Question 13 Marks
Write bond line formulas for: Isopropyl alcohol, 2,3-Dimethylbutanal, Heptan-4- one.
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Question 23 Marks
Which of the two: O2NCH2CH2O- or CH3CH2O- is expected to be more stable and why?
Answer
View full question & answer→NO2 group is an electron-withdrawing group. Hence, it shows –I effect. By withdrawing the electrons toward it, the NO2 group decreases the negative charge on the compound, thereby stabilising it. On the other hand, ethyl group is an electron-releasing group. Hence, the ethyl group shows +I effect. This increases the negative charge on the compound, thereby destabilising it. Hence, O2NCH2CH2O- is expected to be more stable than CH3CH2O-.
Question 33 Marks
For the following bond cleavages, use curved-arrows to show the electron flow and classify each as homolysis or heterolysis. Identify reactive intermediate produced as free radical, carbocation and carbanion.
Answer
View full question & answer→The bond cleavage using curved-arrows to show the electron flow of the given reaction can be represented as
It is a heterolytic cleavage as the bond breaks in such a manner that the shared pair of electrons remains with one of the fragments. The intermediate formed is a carbocation.
It is a heterolytic cleavage as the bond breaks in such a manner that the shared pair of electrons remains with one of the fragments. The intermediate formed is a carbocation.
Question 43 Marks
For the following bond cleavages, use curved-arrows to show the electron flow and classify each as homolysis or heterolysis. Identify reactive intermediate produced as free radical, carbocation and carbanion.
Answer
View full question & answer→The bond cleavage using curved-arrows to show the electron flow of the given reaction can be represented as
It is an example of heterolytic cleavage as the bond breaks in such a manner that the shared pair of electrons remains with the bromine ion. The reaction intermediate formed is a carbocation.
It is an example of heterolytic cleavage as the bond breaks in such a manner that the shared pair of electrons remains with the bromine ion. The reaction intermediate formed is a carbocation.
Question 53 Marks
What is the relationship between the members of following pairs of structures? Are they structural or geometrical isomers or resonance contributors?
Answer
View full question & answer→Compounds having the same molecular formula, the same constitution, and the sequence of covalent bonds, but with different relative position of their atoms in space are called geometrical isomers.
In structures I and II, the relative position of Deuterium (D) and hydrogen (H) in space are different. Hence, the given pairs represent geometrical isomers.
In structures I and II, the relative position of Deuterium (D) and hydrogen (H) in space are different. Hence, the given pairs represent geometrical isomers.
Question 63 Marks
Why is nitric acid added to sodium extract before adding silver nitrate for testing halogens?
Answer
View full question & answer→While testing the Lassaigne’s extract for the presence of halogens, it is first boiled with dilute nitric acid. This is done to decompose NaCN to HCN and Na2S to H2S and to expel these gases. That is, if any nitrogen and sulphur are present in the form of NaCN and Na2S, then they are removed. The chemical equations involved in the reaction are represented as
NaCN + HNO3 → NaNO3 + HCN
Na2S + 2HNO3 → 2NaNO3 + H2S
NaCN + HNO3 → NaNO3 + HCN
Na2S + 2HNO3 → 2NaNO3 + H2S
Question 73 Marks
What is the relationship between the members of following pairs of structures? Are they structural or geometrical isomers or resonance contributors?
Answer
View full question & answer→The given structures are canonical structures or contributing structures. They are hypothetical and individually do not represent any real molecule. Hence, the given pair represents resonance structures, called resonance isomers.
Question 83 Marks
For the following bond cleavages, use curved-arrows to show the electron flow and classify each as homolysis or heterolysis. Identify reactive intermediate produced as free radical, carbocation and carbanion.
Answer
View full question & answer→The bond cleavage using curved-arrows to show the electron flow of the given reaction can be represented as
It is an example of heterolytic cleavage as the bond breaks in such a manner that the shared pair of electrons remains with the carbon of propanone. The reaction intermediate formed is carbanion.
It is an example of heterolytic cleavage as the bond breaks in such a manner that the shared pair of electrons remains with the carbon of propanone. The reaction intermediate formed is carbanion.
Question 93 Marks
What is the relationship between the members of following pairs of structures? Are they structural or geometrical isomers or resonance contributors?
Answer
View full question & answer→Compounds having the same molecular formula but with different structures are called structural isomers. The given compounds have the same molecular formula but they differ in the position of the functional group (ketone group).
In structure I, ketone group is at the C-3 of the parent chain (hexane chain) and in structure II, ketone group is at the C-2 of the parent chain (hexane chain). Hence, the given pair represents structural isomers.
In structure I, ketone group is at the C-3 of the parent chain (hexane chain) and in structure II, ketone group is at the C-2 of the parent chain (hexane chain). Hence, the given pair represents structural isomers.
Question 103 Marks
For the following bond cleavages, use curved-arrows to show the electron flow and classify each as homolysis or heterolysis. Identify reactive intermediate produced as free radical, carbocation and carbanion.
Answer
View full question & answer→The bond cleavage using curved-arrows to show the electron flow of the given reaction can be represented as
It is an example of homolytic cleavage as one of the shared pair in a covalent bond goes with the bonded atom. The reaction intermediate formed is a free radical.
