Suppose n-butyraldehyde-1- \({ }^{2} \mathrm{H}\left(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CDO}\right)\) were reduced with aluminum tri-s-butoxide made from optically active s-buty 1 alcohol. Assuming the cyclic reduction mechanism, would you expect the first 10 per cent of n-butanol-1- \({ }^{2} \mathrm{H}\) to be optically active? Explain. What would be the products at equilibrium?

First, we'll explore the cyclic reduction mechanism. In this reaction, the aluminum tri-s-butoxide made from optically active s-butyl alcohol would chelate the carbonyl group of n-butyraldehyde-1-\({ }^{2}\mathrm{H}\), leading to the formation of a transition state. This chelation process can occur from either the Re side or the Si side of the carbonyl group, which means it can lead to two diastereomeric transition states. The kicker here is that one of the diastereomeric transition states will be favored in the energetically more favorable state due to the steric factors, and since we are using an optically active s-butyl alcohol, this will result in a preferential enantiomeric chirality of the product.

As the cyclic reduction mechanism progresses, the reaction is expected to proceed towards a diastereomeric imbalance, which means we should expect an enantiomeric preference in the first 10% of n-butanol-1-\({ }^{2}\mathrm{H}\). As a result, the first 10% of n-butanol-1-\({ }^{2}\mathrm{H}\) would indeed be optically active. However, as the reaction progresses, the remaining 90% of the n-butanol-1-\({ }^{2}\mathrm{H}\) will be produced in an equal mixture of enantiomers, effectively rendering it racemic.

At the equilibrium, we have two possible products: the optically active n-butanol-1-\({ }^{2}\mathrm{H}\) and the racemic n-butanol-1-\({ }^{2}\mathrm{H}\). To summarize, the products at equilibrium would be a mixture of optically active and racemized n-butanol-1-\({ }^{2}\mathrm{H}\). The first 10% of n-butanol-1-\({ }^{2}\mathrm{H}\) will be optically active, while the remaining 90% will be racemized. Thus, in the presence of an optically active aluminum tri-s-butoxide made from optically active s-butyl alcohol and following the cyclic reduction mechanism, the first 10% of n-butanol-1-\({ }^{2}\mathrm{H}\) produced will indeed be optically active. The products at equilibrium will be a mixture of optically active and racemized n-butanol-1-\({ }^{2}\mathrm{H}\).