Question

p-Nitrobromobenzene can be converted to p-nitroaniline by using NaNH2. The reaction proceeds through the intermediate named

Answer 1

Hint: NaNH₂ is an effective base. When its strong basicity doesn't have negative effects, it can occasionally be a great nucleophile. It is employed in elimination reactions as well as the deprotonation of weak acids. Alkynes, alcohols, and a variety of other functional groups with acidic protons, such as esters and ketones, will all be deprotonated by NaNH₂.

Complete step by step solution:
Firstly, when NaNH₂ reacts with p-Nitrobromobenzene, NaNH₂ being a base will attack the most acidic hydrogen of p-Nitrobromobenzene. The carbon attached to Br will leave producing a carbanion.

In the next step, the bromine-carbon bond of p-Nitrobromobenzene will break and attach to the Na+ of NaNH₂, leaving a positive charge on that carbon i.e, an empty orbital.

Now, two adjacent carbon having a positive and a negative charge creates instability and thus the carbon containing an electron pair will donate its electron pair to the adjacent carbon containing an empty orbital.

After this, a triple bond is formed between the two carbon atoms. The intermediate now formed is called benzyne. The nomenclature of the compound formed is 3,4-aryne.

The correct answer is Benzyne.

Note: Because bases give up their electrons to form new covalent bonds, they are less likely to share them if they are more attracted to them. As a result we see that electronegativity is related to basicity. The greater the electronegativity of an atom, the less it is willing to share its electrons. Electronegativity is important to consider when comparing atoms in the same row of the periodic table.