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I want to do intramolecular aldol condensation of 7-oxooctanal:

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I am talking about the major product. Clearly, a 6-membered ring would be a major product, but there are 2 possibilities: taking hydrogen from $\ce{C^2}$ or $\ce{C^6}$. Which one would be preferred?

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3 Answers 3

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If in doubt, draw:

aldol condensation

Deprotonation at C-6 under thermodynamic control results in the more stable enolate (1c) and leaves the aldehyde (= the centre with the higher carbonyl activity) intact for the aldol reaction.

As already outlined by Uncle Al, the product is likely to undergo elimination to furnish acetylcyclohexene.

However, if that is your target molecule, addition of acetylene to cyclohexanone, followed by an acid-catalyzed rearrangement of the propargylic alcohol would be an alternative route.

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    $\begingroup$ Curious: Why did you not consider the other two dehydration steps for the first and second path? $\endgroup$
    – Jan
    Commented Sep 23, 2017 at 4:27
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The more stable enol would involve $\ce{C^6}$, with the aldol condensation you would get a six member ring with $\ce{-COCH3}$ attached to the ring.

With the enol involving $\ce{C^2}$, the product would have a methyl and $\ce{-CHO}$ attached to the ring.

I don't know the distribution between the two products; it could be 51:49 or as much as 99:1. You should do the reaction to determine this. Factors that may effect the product ratio: temperature, solvent, reaction time, concentration, etc.

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  • $\begingroup$ How do you say that first enol is more stable? $\endgroup$
    – evil999man
    Commented Mar 29, 2014 at 14:37
  • $\begingroup$ The C8 enol would be less stable than the C6 enol. With the hydrogen on C1, I would think the C2 enol would be less stable than the C6 enol. But like I said, I don't know what the distribution actually is. $\endgroup$
    – LDC3
    Commented Mar 29, 2014 at 14:44
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We expect the aldehyde to be more reactive toward nucleophilic attack for its smaller steric hindrance (kinetics). The product ratio may vary with time, kinetics vs. thermodynamics at equilibrium. Attack followed by elimination helps lock in the product.

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  • $\begingroup$ Am I correct in thinking that major product would be that of 6 membered ring not attack of $C_8$? $\endgroup$
    – evil999man
    Commented Mar 29, 2014 at 14:03
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    $\begingroup$ I expect cyclohexene with a conjugated methyl ketone hanging off. NMR product. Methyl ketone versus aldehyde is obvious for the methyl group around 2 versus an aldehyde proton around 9.5 or lower (conjugation). I vote for the kinetic product. The tetrasubstituted endocyclic olefin is probably the thermodynamic product. $\endgroup$
    – Uncle Al
    Commented Mar 29, 2014 at 15:52

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