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Reductions Involving Aldehydes or Ketones

Aldehydes and...

Question

Aldehydes and ketones undergo reduction. The products formed depend on the reducing agent.
$C H_{3} - C H = C H - C H O L i A l H_{4} - --- \to (P)$
$C_{6} H_{5} - C H O N a, C_{2} H_{5} O H - ------- \to (Q)$
The compounds (P) and (Q) respectively are :

C H_{3} - C H_{2} - C H_{2} - C H O

a n d

C_{6} H_{5} - C H_{3}

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C H_{3} - C H_{2} - C H_{2} - C H O

a n d

C_{6} H_{5} - C H_{2} O H

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C H_{3} - C H = C H - C H_{2} O H

a n d

C_{6} H_{5} - C H O H - C H O H - C_{6} H_{5}

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C_{6} H_{5} - C H_{3}

and

C H_{3} - C H_{2} - C H_{2} - C H_{3}

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Solution

The correct option is C $C H_{3} - C H = C H - C H_{2} O H$ $a n d$ $C_{6} H_{5} - C H O H - C H O H - C_{6} H_{5}$
When the reducing agent is lithium aluminium hydride, the aldehyde group is reduced to primary alcohol but the carbon-carbon double bond is unaffected.
$C H_{3} - C H = C H - C H O L i A l H_{4} ⟶ C H_{3} - C H = C H - C H_{2} O H (P)$
When the reducing agent is sodium in ethanol, the aldehyde group is reduced to alcohol and the product formed is obtained by coupling reaction.
$2 C_{6} H_{5} - C H O N a, C_{2} H_{5} O H ⟶ C_{6} H_{5} - C H O H - C H O H - C_{6} H_{5} (Q)$

Therefore, the correct option is $C$ .

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I_{2} / N a O H

C H_{3} - C O O H

C H_{2} = C H - C O O H

C_{6} H_{5} - C O O H

C_{6} H_{5} - C H_{2} - C O O H

C H_{3} - C H_{2} - C O O H

C H_{3} - C H | O H - C O O H

C H_{3} - C O - C H_{3} \to C H_{3} - C H - C H_{3}

O H

C_{6} H_{5} - C H_{2} - C H_{3} \to C_{6} H_{5} - C O O^{-} K^{+}

E_{1}

C H_{3} - H O | C H - C H_{3}

C H_{3} - H O | C | C H_{3} H - C H_{3}

C_{6} H_{5} - H O | C | C H_{3} H - C H_{3}

C H_{3} - C H_{2} - C H_{2} O H

C H_{2} = C H - ⨁ C H_{2}

C_{6} H_{5} - ⨁ C H_{2}

C H_{3} - ⨁ C H - C H_{3}

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