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(D) The reaction of benzyl phenyl ether with $HI$ involves the protonation of the ether oxygen atom. Following protonation,the $C-O$ bond between the

Vedclass · Accepted Answer

$A = C_6H_5CH_2I$ and $B = C_6H_5OH$

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(D) The reaction of benzyl phenyl ether with $HI$ involves the protonation of the ether oxygen atom. Following protonation,the $C-O$ bond between the benzyl carbon and the oxygen atom breaks to form a stable benzyl carbocation $(C_6H_5CH_2^+)$ and phenol $(C_6H_5OH)$. The benzyl carbocation is stabilized by resonance with the phenyl ring. Finally,the iodide ion $(I^-)$ attacks the benzyl carbocation to form benzyl iodide $(C_6H_5CH_2I)$. Thus,the products are $A = C_6H_5CH_2I$ and $B = C_6H_5OH$.

Consider the following reaction. Identify products $A$ and $B$ :-

Similar Questions

Consider the reaction sequence:
Dimethyl ketone $\xrightarrow[(ii) H_2O]{(i) CH_3MgCl} X$ $\xrightarrow[(ii) CH_3Br]{(i) Na} Y$
How many $sp^3$ carbons are present in $Y$?

Diethyl ether absorbs oxygen to form

Assertion : Iodide ion combines with smaller group to avoid steric hindrance.
Reason : With $HI$,anisole gives iodobenzene and methyl alcohol.

Complete the following reaction sequence.
$CH_3-CO-CH_3$ $\xrightarrow[(ii) H_2O]{(i) CH_3MgBr} [A]$ $\xrightarrow{Na, \text{Ether}} [B]$ $\xrightarrow{CH_3-Br} [C]$

In the following reaction,$C_2H_5OC_2H_5 + 4[H] \xrightarrow{\text{Red } P + HI} 2X + H_2O$,$X$ is

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