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Amines · Question 53

AIPMT 2005MCQMedium📈 Topic in 0% papers+4−1
1

Aniline in a set of reactions yielded a product D. [figure] The structure of the product D would be

  1. C6H5NHOH
  2. C6H5NHCH2CH3
  3. C6H5CH2NH2
  4. C6H5CH2OH

Answer

(4) C6H5CH2OH

Solution

C6H5NH2 [HCl, 273-278 K]NaNO2 C6H5N2+Cl- (A, benzenediazonium chloride)

C6H5N2+Cl- CuCN C6H5CN (B, benzonitrile; Sandmeyer reaction)

C6H5CN [Ni]H2 C6H5CH2NH2 (C, benzylamine)

C6H5CH2NH2 HNO2 C6H5CH2OH + N2 + H2O (D)

As per NCERT, primary aliphatic amines react with nitrous acid to form aliphatic diazonium salts which, being unstable, liberate nitrogen gas quantitatively and give alcohols. Benzylamine behaves as a primary aliphatic amine (-NH2 is not on the ring).

D is benzyl alcohol, C6H5CH2OH.

Answer: (4)

AIPMT 2005

Keep practising — Amines

2

In a reaction of aniline a coloured product C was obtained. [figure] The structure of C would be

  1. [figure]
  2. [figure]
  3. [figure]
  4. [figure]
AIPMT 2004
3

The final product C, obtained in this reaction [figure] would be

  1. [figure]
  2. [figure]
  3. [figure]
  4. [figure]
AIPMT 2003
4

CH3CH2Cl NaCN X Ni/H2 Y acetic anhydride Z Z in the above reaction sequence is

  1. CH3CH2CH2NHCOCH3
  2. CH3CH2CH2NH2
  3. CH3CH2CH2CONHCH3
  4. CH3CH2CH2CONHCOCH3
AIPMT 2002
5

Intermediates formed during reaction of [figure] with Br2 and KOH are

  1. RCONHBr and RNCO
  2. RNHCOBr and RNCO
  3. RNH-Br and RCONHBr
  4. RCONBr2
AIPMT 2001
6

A reduction B CHCl3/KOH C reduction N-methylaniline then A is

  1. [figure]
  2. [figure]
  3. CH3NH2
  4. [figure]
AIPMT 2000

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