Question

Question: When alkyne a is treated with ${\mathbf{NH}}_{\mathbf{2}}$ followed by ${\mathbf{CH}}_{\mathbf{3}}$ , a product having the molecular formula ${\mathbf{C}}_{\mathbf{6}}{\mathbf{H}}_{\mathbf{10}}\mathbf{O}$ is formed, but it is not compound B. What is the structure of the product and why is it formed?

Short answer

Answer

Compound B is not formed because deprotonation of alcohol occurs faster than the deprotonation of an alkyne.

Step-by-step solution

Step-by-Step SolutionStep 1: Deprotonation 

The removal of the most loosely held proton from a molecule to form the corresponding conjugate base is called deprotonation.

When sodium amide reacts with pent-4-yn-1-ol, the hydroxy group undergoes deprotonation as an electronegative atom, that is, oxygen, abstracts the proton.

Alkylation reaction

The organic reaction involving the transfer of an alkyl group to an intermediate ion is known as the alkylating reaction.

The conjugate base undergoes methylation to form the corresponding ether as shown below:

Formation of product