Q.3,3-dimethylbutan-2-ol on treatment with conc. H2SO4 gives tetramethylethylene (2,3-dimethylbut-2-ene) as the major product. Suggest a suitable mechanism.

Step 1. Step 1 (protonation). 3,3-Dimethylbutan-2-ol, (CH3)3C-CH(OH)-CH3, has its -OH protonated by H2SO4 to the good leaving group -OH2+.

Step 2. Step 2 (ionisation). Loss of water gives a SECONDARY carbocation at C2, flanked immediately by the bulky, fully-substituted C3 (the former tert-butyl carbon).

Step 3. Step 3 (1,2-methyl shift). One of the three methyl groups on the adjacent C3 migrates, with its bonding electron pair, to the cationic C2 -- converting the secondary C2 cation into a more stable TERTIARY carbocation now centred at C3 (which is flanked by two methyls from its own original substitution plus the newly-arrived methyl and the remaining C2-CH3 group). …

Write out all four E1 steps explicitly, making the 1,2-methyl shift from the secondary to the more stable tertia …

• Drawing only the direct, unrearranged elimination product from the initial secondary cation -- the METHYL SHIFT to the more stable tertiary cation is the essential step that leads to the major, fully-substituted alkene actually observed. …