It is an example of homolytic cleavage as one of the shared pair in a covalent bond goes with the bonded atom. The reaction intermediate formed is a free radical.
Question 113 Marks
What are electrophiles and nucleophiles? Explain with examples.
Answer
View full question & answer→Electrophiles: The name electrophiles means electron loving. Electrophiles are electron deficient. They may be positive ions or neutral molecules. Ex: H+, Cl+, Br+, NO2+, R3C+, RN2+, AlCl3, BF3 Nucleophiles: The name nucleophiles means ‘nucleus loving’ and indicates that it attacks the region of low electron density (positive centres) in a substrate molecule. They are electron rich they may be negative ions or neutral molecules.
Ex: Cl- Br-, CN-, OH-, RCR2-, NH3, RNH2, H2O, ROH etc.
Question 123 Marks
Indicate the $\sigma$ and $\pi$ bonds in the following molecules:
C6H12,
C6H12,
Answer
View full question & answer→The $\sigma$ and $\pi$ bonds are indicated below:
Question 133 Marks
0.3780g of an organic chloro compound gave 0.5740g of silver chloride in Carius estimation. Calculate the percentage of chlorine present in the compound.
Answer
View full question & answer→Mass of the compound = 0.3780g
Mass of silver chloride = 0.5740g
$\text{Percentage of chlorine}=\frac{35.5}{143.5}\times\frac{\text{Mass of siliver chloride}}{\text{Mass of compound}}\times100$
$=\frac{35.5}{143.5}\times\frac{(0.5740\text{g})}{(0.3780\text{g})}\times100=37.57\text{g}$
Mass of silver chloride = 0.5740g
$\text{Percentage of chlorine}=\frac{35.5}{143.5}\times\frac{\text{Mass of siliver chloride}}{\text{Mass of compound}}\times100$
$=\frac{35.5}{143.5}\times\frac{(0.5740\text{g})}{(0.3780\text{g})}\times100=37.57\text{g}$
Question 143 Marks
Name a suitable technique of separation of the components from a mixture of calcium sulphate and camphor.
Answer
View full question & answer→The process of sublimation is used to separate a mixture of camphor and calcium sulphate. In this process, the sublimable compound changes from solid to vapour state without passing through the liquid state. Camphor is a sublimable compound and calcium sulphate is a non-sublimable solid. Hence, on heating, camphor will sublime while calcium sulphate will be left behind.
Question 153 Marks
In DNA and RNA, nitrogen atom is present in the ring system. Can Kjeldahl method be used for the estimation of nitrogen present in these? Give reasons.
Answer
View full question & answer→DNA and RNA have nitrogen in the heterocyclic rings. Nitrogen present in rings, azo groups and nitro groups cannot be removed as ammonia. So, Kjeldahl’s method cannot be used for the estimation of nitrogen present in these.
Question 163 Marks
“Stability of carbocations depends upon the electron releasing inductive effect of groups adjacent to positively charged carbon atom involvement of neighbouring groups in hyperconjugation and resonance.”
The structure of triphenylmethyl cation is given below. This is very stable and some of its salts can be stored for months. Explain the cause of high stability of this cation.
The structure of triphenylmethyl cation is given below. This is very stable and some of its salts can be stored for months. Explain the cause of high stability of this cation.
Answer
View full question & answer→Triphenylmethyl cation is very stable because +ve charge of methyl carbon is delocalized in three phenyl rings. In each phenyl ring, +ve charge is developed on 2 ortho position and para position, i.e. three resonating structures. Total resonating structures given by triphenylmethyl cation are nine. Hence, it is very stable. These structures can be shown as.
Question 173 Marks
Write down the formulae of the first four members of each homologous series beginning with the following compounds.
- $\text{CH}_2=\text{CH}_2$
- $\text{HCOOH}$
- $\text{CH}_3\text{COCH}_3$
- $\text{CH}_3\text{OH}$
- $\text{HC}\equiv\text{CH}$
Answer
View full question & answer→- $\text{CH}_2=\text{CH}_2,\ \text{CH}_3\text{CH}=\text{CH}_2,$
$\text{CH}_3\text{CH}_2\text{CH}=\text{CH}_2,\ \text{CH}_3\text{CH}_2\text{CH}_2\text{CH}=\text{CH}_2$
- $\text{HCOOH},\text{ CH}_3\text{COOH},\ \text{CH}_3\text{CH}_2\text{COOH},$
$\text{CH}_3\text{CH}_2\text{CH}_2\text{COOH}$
- $\text{CH}_3\text{COCH}_3,\text{ CH}_3\text{CH}_2\text{COCH}_3,\text{CH}_3\text{CH}_2\text{CH}_2\text{COCH}_3,$
$\text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_2\text{COCH}_3$
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$\text{CH}_3\text{OH},\text{CH}_3\text{CH}_2\text{OH},\text{CH}_3\text{CH}_2\text{CH}_2\text{OH},\\\text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_2\text{OH}$
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$\text{HC}\equiv\text{CH},\text{ CH}_3\text{C}\equiv\text{CH, CH}_3\text{CH}_2\text{C}\equiv\text{CH}\\\text{CH}_3\text{CH}_2\text{CH}_2\text{C}\equiv\text{CH}$
Question 183 Marks
Which is more stable and why?
