2-Bromo-2-methylbutane gives 2-methylbut-2-ene (70%) and 2-methylbut-1-ene (30%) with $NaOEt$ , but 2-methylbut-2-ene (28%) and 2-methylbut-1-ene (72%) with $(CH X_{3}) X_{3} COK$ . The ratio (2-methylbut-1-ene : 2-methylbut-2-ene) with $(CH X_{3}) X_{3} COK$ , to two decimals, is:

Answer: D

The ratio is

72/28 = 2.57

. With ethoxide it is

30/70 = 0.43

, so the bulkier base raises the Hofmann : Saytzeff ratio about six-fold. Inverting the ratio gives 0.39.