Ph - mathop Olimits^{18} - CH_3 xrightarrow{c | vedclass

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

Question

(A) The cleavage of ethers with $conc. HI$ follows an $S_N2$ mechanism for primary alkyl groups. In the case of anisole $(Ph-O-CH_3)$,the $C-O$ bond b

Vedclass · Accepted Answer

$Ph - \mathop O\limits^{18} H + CH_3 - I$

Vedclass · Answer

(A) The cleavage of ethers with $conc. HI$ follows an $S_N2$ mechanism for primary alkyl groups. In the case of anisole $(Ph-O-CH_3)$,the $C-O$ bond between the methyl group and oxygen is weaker than the $C-O$ bond between the phenyl group and oxygen due to the partial double bond character of the $Ph-O$ bond. Therefore,the $HI$ attacks the methyl group,leading to the formation of $CH_3I$ and phenol $(Ph-OH)$. Since the oxygen atom in the ether is labeled with $^{18}O$,the label will remain with the phenol product $(Ph - \mathop O\limits^{18} H)$. Thus,the products are $Ph - \mathop O\limits^{18} H$ and $CH_3 - I$.

$Ph - \mathop O\limits^{18} - CH_3 \xrightarrow{conc. HI} \text{products}$

Similar Questions

Explain: In alkoxybenzene,the alkoxy group is ortho-para directing,but why does electrophilic substitution occur mainly at the para-position and to a lesser extent at the ortho-position?

Among the following ethers,which one will produce methyl alcohol on treatment with hot concentrated $HI$?

Which reagent is used for the electrophilic aromatic bromination of methyl phenyl ether (anisole)?

Reactant $(A)$ is:

Product $(B)$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module