Consider the following reaction sequence:4tex | vedclass

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(B) Step $1$: The reduction of $4	ext{-aminobenzonitrile}$ with $DIBAL-H$ $(AlH(i-Bu)_2)$ followed by hydrolysis yields $4	ext{-aminobenzaldehyde}$

Vedclass · Accepted Answer

$4	ext{-aminocinnamaldehyde}$

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(B) Step $1$: The reduction of $4	ext{-aminobenzonitrile}$ with $DIBAL-H$ $(AlH(i-Bu)_2)$ followed by hydrolysis yields $4	ext{-aminobenzaldehyde}$ as product '$A$'.Step $2$: The reaction of $4	ext{-aminobenzaldehyde}$ with acetaldehyde $(CH_3CHO)$ in the presence of dilute $NaOH$ and heat is a cross-aldol condensation reaction.Step $3$: The aldehyde group of $4	ext{-aminobenzaldehyde}$ reacts with the $\alpha	ext{-hydrogen}$ of acetaldehyde to form $4	ext{-aminocinnamaldehyde}$ as the major product '$B$'.

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?

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Product $(C)$ is

In the given reaction $CH_3-CH_2-C(=O)-CH_2-COOC_2H_5$ $\xrightarrow{[X]} (A)$ $\xrightarrow{(i) LiAlH_4, (ii) H_2O/H^{+}} CH_3-CH_2-C(=O)-CH_2-CH_2OH + C_2H_5OH$,$[X]$ will be:

$CH_3-CH=CH-CH_2OH \xrightarrow{PCC} CH_3-CH=CH-CHO$
What is the change in hybridisation at $C-1$ in the above reaction?

The above conversion can be carried out by,

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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?