CBSE 12th Chemistry Board Exam 2020: Important Questions & Answers from Chapter 13 - Amines
Check the important questions and answers from the chapter Amines for the upcoming CBSE Class 12th Chemistry Examination 2020.
The students appearing for the CBSE Class 12th Chemistry Examination 2020 can go through the below-mentioned list of questions for the revision purpose at the last minute. These questions are strictly based on the latest CBSE pattern.
Key Points to be mentioned while writing the answers to the below mentioned important questions:
Question 1- Give the chemical tests to distinguish between:
(i) Ethylamine and Aniline
(ii) Aniline and Benzylamine
(iii) Methylamine and Dimethylamine
(iv) Aniline and N-methylaniline
Answer: (i) Through Azo dye test, the reaction of any aromatic primary amine with HNO2(NaNO2 + dil. HCl) at 273-278 K followed by treatment with an alkaline solution of 2-naphthol when a brilliant yellow, orange or red coloured dye is obtained.
(ii) Through Nitrous acid test, Benzylamine reacts with HNO2 to form a diazonium salt which being unstable even at low temperature decomposes with evolution of N2 gas.
(iii) Methylamine is a primary amine, thus it gives Carbylamine test whereas Dimethylamine doesn’t give this test because it is a secondary amine.
(iv) Aniline is a 1° aromatic amine thus gives carbylamine test while N-methylaniline is a secondary aromatic amine and doesn’t give this test.
Question 2- Arrange the following in increasing order of their basic strength:
(i) C6H5NH2, C2H5NH2, (C2H5) 2NH, NH3
(ii) C2H5 NH2 , C6 H5NH2, NH3, C6H5CH2NH2 and (C2H5) 2NH
(iii) CH3NH2, (CH3) 2NH, (CH3) 3N, C6H5NH2, C6H5CH2NH2
Answer: (i) C6H5NH2 < NH3 < C2H5NH2< (C2H5) 2NH
(ii)C6H5CH2NH2 < NH3 <C6H5CH2NH2 < C2H5 NH2 < (C2H5) 2NH
(iii) C6H5NH2 < C6H5CH2NH2 < (CH3) 3N < CH3NH2 <(CH3) 2NH
Question 3- Write structures along with the IUPAC names:
(i) the amide which gives propanamine by Hoffmann bromamide reaction.
(ii) the amine produced by the Hoffmann degradation of benzamide.
Answer: (i) Propanamine contains three carbons. Hence, the amide molecule must contain four carbon atoms. Therefore, the IUPAC name will be Butanamide.
(ii) Benzamide is an aromatic amide containing seven carbon atoms. Hence, the amine formed from benzamide is aromatic primary amine-containing six carbon atoms. Therefore, the IUPAC name will be either Aniline or benzenamine.
Question 4- Write chemical equations for:
(i) CH3–CH2–Cl into CH3–CH2–CH2–NH2
(ii) C6H5–CH2–Cl into C6H5–CH2–CH2–NH2
Question 5- Define:
(i) Sandmeyer reaction
(ii) Coupling reaction
(iii) Gabriel phthalimide synthesis
(iv) Hoffmann bromamide degradation reaction
Answer: (i) Sandmeyer reaction: The Cl–, Br– and CN– nucleophiles can easily be introduced in the benzene ring in the presence of Cu(I) ion. This reaction is called the Sandmeyer reaction.
(ii) Coupling reaction: The azo products obtained have an extended conjugate system having both the aromatic rings joined through the –N=N– bond. These compounds are often coloured and are used as dyes. Benzene diazonium chloride reacts with phenol in which the phenol molecule at its para-position is coupled with the diazonium salt to form p-hydroxyazobenzene. This type of reaction is known as coupling reaction. Similarly the reaction of the diazonium salt with aniline yields p-aminoazobenzene. This is an example of an electrophilic substitution reaction.
(iii) Gabriel phthalimide synthesis: It is used for the preparation of primary amines. Phthalimide on treatment with ethanolic potassium hydroxide forms potassium salt of phthalimide which on heating with alkyl halide followed by alkaline hydrolysis produces the corresponding primary amine. Aromatic primary amines cannot be prepared by this method because aryl halides do not undergo nucleophilic substitution with the anion formed by phthalimide.
(iv) Hoffmann bromamide degradation reaction: Hoffmann developed a method for preparation of primary amines by treating an amide with bromine in an aqueous or ethanolic solution of sodium hydroxide. In this degradation reaction, migration of an alkyl or aryl group takes place from carbonyl carbon of the amide to the nitrogen atom. The amine so formed contains one carbon less than that present in the amide.
Question 6- State Reasons:
(a) Aniline is a weaker base than cyclohexylamine.
(b) It is difficult to prepare pure amines by ammonolysis of alkyl halides.
(c) Electrophilic substitution in aromatic amines takes place more readily than benzene.
Answer: (a) Aniline is a weaker base than cyclohexylamine because the lone pair of electrons on the N-atom is delocalised over the benzene ring in aniline which results in the decrease in electron density on Nitrogen. In cyclohexylamine, the lone pair of the electrons on N-atom is readily available due to absence of π-electrons.
(b) It is difficult to prepare pure amines by ammonolysis of alkyl halides as the primary amine formed by ammonolysis acts as a nucleophile. It further produces 2° and 3° alkyl amine.
(c) Electrophilic substitution in aromatic amines takes place more readily than benzene as aromatic amines undergo electrophilic substitution reactions readily than benzene.
The above-mentioned questions are based on previous year questions, sample papers and the NCERT textbook. The students can also go through the below-mentioned links for the upcoming CBSE Class 12th Examination 2020: