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Amines · Question 55

AIPMT 2003MCQMedium📈 Topic in 62% papers+4−1
1

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

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

Answer

(3) [figure]

Solution

Step 1 (acetylation): Ac2O converts -NH2 into -NHCOCH3 A = 4-methylacetanilide. As per NCERT, this moderates the activating effect of -NH2 so that only monobromination occurs.

Step 2 (bromination): -NHCOCH3 is a stronger o/p-director than -CH3; the para position is occupied Br enters ortho to -NHCOCH3 B = 2-bromo-4-methylacetanilide.

Step 3 (acid hydrolysis): -NHCOCH3 H2O/H+ -NH2 C = 2-bromo-4-methylaniline.

Answer: (3)

AIPMT 2003

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2

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
3

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
4

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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