CH equiv CH xrightarrow{text{Red hot Fe tube} | vedclass

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(B) $1$. The reaction of acetylene $(CH \equiv CH)$ with a red hot iron tube leads to the formation of benzene $(A)$.$2$. Benzene undergoes Friedel-Cr

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$1-$Phenylethanol

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(B) $1$. The reaction of acetylene $(CH \equiv CH)$ with a red hot iron tube leads to the formation of benzene $(A)$.$2$. Benzene undergoes Friedel-Crafts acylation with acetyl chloride $(CH_3COCl)$ in the presence of $AlCl_3$ to form acetophenone ($B$,$C_6H_5COCH_3$).$3$. Reduction of acetophenone with $LiAlH_4$ followed by acid hydrolysis $(H_3O^\oplus)$ reduces the ketone group to a secondary alcohol,yielding $1-$phenylethanol ($C$,$C_6H_5CH(OH)CH_3$).

$CH \equiv CH$ $\xrightarrow{\text{Red hot Fe tube}} A$ $\xrightarrow{CH_3COCl / AlCl_3} B$ $\xrightarrow{(1) LiAlH_4 / (2) H_3O^\oplus} C$
$C$ will be

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$CH \equiv CH$ $\xrightarrow{\text{Red hot Fe tube}} A$ $\xrightarrow{CH_3COCl / AlCl_3} B$ $\xrightarrow{(1) LiAlH_4 / (2) H_3O^\oplus} C$ $C$ will be