The optically active amide shown is treated with $\mathrm{Br}{}_2$/aqueous $\mathrm{NaOH}$. Which structure is the major product?

The amide shown is (S)-2-methylbutanamide. In the Hofmann rearrangement the migrating carbon moves with retention of configuration, so the amine has the same spatial arrangement. The substituent priorities stay N (replacing C=O) > ethyl > methyl > H, so the product is (S)-butan-2-amine, not the inverted or racemic form.