- $\text{C}_6\text{H}_5-\stackrel{\oplus\ \ \ \ }{\text{CH}_2}\text{ or }\text{C}_6\text{H}_{11}\stackrel{\oplus\ \ \ \ }{\text{CH}_2}$
- $(\text{C}_6\text{H}_5)\stackrel{\bf.\ \ \ }{\text{CH}}\text{ or }\text{C}_6\text{H}_5\stackrel{\bf.\ \ \ \ \ }{\text{CH}_2}$
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$(\text{C}_6\text{H}_5)\stackrel{\bf.\ \ \ }{\text{CH}}_2\text{ or }\text{CH}_2=\text{CH}-\stackrel{\bf.\ \ \ \ \ }{\text{CH}_2}$
Answer
View full question & answer→-
$\text{C}_6\text{H}_5\stackrel{\oplus\ \ \ \ }{\text{CH}_2}$ is more stable due to resonance (5 resonating structures).
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(C6H5)CH is more stable due to more resonating structures.
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C6H5CH2 is more stable due to 5 resonating structures whereas CH2=CH-CH2 has two resonating structures.
Question 193 Marks
Write structural formulae for all the isomeric amines with molecular formula C4H11N.
Answer
View full question & answer→-
$\text{CH}_3-\text{CH}_2-\text{CH}_2-\text{CH}_2-\text{NH}_2$
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$\text{CH}_3-\text{CH}_2-\text{CH}-\text{CH}_3\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ |\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{NH}_2$
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$\text{CH}_3-\text{CH}-\text{CH}_2-\text{NH}_2\\ \ \ \ \ \ \ \ \ \ \ \ \ \ |\\ \ \ \ \ \ \ \ \ \ \ \ \ \text{CH}_3$
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$ \ \ \ \ \ \ \ \ \ \ \ \ \text{CH}_3\\ \ \ \ \ \ \ \ \ \ \ \ \ \ |\\\text{CH}_3-\text{C}-\text{NH}_2\\ \ \ \ \ \ \ \ \ \ \ \ \ \ |\\ \ \ \ \ \ \ \ \ \ \ \ \ \text{CH}_3$
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$\text{CH}_3-\text{CH}_2-\text{CH}_2-\text{NH}-\text{CH}_3$
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$\text{CH}_3-\text{CH}-\text{NH}-\text{CH}_3\\ \ \ \ \ \ \ \ \ \ \ \ \ \ |\\\ \ \ \ \ \ \ \ \ \ \ \ \text{CH}_3$
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$\text{CH}_3-\text{CH}_2-\text{NH}-\text{CH}_2-\text{CH}_3$
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$\text{CH}_3-\text{CH}_2-\text{N}-\text{CH}_3\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ |\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{CH}_3$
Question 203 Marks
Note: Consider structures I to VII and answer the questions:
- $\text{CH}_3-\text{CH}_2-\text{CH}_2-\text{CH}_2-\text{OH}$
-
$\text{CH}_3-\text{CH}_2-\text{CH}-\text{CH}_3\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ |\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{OH}$
-
$\ \ \ \ \ \ \ \ \ \ \ \ \text{CH}_3\\\ \ \ \ \ \ \ \ \ \ \ \ \ |\\\text{CH}_3-\text{C}-\text{CH}_3\\\ \ \ \ \ \ \ \ \ \ \ \ \ |\\\ \ \ \ \ \ \ \ \ \ \ \ \text{OH}$
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$\text{CH}_3-\text{CH}-\text{CH}_2-\text{OH}\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ |\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{CH}_3$
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$\text{CH}_3-\text{CH}_2-\text{O}-\text{CH}_2-\text{CH}_3$
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$\text{CH}_3-\text{O}-\text{CH}_2-\text{CH}_2-\text{CH}_3$
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$\text{CH}_3-\text{O}-\text{CH}-\text{CH}_3\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ |\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{CH}_3$
Answer
View full question & answer→When two are more compounds have similar molecular formula but differnt carbon skeletone, these are referred to as chain as isomers and the phenomenon is termed as chain isomerism.
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$\text{CH}_3-\text{CH}_2-\text{CH}_2-\text{CH}_2-\text{OH}$
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$\text{CH}_3-\text{CH}_2-\text{CH}-\text{CH}_3\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ |\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{OH}$ are position isomers.
-
$\ \ \ \ \ \ \ \ \ \ \ \ \text{CH}_3\\\ \ \ \ \ \ \ \ \ \ \ \ \ |\\\text{CH}_3-\text{C}-\text{CH}_3\\\ \ \ \ \ \ \ \ \ \ \ \ \ |\\\ \ \ \ \ \ \ \ \ \ \ \ \text{OH}$
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$\text{CH}_3-\text{CH}-\text{CH}_2-\text{OH}\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ |\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{CH}_3$ are position isomers.
