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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

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:

Similar Questions

Product $(A)$ is

Which of the following are not aromatic?

The electrophile involved in the sulphonation of benzene is

The number of chlorine atoms present in the organic products $X$ and $Y$ of the following reactions, respectively, are:
$\text{Benzene} + 6Cl_2 \xrightarrow[\text{dark, cold}]{\text{Anhydr. } AlCl_3} X$
$\text{Benzene} + 3Cl_2 \xrightarrow{UV, 500 \text{ K}} Y$

The hybridization involved in the six carbon atoms of benzene is

Vedclass Test Series

Exam Paper Generator

Online Exam Module