0.396 gm of a bromoderivative of a hydrocarbon (A) when vapourized occupied 67.2 ml at NTP. On reaction with aqueous NaOH, (A) gives (B). (B) when passed over alumina at 250^(@)C gives a neutral compound (C ), while at 350^(@)C it gives a hydrocarbon (D): (D) when heated with HBr gives an isomer of (A). When (D) is treat with conc. H_(2)SO_(4) and the product is diluted with water and distilled, (E) is obtained. Identify (A) to (E) and explain the reaction involved.

0.396 gm of a bromoderivative of a hydrocarbon (A) when vapourized occ

1.	0.396 gm of a bromoderivative of a hydrocarbon (A) when vapourized occupied 67.2 ml at NTP. On reaction with aqueous NaOH, (A) gives (B). (B) when passed over alumina at 250^(@)C gives a neutral compound (C ), while at 350^(@)C it gives a hydrocarbon (D): (D) when heated with HBr gives an isomer of (A). When (D) is treat with conc. H_(2)SO_(4) and the product is diluted with water and distilled, (E) is obtained. Identify (A) to (E) and explain the reaction involved.
Answer» Solution :Molecular WEIGHT of `A=(0.369xx22.4xx1000)/(67.2)=123` (A) is a bromoderivative, so it may be written as R - Br. `M_(RBr)=123`, So, R = 43 `because` Br = 80 (Atomic weight) i.e. `C_(n)H_(2n+1)=43` or `12n+2n+1=43` or n = 3 `THEREFORE A=C_(3)H_(7)Br` So, the POSSIBLE structures of (A) are `CH_(3)CH_(2)CH_(2)BrorCH_(3)-UNDERSET(Br)underset(\|)(CH)-CH_(3)` The reactions are as follow. Since (D) gives an isomer of (A) on HBr addition, hence (A) = `CH_(3)CH_(2)CH_(2)Brand(E)=CH_(3)underset(OH)underset(\|)(CH)CH_(3)`