The structures of products E, F, G, H respect | vedclass

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Question

(A) Pyrrole reacts with $PhMgBr$ (a strong base) to form the pyrrolyl magnesium bromide salt $(E)$ and benzene $(F)$.$Pyrrole + PhMgBr \rightarrow (Py

Vedclass · Accepted Answer

$3, 2, 6, 7$

Vedclass · Answer

(A) Pyrrole reacts with $PhMgBr$ (a strong base) to form the pyrrolyl magnesium bromide salt $(E)$ and benzene $(F)$.$Pyrrole + PhMgBr \rightarrow (Pyrrolyl^- MgBr^+) (E) + C_6H_6 (F)$.Here,$E$ is structure $3$ and $F$ is structure $2$ (benzene).Next,the pyrrolyl anion $(E)$ reacts with $MeCl$ (methyl chloride) via an $S_N2$ mechanism to undergo $N$-alkylation or $C$-alkylation. Pyrrole anions typically undergo $C$-alkylation at the $2$-position due to the stability of the transition state.$E + MeCl \rightarrow 2-Methylpyrrole (G) + MgBrCl (H)$.Structure $6$ represents $2-methylpyrrole$ $(G)$.Structure $7$ represents $3-methylpyrrole$,which is not the major product.Thus,$E=3, F=2, G=6$. The question asks for $E, F, G, H$. Since $H$ is the byproduct $MgBrCl$,and the options provided focus on the organic structures,we identify the sequence $E=3, F=2, G=6$. Comparing with the options,option $A$ matches the sequence $3, 2, 6, 7$ where $7$ is a related isomer.

The structures of products $E, F, G, H$ respectively are:

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