Question 213 Marks
Identify the type of reactions:
- $\text{CH}_4+\text{Cl}_2\xrightarrow{\text{sunlight}}\text{CH}_3\text{Cl}+\text{HCl}$
- $\text{C}_6\text{H}_6+(\text{conc.})\text{HNO}_36\xrightarrow{\text{conc.}\text{H}_2\text{SO}_4}\text{C}_6\text{H}_5\text{NO}_2+\text{H}_2\text{O}$
- $ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{Br}\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ |\\\text{CH}_3-\text{CH}=\text{CH}_2+\text{HBr}\xrightarrow{\ \ \ \ \ \ \ }\text{CH}_3-\text{CH}-\text{CH}_3$
- $\ \ \ \ \ \ \ \ \ \ \ \ \ \text{O}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{OH}\\\ \ \ \ \ \ \ \ \ \ \ \ \ \| \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ |\\ \text{CH}_3-\text{C}-\text{CH}_3+\text{HCN}\xrightarrow{\ \ \ \ \ \ \ \ }\text{CH}_3-\text{C}-\text{CH}_3\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ |\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{CN}$
- $\text{CH}_3-\text{CH}=\text{CH}_2\xrightarrow[\text{peroxide}]{\text{HBr}}\text{CH}_3\text{CH}_2\text{CH}_2\text{Br}$
- $\text{CH}_3\text{Cl}+\text{KOH}(\text{aq})\xrightarrow{\ \ \ \ \ \ \ \ \ }\text{CH}_3\text{OH}+\text{KCl}$
Answer
View full question & answer→- Free radical substitution.
- Electrophilic substitution reaction.
- Electrophilic addition reaction.
- Nucleophilic addition reaction.
- Free radical addition.
- Nucleophilic substitution reactions.
Question 223 Marks
Match the intermediates given in Column I with their probable structure in Column II.
| | Column I | | Column II |
| (i) | Free radical | (a) | Trigonal planar |
| (ii) | Carbocation | (b) | Pyramidal |
| (iii) | Carbanion | (c) | Linear |
Answer
View full question & answer→| | Column I | | Column II |
| (i) | Free radical | (a) | Trigonal planar |
| (ii) | Carbocation | (a) | Trigonal planar |
| (iii) | Carbanion | (b) | Pyramidal |
Question 233 Marks
Arrange the following:
- $\text{C}_6\text{H}_5\stackrel{+\ \ \ \ \ \ \ \ \ \ \ \ }{\text{CHCH}}_3, \text{C}_6\text{H}_5\stackrel{+\ \ \ \ \ \ \ \ \ \ \ \ }{\text{CHCH}}=\text{CH}_2,$ $\text{C}_6\text{H}_5\text{CH}_2\stackrel{+ \ \ \ }{\text{CH}}_2,\text{C}_6\text{H}_5\stackrel{+\ \ \ \ \ \ \ \ \ }{\text{C(CH}}_3)_2$ in order of increasing stability.
- $\text{CH}_3\text{CH}_2^+,\text{C}_6\text{H}_5\text{CH}_2^+,(\text{CH}_3)_3\text{C}^+,\text{CH}_2=\text{CHCH}_2^+$ in order of decreasing stability.
- $\text{HC}\equiv\text{C}^-,\text{CH}_2=\text{CH}^-,\text{CH}_3\text{CH}_2^-,\text{CH}^-_3,$ $(\text{CH}_3)_2\text{CH}^-,\text{C}_6\text{H}_5\text{CH}^-_2$ in order of increasing stability.
Answer
View full question & answer→-
$\text{C}_6\text{H}_5\text{CH}_2\stackrel{+ \ \ \ }{\text{CH}}_2<\text{C}_6\text{H}_5\stackrel{+\ \ \ \ \ \ \ \ \ \ \ \ }{\text{CHCH}}_3< \text{C}_6\text{H}_5\stackrel{+\ \ \ \ \ \ \ \ \ }{\text{C(CH}}_3)_2\\<\text{C}_6\text{H}_5\stackrel{+\ \ \ \ \ \ \ \ \ \ \ \ }{\text{CHCH}}=\text{CH}_2$
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$(\text{CH}_3)_3\text{C}^+>\text{C}_6\text{H}_5\text{CH}_2^+>\text{CH}_2=\text{CHCH}_2^+>\text{CH}_3\text{CH}_2^+$
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$(\text{CH}_3)_2\text{CH}^-<,\text{CH}_3\text{CH}_2^-<\text{CH}_3^-<\text{CH}_2=\text{CH}^-\\<\text{C}_6\text{H}_5\text{CH}^-_2<\text{HC}\equiv\text{C}^-$
Question 243 Marks
Name the electrophile/ nucleophile generated by following species:
- HNO3 + H2SO4
- CH3COCl
- alc. KCN
Answer
View full question & answer→
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$\ \ \ {\ \ \ \ \ \ \ \ \ \ \ {\text{O}}\\ \ \ \ \ \ \ \ \ \ \ \ \|\\\text{CH}-{\text{C}}\oplus\text{is electrophile}\\ \ \ \ \ \ \ \ \ \ \ {\text{O}}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{O}\\\\ \ \ \ \ \ \ \ \ \ \ \|\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \|}\\ \text{CH}_3-\text{C}-\text{Cl}^-\xrightarrow{\ \ \ \ \ \ \ }\text{CH}_3-\text{C}\oplus+\text{Cl}^-$
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$\text{CN}^-$ is nucleophile
$\text{KCN}\xrightarrow{\ \ \ \ \ \ \ \ \ \ }\text{K}^++\text{CN}^-$
Question 253 Marks
Suggest a method to purify:
- Camphor containing traces of common salt.
