CBSE Class 12 · Haloalkanes and HaloarenesShort Answer

(i) Out of (CH

_{3}

)

_{3}

C–Br and (CH

_{3}

)

_{3}

C–I, which one is more reactive towards S

_{N}

1, and why?
(ii) Write the product formed when p-nitrochlorobenzene is heated with aqueous NaOH at 443 K, followed by acidification.
(iii) Why are the dextro- and laevo-rotatory isomers of butan-2-ol difficult to separate by fractional distillation?

Solution

(i) \textbf{(CH

_{3}

)

_{3}

C–I}: the C–I bond is the weakest and I

^{-}

the best leaving group, so ionisation (the S

_{N}

1 slow step) is fastest.
(ii) The o/p-nitro group activates the haloarene toward nucleophilic substitution:

p - O_{2} N - C_{6} H_{4} - Cl NaOH (aq) 443 K p - O_{2} N - C_{6} H_{4} - ONa H^{+} p-nitrophenol

(iii) The d- and l-forms of butan-2-ol are enantiomers: every scalar physical property — boiling point included — is identical; they differ only in the sign of optical rotation, so fractional distillation cannot tell them apart.

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CBSE Class 12 · Haloalkanes and HaloarenesShort Answer

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