Which of the following reaction(s) can be used for the preparation of alkyl halides ?
(I) $C H_{3} C H_{2} O H + H C l a n h . Z n C l_{2} - ----- \to$
(II) $C H_{3} C H_{2} O H + H C l \to$
(III) $(C H_{3})_{3} C O H + H C l \to$
(IV) $(C H_{3})_{2} C H O H + H C l a n h . Z n C l_{2} - ----- \to$

(IV) only

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(III) and (IV) only

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(I), (III) and (IV) only

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(I) and (II) only

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Solution

The correct option is C (I), (III) and (IV) only
Luca's reagent $(H C l + Z n C l_{2})$ can be used for $1^{0} / 2^{0} / 3^{0}$ -alcohols, but if $Z n C l_{2}$ is absent then only $3^{0}$ alcohol can be used. This is because it is reactive due to the formation of a stable $3^{0}$ carbocation.

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Similar questions

C H_{3} C H_{2} O H + H C l a n h . Z n C l_{2} - ----- \to

C H_{3} C H_{2} O H + H C l \to

(C H_{3})_{3} C O H + H C l \to

(C H_{3})_{2} C H O H + H C l a n h . Z n C l_{2} - ----- \to

C H_{3} C H_{2} O H + {C l}_{2} ⟶ C H_{3} C H O + H C l

C H_{3} C H O + {C l}_{2} ⟶ C {C l}_{3} C H O + H C l

C H_{4} + {C l}_{2} ⟶ C H_{3} C l + H C l

C H_{3} C H_{2} O H + C l_{2} \to C H_{3} C H O + H C l

C H_{3} C H O + C l_{2} \to C C l_{3} C H O + H C l

C H_{4} + C l h v - \to C H_{3} C l + H C l

C H_{3} C H_{2} O H 1

(C H_{3})_{2} C H O H 2

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

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