Question

Draw the alkyne formed when 3,4-dichloroheptane is treated with an excess of strong base such as sodium amide.

Answer 1

Concepts and reason

Preparation of alkyne from 3,4-dichloroheptane and excess of sodium amide.

Fundamentals

Sodium amide ${\rm{(NaN}}{{\rm{H}}_2})$ is strong base and it is used as dehydrohalogenating agent. Sodium amide eliminates HCl from dichloro substituted compound to form chloro substituted alkene. another molecule of HCl is eliminated by sodium amide from chloro substituted alkene to form alkyne as product. So, to prepare alkyne from dichloro substituted compound excess sodium amide ${\rm{(NaN}}{{\rm{H}}_2})$ is required.

The structure of 3,4-dichloroheptane is,

The reaction is,

Ans:

Thus, the structure of alkyne is,