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Identify the major products P, Q and R in the following sequence of reactions:
Explanation
Friedel–Crafts alkylation with rearrangement gives cumene, which is converted to phenol and acetone.
Detailed Solution
$CH_3CH_2CH_2-Cl + AlCl_3 \rightarrow CH_3CH_2\overset{\delta+}{C}H_2\cdots Cl\cdots\overset{\delta-}{AlCl_3}$ (incipient carbocation)
1,2-H shift gives the more stable secondary carbocation $CH_3-\overset{+}{C}H-CH_3$.
Benzene $+ CH_3-\overset{+}{C}H-CH_3 \rightarrow C_6H_5CH(CH_3)_2$ (cumene, P)
$C_6H_5CH(CH_3)_2 \xrightarrow{O_2} C_6H_5C(CH_3)_2-O-O-H$ (cumene hydroperoxide)
$C_6H_5C(CH_3)_2OOH \xrightarrow[\text{rearrangement}]{H^+/H_2O} C_6H_5OH\ (Q) + CH_3COCH_3\ (R)$
So P = cumene, Q = phenol and R = acetone.
1,2-H shift gives the more stable secondary carbocation $CH_3-\overset{+}{C}H-CH_3$.
Benzene $+ CH_3-\overset{+}{C}H-CH_3 \rightarrow C_6H_5CH(CH_3)_2$ (cumene, P)
$C_6H_5CH(CH_3)_2 \xrightarrow{O_2} C_6H_5C(CH_3)_2-O-O-H$ (cumene hydroperoxide)
$C_6H_5C(CH_3)_2OOH \xrightarrow[\text{rearrangement}]{H^+/H_2O} C_6H_5OH\ (Q) + CH_3COCH_3\ (R)$
So P = cumene, Q = phenol and R = acetone.