I am supposed to find the product of the following reaction:

enter image description here

I know that six-membered rings are more stable than five-membered rings, but the C-D bonds are stronger, so it confuses me where will the attack happen.

  • 5
    $\begingroup$ C-D are only marginally stronger, the reaction is reversible and 6-member rings are very much favoured over 8-member. $\endgroup$
    – Waylander
    Jun 18, 2018 at 15:52

2 Answers 2


There are two factors to take into consideration.

  1. The acidity of the deprontonated $\alpha-\ce{H}$.
  2. The electrophilicity of the carbonyl being attacked.

If 1 dominates, then the reaction is under kinetic control. While if 2 dominates then the reaction is under thermodynamic control. (Detailed arguments are omitted here.) Since addition to carbonyls is generally highly reversible, the reaction should generally be under theromodynamic control and yield 1-(2-hydroxycyclohexyl)ethan-1-one:

enter image description here

There are existing experimental results on aldol condensation of 7-oxo-octanal.

Ghobril et al. (Eur. J. Org. Chem., 2008: 4104-4108) used TBD as base and obtained the thermodynamic product, while Pidathala et al. (Angew. Chem. Int. Ed., 42: 2785-2788) used proline and obtained the kinetic product. Ghobril et al. also pointed out that

When unsymmetrical ketoaldehydes such as 3 are submitted to intramolecular aldol reactions, the ketone usually acts as the CH-acidic component, whereas the aldehyde plays the role of the carbonyl-active counterpart to afford regioisomer 4. This regiochemical outcome is also favoured when thermodynamic conditions are used.


In both cases, it would form a 6-membered ring (I can't see where you spotted a 5-membered ring product). The alpha hydrogen of the aldehyde (pKa around 17) is more acidic than the alpha deuterium of the ketone (pKa around 20) since neighboring alkyl groups tend to destabilize carbanions by donating electron density. Furthermore, as you mentioned, C-D bonds are slightly stronger than C-H so it is safe to say that the enolate will form on the alpha carbon at the aldehyde side. (I'm sorry I wrote the carbanion attacks the carbonyl, to be more correct I should've written the enol double bond acting as the nucleophile, not the carbanion. For practical purposes, it makes no difference though.) enter image description here

Information on the acidity of alpha hydrogens: http://www.chem.ucalgary.ca/courses/350/Carey5th/Ch21/ch21-2.html

  • 2
    $\begingroup$ Both enolates will reversibly form and the selectivity of the reaction should not be determined by which enolate forms but rather by thermodynamic stability of the products. In this case it also seems highly likely that a subsequent E1cb reaction will furnish the enone instead of the hydroxyketone. $\endgroup$ Jun 18, 2018 at 19:56
  • $\begingroup$ I thought about continuing to the enone formation but the conditions were not specified and I wasn't sure which product they were hoping to obtain. $\endgroup$ Jun 18, 2018 at 20:13

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