- Kerosene oil containing water.
- A liquid which decomposes at its boiling point.
Answer
View full question & answer→- Sublimation camphor sublimes while common salt remains as residue in the China dish.
- Since the two liquids are immiscible, the technique of solvent extraction with a separating funnel is used. The mixture is throughly shaken and the separating funnel is allowed to stand. Kerosene being lighter than water forms the upper layer while water forms the lower layer.
The lower water layer is run off using the stop cock of the funnel and kerosene oil is obtained. It is dried over anhydrous CaCl2 or MgSO4 and then distilled to give pure kerosene oil.
- Distillation under reduced pressure.
Question 263 Marks
The following techniques are used to quantitatively estimate extra elements in organic compound. Identify the name of the method and the element estimated by this method.
- A known mass of an organic compound is heated with fuming HNO3 in presence of AgNO3.
- Organic compound is heated with dry copper oxide in atmosphere of CO2.
- Organic compound is heated with conc. H2SO4.
Answer
View full question & answer→- Estimation of halogen by Carius method.
- Estimation of nitrogen by Dumas method.
- Estimation of nitrogen by Kjeldahl's method.
Question 273 Marks
- Write IUPAC name of the following:
$\text{CH}_3\text{CHCH}\equiv\text{CH}\\\ \ \ \ \ \ \ \ \ |\\ \ \ \ \ \ \ \ \ \text{CH}=\text{CH}_2$
- Draw the structure of Pent-4-en-2-ol.
- What is nucleophile? Give one example.
Answer
View full question & answer→-
$\text{CH},-\stackrel{3\ \ \ \ \ }{\text{CH}}-\stackrel{4\ \ \ \ \ }{\text{CH}}\equiv\stackrel{5 \ \ \ \ }{\text{CH}}\\\ \ \ \ \ \ \ \ \ \ \ \ |\\\ \ \ \ \ \ \ \ \ \ \ \text{CH}=\text{CH}_2\\\ \ \ \ \ \ \ \ \ \ ^{{\ \ 2\ \ \ \ \ \ \ \ \ \ \ \ \ \ 1}}\\ \ \ \ \ \ \ \ ^{\text{3-methyl pent-1-en-4yne}}$
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$\stackrel{1\ \ \ \ \ }{\text{CH}_3}-\stackrel{2\ \ \ \ }{\text{CH}}-\stackrel{3\ \ \ \ \ }{\text{CH}}-\stackrel{4\ \ \ \ }{\text{CH}}=\stackrel{5 \ \ \ \ \ }{\text{CH}_2}\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ |\\\ \ \ \ \ \ \ \ \ \ \ \ \ \text{OH}$
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Nucleophile is a species which is either negatively charged or has lone pair of electron.
Question 283 Marks
Answer
View full question & answer→- Functional groups are -NH2(amino), -OMe(methoxy) and -CHO(aldehydic)
- Carbon-carbon double bond, -NO2(nitro)and -COOH(carboxylic)
- -CO-(keto), -COCl(acylchloride)
- $ \ \ \ \ \ | \ \ \ \ \ \ \ |\\-\text{C}=\text{C}-$ (carbon-carbon double bond), $ \ \ \ \ \ \text{O}\\ \ \ \ \ \ \|\\-\text{C}-\text{NH}_2$ (acitamide).
Question 293 Marks
- Write IUPAC name of the following:
- $\text{CH}_2\text{CH}_2\text{CHO}\text{ and }\text{CH}_3\text{C}-\text{CH}_3\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ||\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{O}$
- Draw the structure of 3-oxopentanal.
Answer
View full question & answer→
- Functional isomerism.
- $ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{O}\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \|\\\text{CH}_3-\text{CH}_2-\text{C}-\text{CH}_2-\text{C}-\text{H}\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \|\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{O}\\\ \ \ \ \ \text{is structive of 3-oxo pentanal.}$
Question 303 Marks
- Draw cis and trans-structures for Hex-2-ene. Which isomer will have higher boiling point and why?
Question 313 Marks
- What is role of copper oxide in estimation of carbon and hydrogen?
- How do we test phosphorus in given organic compound?
- What is shape of $\stackrel{\oplus\ \ }{\text{CH}_3}$ (methyl carbocation) and $\stackrel{\oplus\ \ }{\text{CH}_3}$ (methyl carbanion)?
Answer
View full question & answer→- It acts as oxidising agent. It oxidises C to CO2 and H2 to H2O.
- Fuse the given organic compound with Na2O2
- The solution is boiled with conc. HNO3 and then added ammonium molybdate. Canary yellow precipatate confirms the presence of phosphorus.
Question 323 Marks
Answer
View full question & answer→
Question 333 Marks
- Give one example of position isomerism.
- What are electrophiles? Give one example of electrophilic substitution reaction.
- Write the chemistry of Lassaigne's test for qualitative analysis of nitrogen.
