C' in the above reaction is : $C_{6}H_{5}CH_{2}OH\xrightarrow{HBr}A\xrightarrow{KCN}B\xrightarrow{H_3O^+,\Delta}C$

Question

Accepted Answer

To understand why Option 2, "Phenyl ethanoic acid," is the correct answer, let's analyze the reaction sequence step by step:

1. **First Step: $C_{6}H_{5}CH_{2}OH \xrightarrow{HBr} A$**

- The starting compound is benzyl alcohol ($C_{6}H_{5}CH_{2}OH$). When treated with hydrobromic acid (HBr), the hydroxyl group (-OH) is replaced by a bromine atom, resulting in benzyl bromide ($C_{6}H_{5}CH_{2}Br$). This is compound A.

2. **Second Step: $A \xrightarrow{KCN} B$**

- Benzyl bromide ($C_{6}H_{5}CH_{2}Br$) reacts with potassium cyanide (KCN) in a nucleophilic substitution reaction. The bromine atom is replaced by a cyano group (-CN), forming benzyl cyanide ($C_{6}H_{5}CH_{2}CN$). This is compound B.

3. **Third Step: $B \xrightarrow{H_3O^+, \Delta} C$**

- Benzyl cyanide ($C_{6}H_{5}CH_{2}CN$) undergoes hydrolysis in the presence of acid (H₃O⁺) and heat (Δ). The cyano group (-CN) is converted into a carboxylic acid group (-COOH), resulting in phenyl ethanoic acid ($C_{6}H_{5}CH_{2}COOH$). This is compound C.

Now, let's justify why the other options are incorrect:

1. **Option 1: Benzoic acid**

- Benzoic acid has the formula $C_{6}H_{5}COOH$. In the reaction sequence, the cyano group is attached to the methylene group ($CH_2$) of benzyl cyanide, not directly to the benzene ring. Therefore, the product is phenyl ethanoic acid, not benzoic acid.

2. **Option 3: Phenyl ethylamine**

- Phenyl ethylamine would require the conversion of the cyano group to an amine group (-NH₂). However, the reaction conditions given (acidic hydrolysis) convert the cyano group to a carboxylic acid, not an amine.

3. **Option 4: Hexane-1, 6-dioic acid**

- Hexane-1, 6-dioic acid is a dicarboxylic acid with six carbon atoms in a straight chain. The reaction sequence involves a benzene ring and does not lead to the formation of a straight-chain dicarboxylic acid.

In conclusion, the reaction sequence leads to the formation of phenyl ethanoic acid, which is consistent with Option 2. The other options do not match the structure or the reaction pathway described.