The major organic product in the reaction is: | vedclass

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

Question

(A) In the reaction of an unsymmetrical ether with $HI$,if the alkyl groups are primary or secondary,the reaction follows the $S_N2$ mechanism.The iod

Vedclass · Accepted Answer

$CH_3I + (CH_3)_2CHOH$

Vedclass · Answer

(A) In the reaction of an unsymmetrical ether with $HI$,if the alkyl groups are primary or secondary,the reaction follows the $S_N2$ mechanism.The iodide ion $(I^{-})$ attacks the smaller (less sterically hindered) alkyl group.In $CH_3-O-CH(CH_3)_2$,the methyl group $(-CH_3)$ is smaller than the isopropyl group $(-CH(CH_3)_2)$.Therefore,the $I^{-}$ ion attacks the methyl group,leading to the formation of $CH_3I$ and $(CH_3)_2CHOH$ as the major products.

The major organic product in the reaction is: $CH_3-O-CH(CH_3)_2 + HI \to$ products

Similar Questions

$C_6H_5OC_2H_5 + HI \rightarrow X + Y$. Here $X$ and $Y$ are . . . . . . .

The product of the reaction $CH_3O-CH_2-CH(OCH_3)-CH_2-OCH_3 \xrightarrow{HI(excess)}$ is:

What is $X$ in the following reaction?

What are $A$ and $B$ in the following reaction?
$CH_3-CH_2-CH_2-O-C(CH_3)_2-CH_2-CH_3 \xrightarrow{HI} A + B$

Assertion : The major products formed by heating $C_6H_5CH_2OCH_3$ with $HI$ are $C_6H_5CH_2I$ and $CH_3OH$.
Reason : Benzyl cation is more stable than methyl cation.

Vedclass Test Series

Exam Paper Generator

Online Exam Module