We have seen that the reduction of lactones with LiAlH₄ proceeds through ring opening and ultimately gives diols after two consecutive hydride additions. A very similar hydride reduction can also be applied to lactams, although the outcome and key mechanistic details are different. The main difference is that the product of this reaction is the corresponding cyclic amine.
An important intermediate in this reaction is the formation of the iminum intermediate, which, being very electron-deficient, is attacked by another hydride ion, forming the amine.
Once again, unlike the reduction of lactones, there is no ring opening because the conjugate base of the amino fragment is a lot stronger than the akoxide. So, when the tetrahedral intermediate is broken, the Al-O fragment is the leaving group, not the conjugate base of the amine.
We have a dedicated post on the reactions of Lactones and Lactams with lots of practice problems here.
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