The reaction of (+) 2-iodobutane and NaI * (I* is radioactive isotope of iodine) in acetate was studied by… · ABC26OR0833

Subject: Organic Chemistry · Chapter: Reagents, Reaction Mechanisms & Named Reactions · Topic: Reagents, Reaction Mechanisms & Named Reactions – General · Exam: JAM 2005-2022 · Marks: 2 · Difficulty: Medium

The reaction of $(+)$ 2-iodobutane and $\mathrm{NaI} *$ (I* is radioactive isotope of iodine) in acetate was studied by measuring the rate of racemization $\left(\mathrm{k}_{\mathrm{r}}\right)$ and the rate of incorporation of $\mathrm{I}^{*}\left(\mathrm{k}_{\mathrm{i}}\right)$.
\[ (+) \mathrm{CH}_{3} \mathrm{CH}(\mathrm{I}) \mathrm{CH}_{2} \mathrm{CH}_{3}+\mathrm{NaI}^{*} \rightarrow \mathrm{CH}_{3} \mathrm{CH}\left(\mathrm{I}^{*}\right) \mathrm{CH}_{2} \mathrm{CH}_{3}+\mathrm{NaI} \]
For the reaction, the relationship between $\mathrm{k}_{\mathrm{r}}$ and $\mathrm{k}_{\mathrm{i}}$ is:
(a)$\mathrm{k}_{\mathrm{i}}=2 \times \mathrm{k}_{\mathrm{r}}$
(b)$\mathrm{k}_{\mathrm{i}}=(1 / 2) \times \mathrm{k}_{\mathrm{r}}$
(c)$\mathrm{k}_{\mathrm{i}}=\mathrm{k}_{\mathrm{r}}$
(d)$\mathrm{k}_{\mathrm{i}}=(1 / 3) \times \mathrm{k}_{\mathrm{r}}$
Answer
Answer: B ✓ checked by 4AB · confidence high
Explanation
Each S_N2 exchange inverts one molecule, destroying the rotation of two molecules' worth: k_r = 2k_i, so k_i = ½k_r (b).

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