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Why doesn't Iodine liberate in the following reaction?

enter image description here
(A) Inadequate ring size
(B) Two Bromines are in equatorial positions
(C) Steric hindrance of t-butyl group towards attacking $\ce{I-}$

Initially, I thought it would simply undergo a simple finkelstein kind of reaction and since two iodines in vicinal carbons aren't stable, it would lose $\ce{I2}$ to give a double bond according to the following mechanism:

Orgainc reaction

But my book said that $\ce{I2}$ would not be liberated.

I had a thought that this might be related to the chair conformation and the bromines being at equatorial positions since the options provided point towards this,
so I felt that's what the exercise wants me to do but I can't get to the reasoning.

For reference, the answer given in my book is:

(B) Two Bromines are in equatorial positions

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  • $\begingroup$ To loose $\ce{I2}$, the oxidation reaction $\ce{2I- -> I2 + 2e-}$ should happen. But there is no oxidizer in this reaction. $\endgroup$
    – Mathew Mahindaratne
    Jun 14, 2021 at 15:24
  • $\begingroup$ @MathewMahindaratne Oh okay but the reason given in my book says that its because the two bromines are in equatorial positions. Maybe the exercise wanted me to look at this factor rather than that? Can you help me reason this out? $\endgroup$
    – Prajwal Tiwari
    Jun 14, 2021 at 16:08
  • $\begingroup$ @MathewMahindaratne The OP most probably is talking about homolytic clevage of the vicinal iodines to form iodine and an alkene and not direct oxidation of the iodide ion. $\endgroup$
    – Nisarg Bhavsar
    Jun 14, 2021 at 16:26
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    $\begingroup$ @PrajwalTiwari See the edit I made. Is that what you meant to say? If it's not than feel free to rollback. $\endgroup$
    – Nisarg Bhavsar
    Jun 14, 2021 at 16:41
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    $\begingroup$ How do you know whether the bromine substituents, or the iodine ones, are equatorial? The stereochemistry isn’t marked, so what enables you to make that assumption? I actually think there is a good reason to say so, but you haven’t provided it. $\endgroup$
    – orthocresol
    Jun 14, 2021 at 17:50

1 Answer 1

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Why should we assume that the halogens are in equatorial position?

As @Orthocresol pointed out there is probably enough reason (it should be still mentioned as any other conformer may still exist even if unstable) for both the halogens to be in equatorial position. The t-butyl group attached provides massive steric repulsion to both the halides so they are locked in equatorial position. This is a good proof of tert-butyl's massive steric hindrance which results in cyclohexane ring ditching it's usual chair conformer for a twist-boat conformation for relief.

OK, but why is the configuration of the $\ce{C-Br}$ bonds relevant to the question?

The reaction mentioned here is the well-known finkelstein reaction which prefers $S_{N}2$ mechanism (mainly due to solvent effects). As we know the rate of $S_{N}2$ reaction depends upon it's transition state's stability, whose stability reduces as steric inhibition increases. ($\ce{I^{-}}$ is not exactly the smallest nucleophile out there.)

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The problem doesn't end here. The mechanism depends upon nucleophilic attack on $\ce{C-X}$ bond as a back attack. Now we can't expect the to put Br in axial position because of high repulsion of tert-butyl group and in equatorial position the tert-butyl group again interferes directly in approach of nucleophile. As a result this reaction cannot proceed and no $\ce{I_2}$ is liberated

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