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Preparations and Uses of Caustic Soda

S N I reactio...

Question

$S_{N}$ I reaction undergoes through a carbocation intermediate as follows :
$R - X (a q .) S l o w ⇌ R * (a q .) + X^{-} (a q .) H_{2} O - - \to f a s t R O H (a q .) + H^{+} (a q .)$
[R = t-Bu, iso-Pr, Et, Me] (X = Cl, Br, I)
The correct statements are
(I) The decreasing order of rate of S_N 1 reaction is t-BuX > iso-PrX > EtX > MeX
(II) The decreasing order of ionization energy is MeX > EtX > iso-PrX> t-BuX
(III) The decreasing order of energy of activation is t-BuX > iso-PrX > EtX > MeX

I and II are correct

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I and III are correct

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II and III are correct

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I, II and III are correct

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Solution

The correct option is D I, II and III are correct
Rate of $S_{N} 1$ reaction is a
t – Bu X > iso – PrX > EtX > MeX
So order of activation energy
t – Bu X > iso – PrX > EtX > MeX
and order of ionization energy
t – BuX < iso – PrX < EtX < MeX

Suggest Corrections

Similar questions

S_{N} 1

[R = t - B u, i s o - P r, E t, M e] (X = C l, B r, I)

S N_{1}

t - B u X > i s o - P r X > E t X > M e X

M e X > E t X > i s o - P r X > t - B u X

t - B u X > i s o - P r X > E t X > M e X

[R = t-Bu, iso-Pr, Et, Me]

(X = C l, B r, I)

The correct statements are:
I. The decreasing order of rate of SN1reaction is

t - B u X > i s o - P r X > E t X > M e X

II. The decreasing order of ionization energy is

M e X > E t X > i s o - P r X > t - B u X

III. The decreasing order of energy of activation is

t - B u X > i s o - P r X > E t X > M e X

S_{N} 1

R - X (a q) s l o w ⇌ R^{+} (a q) + X^{-} (a q) H_{2} O - -- \to F a s t R O H (a q) + H^{+} (a q)

[R = t - B u, i s o - P r, E t, M e]

I

I V

Methylformate I

Ethylformate I I

Iso-propylformate I I I

n-Propylformtae I V

N_{2} O

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