When alcohol reacts with concentrated H_2SO_4 | vedclass

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(A) The correct answer is $A$. In the dehydration of alcohols using concentrated $H_2SO_4$,the reaction proceeds via the formation of a carbonium ion

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Carbonium ion

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(A) The correct answer is $A$. In the dehydration of alcohols using concentrated $H_2SO_4$,the reaction proceeds via the formation of a carbonium ion (carbocation) intermediate.Step $1$: Protonation of the alcohol group: $R-CH_2-CH_2-OH + H^+ \rightarrow R-CH_2-CH_2-OH_2^+$Step $2$: Loss of water molecule to form a carbonium ion: $R-CH_2-CH_2-OH_2^+ \rightarrow R-CH_2-CH_2^+ + H_2O$Step $3$: The carbonium ion may undergo $1, 2-$ hydride or $1, 2-$ methyl shifts to form a more stable carbonium ion before losing a proton to form an alkene.

Column-$I$ (Reagents)	Column-$II$ (Products)
$A$. $HCl + ZnCl_2$	$i$. $CH_3CH=CHCH_3$
$B$. $HBr$	$ii$. $CH_3CH_2CH_2CH_2Cl$
$C$. $SOCl_2$	$iii$. $CH_3CH_2CH_2CH_2Br$
$D$. $H_2SO_4, \Delta$	$iv$. $CH_3CH_2CH_2CH_2Cl$

When alcohol reacts with concentrated $H_2SO_4$,the intermediate compound formed is:

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