The product B formed in the following reactio | vedclass

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

Question

(D) Step $1$: The reaction of $1-(4-methylphenyl)prop-1-ene$ with $HCl$ follows Markovnikov's rule. The electrophilic addition of $H^+$ to the double

Vedclass · Accepted Answer

$1-$($4$-methylphenyl)propyl isocyanide

Vedclass · Answer

(D) Step $1$: The reaction of $1-(4-methylphenyl)prop-1-ene$ with $HCl$ follows Markovnikov's rule. The electrophilic addition of $H^+$ to the double bond forms a stable benzylic carbocation at the carbon adjacent to the benzene ring. Then,$Cl^-$ attacks this carbocation to form $1-(4-methylphenyl)-1-chloropropane$ as the major product $(A)$.Step $2$: The reaction of the alkyl halide $(A)$ with $AgCN$ is a nucleophilic substitution reaction. Since $AgCN$ is a covalent compound,the nitrogen atom acts as the nucleophile,leading to the formation of an isocyanide (isonitrile) as the major product $(B)$.Thus,the final product is $1-(4-methylphenyl)propyl$ isocyanide.

Reaction	Product
$I$. $RX + AgCN$	$RNC$
$II$. $RX + KCN$	$RCN$
$III$. $RX + KNO_2$	$RONO$
$IV$. $RX + AgNO_2$	$RNO_2$

The product $B$ formed in the following reaction sequence is:

Similar Questions

Which of the following pairs is/are correctly matched?

Reaction Product

$I$. $RX + AgCN$ $RNC$

$II$. $RX + KCN$ $RCN$

$III$. $RX + KNO_2$ $RONO$

$IV$. $RX + AgNO_2$ $RNO_2$

Which of the following chloro-compounds shows complete inversion of configuration during a reaction?

The rate of $S_{N}1$ reaction for the given compounds is:

Grignard reagent is not prepared in aqueous medium but prepared in ether medium because the reagent

$C_2H_5I$ $\xrightarrow{\text{alc. } KOH} X$ $\xrightarrow{Br_2} Y$ $\xrightarrow{KCN} Z$; $Z = \dots$

Vedclass Test Series

Exam Paper Generator

Online Exam Module