The products P and Q in the following sequenc | vedclass

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(A) $1$. The first step is a Birch reduction of $p$-methylanisole. The electron-donating $-OCH_3$ group directs the reduction to give $1$-methoxy-$4$-

Vedclass · Accepted Answer

$P$ = $1-$methoxy$-4-$methylcyclohexa$-1,4-$diene,$Q$ = $5-$oxohexanal

Vedclass · Answer

(A) $1$. The first step is a Birch reduction of $p$-methylanisole. The electron-donating $-OCH_3$ group directs the reduction to give $1$-methoxy-$4$-methylcyclohexa-$1,4$-diene as product $P$.$2$. The second step is ozonolysis of $P$ with $1$ equivalent of $O_3$. The more electron-rich double bond (the one conjugated with the $-OCH_3$ group) reacts with $O_3$.$3$. Cleavage of the $C1=C2$ double bond in $1$-methoxy-$4$-methylcyclohexa-$1,4$-diene yields a dicarbonyl compound. Specifically,the $C1$ becomes a ketone (as part of an ester/enol ether derivative) and $C2$ becomes an aldehyde. The resulting structure is $5$-oxohexanal.

The products $P$ and $Q$ in the following sequence of reactions are:
$p$-Methylanisole $\xrightarrow{Li, EtOH, NH_3(l)} P$ $\xrightarrow[(ii) -78^{\circ}C, Me_2S]{(i) O_3 (1 \text{ equiv.})} Q$

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The products $P$ and $Q$ in the following sequence of reactions are:$p$-Methylanisole $\xrightarrow{Li, EtOH, NH_3(l)} P$ $\xrightarrow[(ii) -78^{\circ}C, Me_2S]{(i) O_3 (1 \text{ equiv.})} Q$