Consider the following reaction sequence:C_6H | vedclass

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(D) $1$. The starting material is cinnamaldehyde $(C_6H_5-CH=CH-CHO)$.$2$. Treatment with $LiAlH_4$ at $35^{\circ}C$ reduces both the aldehyde group a

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$CH_4$

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(D) $1$. The starting material is cinnamaldehyde $(C_6H_5-CH=CH-CHO)$.$2$. Treatment with $LiAlH_4$ at $35^{\circ}C$ reduces both the aldehyde group and the carbon-carbon double bond to give $3-$phenylpropan$-1-$ol $(A = C_6H_5-CH_2-CH_2-CH_2OH)$.$3$. When $A$ is treated with excess $CH_3MgBr$,the Grignard reagent acts as a base and reacts with the acidic hydroxyl group $(-OH)$ of the alcohol to form an alkoxide and methane gas $(CH_4)$.$4$. The reaction is: $C_6H_5-CH_2-CH_2-CH_2OH + CH_3MgBr \rightarrow C_6H_5-CH_2-CH_2-CH_2OMgBr + CH_4$.$5$. Thus,$B$ is $CH_4$.

Consider the following reaction sequence:
$C_6H_5-CH=CH-CHO$ $\xrightarrow[35^{\circ}C]{LiAlH_4} A$ $\xrightarrow[Excess]{CH_3MgBr} B$
Identify $B$ in the reaction.

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Consider the following reaction sequence:$C_6H_5-CH=CH-CHO$ $\xrightarrow[35^{\circ}C]{LiAlH_4} A$ $\xrightarrow[Excess]{CH_3MgBr} B$Identify $B$ in the reaction.