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Hydrocarbons · Question 55

AIPMT 2002MCQEasy📈 Topic in 0% papers+4−1
1

When CH3CH2CHCl2 is treated with NaNH2, the product formed is

  1. CH3-CH=CH2
  2. CH3-C CH
  3. [figure]
  4. [figure]

Answer

(2) CH3-C CH

Solution

CH3CH2CHCl2 is a gem-dihalide (1,1-dichloropropane).

As per NCERT, a strong base such as NaNH2 removes two molecules of HCl (dehydrohalogenation) from a dihalide to give an alkyne.

CH3CH2CHCl2 [-HCl]NaNH2 CH3CH=CHCl [-HCl]NaNH2 CH3-C CH

∴ The product is propyne.

Answer: (2)

AIPMT 2002

Keep practising — Hydrocarbons

2

Which alkene on ozonolysis gives CH3CH2CHO and CH3COCH3?

  1. [figure]
  2. CH3CH2CH=CHCH2CH3
  3. CH3CH2CH=CHCH3
  4. [figure]
AIPMT 2001
3

Increasing order of electrophilic substitution for following compounds [figure]

  1. IV < I < II < III
  2. III < II < I < IV
  3. I < IV < III < II
  4. II < III < I < IV
AIPMT 2000
4

In Friedel-Crafts reaction, toluene can be prepared by

  1. C6H6 + CH3Cl
  2. C6H5Cl + CH4
  3. C6H6 + CH2Cl2
  4. C6H6 + CH3COCl
AIPMT 2000

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