I am supposed to find the product of the following reaction:
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.
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Sign up to join this communityI am supposed to find the product of the following reaction:
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.
There are two factors to take into consideration.
- The acidity of the deprontonated $\alpha-\ce{H}$.
- 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:
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.)
Information on the acidity of alpha hydrogens: http://www.chem.ucalgary.ca/courses/350/Carey5th/Ch21/ch21-2.html