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(C) Step $1$: Benzene reacts with acetyl chloride $(CH_3COCl)$ in the presence of anhydrous $AlCl_3$ (Friedel-Crafts acylation) to form acetophenone $

Vedclass · Accepted Answer

$C_9H_{12}O$

Vedclass · Answer

(C) Step $1$: Benzene reacts with acetyl chloride $(CH_3COCl)$ in the presence of anhydrous $AlCl_3$ (Friedel-Crafts acylation) to form acetophenone $(P)$,which is $C_6H_5COCH_3$.Step $2$: Acetophenone $(P)$ reacts with methyl magnesium bromide $(CH_3MgBr)$ followed by hydrolysis $(H_2O)$ to form $2$-phenylpropan-$2$-ol $(Q)$.The reaction is: $C_6H_5COCH_3 + CH_3MgBr$ $\rightarrow C_6H_5C(OMgBr)(CH_3)_2$ $\xrightarrow{H_2O} C_6H_5C(OH)(CH_3)_2$.The structure of $Q$ is $C_6H_5C(OH)(CH_3)_2$,which contains $9$ carbon atoms,$12$ hydrogen atoms,and $1$ oxygen atom.Thus,the molecular formula of $Q$ is $C_9H_{12}O$.

Benzene on reaction with acetyl chloride in the presence of anhydrous $AlCl_3$ gave the product $P$. The product $P$ on reaction with methyl magnesium bromide followed by treatment with water furnished the product $Q$. The molecular formula of $Q$ is

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