Q.A carbonyl compound A having molecular formula C5H10O forms a crystalline precipitate with sodium bisulphite and gives a positive iodoform test. A does not reduce Fehling's solution. Identify A.

Step 1. A crystalline precipitate with sodium bisulphite (Section 12.5.A.2) restricts A to an ALDEHYDE or a METHYL KETONE -- only these are small/unhindered enough at the carbonyl carbon to add the bulky bisulphite ion cleanly.

Step 2. A does NOT reduce Fehling's solution (Section 12.6), which rules OUT any aldehyde (an aldehyde would reduce Fehling's readily) -- so A must specifically be a METHYL ketone, not an aldehyde.

Step 3. A positive iodoform test (Section 12.5.D) confirms the presence of a CH3-CO- fragment directly -- fully consistent with A being a methyl ketone. …

Use each clue to narrow the functional-group family in turn: bisulphite addition means aldehyde-or-methyl-ketone; a negative Fehling's test eliminates the aldehyde possibility; a positive iodoform test conf …

• Picking an aldehyde such as pentanal, since it too could plausibly give a bisulphite adduct and a (weaker) iodoform-adjacent signal via its own alpha-methyl-bearing chain -- the explicitly NEGATIVE Fehling's result is t …