CH3 CH2 OH - HCl CH3 CH2 Cl- H2O In the above reaction- the leaving group is -

Zinc chloride is a Lewis acid and coordinates with the alcoholic group. R ....O + ZnCl_2 → R H+O Z ⊖nCl_2 → R^+ + HO Z ⊖nCl_2 R^+ + Cl^ (from HCl) → RC ...

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Standard XI

Chemistry

Chemical Properties of Alcohol

CH3 CH2 OH + ...

Question

$C H_{3} C H_{2} O H + H C l Z n C l_{2} - --- \to C H_{3} C H_{2} C l + H_{2} O$ In the above reaction, the leaving group is ?

$O H^{-}$

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$H_{2} O$

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$H O - Z ⊖ n C l_{2}$

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$H_{3} O^{+}$

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Solution

The correct option is C
$H O - Z ⊖ n C l_{2}$

Zinc chloride is a Lewis acid and coordinates with the alcoholic group.
$R - . . O . . + Z n C l_{2} \to R - + O H - Z ⊖ n C l_{2} \to R^{+} + H O - Z ⊖ n C l_{2}$
$R^{+} + C l^{-} (f r o m H C l) \to R C l + H^{+}$
$H^{+} + H O^{-} Z n C l_{2} \to H_{2} O + Z n C l_{2}$

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Similar questions

$C H_{3} C H_{2} O H + H C l Z n C l_{2} - --- \to C H_{3} C H_{2} C l + H_{2} O$ In the above reaction, the leaving group is ?

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

In the above reaction, the leaving group is:

Q. In the preparations of ethyl chloride by Grooves’ method,

C H_{3} C H_{2} O H + H C l A n h y d . Z N C l_{2} - ------- \to C H_{3} C H_{2} C l + H_{2} O

the main reason for using anhydrous

Z n C l_{2}

as catalyst is that

Q. Complete the following reaction:

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

+ H_{2} O

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CH3CH2OH+HClZnCl2−−−−→CH3CH2Cl+H2O In the above reaction, the leaving group is ?

The correct option is C HO−Z⊖nCl2 Zinc chloride is a Lewis acid and coordinates with the alcoholic group. R−..O..+ZnCl2→R−+OH−Z⊖nCl2→R++HO−Z⊖nCl2 R++Cl−(from HCl)→RCl+H+ H++HO−ZnCl2→H2O+ZnCl2

$C H_{3} C H_{2} O H + H C l Z n C l_{2} - --- \to C H_{3} C H_{2} C l + H_{2} O$ In the above reaction, the leaving group is ?