Today I was told:

When 2-chlorobutane is reacted with 2-chloropropane in the presence of sodium and dry ether the following products are formed:

  • 2,3-dimethylpentane
  • 3,4-dimethylhexane
  • 2,3-dimethylbutane

3,4-dimethylhexane, being most symmetric, is the major product.

I'm not satisfied with the reason provided. Can anyone please explain the fact in a better manner?

What I'm unable to understand is that both 3,4-dimethylhexane and 2,3-dimethylbutane and equally symmetric since both have a plane of symmetry, don't they? Then how are we using symmetry to compare the yield?

  • $\begingroup$ What part of it didn't you understand? You may get better answers if you exactly point that out. $\endgroup$ – M.A.R. ಠ_ಠ Jun 18 '15 at 20:08
  • $\begingroup$ "2,3-dimethylhexane, being most symmetric, is the major product." 2,3-dimethylhexane is not an expected product, do you mean 2,3-dimethylbutane? Were the starting materials present in equimolar amounts? $\endgroup$ – ron Jun 18 '15 at 20:10
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    $\begingroup$ @ron I think OP meant 2,3-dimethylpentane, which would statistically be the major product if the starting materials are present in equimolar amounts. (That has nothing to do with symmetry... so I don't know what OP was told.) $\endgroup$ – orthocresol Jun 18 '15 at 20:13
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    $\begingroup$ @ron the OP had it right then did an edit. $\endgroup$ – user1945827 Jun 18 '15 at 20:17
  • $\begingroup$ Sorry for the confusion here, it's actually 3,4 and not 2,3. :) $\endgroup$ – Harshal Gajjar Jun 19 '15 at 18:10

I think that there should be a 'statistical product distribution', if I'm correct the products should be in the ratios 2:1:1.

You get to this conclusion as follows, call 2-chlorobutane A and 2-chloropropane B.

A can react with A to give AA, and A can react with B to give AB, while B can react with A to give BA, while B can react with B to give BB.

When you realise that AB and BA are the same molecule you get AA:2AB:BB; that is, 1:2:1.

  • $\begingroup$ +1 Then, is symmetry responsible for anything here? $\endgroup$ – Harshal Gajjar Jun 19 '15 at 18:18
  • $\begingroup$ Not as far as I can see. $\endgroup$ – user1945827 Jun 19 '15 at 19:11

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