Work backwards: ozonolysis of B gives benzaldehyde and butane-2,3-dione, so B is a diene formed by Grignard addition to a methyl ketone (iodoform positive) followed by dehydration.
The figure shows (A) giving a positive iodoform test, then (i) \(\mathrm{CH_3MgBr}\), (ii) \(\mathrm{H^+/H_2O}\), (iii) conc. \(\mathrm{H_2SO_4/\Delta}\) to give (B), and then \(\mathrm{O_3/Zn,\ H_2O}\) giving benzaldehyde (\(\mathrm{C_6H_5CHO}\)) and butane-2,3-dione (\(\mathrm{CH_3COCOCH_3}\)). Working backwards, the diene B must be \(\mathrm{C_6H_5CH=C(CH_3)-C(CH_3)=CH_2}\): ozonolysis gives \(\mathrm{C_6H_5CHO}\), \(\mathrm{CH_3COCOCH_3}\) and \(\mathrm{HCHO}\). B arises by dehydration of the tertiary alcohol \(\mathrm{C_6H_5CH=C(CH_3)-C(OH)(CH_3)_2}\), which comes from \(\mathrm{CH_3MgBr}\) adding to the methyl ketone A = \(\mathrm{C_6H_5CH=C(CH_3)-CO-CH_3}\) (3-methyl-4-phenylbut-3-en-2-one). A contains the \(\mathrm{CH_3CO-}\) group, so it gives a positive iodoform test. Check of the other options: option (2) would give acetophenone + butane-2,3-dione, option (3) would give benzaldehyde + methylglyoxal, option (4) would give acetophenone + methylglyoxal; none matches the products drawn. Answer: (1) \(\mathrm{C_6H_5CH=C(CH_3)COCH_3}\) (first structure)
