In solvolysis of 1,2-dimethylpropyl p-toluene | vedclass

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

Question

(D) The solvolysis of a secondary substrate like $1,2$-dimethylpropyl $p$-toluenesulfonate in a polar protic solvent such as acetic acid proceeds via

Vedclass · Accepted Answer

$S_N1, E_1$

Vedclass · Answer

(D) The solvolysis of a secondary substrate like $1,2$-dimethylpropyl $p$-toluenesulfonate in a polar protic solvent such as acetic acid proceeds via the formation of a carbocation intermediate.Since the solvent is polar protic and the substrate is secondary,the rate-determining step is the ionization of the leaving group ($p$-toluenesulfonate) to form a carbocation.This carbocation intermediate can then undergo nucleophilic attack by the solvent to form substitution products $(S_N1)$ or lose a proton to form alkenes $(E_1)$.Both processes follow unimolecular kinetics,hence the mechanism is $S_N1$ and $E_1$.

In solvolysis of $1,2$-dimethylpropyl $p$-toluenesulfonate in acetic acid at $75^\circ C$,(alkene + substitution products) will be formed by which mechanism?

Similar Questions

In the reaction shown,the major product obtained is:

$2-$chloro$-2-$methylpentane on reaction with sodium methoxide in methanol yields:
$(1)\ C_2H_5-CH_2-C(CH_3)_2-OCH_3$
$(2)\ C_2H_5-CH_2-C(CH_3)=CH_2$
$(3)\ C_2H_5-CH=C(CH_3)_2$

In the following pairs of halogen compounds,which compound undergoes faster $S_N1$ reaction?
$(i)$ $2$-chloro-$2$-methylpropane and $3$-chloropentane
(ii) $2$-chloroheptane and $1$-chlorohexane

Out of $C_{6}H_{5}CH_{2}Cl$ and $C_{6}H_{5}CHClC_{6}H_{5},$ which is more easily hydrolysed by aqueous $KOH$?

Alkyl halides react with $Mg$ in dry ether to form

Vedclass Test Series

Exam Paper Generator

Online Exam Module