The reaction of 4-bromobenzyl chloride with NaCN in ethanol leads to

4-Bromobenzyl cyanide

The correct answer is option 1. 4-Bromobenzyl cyanide.

The main reactant that has been given to us is of 4 – bromobenzyl chloride. But before we make the molecular structure of this compound, let us understand how to step by step decode the IUPAC name to form the molecular structure. The parent chain in this form of 4 – bromobenzyl chloride is evidently benzyl chloride. The molecular structure for the benzyl group is basically a benzene with a methyl group at position 1 of the ring. Hence, benzyl chloride is formed when a chlorine atom gets attached on the methyl group of the benzyl molecule. Now, 4 – bromobenzyl chloride is formed when a bromine atom is bonded with a carbon atom at position 4 in the benzene ring. Hence, the molecular structure of 4 – bromobenzyl chloride can be given as:

Now this compound is made to react with a compound known as sodium cyanide or NaCN. Another compound used here is ethanol or ethyl alcohol. Ethyl alcohol is a polar aprotic compound, which means that it does not give away its hydrogen atom when dissolved in a solvent. On the other hand, sodium cyanide is a nucleophilic compound. This means that CN⁻ acts like the nucleophile. This Chlorine present on the benzyl chloride molecule will act as a leaving group and this results in the substitution of Cl⁻ with CN⁻. This reaction proceeds with S_N2 substitution reaction. hence, the chemical reaction can be given as: