- ACH3CHBrCH2CH3
- B
- C
- DAll of the above
- All of the above
188 questions · self-marked practice — reveal the answer and mark yourself.
Explanation:
Both HCHO and HCOOH react with AgNO3/NH4OH
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 + NaHCO3 $\rightarrow$ HCOONa + H2O + CO2 $\uparrow$
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.
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.
| | %C | %Н |
| (a) | 21.95 | 4.58 |
| (b) | 21.95 | 5.58 |
| (c) | 11.95 | 5.58 |
| (d) | 11.95 | 4.58 |
| | %C | %Н |
| (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
$= \frac{2\times 0.1014 \times100}{18 \times 0.246}= 4.58\%$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.
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.
Explanation:
The common name of ethanol is ethyl alcohol.


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.
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.
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.
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.
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.

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.




Explanation:
The correct bond-line formula for

Explanation:
Nucleophiles are the electron rich species.
Nitrene is a nucleophile due to the lone pair electrons of Nitrogen.
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.
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.
$4\text{FeCl}_3+3\text{Na}_4[\text{Fe(CN)}_6]\rightarrow\text{Fe}_4[\text{Fe(CN)}_6]_3+12\text{NaCl}$
Explanation:
It is least stable because $\text{C}^\ominus$ is attached to three electron releasing CH3 groups.
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.
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.
Explanation:
Cu2+ and Hg2+ lie in IInd group of qualitative analysis H2S in acid medium is reagent for it.
HCl $\rightarrow$ H+ + Cl−
H2S $\leftrightharpoons$ 2H+ + S2−
Explanation:
Alkanes have the general formula CnH2n+2.
Thus, C3H4 does not belong to homologous series of alkanes.

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.


Explanation:
+R-effect of OH > -CH3
Hence, 2-methyl phenol is more reattive towards electrophilic reagent.
Explanation:
-COOH is given more preference over keto, prefix for ketone group is oxo.
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 $\text{HO(CH}_2)_3\text{CHCH}_3\text{CH(CH}_3)_2$
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.
Explanation:
$\text{Na}_2\text{S}+\text{Na}_2\big[\text{Fe}(\text{CN})_5\text{NO}\big]\xrightarrow{\ \ \ \ }\text{Na}_4\big[\text{Fe}(\text{CN})_5\text{NOS}\big]\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{Violet coloour}$

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.
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 C2H6 are homologous series. Its molar mass is 16 and 30gm/mole respectively.
Hence, it is differed by unit 14.
Explanation:
Electronegativity increases as the state of hybridization changes from sp3 to sp2 and sp2 to sp. Thus, sp hybridized carbon has the highest electronegativity.
Explanation:
Acetaldehyde i.e CH3CHO and propionaldehyde i.e CH3CH2CHO belong from the aldehyde family hence they are homologous series.
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.
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
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 CuSO4.5H2O (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.
Explanation:
Color of the substance is the qualitative property. Qualitative properties are the properties which can be observed without making any measurements or calculations.
Explanation:
Since CdS belongs to IIA group and As2S3 belongs to IIB group, they can be separated with yellow ammonium sulphide as only IIB group sulphides dissolve in yellow ammoium sulphide.
Explanation:
$\frac{\text{P}_1\text{V}_1}{\text{T}_1}=\frac{\text{P}_2\text{V}_2}{\text{T}_2}\Rightarrow \frac{40\times760}{300}=\frac{\text{V}_2\times760}{273}$
$\Rightarrow\text{V}_2=\frac{28\times273}{76\times3}$ $[\text{P}=725-25=700\text{mm}]$
$\text{V}_2=\frac{7644}{228}=33.53\text{mL}$
$\%\text{ of }\text{N}=\frac{\text{V}_2}{8\text{w}}=\frac{33.53}{8\times0.25}=16.76\%$
Explanation:
Two types of adsorption chromatography techniques are column chromatography and thin layer chromatography.
Explanation:
$\ \ \ \ \ \ \ \ \text{sp}\ \ \ \ \text{sp}\ \ \ \ \text{sp}\ \ \ \ \text{sp}\\\text{H}-\text{C}\equiv\text{C}-\text{C}\equiv\text{C}-\text{H}$
$\text{sp}^3\ \ \ \ \ \ \ \text{sp}\ \ \ \ \text{sp}\ \ \ \ \text{sp}^3\\\text{CH}_3-\text{C}\equiv\text{C}-\text{CH}_3$
$\text{sp}^2\ \ \ \ \ \ \ \text{sp}\ \ \ \ \ \text{sp}^2\\\text{CH}_2=\text{C}=\text{CH}_2$
$\text{sp}^2\ \ \ \ \ \ \ \text{sp}^2\ \ \ \ \ \ \text{sp}^2\ \ \ \ \ \text{sp}^2\\\text{CH}_2=\text{CH}-\text{CH}=\text{CH}_2$
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: CH3OH,C2H5OH,C3H7OH,C4H9OH.
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.
Explanation:
1-heptyne is alkyne. The general formula for alkyne is CnH2n−2.
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.
Explanation:
Carbonyl group is −C = O group, which present in aldehydes and ketones but not present in alcohols.
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.
$\ \ \ \ \ \ \ \ \text{H}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{H}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{H}\\ \ \ \ \ \ \ \ \ \ | \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{H}^\oplus \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ |\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ |\\\text{H}-\text{C}-\text{CH}=\text{CH}_2\leftrightarrow\text{H}-\text{C}=\text{CH}-\stackrel{{\ominus}}{\hbox{C}\text{}}\text{H}_2\leftrightarrow\text{H}^\oplus\text{C}=\text{CH}-\stackrel{{\ominus}}{\hbox{C}}\text{H}_2\leftrightarrow\text{H}-\text{C}-\text{CH}-\stackrel{{\ominus}}{\hbox{C}}\text{H}_2\\\ \ \ \ \ \ \ \ \ |\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ |\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ |\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{H}^\oplus\\\ \ \ \ \ \ \ \ \text{H}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{H}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{H}$
Explanation:
Structural isomers of C3H9N are

