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(B) The reaction involves a nucleophilic substitution $(S_N2)$ on benzyl chloride $(C_6H_5CH_2Cl)$.There are two nucleophiles present: phenoxide ion $

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Benzyl ethyl ether $(C_6H_5CH_2OC_2H_5)$

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(B) The reaction involves a nucleophilic substitution $(S_N2)$ on benzyl chloride $(C_6H_5CH_2Cl)$.There are two nucleophiles present: phenoxide ion $(C_6H_5O^-)$ and ethoxide ion $(C_2H_5O^-)$.Ethoxide ion $(C_2H_5O^-)$ is a stronger nucleophile than the phenoxide ion $(C_6H_5O^-)$ because the negative charge on the phenoxide ion is delocalized over the benzene ring,making it less available for attack.Therefore,the ethoxide ion will preferentially attack the electrophilic carbon of benzyl chloride to form benzyl ethyl ether $(C_6H_5CH_2OC_2H_5)$ as the major product.

Column-$I$	Column-$II$
$(A)$ $CH_3-O-CH_3$	$(1)$ $C_6H_5OH + CH_3I$
$(B)$ $(CH_3)_2CH-O-CH_3$	$(2)$ $(CH_3)_3C-I + CH_3OH$
$(C)$ $(CH_3)_3C-O-CH_3$	$(3)$ $C_6H_5I + CH_3OH$
$(D)$ $C_6H_5-O-CH_3$	$(4)$ $CH_3OH + CH_3I$
	$(5)$ $(CH_3)_2CHOH + CH_3I$
	$(6)$ $(CH_3)_2CH-I + CH_3OH$
	$(7)$ $(CH_3)_3COH + CH_3I$

The major product in the given reaction will be:
(One mole of each)
$C_6H_5O^-Na^+ + C_6H_5CH_2Cl + C_2H_5OK^+ \xrightarrow{\text{Major Product}}$

Similar Questions

The ether $C_6H_5-O-CH_2-C_6H_5$ when treated with $HI$ produces:

What are $X$ and $Y$ in the following reaction?
$(CH_3)_3COCH_3 + HI \rightarrow X + Y$

Write the equations for the reaction of hydrogen iodide with:
$(i)$ $1-propoxypropane$
$(ii)$ $methoxybenzene$
$(iii)$ $benzyl \ ethyl \ ether$

$Ph-O-CH_2-Ph \xrightarrow{HI} A + B$ will be

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The major product in the given reaction will be:(One mole of each)$C_6H_5O^-Na^+ + C_6H_5CH_2Cl + C_2H_5OK^+ \xrightarrow{\text{Major Product}}$