The product of the reaction is:

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

Question

(A) The reaction involves the nucleophilic ring opening of an epoxide by a Grignard reagent $(RMgX)$.In an unsymmetrical epoxide,the nucleophile $(R^-

Vedclass · Accepted Answer

The product is $Ph-CH_2-CH(OH)-CH(R)-CH_3$

Vedclass · Answer

(A) The reaction involves the nucleophilic ring opening of an epoxide by a Grignard reagent $(RMgX)$.In an unsymmetrical epoxide,the nucleophile $(R^-)$ attacks the less sterically hindered carbon atom.The epoxide is $1-phenyl-2,3-epoxybutane$ (or $2-methyl-3-benzyloxirane$).The carbon atom attached to the $CH_3$ group is less sterically hindered compared to the carbon atom attached to the $PhCH_2$ group.Therefore,the $R$ group attacks the carbon atom bearing the $CH_3$ group.After the attack,the $O^-$ is protonated by $H_3O^+$ to form an alcohol.The resulting product is $1-phenyl-3-substituted-butan-2-ol$,which is $Ph-CH_2-CH(OH)-CH(R)-CH_3$.

Similar Questions

Which of the following gives a ketone on oxidation?

Which product is not obtained when $\text{CH}_3\text{CH}_2\text{OH}$ reacts with $\text{H}_2\text{SO}_4$?

Based on which rule is $2$-butene obtained as the major product from the dehydration of $2$-butanol?

In the following sequence of reaction,$Z$ $\xrightarrow{PCl_5} X$ $\xrightarrow[\Delta]{\text{Alc. KOH}} Y$ $\xrightarrow{H_2O/H^{+}} Z$ (Major),$Z$ is:

$HO-CH_2-C(=O)-CH(OH)-CH_2-OH \xrightarrow{HIO_4} \text{Product obtained is}$

Vedclass Test Series

Exam Paper Generator

Online Exam Module