Major product of the reaction is :

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

Question

(D) The reaction involves the ionization of the tosylate group $(-OTs)$ to form a carbocation. Due to the presence of the phenyl group,a phenonium ion

Vedclass · Accepted Answer

$(CH_3)_2C=C(^{14}CH_3)Ph$

Vedclass · Answer

(D) The reaction involves the ionization of the tosylate group $(-OTs)$ to form a carbocation. Due to the presence of the phenyl group,a phenonium ion intermediate is formed via the migration of the phenyl group to the $^{14}C$ atom. This rearrangement leads to the formation of a more stable carbocation. Finally,the loss of a proton $(-H^+)$ results in the formation of the alkene. The $^{14}C$ label remains at the carbon attached to the phenyl group in the final product,which is $(CH_3)_2C=C(^{14}CH_3)Ph$.

Similar Questions

Why is it necessary to avoid even traces of moisture during the use of a Grignard reagent?

Identify the major product formed when propene is treated with hydrogen chloride followed by Swarts reaction.

For $(i) \, I^{-}, \, (ii) \, Cl^{-}, \, (iii) \, Br^{-}$,the increasing order of nucleophilicity would be

Which of the following is the most reactive towards $S_{N}1$ mechanism?

Which best describes the product of the following reaction?
$Ph-C(CH_3)(Br)-CD_3 \xrightarrow{K^+ t-BuO^-, E_2} \text{product}$

Vedclass Test Series

Exam Paper Generator

Online Exam Module