydrogen in a carbon chain.

                        Example: CH3​OH, C2​H5​OH, C3​H7​OH.

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                        MCQ 1471 Mark
                        Electrophiles are electron seeking species. Which of the following groups contain only electrophiles?
                        • A
                          $\text{BF}_3,\ \text{NH}_3,\ \text{H}_2\text{O}$
                        • B
                          $\text{AlCl}_3,\ \text{SO}_3,\ \text{NO}_2^+$
                        • C
                          $\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ +\\\text{NO}_2^+,\ \text{CH}_3^+,\ \text{CH}_3-\text{C}=0$
                        • D
                          $\text{C}_2\text{H}_5^-,\ \dot{\text{C}}_2\text{H}_5,\ \text{C}_2\text{H}_5^+$
                        Answer
                        1. $\text{AlCl}_3,\ \text{SO}_3,\ \text{NO}_2^+$

                        2. $\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ +\\\text{NO}_2^+,\ \text{CH}_3^+,\ \text{CH}_3-\text{C}=0$

                        Explanation:

                        In (a), NH3 and H2O are nucleophiles but BF3 is an electrophile. in (d), $\text{C}_2\text{H}^-_5​​​​​​​$ are electrophiles.

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                        MCQ 1481 Mark
                        IUPAC name of CH3​OC2​H5​ is:
                        • A
                          Ethoxy methane
                        • B
                          Methoxy ethane
                        • C
                          Both (a) and (b)
                        • D
                          None of these
                        Answer
                        1. Methoxy ethane

                        Explanation:

                        According to IUPAC nomenclature system, ethers are known as alkoxy alkane.

                        The ether oxygen is taken with a smaller alkyl group and creates the alkoxy group while larger alkyl group is considered as a alkyl group.

                        So the name will be methoxy ethane. 

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                        MCQ 1491 Mark
                        The absorption of hydrogen by platinum is known as:
                        • A
                          Hydrogenation
                        • B
                          Reduction
                        • C
                          Dehydrogenation
                        • D
                          Occlusion
                        Answer
                        1. Occlusion

                        Explanation:

                        A huge amount hydrogen gas generally loosely bounds with the surface of the platinum metal. This is known as occlusion and the hydrogen is known as occluded hydrogen.

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                        MCQ 1501 Mark
                        During the fusion of an organic compound with sodium metal, nitrogen of the organic compound is converted into:
                        • A
                          NaNO2
                        • B
                          NaNH2
                        • C
                          NaCN
                        • D
                          NaNC
                        Answer
                        1. NaCN
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                        MCQ 1511 Mark
                        Which of the following would be expected to be most highly ionised in water?
                        • A
                          CH3​CH2​CH2​COOH
                        • B
                          CH3CHClCH2​COOH
                        • C
                          CH3​CH2​CCl2​COOH
                        • D
                          CH3​CH2​CHClCOOH
                        Answer
                        1. CH3​CH2​CCl2​COOH

                        Explanation:

                        Fast ionization is shown by strong acids. Chlorine atom has high -I effect so when attached to nearly carbon, it increases acidity thus is more acidic.

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                        MCQ 1521 Mark
                        Identify the correct statement.
                        • A
                          Reaction mechanisms are studied using isotopic labelling.
                        • B
                          Isolation of reactive intermediates is a method to establish reaction mechanism.
                        • C
                          Both a and b are correct.
                        • D
                          Neither a nor b is correct.
                        Answer
                        1. Both a and b are correct.

                        Explanation:

                        Both a and b are correct.

                        1. Reaction mechanisms are studied using isotopic labeling. One of the atoms of one reactant/ reagent is replaced with its isotope. Then, the product in which this labelled atom is present is determined. From this information, we can decide which atom of reactant goes to which product.

                        CH3​−CO−O*H + CH3​−OH $\xrightarrow{\text{H}^+}$ ​CH3​−COO−CH3​ + H2​O*

                        In the above reaction, O atom of −OH group of acetic acid is labelled. It is then observed that labelled O atom is found in water. Hence, we can conclude that

                        −OH group of acetic acid combines with H atom of methanol.

