An organic compound A with molecular formula C_4H_9 Br on treatment with alcoholic KOH gave two isomeric compounds B and C with the formula C_(4)H_(8) On ozonolysis, B gave only one product CH_3CHO while C gave two different products. Find the sum of carbon atoms in compound A, B and C.

An organic compound A with molecular formula C_4H_9 Br on treatment wi

1.	An organic compound A with molecular formula C_4H_9 Br on treatment with alcoholic KOH gave two isomeric compounds B and C with the formula C_(4)H_(8) On ozonolysis, B gave only one product CH_3CHO while C gave two different products. Find the sum of carbon atoms in compound A, B and C.
Answer» Solution : The compound A is a haloalkane which UNDERGOES dehydrohalogenation with alcoholic KOH to form two isomeric alkenes B and C. SINCE B upon ozonolysis gave only ONE product, i.e., `CH_3CHO`, B is expected to be 2-butene `(CH_3CH = CHCH_3)`. Since C gave different products on ozonolysis, it must be a position isomer of 2-butene, i.e., 1-butene. The entire sequence of reaction is as follows: `underset("2-Bromobutane (A)")(CH_(3)-underset(Br)underset(\|)CH-CH_(2)-CH_(3) overset("alc. KOH") to``overset(CH_(3)CH=CHCH_(3))overset(2-"Butene (B)")underset("1 product")underset(2CH_(3)CHO)(darr)(ozonolysis)` + `overset(CH_(3)CH_(2)CH=CH_(2))overset("1-Butene")underset(2-"products")underset(CH_(3)CH_(2)CHO + HCHO)(darr) ("ozonolysis")` Thus, compound A, B & C contains total 12 carbon atoms.