(i) Out of ${(C H_{3})}_{3} C - B r$ and ${(C H_{3})}_{3} C - I$ , 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 443K followed by acidification?
(iii) Why dextro and laevo-rotatory isomers of Butan-2-ol are difficult to separate by fractional distillation?

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Solution

(i) The reactivity of $S_{N} 1$ reactions depends upon the ease qith which the leaving group leaves the substrate and the stability of the intermediate formed. The intermediate carbocation is same in both cases. Iodide ion is a better leaving group than bromide ion and so ${(C H_{3})}_{3} C C - B I$ will be more reactive towards $S_{N} 1$ reactions.
(ii) When p-nitrochlorobenzene is heated with aqueous NaOH at 443K followed by acidification, $O H^{-}$ group will replace $C l^{-}$ and 4-Nitro-phenol is formed in acidic medium.
(iii) Since dextro and laevo-rotatory isomers are enantiomers they have the same boiling point. So it is difficult to separate them by fractional distillation.

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{(C H_{3})}_{3} C - B r

{(C H_{3})}_{3} C - I

S_{N} 1

(C H_{3})_{3} C - O - C H_{3}

(C H_{3}) C - I

C H_{3} - O H

(C H_{3})_{3} C - O H

C H_{3} - I

Which of the following alkyl halides will undergo $S_{N} 1$ reaction most readily?

(a) $(C H_{3})_{3} C - F$ (b) $(C H_{3})_{3} C - C l$ (c) $(C H_{3})_{3} C - B r$ (d) $(C H_{3})_{3} C - I$

C H_{3} - C H O, C_{6} H_{5} C O C H_{3}, H C H O

p K_{a}

C l - C H_{2} - C O O H

C H_{3} C O O H

C H_{3} C H_{2} C H = C H - C H_{2} C N 1. (i - B u)_{2} A I H - -------- \to 2. H_{2} O

C_{3} H_{6} O

I_{2}

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