Explanation:
CH3CH2CH2CH3 (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.
Explanation:
Because it does not have carbon.

Explanation:
Among a, b and c electron donating ability order is −OCH3 > −CH3 > −CH2OCH3 more the electron donating ability of substituent more is the stability of carbocation.

Explanation:
Molecular formula of the compound is C6H14. 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.
Explanation:
Esters have a fruity smell due to their volatile nature caused by their chemical composition and conformations.
Explanation:
Tube in which a known mass of an organic compound is heated for the quantitative analysis of S is known as Carius tube.
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.
Explanation:
Petroleum and coaltar are the main sources of organic compounds.
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.
Explanation:
(d) is the correct answer as the group associated with it is carboxylic acid group.
Explanation:
A homologous series is a family of hydrocarbons with similar chemical properties that share the same general formula.
Both given compound HCOOCH3 and CH3COOCH3 are homologous series of a simple ester group RCOOH, where a successive methyl group −CH3 added to previous one.
Explanation:
Amides > Nitriles > Aldehydes > ketones
Explanation:
CH3CHO − Acetaldehyde
No of atoms of H:O:C = 4:1:2
Explanation:
Due to common ion effect, the ionisation of H2S is suppressed and at low concentration of S2- ions, the second group sulphides are precipitated.

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.





Explanation:
It is aromatic $\because$ it has $(4\text{n}+2)\pi,$ i.e., $6\pi$ electrons which are delocalised.
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.
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.
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.
Explanation:
The functional group present in CH3−O−CH2CH3 is an ether group −O−.
Note: In general, ethers can be represented as R−O−R′.
Explanation:
Dimethyl ether is CH3OCH3.
It's IUPAC name is methoxymethane.
Explanation:
$\text{cl}-\text{cl}\xrightarrow{\ \ \text{hv}\ \ \ }2\text{c}\dot{\text{l}}$
Explanation:
From the image we can see that there is only 1 pi-bond present in phosgene (cabonyl chrolide).


Explanation:
Arrow denotes the direction of movement of electrons

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

B → Organic compound in solvent layer.

Explanation:
Aldehyde group is preferred over keto group.

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

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.

Explanation:

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

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}$

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.
Explanation:
Essential oils are insoluble in water, soluble in steam and have high vapour pressure. Therefore, they can be separated by steam distillation.
Explanation:
Alkyl halides do not show functional isomerism. Alcohols and ethers; aldehydes and ketones; cyanides and isocyanides are functional isomers.
Explanation:
The functional group present every organic acid is −COOH as it dissociate in water as −COO− and H+.
Explanation:
Glycerol decomposes at its boiling point is separated from spent-lye by using distillation under reduced pressure.

Explanation:
Generally transition state tries to bring stability with less energy difference. From the figure it is evident that energy difference between TS2 and intermediate 2 is least.
So it is most likely that TS2 and intermediate 2 have similar structures.
Explanation:
−OH represents an alcoholic group. It represents alcohols such as ethanol CH3, −CH2, −OH. The general formula of alcohols is R−OH.
Explanation:
Ethane have a higher boiling point than methane because molecules of ethane (C2H6) 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.
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)
$\text{C}+ 2\text{CuO} \xrightarrow[]{\Delta}2\text{Cu + CO}_{2}\uparrow$
$\text{2H} + \text{CuO}\xrightarrow[]{\Delta} \text{Cu} + \text{H}_{2}\text{O}$
$\text{CO}_{2} + \text{Ca}(\text{OH})_{2} \xrightarrow{\ \ \ \ \ }\text{CaCO}_{3} \downarrow + \text{H}_{2}\text{O}\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{White turbidity}$
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.
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.
Explanation:
$\pi$ bond is formed by lateral (side ways) overlap of either p−p, d−d or d−p orbitals.
Explanation:
Methane and ethane are alkanes. These are saturated carbon compounds. They have the general formula CnH2n+2.
All the alkanes differ by CH2 group successively in the series.
Explanation:
Preceeeding homologue is C9H20.
Succeeding homologue is C11H24.
Explanation:
Members of Homologous series have same chemical properties.

Explanation:
Structure of chlorocyclohexane is:

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 CuSO4.5H2O (by water vapor) then the compound contains hydrogen. If lime water is turned milky by CO2, then the compound contains carbon.
Explanation:
As %s character increases, electronegativity of orbitals increases which means its tendency to pull electron towards itself also increases hence -I effect increases.
Explanation:
A sigma bond is a single bond formed by linear combination of hybridised orbitals.
Explanation:

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.

Explanation:
Marsh gas is methane, CH4.
It is called so because it is often obtained from marshy places.
Explanation:
General formula of homologous series is CnH2n+2.
First member is methane i.e. CH4
Second member is ethane i.e. C2H6
Third member is propane i.e. C3H8
Forth member is butane i.e. C4H10
Similarly, fifth member is pentane i.e
Explanation:
A homologation reaction is a chemical process that converts one member of a homologous series to the next member of the homologous series.
Explanation:
Compounds within a homologous series typically have a fixed set of functional groups that gives them similar chemical and physical properties.


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.

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,
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.
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.

Explanation:
Since, the given compounds contain benzene nucleus so, naphthalene is a benzenoid aromatic compound.
Explanation:
Phenoxide is stabilized by resonance.
Explanation:
Formation of sodium thionitroprusside (purple) shows the presence of sulphur in organic compound.
$\text{Na}_{2}\text{S} \ \ \ \ + \ \ \ \ \ \text{Na}_{2}[\text{Fe}(\text{NO})(\text{CN})_{5}] \xrightarrow{\ \ \ \ \ \ }\text{Na}_{4}[\text{Fe}(\text{CN})_{5}\text{NOS}]\\\text{Sodium extract}$
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 $\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{CH}_3\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ |\\\text{CH}_3-\text{CH}_2-\text{CH}_2-\text{C}-\text{CH}-\text{CH}_2-\text{CH_3}\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ |\ \ \ \ \ \ \ |\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{CH}\ \ \text{C}_2\text{H}_5$
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.
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 CnH2n+2 and for alkenes it is CnH2n etc.
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.
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.
Explanation:
Cyanogen, $\text{N}\equiv\text{C}-\text{C}\equiv\text{N}$has three sigma and four 'pi' bonds.

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

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.
Explanation:
Acidified sodium fusion extract on addition of ferric chloride solution gives blood red colouration, which confirms the presence of N and S both.
$\text{Na} + \text{C} + \text{N} + \text{S} \xrightarrow{\ \ \ \ \ \ } \text{NaSCN}$
$\text{Fe}^{+3} + \text{SCN}^{-} \xrightarrow{\ \ \ \ \ \ } [\text{Fe}(\text{SCN})]^{2+}\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{Blood red}$
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)

Explanation:
Methane, ethane, and propane-CH4, C2H6, C3H8−CnH2n+2 belong to alkane family with same general formula except C4H6

