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Topic: Stability of isomers : Ortho dimethyl cyclohexadiene  (Read 4190 times)

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Offline OccamsAftershave

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Stability of isomers : Ortho dimethyl cyclohexadiene
« on: October 14, 2012, 11:51:53 AM »
I am not sure about this. Which isomer is stablest ? The one where the double bonds are parallel to each other ? Or the one where a double bond attaches itself to one carbon attached to the external methyl group and the other double bond does the same with the other carbon attached to external methyl group ?

If the parallel bonds one, then which configuration of parallel bonds ?

Offline discodermolide

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Re: Stability of isomers : Ortho dimethyl cyclohexadiene
« Reply #1 on: October 14, 2012, 11:56:29 AM »
Double bonds prefer to be in the ring where they have the greatest number of substituents.
But I think it would be better if you drew your idea out so we can see exactly what you mean.
You can easily attach files to your post.
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Offline OccamsAftershave

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Re: Stability of isomers : Ortho dimethyl cyclohexadiene
« Reply #2 on: October 14, 2012, 12:09:08 PM »
Here are the 2 isomers via crude MSpaint :



Offline discodermolide

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Re: Stability of isomers : Ortho dimethyl cyclohexadiene
« Reply #3 on: October 14, 2012, 12:20:13 PM »
Thank you.
I would say the right hand structure is probably the more stable form.
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Offline OccamsAftershave

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Re: Stability of isomers : Ortho dimethyl cyclohexadiene
« Reply #4 on: October 14, 2012, 12:23:26 PM »
I see. Can you explain why ? What mechanism causes the stability ?

I myself went with the right hand one ( this was in a test ) as I thought that the left one, having double bonds separated by 2 C atoms in both directions wouldn't possess any aromatic nature at all, but the right one would have it along 1 direction, hence some extra stability would occur. Not a very solid reason, IMO, and I couldn't really explain why the right one would be stabler than a similar config of double bonds separated by 1 carbon along one side and 3 on the other but having the double bonds along different carbons in the ring.

Offline discodermolide

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Re: Stability of isomers : Ortho dimethyl cyclohexadiene
« Reply #5 on: October 14, 2012, 12:30:00 PM »
Here is a good explanation.
http://www.bluffton.edu/~bergerd/classes/CEM222/Handouts/cyclohexadiene_orbitals.pdf

makes a better job of it than I can ;D
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Offline OccamsAftershave

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Re: Stability of isomers : Ortho dimethyl cyclohexadiene
« Reply #6 on: October 14, 2012, 01:09:54 PM »
The link does say that 1,3 cyclohexadiene is slightly more stable than 1,4 cyclohexadiene according to the table at the end, which'd work...but isn't that a negligible amount of energy ? And how do the methyl groups interact with this ?

Also wiki puts 1,4 as more stable than 1,3 - go figure. This is the source :

http://webbook.nist.gov/cgi/cbook.cgi?ID=C628411&Units=SI&Mask=8#Thermo-React

The enthalpy value is indeed negative. Which is correct ?

Offline curiouscat

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Re: Stability of isomers : Ortho dimethyl cyclohexadiene
« Reply #7 on: October 14, 2012, 01:27:14 PM »
I don't know about the substituted cases; but wouldn't conjugated double bonds (1,3-case) be generally more stable in dienes?

Offline OccamsAftershave

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Re: Stability of isomers : Ortho dimethyl cyclohexadiene
« Reply #8 on: October 14, 2012, 01:52:23 PM »
@Curiouscat : Exactly, that was my initial reasoning as well. But why that particular arrangement of double bonds separated by 1 carbon along 1 direction and 3 carbons along the other ? I mean, it's possible to think of another pair of double bonds similarly separated but disconnected to the carbons bearing the external methyl groups. Which arrangement of double bonds then is stabler ?

Offline curiouscat

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Re: Stability of isomers : Ortho dimethyl cyclohexadiene
« Reply #9 on: October 14, 2012, 02:06:43 PM »
@Curiouscat : Exactly, that was my initial reasoning as well. But why that particular arrangement of double bonds separated by 1 carbon along 1 direction and 3 carbons along the other ? I mean, it's possible to think of another pair of double bonds similarly separated but disconnected to the carbons bearing the external methyl groups. Which arrangement of double bonds then is stabler ?

Then the more pertinent stability comparison is between these 4 :




#2 and #4 (counting from left; pardon my messily aligned sketch) look sterically hindered; the rest I've no clue.

If I were to hazard a guess I'd say #1 is more stable than #3 though I'd be hard pressed to explain why! (tert-C in its formation? Dunno!)
« Last Edit: October 14, 2012, 02:17:23 PM by curiouscat »

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