(R)-2 -Iodobutane is treated with NaI in acetone and allowed to stand for a long time.The product eventually formed is:

(R) - 2 - i o d o b u t a n e

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(S) - 2 - i o d o b u t a n e

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(\pm) - 2 - i o d o b u t a n e

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(\pm) - 1, 2 - d i i o d o b u t a n e

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Solution

The correct option is C $(\pm) - 2 - i o d o b u t a n e$
$I^{-}$ is a good nucleophile as well as a good leaving group. Therefore, if (R)-2-iodobutane is treated with NaI, repeated $S_{N} 2$ reactions occur. As a result , eventually a racemic mixture of $(\pm)$ -2 iodobutane is obtained.

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Similar questions

(R) - 2 -

[α] = {15.9}^{\circ}

1.0

- {15.58}^{\circ},

Assertion : Optically active 2-iodobutane on treatment with NaI in acetone undergoes recemization.

Reason : Repeated Walden inversions on the reactant and its product eventually gives a racemic mixture.

2

N a I

2

Which is the correct increasing order of boiling points of the following compounds?

1-iodobutane , 1-bromobutane, 1-chlorobutane, Butane

(a) Butane < 1-chlorobutane < 1-bromobutane < 1-iodobutane

(b) 1-iodobutane < 1-bromobutane < 1-chlorobutane < Butane

(d) Butane < 1-chlorobutane < 1-iodobutane < 1-bromobutane

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