Question
How will you bring about the following conversion in not more than two step?
Benazaldehyde to $\alpha$-Hydroxyphenylacetic acid.
✓
Answer
Benazaldehyde to $\alpha$-Hydroxyphenylacetic acid:
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{OH}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{OH}\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ |\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ |\\ \text{C}_6\text{H}_5\text{CHO}\xrightarrow[\text{pH. 9-10}]{\text{HCN}}\text{C}_6\text{H}_5\text{CH}-\text{CN}\xrightarrow{\text{H}^+/\text{H}_2\text{O}}\text{C}_6\text{H}_5-\text{CH}-\text{COOH}$
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