                        1. Isolation of reactive intermediates is a method to establish reaction mechanism.

                        Consider the conversion A → D.

                        It can proceed through following two different mechanisms.

                        A → B → D.

                        A → C → D.

                        Suppose that we are able to isolate intermediate B but we cannot isolate intermediate C. Hence, we can say that the reaction mechanism is A → B → D.

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                        MCQ 1531 Mark
                        Choose the correct order of stability of carbocation using the concept of hyperconjugation.
                        $\begin{matrix}{\ \ \ \ \ \ \ \ \ \ \ \ \text{CH}_3\\ \ \ \ \ \ \ \ \ \ \ \ \ \ |\\\text{CH}_3-\text{C}^\oplus\\ \ \ \ \ \ \ \ \ \ \ \ \ \ |\\\ \ \ \ \ \ \ \ \ \ \ \ \text{CH}_3} & { \ \ \ \ \ \ \ \ \ \ \ \ \text{CH}_3\\ \ \ \ \ \ \ \ \ \ \ \ \ \ |\\\text{CH}_3-\text{C}^\oplus\\ \ \ \ \ \ \ \ \ \ \ \ \ \ |\\ \ \ \ \ \ \ \ \ \ \ \ \ \text{H} } & {\text{CH}_3\stackrel{{\oplus} }{\hbox{CH}_2}} & \stackrel{{\oplus} }{\hbox{CH}_3} \\ \ \ \ \ \ \ \ \ \text{I} & \ \ \ \ \ \ \ \text{II} & \ \ \text{III} & \text{IV} \end{matrix}$
                        • A
                          I < II < III < IV
                        • B
                          IV < III < II < I
                        • C
                          III < IV < II < I
                        • D
                          None of these.
                        Answer
                        1. IV < III < II < I

                        Explanation:

                        Greater the number of alkyl groups attached to a positively charged carbon atom, the greater is the hyperconjugation interaction and stabilisation of the cation. Thus, we have the following relative stability of carbocation:

                        $ \ \ \ \ \ \ \ \ \ \ \ \ \text{CH}_3\\ \ \ \ \ \ \ \ \ \ \ \ \ \ | \\\text{CH}_3-\stackrel{{+} }{\hbox{C}}>(\text{CH}_3)_2\stackrel{{+ \ \ \ \ } }{\hbox{CH}}>\text{CH}_3\stackrel{{+ \ \ \ \ \ \ } }{\hbox{CH}_2}>\stackrel{{+ \ \ \ \ \ } }{\hbox{CH}}_3\\ \ \ \ \ \ \ \ \ \ \ \ \ \ |\\ \ \ \ \ \ \ \ \ \ \ \ \ \text{CH}_3$

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                        MCQ 1541 Mark
                        Which functional group is present in a molecule of CH3​Cl?
                        • A
                          Amine
                        • B
                          Halide
                        • C
                          Ether
                        • D
                          Ketone
                        Answer
                        1. Halide

                        Explanation:

                        Methyl chloride, CH3​Cl is an alkyl chloride. The functional group present is chlorine atom −Cl.

                        Note: In general we can write alkyl halides as R-X where X is halogen atom (Cl, Br, I or F).

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                        MCQ 1551 Mark
                        Dumas method is used for estimation of:
                        • A
                          Carbon
                        • B
                          Nitrogen
                        • C
                          Oxygen
                        • D
                          Sulphur
                        Answer
                        1. Nitrogen

                        Explanation:

                        The Dumas method in analytical chemistry is a method for the quantitative determination of nitrogen in chemical substances based on a method first described by Jean-Baptiste Dumas in 1826.

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                        MCQ 1561 Mark
                        The elements always present in organic compounds are C and H. In addition to these, name the other elements that may be present in it:
                        • A
                          Oxygen and nitrogen.
                        • B
                          Sulphur and halogens.
                        • C
                          Phosphorus.
                        • D
                          All of these.
                        Answer
                        1. All of these.

                        Explanation:

                        The elements present in organic compounds are C and H, other elements are O, NS, X (halogen).

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                        MCQ 1571 Mark
                        The product will be:

                        • A
                          Cis - 2 - butene
                        • B
                          Trans - 2 - butene
                        • C
                          2 - butyne
                        • D
                          Buta - 1 , 3 - diene
                        Answer
                        1. Trans - 2 - butene

                        Explanation:

                        Since elimination is trans-in nature so meso - compound will give generally trans - alkene as major.

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                        MCQ 1581 Mark
                        The dark turquoise-green solid copper (II) carbonate dissolves in hydrochloric acid to form a:
                        • A
                          Green-blue solution of copper (II) chloride.
                        • B
                          Red solution of copper (II) chloride.
                        • C
                          Green-blue solution of copper (II) chlorinate.
                        • D
                          Green-blue solution of carbon dioxide.
                        Answer
                        1. Green-blue solution of copper (II) chloride.

                        Explanation:

                        The dark turquoise-green solid copper (II) carbonate dissolves in hydrochloric acid to form a green-blue solution of copper (II) chloride and effervescence from carbon dioxide formed.

                        CuCO3 ​+ 2HCl → CuCl​+ H2​O + CO2

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                        MCQ 1591 Mark
                        Groups that can show both +M and -M effect:
                        • A
                          vinyl
                        • B
                          phenyl
                        • C
                          -NO
                        • D
                          All of these
                        Answer
                        1. All of these

                        Explanation:

                        CH2​ = CH− and phenyl both contain double bonds, Since we know -NO, C=C both can show +M and -M effect both hence all of the above options are correct.

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                        MCQ 1601 Mark
                        The increasing order of stability of the following free radicals:
                        • A
                           $(\text{C}_6\text{H}_5)_3\stackrel{\bf.}{\text{C}}<(\text{C}_6\text{H}_5)_2\stackrel{\bf.\ \ \ }{\text{CH}}<(\text{CH}_3)\stackrel{\bf.}{\text{C}}<(\text{CH}_3)_2\stackrel{\bf.\ \ \ }{\text{CH}}$
                        • B
                          $(\text{C}_6\text{H}_5)_2\stackrel{\bf.\ \ \ \ \ }{\text{CH}}<(\text{C}_6\text{H}_5)_3\stackrel{\bf.\ }{\text{C}}<\text{(CH}_3)_3\stackrel{\bf.}{\text{C}}<(\text{CH}_3)_2\stackrel{\bf.\ \ \ }{\text{CH}}$
                        • C
                          $(\text{CH}_3)_2\stackrel{\bf.\ \ \ \ }{\text{CH}}<(\text{CH}_3)_3\stackrel{\bf.}{\text{C}}<(\text{C}_6\text{H}_5)_3\stackrel{\bf.}{\text{C}}<(\text{C}_6\text{H}_5)_2\stackrel{\bf.\ \ \ \ \ }{\text{CH}}$
                        • D
                          $(\text{CH}_3)_2\stackrel{\bf.\ \ \ \ }{\text{CH}}<(\text{CH}_3)_3\stackrel{\bf.\ }{\text{C}}<(\text{C}_6\text{H}_5)_2\stackrel{\bf.\ \ \ \ \ }{\text{CH}}<(\text{C}_6\text{H}_5)_3\stackrel{\bf.}{\text{C}}$ 
                        Answer
                        1. $(\text{CH}_3)_2\stackrel{\bf.\ \ \ \ }{\text{CH}}<(\text{CH}_3)_3\stackrel{\bf.\ }{\text{C}}<(\text{C}_6\text{H}_5)_2\stackrel{\bf.\ \ \ \ \ }{\text{CH}}<(\text{C}_6\text{H}_5)_3\stackrel{\bf.}{\text{C}}$
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                        MCQ 1611 Mark
                        In paper chromatography, chromatography paper contains water trapped in it, which acts as the:
                        • A
                          Mobile phase.
                        • B
                          Stationary phase.
                        • C
                          Stationary medium.
                        • D
                          None of these.
                        Answer
                        1. Stationary phase.

