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What is the number of possible enantiomers in monochlorination of 2-methybutane?

My approach

When $\ce{Cl2}$ is added to any alkane it prefers $3^\circ > 2^\circ > 1^\circ$. When it is being added to 2-methybutane, final product (major) formed will be 2-chloro-2-methybutane, which will show 0 chiral centre, hence the maximum number of enantiomers will be

$$2^n = 2^0 = 1$$

My book gave me answer $2$, but no explanation is provided.

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    $\begingroup$ You gave a good explanation, the book did not give any - you win! Or they meant the question differently: of all the products that could form, how many are chiral? $\endgroup$
    – Karsten
    Jun 11, 2019 at 1:12
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    $\begingroup$ @Who, I was wondering what happened to your doctoral title (:D). Btw preference of chlorination is highest at tertiary carbons, that's true, but don't you think it will definitely happen even at secondary and primary centers (albeit to a lesser extent). That's why we say "major product" is the tertiary one, but minor products also form. $\endgroup$ Jun 11, 2019 at 1:55
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    $\begingroup$ Which book are you taking this question from? $\endgroup$ Jun 11, 2019 at 2:00
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    $\begingroup$ Chlorine is less selective then Bromine in free radical reactions(at least here).Even though tertiary radical is most stable as in the post , its product product may not be major. $\endgroup$ Jun 11, 2019 at 3:09
  • $\begingroup$ Partly related: What is the major product on chlorination of 2-methylbutane? $\endgroup$
    – user7951
    Nov 8, 2019 at 8:35

1 Answer 1

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Assuming monochlorination of 2-Methyl Butane ,the following products are possible as shown in the figure below. enter image description here

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