MCQ
The best reagent for converting$, 2-$phenylpropanamide into $1-$phenylethanamine is $.........$
  • A
    excess $H_2/ Pt.$
  • $\ce{NaOH/ Br_2}.$
  • C
    $\ce{NaBH_4}/$ methanol.
  • D
    $\ce{LiAlH_4}/$ ether.

Answer

Correct option: B.
$\ce{NaOH/ Br_2}.$
$\text{CH}_3-\text{CH}-\text{CONH}_2\xrightarrow[\text{(Hofmann's bromamide reaction)}]{\text{Br}_2/\text{N}_\text{a}\text{OH}}\text{CH}_3-\text{CH}-\text{NH}_2\\\ \ \ \ \ \ \ \ \ \ \ \ \ |\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ |\\\ \ \ \ \ \ \ \ \ \ \ \ \text{C}_6\text{H}_5\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{C}_6\text{H}_5\\2-\text{Phenylpropanamide}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 1-\text{Phenyletanamine}$

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