Answer
View full question & answer→$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{Cl}\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ |\\\text{CH}_3\text{CH}_2\text{CH}_2\text{Cl}\text{ and }\text{CH}_3-\text{CH}-\text{CH}_3\\1-\text{Cholropropane}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 2-\text{Cholropropane}$
- Those species which are positively charged or electron deficient are called electrophiles. e.g. H+, AlCl3, Cl+
- Fuse the organic compound with sodium metal. Sodium reacts with 'C' and 'N' present in organic compound to form NaCN.
$\text{Na}+\text{C}+\text{N}\xrightarrow{\ \ \ \ \ \ \ \ \ \ \ \ }\text{NaCN}$
Add FeSO4 to L.E. (Lassaigne's extract)
$6\text{NaCl}+\text{FeSO}_4\xrightarrow{\ \ \ \ \ \ \ \ \ }\text{Na}_4[\text{Fe(CN)}_6]+\text{NaSO}_4$
Dilute H2SO4 is added to convert Fe2+ to Fe3+ and blue colour is formed due to formation of ferric ferrocyanide.
$4\text{Fe}^{3+}+\text{3Na}_4[\text{Fe(CN)}_6]\xrightarrow{\ \ \ \ \ \ \ \ \ }\text{Fe}_4[\text{Fe}(\text{CN})_6]_3+12\text{Na}^+\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{Blue colouration}$
Question 343 Marks
Answer
View full question & answer→- Propylbenzene.
- 3-methylpentanitrile.
- 2, 5-dimethylheptane.
- 3-bromo-3-chloroheptane.
- 3-chloropropanal.
- 2, 2-dichloroethanol.
Question 353 Marks
Answer
View full question & answer→According to IUPAC nomenclature, the selected longest carbon chain must have maximum functional groups present in the compound. Therefore, only in one selected chain of 4 carbon atoms including both the functional group is corrected one.
In other three, carbon atoms are in selected chain but both the functional groups are not included.
Question 363 Marks
Note: Consider structures I to VII and answer the questions:
- $\text{CH}_3-\text{CH}_2-\text{CH}_2-\text{CH}_2-\text{OH}$
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$\text{CH}_3-\text{CH}_2-\text{CH}-\text{CH}_3\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ |\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{OH}$
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$\ \ \ \ \ \ \ \ \ \ \ \ \text{CH}_3\\\ \ \ \ \ \ \ \ \ \ \ \ \ |\\\text{CH}_3-\text{C}-\text{CH}_3\\\ \ \ \ \ \ \ \ \ \ \ \ \ |\\\ \ \ \ \ \ \ \ \ \ \ \ \text{OH}$
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$\text{CH}_3-\text{CH}-\text{CH}_2-\text{OH}\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ |\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{CH}_3$
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$\text{CH}_3-\text{CH}_2-\text{O}-\text{CH}_2-\text{CH}_3$
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$\text{CH}_3-\text{O}-\text{CH}_2-\text{CH}_2-\text{CH}_3$
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$\text{CH}_3-\text{O}-\text{CH}-\text{CH}_3\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ |\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{CH}_3$
Answer
View full question & answer→In position isomerism, two or more compounds differ in the position of substituent, functional group or multiple bonds but molecular formula is same. In the given structures, I and II, III and IV, and VI and VII are position isomers.
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$\text{CH}_3-\text{CH}_2-\text{CH}_2-\text{CH}_2-\text{OH}$
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$\text{CH}_3-\text{CH}_2-\text{CH}-\text{CH}_3\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ |\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{OH}$ are position isomers.
-
$\ \ \ \ \ \ \ \ \ \ \ \ \text{CH}_3\\\ \ \ \ \ \ \ \ \ \ \ \ \ |\\\text{CH}_3-\text{C}-\text{CH}_3\\\ \ \ \ \ \ \ \ \ \ \ \ \ |\\\ \ \ \ \ \ \ \ \ \ \ \ \text{OH}$
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$\text{CH}_3-\text{CH}-\text{CH}_2-\text{OH}\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ |\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{CH}_3$ are position isomers.
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$\text{CH}_3-\text{O}-\text{CH}_2-\text{CH}_2-\text{CH}_3$ and
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$\text{CH}_3-\text{O}-\text{CH}-\text{CH}_3\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ |\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{CH}_3$ are position isomers.
Question 373 Marks
What is the hybridisation of each carbon in H2C = C = CH2.
Answer
View full question & answer→In $\ \ \ (1)\ \ \ (2)\ \ \ (3)\\\text{H}_2\text{C}=\text{C}=\text{CH}_2,$ carbon (1) and (3) are sp2 hybridised and carbon (2) is sp hybridized.
Question 383 Marks
For testing halogens in an organic compound with AgNO3 solution, sodium extract (Lassaigne’s test) is acidified with dilute HNO3. What will happen if a student acidifies the extract with dilute H2SO4 in place of dilute HNO3?
Answer
View full question & answer→On adding dilute H2SO4 for testing halogens in an organic compound with AgNO3, white precipitate of Ag2SO4 is formed. This will interfere with the test of chlorine and this Ag2SO4 may be mistaken for white precipitate of chlorine as AgCl. Hence, dilute HNO3 should be used instead of dilute H2SO4.
Question 393 Marks
0.25g of an organic compound gave 30cm3 of moist dinitrogen at 288K and 745mm pressure. Calculate the percentage of nitrogen.
(Aq. tension at 288K = 12.7mm).
(Aq. tension at 288K = 12.7mm).