                        Explanation:

                        In paper chromatography, chromatography paper contains water trapped in it, which acts as the stationary phase.

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                        MCQ 1621 Mark
                        What is not true about homologous series?
                        • A
                          All members have similar chemical properties.
                        • B
                          They have identical physical properties.
                        • C
                          They can be represented by a general formula.
                        • D
                          None of the above.
                        Answer
                        1. They have identical physical properties.

                        Explanation:

                        Homologous series differ by a CH2​ group. They have similar chemical properties as they have a common functional group.

                        They can be represented by a general formula for alkanes, alkenes, alkynes, and other functional groups.

                        However, as the molecular mass increases in a homologous series, a gradation in physical properties is observed. So, they do not have identical physical properties.

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                        MCQ 1631 Mark
                        The compound formed in the positive test for nitrogen and lassaigne's solution of an organic compound is:
                        • A
                          Fe4[Fe(CN)6]3
                        • B
                          Na[Fe(CN)6]
                        • C
                          Fe(CN)3
                        • D
                          Na4[Fe(CN)5NOS]
                        Answer
                        1. Fe4[Fe(CN)6]3

                        Explanation:

                        4Fe3+ + 3Na[Fe(CN)6] → Fe4[Fe(CN)6]3 + 12 Na+

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                        MCQ 1641 Mark
                        Glycerol is purified by:
                        • A
                          Vacuum distillation.
                        • B
                          Simple distillation.
                        • C
                          Steam distillation.
                        • D
                          Fractional distillation.
                        Answer
                        1. Vacuum distillation.

                        Explanation:

                        Because glycerol decomposes at its normal boiling point.

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                        MCQ 1651 Mark
                        Which one of these shows least -I effect:
                        • A
                          $−\text{OH}_2^+​$
                        • B
                          $−\text{NH}_3^-​$
                        • C
                          $−\text{NO}_2​$
                        • D
                          $−\text{F}$
                        Answer
                        1. $−\text{F}$

                        Explanation:

                        Since + charge is more electronegative than neutral atom therefore correct order is

                        $−\text{OH}_2^+​$​>$−\text{NH}_3^-​$>$−\text{NO}_2​$>$−\text{F}$. Hence least $−\text{I}$ effect is shown by $−\text{F}$.

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                        MCQ 1661 Mark
                        Which of the following best describe the given statement "an atom or group of atoms joined in a specific manner which is responsible for the characteristic chemical properties of the organic compounds?
                        • A
                          Heteroatom.
                        • B
                          Functional group.
                        • C
                          Aromatic group.
                        • D
                          Non-aromatic group.
                        Answer
                        1. Functional group.
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                        MCQ 1671 Mark
                        All members of a homologous series differ by a: 
                        • A
                          Methyl group
                        • B
                          Methylene group
                        • C
                          Methane group
                        • D
                          None of these
                        Answer
                        1. Methylene group

                        Explanation:

                        Homologous series differ by a CH2​ group known as the methylene group.

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                        MCQ 1681 Mark
                        What is qualitative analysis?
                        • A
                          Identification of unknown substance by direct observation.
                        • B
                          Identification of unknown substance by chemical tests.
                        • C
                          Both a & b
                        • D
                          None of the above
                        Answer
                        1. Identification of unknown substance by chemical tests.

                        Explanation:

                        Qualitative analysis includes identification of unknown substance by chemical tests.

                        Identification of substances by their physical and chemical properties.

