Answer: A, B, C
The compound is 2-chloro-2-methylbutane. A tertiary halide in a polar protic solvent without strong base reacts by SN1/E1, in which slow ionisation to the carbocation is rate-determining and the rate is first order in halide only. Loss of different protons gives both alkenes. Attack on the planar cation occurs from both faces, so inversion is not complete.