The product B formed in the following reactio | vedclass

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(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

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(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.

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

Similar Questions

Identify $A$ and $B$ in the following reaction:
$C_2H_5Cl \xrightarrow{A} C_2H_5OH \xleftarrow{B} C_2H_5Cl$

The correct reactivity order for $E_2$ reaction with alcoholic $KOH$ will be:
$(i)$ $3-$chlorocyclohexene derivative (conjugated system)
(ii) $1-$chlorodecalin derivative (with double bond)
(iii) $1-$chlorodecalin (saturated)

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

The product of the reaction is:

Arrange the following halides in decreasing order of $S_N1$ reactivity:
$(I)$ $CH_3CH_2CH_2Cl$
$(II)$ $CH_2=CHCHClCH_3$
$(III)$ $CH_3CH_2CHClCH_3$

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