Explanation:
The most common characteristic of homologous series is increase in the number of methylene (−CH2−) units in saturated chain within the molecule.
Explanation:
In CH3—CH2—Cl, the carbon marked with asterisk has the greatest positive charge due to high electronegativity of Cl.
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.
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: CH3OH, C2H5OH, C3H7OH.
$\text{AlCl}_3,\ \text{SO}_3,\ \text{NO}_2^+$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ +\\\text{NO}_2^+,\ \text{CH}_3^+,\ \text{CH}_3-\text{C}=0$
Explanation:
In (a), NH3 and H2O are nucleophiles but BF3 is an electrophile. in (d), $\text{C}_2\text{H}^-_5$ are electrophiles.
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.
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.
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.
Explanation:
Both a and b are correct.
CH3−CO−O*H + CH3−OH $\xrightarrow{\text{H}^+}$ CH3−COO−CH3 + H2O*
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.
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.
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$
Explanation:
Methyl chloride, CH3Cl 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).
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.
Explanation:
The elements present in organic compounds are C and H, other elements are O, NS, X (halogen).

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

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 → CuCl2 + H2O + CO2
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.
Explanation:
In paper chromatography, chromatography paper contains water trapped in it, which acts as the stationary phase.
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.
Explanation:
4Fe3+ + 3Na[Fe(CN)6] → Fe4[Fe(CN)6]3 + 12 Na+
Explanation:
Because glycerol decomposes at its normal boiling point.
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 $−\text{F}$.
Explanation:
Homologous series differ by a CH2 group known as the methylene group.
Explanation:
Qualitative analysis includes identification of unknown substance by chemical tests.
Identification of substances by their physical and chemical properties.
Explanation:
A homologation reaction is a chemical process that converts one member of a homologous series to the next member.
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{O}\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ||\\\text{CH}_3-\text{CH}_2-\text{CH}_2-\text{CH}_2-\text{C}-\text{H}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{O}\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ||\\\text{CH}_3-\text{CH}_2-\text{CH}_2\text{CH}_2-\text{C}-\text{CH}_3$
$\text{CH}_3-\text{CH}_2-\text{C}-\text{CH}_2-\text{CH}_3\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ||\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{O}$
$\text{CH}_3-\text{CH}-\text{CH}_2-\text{C}-\text{H}\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ |\ \ \ \ \ \ \ \ \ \ \ ||\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{CH}_3\ \ \ \ \ \text{O}$
Which of the following pairs are not functional group isomers?
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
Explanation:
NH4NO3 $\xrightarrow{\text{onheating}}$ 2H2O + N2O (NItrous oxide).
Explanation:
The IUPAC name of the given compound is methoxy isopropane.

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.
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{O}\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ||\\\text{CH}_3-\text{CH}_2-\text{CH}_2-\text{CH}_2-\text{C}-\text{H}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{O}\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ||\\\text{CH}_3-\text{CH}_2-\text{CH}_2\text{CH}_2-\text{C}-\text{CH}_3$
$\text{CH}_3-\text{CH}_2-\text{C}-\text{CH}_2-\text{CH}_3\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ||\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{O}$
$\text{CH}_3-\text{CH}-\text{CH}_2-\text{C}-\text{H}\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ |\ \ \ \ \ \ \ \ \ \ \ ||\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{CH}_3\ \ \ \ \ \text{O}$
Which of the following pairs are position isomers?
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
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 H2O which is polar protic solvent.
In reaction 2 polar solvent (crown ether) is used.
Thus reaction 2 is faster than reaction 1.
Explanation:
The correct IUPAC name of the following structure is

$\ \ \ \ \ \ \ \ \ \ \ +\\\text{CH}_3-\text{CH}-\text{OCH}_3$
$\ \ \ \ \ \ \ \ \ \ \ +\\\text{CH}_3-\text{CH}-\text{OCH}_3$
$\ \ \ \ \ \ \ \ \ \ \ +\\\text{CH}_3-\text{CH}-\text{CH}_2-\text{OCH}_3$
Explanation:

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





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.



Hyper conjugation occurs in:
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.

Explanation:
Since double bond is a source of electrons and the charge flows from source of more electron density, therefore, n electrons of the double bond attack the proton.

Explanation:
Box a encloses ester group, b encloses ketone group while c encloses amide group.
Here ether group is not present.

Explanation:
Benzyl Carbocation is most stable due to five resonating structures, I (Allyl carbocation is more stable than II (n-propyl carbocation) due to resonance.