Trimethyl(1-phenylpropan-2-yl)ammonium hydroxide, $\mathrm{C}{}_6\mathrm{H}{}_5-\mathrm{CH}{}_2-\mathrm{CH}(\mathrm{CH}{}_3)-\mathrm{N}{}^{+}(\mathrm{CH}{}_3){}_3\mathrm{OH}{}^{-}$, is heated. The major alkene is

The Hofmann rule alone would remove a methyl hydrogen to give allylbenzene. However, the benzylic $\mathrm{CH}{}_2$ hydrogens are much more acidic, and the E2 transition state has carbanion character and gains conjugation with the ring. Loss of the benzylic hydrogen therefore dominates and the conjugated 1-phenylprop-1-ene is the major product.