Reaction (A):Ph-CH=CH-CH_3 xrightarrow[H_2O_2 | vedclass

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

Question

(D) Reaction $(A)$ is Hydroboration-Oxidation,which follows Anti-Markownikoff addition of water across the double bond.Reaction $(B)$ is Oxymercuratio

Vedclass · Accepted Answer

In reaction $(B)$,the product is a racemic mixture.

Vedclass · Answer

(D) Reaction $(A)$ is Hydroboration-Oxidation,which follows Anti-Markownikoff addition of water across the double bond.Reaction $(B)$ is Oxymercuration-Demercuration,which follows Markownikoff addition of water across the double bond.In reaction $(B)$,the starting material $CH_3-CH=CH-Ph$ is an unsymmetrical alkene. The addition of $H_2O$ follows the Markownikoff rule,resulting in the formation of a chiral center at the carbon atom attached to the phenyl group.Since the reaction proceeds through a cyclic mercurinium ion intermediate,the attack of water can occur from either side,leading to the formation of a racemic mixture of the product $CH_3-CH(OH)-CH_2-Ph$ (or $CH_3-CH_2-CH(OH)-Ph$ depending on regioselectivity,but the formation of a chiral center leads to racemization).

Reaction $(A)$:
$Ph-CH=CH-CH_3 \xrightarrow[H_2O_2 / OH^-]{B_2H_6 / THF}$ products.
Reaction $(B)$:
$CH_3-CH=CH-Ph \xrightarrow[NaBH_4]{Hg(OAc)_2 / H_2O}$ products.
Correct statement is:

Similar Questions

Which compound produces acetaldehyde upon heating with aqueous $KOH$?

An isomer of $C_3H_6Cl_2$ on boiling with aqueous $KOH$ gives acetone. Hence,the isomer is

An organic compound $A$ $(C_4H_9Cl)$ on reaction with $Na$/dry ether gives a hydrocarbon which on monochlorination gives only one chloro derivative. What is $A$?

Differentiate between the $S_N2$ and $S_N1$ mechanisms of alkyl halides.

Assertion : Benzyl bromide when kept in acetone water,it produces benzyl alcohol.
Reason : The reaction follows $S_{N}2$ mechanism.

Vedclass Test Series

Exam Paper Generator

Online Exam Module

Reaction $(A)$:$Ph-CH=CH-CH_3 \xrightarrow[H_2O_2 / OH^-]{B_2H_6 / THF}$ products.Reaction $(B)$:$CH_3-CH=CH-Ph \xrightarrow[NaBH_4]{Hg(OAc)_2 / H_2O}$ products.Correct statement is: