Hydrocarbons

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)