Which of the following compound can give positive iodoform test when treated with aqueous $$KOH$$ solution followed by potassium hypoiodite.

The iodoform test is highly specific for compounds containing either a methyl ketone group ($$\text{CH}_3\text{--C}(=\text{O})--$$) or a secondary methyl alcohol group ($$\text{CH}_3\text{--CH(OH)}--$$) that can easily oxidize to a methyl ketone.

When starting with a geminal dihalide complex, treatment with a strong base like aqueous $$\text{KOH}$$ causes nucleophilic substitution to replace the halogen atoms with hydroxyl ($$-\text{OH}$$) groups. If this intermediate step generates a stable or unstable gem-diol that eliminates water to form a methyl ketone structure, it will give a positive iodoform reaction upon adding $$\text{KOI}$$.

Step-by-Step Analysis of Compound A:

1. Nucleophilic Substitution with Base:

   Compound A is 2,2-dichlorobutane ($$\text{CH}_3\text{--CH}_2\text{--C(Cl)}_2\text{--CH}_3$$). When treated with aqueous $$\text{KOH}$$, the two chlorine atoms on carbon-2 are substituted by two hydroxyl groups, forming an unstable gem-diol intermediate:

   $$\text{CH}_3\text{--CH}_2\text{--C(Cl)}_2\text{--CH}_3 \xrightarrow{\text{aq. KOH}} \text{CH}_3\text{--CH}_2\text{--C(OH)}_2\text{--CH}_3$$

2. Dehydration to Form a Functional Carbonyl:

   Since two hydroxyl groups cannot comfortably sit on the same carbon atom, the intermediate immediately loses a water molecule ($$-\text{H}_2\text{O}$$) to form 2-butanone:

   $$\text{CH}_3\text{--CH}_2\text{--C(OH)}_2\text{--CH}_3 \xrightarrow{-\text{H}_2\text{O}} \text{CH}_3\text{--CH}_2\text{--C}(=\text{O})\text{--CH}_3$$

3. Iodoform Conversion:

   The resulting 2-butanone contains the required methyl ketone ($$\text{--C}(=\text{O})\text{CH}_3$$) structural component. When potassium hypoiodite ($$\text{KOI}$$) is added, it undergoes oxidative cleavage to produce a bright yellow precipitate of iodoform ($$\text{CHI}_3$$) along with a potassium carboxylate salt:

   $$\text{CH}_3\text{--CH}_2\text{--C}(=\text{O})\text{--CH}_3 \xrightarrow{\text{KOI}} \text{CH}_3\text{--CH}_2\text{--COOK} + \text{CHI}_3 \downarrow$$

Conclusion:

As clearly depicted in the reaction scheme, 2,2-dichlorobutane converts directly into a methyl ketone intermediate under basic conditions, making it perfectly suited to deliver a positive iodoform response.

Answer: Option A