# Reduction of Carbonyl Compounds with LiAlH4 [duplicate]

The following appeared in the October/November CIE Camebridge A level chemistry paper 4.

Two isometric aromatic compounds, V and W are shown below.

Draw the structures of the two organic products from the reaction of V and W with $\ce{LiAlH_4}$.

The answer key simply removes the oxygens from the structures. However, I was under the impression that they would be replaced by hydroxyl groups since reducing carbonyl compounds gives one alcohols. Does anyone know why this is the case?

Though, the carbonyls are reduced to alcohols after the treatment with $\ce{LiAlH_4}$,here the $\ce{C=O}$ double bond is part of the Amide($\ce{-CONH_2)}$group, and this group has definitely different chemical functionality than pure carbonyls.So, though both the aldehyde(or, keto) and amides have same $\ce{C=O}$ double bonds, but thir property is different due to extra presence of adjacent of $\ce{NH_2}$ group in amides.
In Aldehydes or ketones, the $\ce{C=O}$ has more double bond character. So, when it is treated with $\ce{LiAlH_4}$, due to the strong bonding between $C$ and $O$, only cleavage of the $\pi$-bond is possible and thus it can possibly form alcohols, because, Nucleophillic attack by $\ce{H^-}$ of $\ce{LiAlH_4}$, can form $C-H$ bonds, and energetically it is not so favourable that it will break two $C-O$ bonds for that.So, that's why, the end products are alcohols in reduction of aldehydes or ketones.
But in Amides, the $\ce{C-O }$, bond has a very less double bond character due to the resonance of carbonyl carbon with the $\ce{-NH_2}$ group, and therefore, the bond in this case becomes very weak, and easy to break.
More Over, if you see the mechanism, for this reduction of amides, in a transition state, that carbonyl carbon forms double bonds with the $\ce{-NH_2}$ group ( $\ce{-C=NH}$), and this becomes energeticaly more favourable ( As, the difference in energy of the $2p$ orbitals of $C$ ans $N$ is very less thaan that of the $O$, so the $\pi$-overlap can form strongly )and thus the $C-O$ bond becomes easier to break totally, which in turn give the amines as the end product in reduction of amides.