Propanal is treated with dilute aqueous NaOH at room temperature and the product is then heated to give X (aldol condensation).

Two propanal molecules combine: the enolate of one (nucleophilic carbon C-2) attacks the carbonyl carbon of the other to give 3-hydroxy-2-methylpentanal; heating eliminates water to the conjugated enal $\mathbf{X}=\mathrm{CH_3CH_2CH{=}C(CH_3)CHO}$, 2-methylpent-2-enal. Formula: $2\,\mathrm{C_3H_6O}-\mathrm{H_2O}=\mathrm{C_6H_{10}O}$; (A) is true. Each alkene carbon bears two different substituents (C2: methyl and CHO; C3: ethyl and H), so E/Z isomerism exists; (B) is true. Cleaving the double bond gives $\mathrm{CH_3CH_2CHO}$ (propanal) and $\mathrm{CH_3COCHO}$ (2-oxopropanal), not ethanal, so (C) is false. There is no $\mathrm{CH_3CO{-}}$ group or $\mathrm{CH_3CH(OH){-}}$ unit, so the iodoform test is negative and (D) is false.