Reaction of ethanol with conc. sulphuric acid | vedclass

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Question

(A) Step $1$: Ethanol $(CH_3CH_2OH)$ undergoes dehydration in the presence of concentrated sulphuric acid $(H_2SO_4)$ at $170^{\circ} C$ to produce et

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$1,2$-dibromoethane

Vedclass · Answer

(A) Step $1$: Ethanol $(CH_3CH_2OH)$ undergoes dehydration in the presence of concentrated sulphuric acid $(H_2SO_4)$ at $170^{\circ} C$ to produce ethene $(CH_2=CH_2)$ gas.$CH_3CH_2OH \xrightarrow{Conc. H_2SO_4, 170^{\circ} C} CH_2=CH_2 + H_2O$Step $2$: The ethene gas is then treated with bromine $(Br_2)$ in carbon tetrachloride $(CCl_4)$,which is an electrophilic addition reaction,resulting in the formation of $1,2$-dibromoethane.$CH_2=CH_2 + Br_2 \xrightarrow{CCl_4} CH_2Br-CH_2Br$Therefore,the major product is $1,2$-dibromoethane.

Reaction of ethanol with conc. sulphuric acid at $170^{\circ} C$ produces a gas which is then treated with bromine in carbon tetrachloride. The major product obtained in this reaction is

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