Why does the reduction of a Weinreb amide give an aldehyde instead of an amine?
Reaction of acyl carbonyl compounds with strong nucleophiles (hydrides, organometallics) is a very common and useful reaction, although it has some limitations. One of the simplest cases is the reaction of an ester with a strong nucleophile. As one would expect, the strong nucleophile attacks the electrophilic carbonyl carbon, giving a tetrahedral intermediate. In most cases, the tetrahedral intermediate breaks down readily to give a new carbonyl compound. In this case, that carbonyl intermediate will be a ketone (from a Grignard, e.g.) or an aldehyde (from LiAlH4). Both of those functional groups are more reactive than the ester starting material, so the next equivalent of nucleophile will react faster with the aldehyde/ketone, eventually giving a primary (from LiAlH4) or tertiary (from a Grignard) alcohol product.



The situation at the tetrahedral intermediate changes slightly when an amide is the electrophile. As cfcief points out, the less electronegative nitrogen will kick out the more electronegative oxygen (stabilized by aluminum) to give an iminium intermediate. Again, this intermediate is more reactive than the starting material, so a second nucleophilc attack takes place.



The key feature of the Weinreb amide is that after the nucleophilic attack at the carbonyl, the tetrahedral intermediate is stabilized as a five-membered cyclic chelate. Without decomposition of the tetrahedral intermediate, the only active electrophile is more of the Weinreb amide starting material. On workup, the strong nucleophile is quenched, the metal washed out, and the tetrahedral intermediate is allowed to decompose to the aldehyde/ketone product.



Although the N-methoxy-N-methylamide was specifically designed by Weinreb to behave this way (ref), it wasn't until more than 20 years later that mechanistic evidence was observed (ref).