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Identify the product in the following reaction: [cyclobutane ring with -Cl substituent] $\xrightarrow{(i)\ KCN\ (ii)\ H_2O/HCl,\Delta\ (iii)\ Br_2/red\ phosphorus\ (iv)\ H_2O}$ Product
A
[add image: cyclobutane with -COBr substituent]
B
[add image: cyclobutane with -COOH and -Br substituents (alpha-bromo acid)]
C
[add image: cyclobutane with -Cl and -Br substituents]
D
[add image: cyclobutane with -Br substituent only]
Detailed Solution
The cyclobutyl chloride reacts with KCN to form the nitrile (substituting Cl with CN), which is hydrolysed with $H_2O/HCl$ on heating to give the carboxylic acid. This acid then undergoes the Hell-Volhard-Zelinsky reaction ($Br_2$/red phosphorus, followed by $H_2O$) to introduce a bromine at the alpha-carbon, giving an alpha-bromo carboxylic acid as the final product.