Q.Which of the following cannot be identified by the iodoform test?

Acetaldehyde and any methyl ketone (containing CH3-CO-) treated with halogen and alkali give the corresponding haloform, CHX3. Base first drives successive, progressively easier alpha-halogenations of the methyl group (each halogen further activating the remaining alpha-hydrogens); the resulting CX3-CO-R is then attacked directly at its now strongly-electrophilic carbonyl by hydroxide, and the tetrahedral intermediate expels the stabilised CX3- carbanion (instantly protonated to CHX3) rather than expelling hydroxide back out. Because the reaction absolutely requires a CH3-CO- fragment (or a CH3-CH(OH)- alcohol precursor that NaOX first oxidises to it in situ), it doubles as the classic IODOFORM TEST for a methyl ketone or a matching secondary alcohol, recognised by the pale yellow, characteristically-smelling CHI3 precipitate. …