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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4$\begingroup$ 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 '18 at 15:52
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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
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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. $\endgroup$ – orthocresol♦ Jun 18 '18 at 19:56
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$\begingroup$ 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. $\endgroup$ – Raul Luciano Jun 18 '18 at 20:13
