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

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

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?

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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.

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    $\begingroup$ (-1) this has little if anything to do with bond strength in the starting material. It is simply that the mechanism allows for loss of OH in only one case. $\endgroup$ – orthocresol Apr 12 '18 at 7:38

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