The major product formed in the following reaction is:
Show Hint
Ask which carbon sits anti to the leaving \(\mathrm{OH}\): here it is a gem-dimethyl carbon that can form a stable tertiary cation, so the oxime fragments (C-C cleavage to a nitrile) instead of doing a normal 1,2-migration, and the resulting cation is captured by the nearby aromatic ring.
This is a Beckmann reaction, but the twist is that the substrate is set up to fragment rather than simply rearrange. Look at what sits next to the oxime carbon: one neighbor is the aromatic ring, and the other neighbor is a carbon carrying two methyl groups. That gem-dimethyl carbon is the key to the whole question.
Normal Beckmann path: if this were an ordinary oxime, protonating the $\mathrm{OH}$ and losing water would trigger the group anti to the leaving $\mathrm{OH}$, here the gem-dimethyl carbon, to migrate onto nitrogen, giving a ring-expanded seven-membered lactam. That is exactly what options (A) and (B) show.
Why that does not happen here: migrating as a full alkyl group is only the low-energy path when there is no better option. Here, breaking off the gem-dimethyl carbon completely gives a tertiary carbocation, stabilized by the two flanking methyls. A tertiary cation is much more stable than the transition state for simple migration, so the molecule fragments instead, the bond between the oxime carbon and the gem-dimethyl carbon breaks fully.
What each fragment becomes: the oxime carbon keeps its bond to the aromatic ring and turns into a nitrile, $\mathrm{Ar{-}C{\equiv}N}$. The gem-dimethyl carbon becomes a free tertiary cation, still attached through three $\mathrm{CH_2}$ groups to the far side of the aromatic ring.
Ring closure: that tertiary cation gets trapped right back onto the same aromatic ring, an intramolecular Friedel-Crafts alkylation, forming a new six-membered ring. The seven-membered ring has effectively contracted to a six-membered ring, with a pendant nitrile group on the aromatic ring and the two methyls sitting on the new ring's quaternary carbon, on the far side of the ring from the nitrile.
That six-membered, nitrile-bearing bicyclic structure is option (C). Option (D) draws almost the same skeleton but puts the gem-dimethyl carbon right next to the nitrile-bearing aromatic carbon instead of across the ring, which is not where the ring-closing carbocation actually lands.
Let's summarize:
A carbon anti to the leaving $\mathrm{OH}$ that can form a stabilized, here tertiary, cation favors Beckmann fragmentation over normal rearrangement.
Fragmentation converts the oxime carbon into a nitrile and releases the other carbon as a cation.
An electron-rich aromatic ring nearby traps that cation intramolecularly, closing a new, smaller ring.
The correct product is the ring-contracted nitrile in option (C).
