Give the reasons for the following:



Give reasons for the following.
(i) Ethyl iodide undergoes 2 reaction faster than ethyl bromide.
(ii) (±) 2-butanol is optically inactive.
(iii) C—X bond length in halobenzene is smaller than C—X bond length
The C—Cl bond length in chlorobenzene is shorter


(i) Iodide is a better leaving group because of its larger size than bromide, therefore, ethyl iodide undergoes SN 2 reaction faster than ethyl bromide.
(ii) (±) 2-butanol is a racemic mixture. It is a mixture which contains two enantiomers in equal proportion and thus, have
zero optical rotation due to internal compensation. Therefore, it is optically inactive.
(iii) In halobenzenes (like chlorobenzene), the lone pair of electrons on halogen atom is delocalised on the benzene ring. As a result, C — X bond (C — Cl bond in case of chlorobenzene) acquires some double bond character while in
$CH _{ 3 }$— X, C—X bond is a pure single bond. Therefore, C—X bond in halobenzene is shorter than in $CH _{ 3 }$—X.