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Consider the following reaction:
Ethanol $\xrightarrow{PBr_3}$ X $\xrightarrow{alc.\ KOH}$ Y $\xrightarrow[(ii)\ H_2O,\ Heat]{(i)\ H_2SO_4,\ room\ temperature}$ Z
The product Z is
Ethanol $\xrightarrow{PBr_3}$ X $\xrightarrow{alc.\ KOH}$ Y $\xrightarrow[(ii)\ H_2O,\ Heat]{(i)\ H_2SO_4,\ room\ temperature}$ Z
The product Z is
A
$CH_3CH_2-OH$
B
$CH_2=CH_2$
C
$CH_3CH_2-O-CH_2CH_3$
D
$CH_3CH_2-O-SO_3H$
Detailed Solution
Step 1: $PBr_3$ replaces –OH by –Br: $3CH_3CH_2OH + PBr_3 \rightarrow 3CH_3CH_2Br + H_3PO_3$. X is ethyl bromide.
Step 2: alcoholic KOH brings about dehydrohalogenation (elimination): $CH_3CH_2Br \xrightarrow{alc.\ KOH} CH_2=CH_2 + KBr + H_2O$. Y is ethene.
Step 3 (i): ethene adds cold concentrated $H_2SO_4$ to give ethyl hydrogen sulphate: $CH_2=CH_2 + H_2SO_4 \rightarrow CH_3CH_2OSO_3H$
Step 3 (ii): on heating with water this is hydrolysed: $CH_3CH_2OSO_3H + H_2O \rightarrow CH_3CH_2OH + H_2SO_4$
Hence Z is ethanol, $CH_3CH_2-OH$ (the starting compound is regenerated).
Step 2: alcoholic KOH brings about dehydrohalogenation (elimination): $CH_3CH_2Br \xrightarrow{alc.\ KOH} CH_2=CH_2 + KBr + H_2O$. Y is ethene.
Step 3 (i): ethene adds cold concentrated $H_2SO_4$ to give ethyl hydrogen sulphate: $CH_2=CH_2 + H_2SO_4 \rightarrow CH_3CH_2OSO_3H$
Step 3 (ii): on heating with water this is hydrolysed: $CH_3CH_2OSO_3H + H_2O \rightarrow CH_3CH_2OH + H_2SO_4$
Hence Z is ethanol, $CH_3CH_2-OH$ (the starting compound is regenerated).