The reaction of $C_{6} H_{5} C H = C H C H O$ with $L i A l H_{4}$ gives.

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

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

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

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C_{6} H_{5} C H_{2} C H O H C H_{3}

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Solution

The correct option is B $C_{6} H_{5} C H_{2} C H_{2} C H O$
$L i A l H_{4}$ does not reduce alkene, but when the alkene is in conjugation with aldehyde and phenyl group is attached to it, then alkene also get reduced.
In amine formation, $p^{H}$ must maintain
$R - C - H ⟶ R - C H = N - R$
At high $p^{H}$ , there wont be enough acid to protanate the $O H$ in the intermediate to allow for removal as $H_{2} O$ . At low $p^{H}$ , most of the amine reactant will be tied up as its ammonium conjugate acid and will become non-nucleophilic.

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Similar questions

C_{6} H_{5} - C H = C H C H O X \to C_{6} H_{5} C H = C H C H_{2} O H

C_{6} H_{5} C H O N H_{2} O H / H^{+} - ------- \to △ (X) P_{2} O_{5} - -- \to △ (Y) H_{2} O / H^{+} - ---- \to (Z)

X

Y

Z

The reaction of $C_{6} H_{5} C H = C H C H_{3}$ with HBr produces :

C_{6} H_{5} C H O + (C H_{3} C O)_{2} O C H_{3} C O O N a - ------- \to 180^{\circ} C X

Predict the order of reactivity of the following compounds in SN1 and SN2 reactions

C6H5CH2Br , C6H5CH(C6H5)Br , C6H5CH(CH3)Br , C6H5C(CH3)(C6H5)

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