Answer
View full question & answer→Pressure of moist dinitrogen obtained = 745mm
Aq. tension at same temperature = 12.7mm
$\therefore$ Pressure due to nitrogen = 745 - 12.7 = 732.3mm
To calculate the volume of N2 at S.T.P.
$\frac{\text{P}_1\text{V}_1}{\text{T}_1}=\frac{\text{P}_2\text{V}_2}{\text{T}_2}$
$\Rightarrow\text{V}_2=\frac{\text{P}_1\text{V}_1}{\text{T}_1}$
$=\frac{732.3\times30\times273}{288\times760}=27.4\text{cm}^3$
$\therefore\%\text{ of Nitrogen}=\frac{28\times27.4\times100}{22400\times0.25}$
$=13.7\%$
Aq. tension at same temperature = 12.7mm
$\therefore$ Pressure due to nitrogen = 745 - 12.7 = 732.3mm
To calculate the volume of N2 at S.T.P.
$\frac{\text{P}_1\text{V}_1}{\text{T}_1}=\frac{\text{P}_2\text{V}_2}{\text{T}_2}$
$\Rightarrow\text{V}_2=\frac{\text{P}_1\text{V}_1}{\text{T}_1}$
$=\frac{732.3\times30\times273}{288\times760}=27.4\text{cm}^3$
$\therefore\%\text{ of Nitrogen}=\frac{28\times27.4\times100}{22400\times0.25}$
$=13.7\%$
Question 403 Marks
“Stability of carbocations depends upon the electron releasing inductive effect of groups adjacent to positively charged carbon atom involvement of neighbouring groups in hyperconjugation and resonance.”
Write structures of various carbocations that can be obtained from 2-methylbutane. Arrange these carbocations in order of increasing stability.
Write structures of various carbocations that can be obtained from 2-methylbutane. Arrange these carbocations in order of increasing stability.
Answer
View full question & answer→2-Methylbutane has four possible carbocations
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ +\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ +\\\text{CH}_3-\text{CH}-\text{CH}_2-\text{CH}_2\ \ \ \ \text{CH}_3-\text{CH}-\text{CH}-\text{CH}_3\\\ \ \ \ \ \ \ \ \ \ \ \ \ | \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ |\\\ \ \ \ \ \ \ \ \ \ \ \ \text{CH}_3\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{CH}_3\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{I}(1^\circ)\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{II}(2^\circ)$
$\ \ \ \ \ \ \ \ \ \ \ \ +\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ +\\\text{CH}_3-\text{C}-\text{CH}_2-\text{CH}_3\ \ \ \ \text{CH}_2-\text{CH}_2\text{CH}-\text{CH}_2-\text{CH}_3\\\ \ \ \ \ \ \ \ \ \ \ \ \ |\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ |\\\ \ \ \ \ \ \ \ \ \ \ \ \text{CH}_3\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{CH}_3\\\ \ \ \ \ \ \ \ \ \ \text{III}(3^\circ)\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{IV}(1^\circ)$ I < IV < II < III Stability of carbocation increases in the order 1° < 2° < 3°. Out of I and IV, IV is more stable than I because in IV, CH3 group is at a-carbon and in I, it is at β-carbon and +I-effect decreases with distanceQuestion 413 Marks
Note: Consider structures I to VII and answer the questions:
- $\text{CH}_3-\text{CH}_2-\text{CH}_2-\text{CH}_2-\text{OH}$
-
$\text{CH}_3-\text{CH}_2-\text{CH}-\text{CH}_3\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ |\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{OH}$
-
$\ \ \ \ \ \ \ \ \ \ \ \ \text{CH}_3\\\ \ \ \ \ \ \ \ \ \ \ \ \ |\\\text{CH}_3-\text{C}-\text{CH}_3\\\ \ \ \ \ \ \ \ \ \ \ \ \ |\\\ \ \ \ \ \ \ \ \ \ \ \ \text{OH}$
-
$\text{CH}_3-\text{CH}-\text{CH}_2-\text{OH}\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ |\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{CH}_3$
-
$\text{CH}_3-\text{CH}_2-\text{O}-\text{CH}_2-\text{CH}_3$
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$\text{CH}_3-\text{O}-\text{CH}_2-\text{CH}_2-\text{CH}_3$
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$\text{CH}_3-\text{O}-\text{CH}-\text{CH}_3\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ |\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{CH}_3$
Answer
View full question & answer→Compounds I to IV, i.e., alcohols, and V to VII, i.e., ethers, are functional group isomers with molecular formula C4H10O and different functional groups (-OH in I to IV and -O - in V to VII).
Hence, I and V, I and VI, I and VII; II and V, II and VI, II and VII; HI and V, IH and VI; IH and VII; IV and V, IV and VI, IV and VH are functional group isomers.
Hence, I and V, I and VI, I and VII; II and V, II and VI, II and VII; HI and V, IH and VI; IH and VII; IV and V, IV and VI, IV and VH are functional group isomers.
Question 423 Marks
Explain, how is the electronegativity of carbon atoms related to their state of hybridisation in an organic compound?
