An organic compound (A) compound of C and H contains 85.71 % C. it shown `M^+ at m//z=42` in the mass spectrum. The compound reacts with HBr in the absence of peroxide to yield an organic compound (B) which is isomeric with the compound ( C) obtained when the compound reacts with HBr in the presence of peroxide.
Identify A, B and C.

Explanation

`C:H = (85.71)/(12):(14.29)/(1) = 7.14:14.29 = 1:2`
Emperical formula = `CH_(2)`, Molar formula `= (CH_(2))_(n)`
According to the problem, `n xx (12 +2) = 42`
`:. n=3`, So, the actual molecular formula is `C_(3)H_(6)`. DBE =1. So, probable structures of `C_(3)H_(6)` are

Structure A is accepted as it can undergo given reactions