Among the following the one that gives positive iodoform test upon reaction with $$I_2$$ and NaOH is

A compound gives a positive iodoform test if it contains either a methyl ketone group, $$-COCH_3$$, or an alcohol that can be oxidized to a methyl ketone, i.e., compounds containing the group $$-CH(OH)CH_3$$.

For option (A), $$\mathrm{CH_3CH_2CH(OH)CH_2CH_3}$$ is pentan-3-ol. On oxidation, it forms pentan-3-one, $$\mathrm{CH_3CH_2COCH_2CH_3}$$, which is not a methyl ketone. Hence, it does not give the iodoform test.

For option (B), $$\mathrm{C_6H_5CH_2CH_2OH}$$ is a primary alcohol. On oxidation, it gives phenylacetaldehyde, $$\mathrm{C_6H_5CH_2CHO}$$, which does not contain the $$-COCH_3$$ group. Hence, it does not give the iodoform test.

For option (D), $$\mathrm{PhCH(OH)CH_3}$$ contains the $$-CH(OH)CH_3$$ group. It is first oxidized by $$\mathrm{I_2/NaOH}$$ to acetophenone, $$\mathrm{PhCOCH_3}$$, which is a methyl ketone. Acetophenone then undergoes the iodoform reaction to give yellow precipitate of $$\mathrm{CHI_3}$$.

Therefore, $$\mathrm{PhCH(OH)CH_3}$$ gives a positive iodoform test and hence option (D) is the correct answer.