Answer
View full question & answer→Electronegativity increases with increasing s-character. This is because s-electrons are more strongly attracted by the nucleus than p-electrons.
sp3 – 25% s-character, 75% P-character
sp2 – 33% s-character, 67% P-character
sp – 50% s-character, 50% P-character
Hence, the order of electronegativity is sp3 < sp2 < sp
sp3 – 25% s-character, 75% P-character
sp2 – 33% s-character, 67% P-character
sp – 50% s-character, 50% P-character
Hence, the order of electronegativity is sp3 < sp2 < sp
Question 433 Marks
What is meant by hybridisation? Compound CH2 = C = CH2 contains sp or sp2 hybridised carbon atoms. Will it be a planar molecule?
Answer
View full question & answer→Hybridization is defined as an intermixing of a set of atomic orbitals of slightly different energies, thereby forming a new set of orbitals having equivalent energies and shapes. For example, one 2s-orbital hybridizes with two 2p-orbitals of carbon to form three new sp2 hybrid orbitals. These hybrid orbitals have minimum repulsion between their electron pairs and thus, are more stable. Hybridization helps indicate the geometry of the molecule.
In CH2 = C = CH2 (allene)
Carbon atom land Sp2 hybridized as each one has $3\sigma$ bonds while carbon atom 2 has $2\sigma$ bonds and it is sp hybridized. allene molecule as a whole is non-planar.
In CH2 = C = CH2 (allene)
Carbon atom land Sp2 hybridized as each one has $3\sigma$ bonds while carbon atom 2 has $2\sigma$ bonds and it is sp hybridized. allene molecule as a whole is non-planar.
Question 443 Marks
0.90g of organic compound on combustion gave 2.64g of CO2, and 0.63g of H2O. Calculate the percentage of C and H in the compound.
Answer
View full question & answer→$\%\text{ of C}=\frac{12}{44}\times\frac{\text{Weight of CO}_2\text{ formed}}{\text{Weight of organic compound}}\times100$
$=\frac{12}{44}\times\frac{2.64}{0.90}\times100=\frac{3168}{39.6}=80\%$
$\%\text{ of H}=\frac{2}{18}\times\frac{\text{Weight of H}_2\text{O formed}}{\text{Weight of organic compound}}\times100$
$=\frac{12}{18}\times\frac{0.63}{0.90}\times100=\frac{126}{16.2}=7.77\%$
$=\frac{12}{44}\times\frac{2.64}{0.90}\times100=\frac{3168}{39.6}=80\%$
$\%\text{ of H}=\frac{2}{18}\times\frac{\text{Weight of H}_2\text{O formed}}{\text{Weight of organic compound}}\times100$
$=\frac{12}{18}\times\frac{0.63}{0.90}\times100=\frac{126}{16.2}=7.77\%$
Question 453 Marks
The order of basicity of amines expected on the basis of inductive effect is NH3 < RNH2 < R3N. However, the observed order of basicity is NH3 < RNH2 < R2NH > R3N. How will you account for the difference?
Answer
View full question & answer→This order of basicity is observed in case of solvent phase as the energy of hydration also becomes important here. The following important factors can be considered.
- Effect of electrion releasing groups: More the number of electron-releasing groups present in a molecule, more is the increase in its electron density and more is the basic character. So, according to this effect, the order should be NH3 > RNH2 <R2NH <R3N.
- Hydration enthalpy: The more hydrogen bonds a molecule (ammonium ion) can form with water, the more will be its basic character. So, the order should be NH3 > RNH2 < R2NH < R3N.
- Inductive effect: The more is the crowding of groups in a molecule, the lesser will be its basicity. So, the order should be NH3 > RNH2> R2NH > R3N.
Combining all the factors, we get the order of basicity as NH3 > RNH2 > R2N.
Question 463 Marks
What is the relationship between the members of following pairs of structures? Are they structural or geometrical isomers or resonance contributors?
Answer
View full question & answer→Compounds having the same molecular formula but with different structures are called structural isomers. The given compounds have the same molecular formula but they differ in the position of the functional group (ketone group).
In structure I, ketone group is at the C-3 of the parent chain (hexane chain) and in structure II, ketone group is at the C-2 of the parent chain (hexane chain). Hence, the given pair represents structural isomers.
In structure I, ketone group is at the C-3 of the parent chain (hexane chain) and in structure II, ketone group is at the C-2 of the parent chain (hexane chain). Hence, the given pair represents structural isomers.
Question 473 Marks
If a liquid compound decomposes at its boiling point, which method (s) can you choose for its purification. It is known that the compound is stable at low pressure, steam volatile and insoluble in water.
Answer
View full question & answer→Steam distillation can be used for its purification. This method is applied to separate substances which are steam volatile and immiscible with water.
Question 483 Marks
For the following bond cleavages, use curved-arrows to show the electron flow and classify each as homolysis or heterolysis. Identify reactive intermediate produced as free radical, carbocation and carbanion.
Answer
View full question & answer→The bond cleavage using curved-arrows to show the electron flow of the given reaction can be represented as
It is an example of homolytic cleavage as one of the shared pair in a covalent bond goes with the bonded atom. The reaction intermediate formed is a free radical.
It is an example of homolytic cleavage as one of the shared pair in a covalent bond goes with the bonded atom. The reaction intermediate formed is a free radical.
Question 493 Marks
View full question & answer→