It is a decarboxylation reaction, which give an alkane with one carbon-atom less, as in CH3CH2COONa i.e. CH3−CH3( ethane ). (iii) C4H10O1 i.e. C4H9OH. When treated with H2SO4 (conc.), it will produce butene-2 as follows: C4H9OH
H2SO4(conc)
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Δ
H3C−HC=CH−CH3 2-butene is an unsaturated compound. Thus, will decolourise bromine water, Hence, A,B,C are respectively CH3CH2COONa−(A) CH3−CH3(ethane )−(B) CH3−CH=CH−CH32 -(butene)−(C) ∴(a) is the correct answer.