Codex @codex  0

 Incoming links: Methine group

protonation of the alcohol converts its OH group into a water leaving group. Departure of water initially gives the secondary carbocation

C H_{3} - C H^{+} - CH (C H_{3})_{2}

. A 1,2-hydride shift from the neighboring methine group transfers its C–H bonding pair to the electron-deficient carbon:

C H_{3} - C H^{+} - CH (C H_{3})_{2} ⟶ C H_{3} C H_{2} - C^{+} (C H_{3})_{2} .

The resulting tertiary carbocation is the tert-amyl electrophile. An aromatic

π

bond of the dimethoxybenzene attacks that carbon, producing a Wheland intermediate. Removal of the proton from the attacked ring carbon restores aromaticity and regenerates acid. Repeating this Friedel-Crafts alkylation at the symmetry-related second site gives the disubstituted product.

The hydride migration explains the new ethyl group and the absence of a benzylic methine group proton. Simply attaching the original alcohol skeleton at its OH-bearing carbon would not explain the twelve-proton NMR singlet.

The product is 1,4-dimethoxy-2,5-bis(2-methylbutan-2-yl)benzene. Each new substituent is

- C (C H_{3})_{2} C H_{2} C H_{3}

, a tert-amyl group, rather than the unrearranged secondary alkyl group of the alcohol. The two substituents are related by the symmetry of the ring.

The proton assignments are:

chemical shift / ppm	Assignment	Reason
$6.75$	Two aromatic H, at positions 3 and 6	symmetry makes them equivalent; neither has an adjacent aromatic H to give a resolved ordinary ortho splitting.
$3.79$	Two methoxy groups, six H	Equivalent $OC H_{3}$ groups give a NMR singlet.
$1.81$	Two ethyl $C H_{2}$ groups, four H	Each couples to a terminal three-proton methyl group, giving a NMR quartet.
$1.33$	Four methyl groups attached to the two quaternary benzylic carbons, twelve H	These methyl groups have no neighboring methine group proton and give a NMR singlet.
$0.65$	Two terminal ethyl methyl groups, six H	Each couples to two methylene group protons, giving a NMR triplet with the same coupling constant as the NMR quartet.

The integrals add to 30 H. The unusually shielded terminal ethyl methyl signal is compatible with an aryl-containing environment; its NMR quartet–NMR triplet partner and integral are the decisive assignment evidence. The molecular formula and degree of unsaturation,

DBE = 4