Question

${C_2}{H_2}\xrightarrow{{NaN{H_2}}}A\xrightarrow{{1 - Bromohep\tan e}}B$
What is B in the above reaction?

Answer 1

Hint : Sodium amide or sodamide ( $NaN{H_2}$ ) is a strong base widely used in organic reactions. It is prepared by the reaction of sodium with ammonia gas. Sodium amide is similar to Lithium diisopropylamide ( $LDA$ ).

Complete Step By Step Answer:
In the question it is given that acetylene reacts with sodamide to yield A which then further reacts with alkyl halide(1-Bromoheptane) to yield B. We have to find B.
Sodium amide $\left( {NaN{H_2}} \right)$ is a strong base which can easily deprotonate alkynes, alcohols, ketones and other functional groups having acidic protons.
In this reaction, Sodium amide will deprotonate acetylene to yield acetylide ion (A) which further reacts with 1-Bromoheptane (alkyl halide) and undergo nucleophilic substitution to yield 1-nonyne (B).

The terminal proton of acetylene is highly acidic and is very easily removed in the presence of sodium amide which is a strong base forming acetylide ion. Bromine is a good leaving group leading to nucleophilic substitution reaction. Also, acetylide ion is a good nucleophile forming 1-nonyne.

Additional Information:
Another important application of sodium amide is the formation of alkynes from geminal or vicinal dihalides. It should be noted that Sodium amide reacts violently with water to produce ammonia and sodium hydroxide hence should be handled with care.

Note :
In order to complete such reactions, we should be aware of the properties and nature of the reagent given. For example; in this question Sodium amide was given which is a good base and hence will abstract the acidic proton using this information we were able to complete the reaction.