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NCERT Exemplar · Class 12 · Chemistry

Chapter 13: Amines

Practice each question, then open the answer or worked solution.

Short Answer Questions

37 questions
Q48Written answer

Under what reaction conditions (acidic/basic), the coupling reaction of aryl diazonium chloride with aniline is carried out?

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Ans: The azo products have an extended conjugate system with both aromatic rings linked by the –N=N- bond. These compounds are frequently colored and used as dyes. Benzene diazonium chloride reacts with phenol to form p-hydroxy azobenzene by coupling the phenol molecule in its para position with the diazonium salt. This is referred to as a coupling reaction.


Q49Written answer

Predict the product of reaction of aniline with bromine in non-polar solvent such as $\mathrm{CS}_{2}$.

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Ans: $\mathrm{CS}_{2}$ is a non-polar solvent which decreases the activating effect of $-\mathrm{NH}_{2}$. As a result, mono substitution occurs only at o- and p- positions giving a mixture of 2-bromoaniline (minor) and 4-bromoaniline (major) as products. 


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Q50Written answer

Arrange the following compounds in increasing order of dipole moment:

$\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{3}, \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{NH}_{2}, \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}$.

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Ans:  $\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{3}<\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{NH}_{2}<\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}$

As oxygen is more electronegative than nitrogen therefore, the O-H bond is more polar than the N-H bond. So ethanol has more dipole moments than ethylamine. Propane is non- polar in nature hence, had the least among all. 


Q51Written answer

What is the structure and IUPAC name of the compound, allyl amine?

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Ans: Prop-2-en-1-amine

$\mathrm{CH}_{2}=\mathrm{CH}-\mathrm{CH}_{2} \mathrm{NH}_{2}$


Q52Written answer

Write down the IUPAC name of


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Ans: N, N-dimethylbenzenanmine 


Q53Written answer

A compound $Z$ with molecular formula $C_{3} H_{9} N$ reacts with $C_{6} H_{5} S O_{2} C l$ to give a solid, insoluble in alkali, identify Z.

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Ans: $Z$ is an aliphatic amine that produces a solid that is insoluble in base. This means that the reaction with $\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{SO}_{2} \mathrm{Cl}$ must produce a product with no replaceable hydrogen attached to nitrogen.

To put it another way, the amine must be a secondary amine.

Z stands for ethylmethylamine.


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Q54Written answer

A primary amine, $\mathrm{RNH}_{2}$ can be reacted with $\mathrm{CH}_{3}-\mathrm{X}$ to get secondary amine, $\mathrm{R}-\mathrm{NHCH}_{3}$ but the only disadvantage is that $3^{\circ}$ amine and quaternary ammonium salts are also obtained as side products. Can you suggest a method where $\mathrm{RNH}_{2}$ forms only $2^{\circ}$ amine?

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Ans: Carbylamine reaction is shown by primary amines only, primary amines react with chloroform in the presence of alcoholic KOH to form isocyanides which upon catalytic reduction gives secondary amines as the final product. 

$\underset{1^{\circ} \mathrm{Amine}}{R-\mathrm{NH}_{2}}+\mathrm{CHCl}_{3}+3 \mathrm{KOH} \rightarrow \underset{\text { Alkyl isocyanide }}{\longrightarrow} \frac{\mathrm{HC}_{2} / \mathrm{Pd}}{(\text { Reduction) })} \underset{2^{\circ} \text { Amine }}{\mathrm{N}}-\mathrm{NH}-\mathrm{CH}_{3}$


Q55Written answer

Complete the following reaction.


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Ans: 


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Q56Written answer

Why is aniline soluble in aqueous HCl?

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Ans: On reaction with hydrochloric acid, aniline forms anilinium chloride ion which is water soluble. Therefore, the aniline solubility in aqueous HCl solution. 


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Q57Written answer

Suggest a route by which the following conversion can be accomplished.


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Ans:


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