CH_3-C(CH_3)=CH_2 xrightarrow[R_2O_2]{HBr} X | vedclass

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

Question

(A) $1$. The reaction of $CH_3-C(CH_3)=CH_2$ with $HBr$ in the presence of peroxide $(R_2O_2)$ follows the Anti-Markovnikov addition (Kharasch effect)

Vedclass · Accepted Answer

$X = CH_3-CH(CH_3)-CH_2-Br, Y = CH_3-CH(CH_3)-CH_2-OCH_3$

Vedclass · Answer

(A) $1$. The reaction of $CH_3-C(CH_3)=CH_2$ with $HBr$ in the presence of peroxide $(R_2O_2)$ follows the Anti-Markovnikov addition (Kharasch effect). The $Br$ atom attaches to the less substituted carbon,yielding $X = CH_3-CH(CH_3)-CH_2-Br$ (isobutyl bromide).$2$. The reaction of the primary alkyl halide $(X)$ with sodium methoxide $(CH_3ONa)$ proceeds via an $S_N2$ mechanism to form an ether,$Y = CH_3-CH(CH_3)-CH_2-OCH_3$ (methyl isobutyl ether).

$CH_3-C(CH_3)=CH_2$ $\xrightarrow[R_2O_2]{HBr} X$ $\xrightarrow{CH_3ONa} Y$; $X$ and $Y$ are:

Similar Questions

Pent-$1$-ene reacts with diborane to form $X$. $X$ on oxidation using $H_2O_2$ in the presence of aqueous $NaOH$ gives $Y$. Compound $Y$ is

Coloration of $Br_2/CCl_4$ will be discharged by

What is the product formed when $CH_2 = CH(CH_2)_2COOH$ reacts with $HBr$?

In which case does the addition of an electrophile $NOT$ take place according to Markovnikov's rule?

Consider the following reactions:
$Ph-CH_2-CH=CH_2 \xrightarrow{H^+/H_2O} P$
$Ph-CH_2-CH=CH_2 \xrightarrow[(ii) NaBD_4]{(i) Hg(OAc)_2, H_2O} Q$
$Ph-CH_2-CH=CH_2 \xrightarrow[(ii) H_2O_2/OH^-]{(i) BD_3, THF} R$
Identify the products $P, Q$ and $R$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module