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                        MCQ 1691 Mark
                        Chemical process that converts one member of a homologous series to the next member is called as:
                        • A
                          Homologation reaction
                        • B
                          Replacement reactions
                        • C
                          Homogolous reaction
                        • D
                          Decomposition reaction
                        Answer
                        1. Homologation reaction

                        Explanation:

                        A homologation reaction is a chemical process that converts one member of a homologous series to the next member.

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                        MCQ 1701 Mark
                        In Duma's method for estimation of nitrogen, 0.3g of an organic compound gave 50 mL of nitrogen collected at 300K temperature and 715 mm pressure. The percentage composition of nitrogen in the compound (Aqueous tension at 300K = 15mm) is:
                        • A
                          15.46%
                        • B
                          17.46%
                        • C
                          18.36%
                        • D
                          28%
                        Answer
                        1. 17.46%
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                        MCQ 1721 Mark
                        Note: Consider the following four compounds for answering questions:

                        1. $\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{O}\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ||\\\text{CH}_3-\text{CH}_2-\text{CH}_2-\text{CH}_2-\text{C}-\text{H}$

                        2. $\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{O}\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ||\\\text{CH}_3-\text{CH}_2-\text{CH}_2\text{CH}_2-\text{C}-\text{CH}_3$

                        3. $\text{CH}_3-\text{CH}_2-\text{C}-\text{CH}_2-\text{CH}_3\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ||\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{O}$

                        4. $\text{CH}_3-\text{CH}-\text{CH}_2-\text{C}-\text{H}\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ |\ \ \ \ \ \ \ \ \ \ \ ||\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{CH}_3\ \ \ \ \ \text{O}$

                        Which of the following pairs are not functional group isomers?

                          Answer
                          1. II and III
                          1. I and IV

                          Explanation:

                          Two or more compounds with same molecular formula but different functional groups are called functional isomers. In the given compounds.

                          I. Aldehydic group.

                          II. Ketonic group.

                          III. Ketonic group.

                          IV. Aldehydic group.

                          Hence II and III, I and IV are not functional isomers

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                          MCQ 1731 Mark
                          Ammonium nitrate on heating gives ___________.
                          • A
                            Nitric oxide
                          • B
                            Nitrogen
                          • C
                            Nitrogen peroxide
                          • D
                            Nitrous oxide
                          Answer
                          1. Nitrous oxide

                          Explanation:

                          NH4​NO3​ $\xrightarrow{\text{onheating}}$​ 2H2​O + N2​O (NItrous oxide).

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                          MCQ 1741 Mark
                          IUPAC name of ​$\text{CH}_3​−\text{O}−\text{CH}−\text{CH}_3 \\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ |\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{CH}_3$ is _____.
                          • A
                            3-Methoxy propane
                          • B
                            2-Methyl methoxy ethane
                          • C
                            1-Methoxy-1-methyl ethane
                          • D
                            Methoxy isopropane
                          Answer
                          1. Methoxy isopropane

                          Explanation:

                          The IUPAC name of the given compound is methoxy isopropane.

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                          MCQ 1751 Mark
                          Which method of estimation is shown in the figure given below?

                          • A
                            Carius method.
                          • B
                            Kjeldahl's method.
                          • C
                            Duma's method.
                          • D
                            None of these.
                          Answer
                          1. Carius method.
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                          MCQ 1761 Mark
                          Nucleophile is a species that should have:
                          • A
                            A pair of electrons to donate.
                          • B
                            Positive charge.
                          • C
                            Negative charge.
                          • D
                            Electron deficient species.
                          Answer
                          1. A pair of electrons to donate.
                          1. Negative charge.

                          Explanation:

                          Nucleophile (nucleus-loving) is a chemical species that donates an, electron pair to an electrophile (electron-loving). Hence, a nucleophile should have either a negative charge or an electron pair to donate.

                          Thus, options (a) r and (c) are correct.

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                          MCQ 1771 Mark
                          Note: Consider the following four compounds for answering questions:

                          1. $\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{O}\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ||\\\text{CH}_3-\text{CH}_2-\text{CH}_2-\text{CH}_2-\text{C}-\text{H}$

                          2. $\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{O}\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ||\\\text{CH}_3-\text{CH}_2-\text{CH}_2\text{CH}_2-\text{C}-\text{CH}_3$

                          3. $\text{CH}_3-\text{CH}_2-\text{C}-\text{CH}_2-\text{CH}_3\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ||\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{O}$

                          4. $\text{CH}_3-\text{CH}-\text{CH}_2-\text{C}-\text{H}\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ |\ \ \ \ \ \ \ \ \ \ \ ||\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{CH}_3\ \ \ \ \ \text{O}$

                          Which of the following pairs are position isomers?

                            Answer
                            1. II and III

                            Explanation:

                            In position isomerism, two or more compounds differ in the position of substituent, functional group or multiple bonds but molecular formula is same. In pentanone-2 and pentanone-3, position of ketonic group is different. CH3-CH2CH2-CO-CH3 and CH3-CH2-CO-CH2-CH3

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                            MCQ 1781 Mark
                            • A
                              Reaction 1
                            • B
                              Reaction 2
                            • C
                              Both have equal rates of reaction.
                            • D
                              Cannot be predicted.
                            Answer
                            1. Reaction 2

                            Explanation:

                            Since given reaction is SN2​ and we know that SN2​ favoured in presence of polar aprotic solvent like crown-ethers, DMSO, DMF, etc not H2​O which is polar protic solvent.

                            In reaction 2 polar solvent (crown ether) is used.

                            Thus reaction 2 is faster than reaction 1.

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                            MCQ 1801 Mark
                            What is the correct order of decreasing stability of the following cations.

                            1. $\ \ \ \ \ \ \ \ \ \ \ +\\\text{CH}_3-\text{CH}-\text{OCH}_3$

                            2. $\ \ \ \ \ \ \ \ \ \ \ +\\\text{CH}_3-\text{CH}-\text{OCH}_3$

                            3. $\ \ \ \ \ \ \ \ \ \ \ +\\\text{CH}_3-\text{CH}-\text{CH}_2-\text{OCH}_3$

                              Answer
                              1. II > I > III

                              Explanation:

                              Thus, the stability of carbocation decreasws in the order: II > I >III.

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                              MCQ 1811 Mark
                              In which of the following representations given below spatial arrangement of group/ atom different from that given in structure ‘A’?

                              • A
                                 
                              • B
                                 
                              • C
                                 
                              • D
                                 
                              Answer

                              Explanation:

                              The spatial arrangement of groups or atoms can be checked by doing two interchange and bringing H below the plane of the paper. Find out the sequence of the remaining groups in a particular order whether clockwise or anticlockwise starting from atom with highest atomic number to atom with lower atomic numbers. Hence, option (ii) has same spatial arrangement as (A) while in rest three is different.

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                              MCQ 1821 Mark
                              Consider the following compounds

                              1.  

                              1.  

                              1.  

                              Hyper conjugation occurs in:

                                Answer
                                1. III

                                Explanation:

                                Because it has 'H' in conjugation with free radical.

                                OR

                                Hyperconjugation occurs through the H- atoms present on the carbon atom next to the double bond i.e alpha hydrogen atoms. There is no alpha -H in the structure I and II. So, hyperconjugation occurs in structure III only ie.

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                                MCQ 1871 Mark
                                The IUPAC name of iso-octane is:
                                • A
                                  2, 2-dimethylpentane.
                                • B
                                  2, 3-dimethylpentane.
                                • C
                                  2, 3, 3-trimethylpentane.
                                • D
                                  2, 2, 4-trimethylpentane.
                                Answer
                                1. 2, 2, 4-trimethylpentane.
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                                MCQ 1881 Mark
                                For the purification, isolation and separation of organic compounds, the latest technique is:
                                • A
                                  Chromatography.
                                • B
                                  Steam distillation.
                                • C
                                  Fractional crystallization.
                                • D
                                  Sublimation.
                                Answer
                                1. Chromatography.
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