A compound (A) (which gives haloform reaction), on reduction with zinc amalgam and conc.HCl gives (B). (B) is also obtained by treating \(1,1,2,2\) - tetrachlorobutane with zinc dust. The compound (A) is (a) C#CC(C)O (b) C=C(C)C=O (c) C=CC(C)O (d) C#CC=O

(a) \(\mathrm{CH_3C\equivCCH_2OH}\): This compound has a terminal alkyne functional group and a primary alcohol group. Alcohols (except for methyl ketones) do not give haloform reactions. So, option (a) can be ruled out. (b) \(\mathrm{CH_3CH=C(CH_3)CHO}\): This compound has an aldehyde functional group and an alkene functional group. Aldehydes with at least one alpha hydrogen can give haloform reactions. In this case, there's one alpha hydrogen present. So, option (b) is a candidate for compound (A). (c) \(\mathrm{CH_3CH=C(CH_3)CH_2OH}\): This compound has an alkene functional group and a primary alcohol functional group, which does not give haloform reactions. So, option (c) can be ruled out. (d) \(\mathrm{CH_3C\equivCCHO}\): This compound has a terminal alkyne functional group and an aldehyde functional group. Like we discussed earlier, aldehydes with one alpha hydrogen can give haloform reactions, which is true for this compound as well. Therefore, option (d) is also a candidate for compound (A). Now, we have two candidates for compound (A): option (b) and option (d).

(b) Reducing \(\mathrm{CH_3CH=C(CH_3)CHO}\) with zinc amalgam and conc. HCl will reduce the aldehyde group to a primary alcohol: \(\mathrm{CH_3CH=C(CH_3)CH_2OH}\). (d) Reducing \(\mathrm{CH_3C\equiv CCHO}\) with zinc amalgam and conc. HCl will reduce the aldehyde group to a primary alcohol: \(\mathrm{CH_3C\equiv CCH_2OH}\).

Treating 1,1,2,2-tetrachlorobutane with zinc dust will reduce two carbon-chlorine bonds to form a double bond: \(\mathrm{CCl_2CHCHCl_2}\) \(\xrightarrow{\mathrm{Zn\ \mathrm{dust}}}\) \(\mathrm{CH_2=CH-CHCl_2}\). Comparing this product with the products obtained by reducing compounds (b) and (d), we see a match between this product and the one obtained by reducing compound (b). Therefore, compound (A) is: (b) \(\mathrm{CH_3CH=C(CH_3)CHO}\).