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(A-IV, B-II, C-III, D-I) . $1,6$-dibromohexane reacts with $Zn$ to undergo intramolecular cyclization to form cyclohexane. Thus,$A-iv$.$B$. Ethanol $(

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(A-IV, B-II, C-III, D-I) . $1,6$-dibromohexane reacts with $Zn$ to undergo intramolecular cyclization to form cyclohexane. Thus,$A-iv$.$B$. Ethanol $(C_2H_5OH)$ reacts with concentrated $H_2SO_4$ at $443 \ K$ to undergo dehydration to form ethene $(H_2C=CH_2)$. Thus,$B-ii$.$C$. Propene reacts with $HBr$ in the presence of peroxide (anti-Markovnikov addition) to form $1$-bromopropane $(CH_3-CH_2-CH_2-Br)$. Thus,$C-iii$.$D$. $1,1$-dibromopropane reacts with $NaNH_2$ (a strong base) to undergo dehydrohalogenation to form propyne $(H_3C-C \equiv CH)$. Thus,$D-i$.Therefore,the correct matching is $A-iv, B-ii, C-iii, D-i$.

List-$I$	List-$II$
$A$. The reaction of $1,6$-dibromohexane with $Zn$.	$i$. $H_3C-C \equiv CH$
$B$. Reaction of ethanol with concentrated $H_2SO_4$ at $443 \ K$.	$ii$. $H_2C=CH_2$
$C$. Major product in the reaction of propene with $HBr$ in the presence of benzoyl peroxide.	$iii$. $CH_3-CH_2-CH_2-Br$
$D$. The reaction of $1,1$-dibromopropane with $NaNH_2$ at $433 \ K$.	$iv$. Cyclohexane

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