An alkene (A) having formula `C_4H_8`, on ozonolysis, gives propanal & methanal. A reacts with HBr to produce a compound of molecular formula, `C_4H_9Br`.
This compound, when heated with alcoholic KOH, produces another alkene (B) which is isomeric with A. Identify the alkenes (A) and (B).

Explanation

The products of ozonolysis of the alkene (A) are propanal `(CH_(3)CH_(2)CHO)` and methanal (HCHO). Hence, the alkene (A) is 1-butene `(CH_(3)CH_(2)CH = CH_(2))`
`underset("Propanal")(CH_(3)CH_(2) - overset(overset(H)(|))(C) = O) + underset("Methanal")(O = CH_(2)) rArr underset("1-butene(A)")(CH_(3)CH_(2)CH = CH_(2))`
1-butene, being an usymmetrical alkene reacts with HBr according to Markownikoff's rule to give 2-bromobutane. Its molecular formula is `C_(4)H_(9)Br`.
`CH_(3)CH_(2)CH = CH_(2) + HBr rarr underset("2-bromobutane")(CH_(3)CH_(2)CHBrCH_(3))`
When 2-bromobutane is heated with alcoholic KOH solution, it undergoes dehydrobromination and according to Saytzeff's rule, 2-butene (an isomer of 1-butene) is obtained as the major product. Therefore, the alkene (B) is 2-butene `(CH_(3)CH = CHCH_(3))`.