Consider the following reaction sequence:
$4\text{-aminobenzonitrile}$ $\xrightarrow[(ii) H_2O]{(i) AlH(i-Bu)_2} 'A'$ $\xrightarrow[dil. NaOH, \Delta]{CH_3CHO} B$
The product $B$ is?

  • A
    $4\text{-aminobenzaldehyde}$
  • B
    $4\text{-aminocinnamaldehyde}$
  • C
    $4\text{-amino-N-ethylidenebenzylamine}$
  • D
    $4\text{-amino-N-formylbenzamide}$

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$HCDO \xrightarrow[{(50\%)}]{{OH^{-}}}$ Product of this Cannizzaro reaction is:

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$(i)$ Benzaldehyde
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