Which of the following on reaction with NaOI will give yellow precipitates?

(A) Phenylethanone
(B) Sec-Butyl alcohol
(C) Phenylethanal
(D) Methyl n-propylketone

Choose the correct answer from the options given below:

(A), (B) and (D) only

The correct answer is Option (1) → (A), (B) and (D) only
Reaction with NaOI (sodium hypoiodite) gives the iodoform test, producing a yellow precipitate of iodoform (CHI₃).
This test is given by:

• Methyl ketones having the group CH₃CO–

• Secondary alcohols containing the group CH₃–CHOH– (which are oxidized to methyl ketones).

(A) Phenylethanone (Acetophenone)
Structure: C₆H₅COCH₃
It contains the CH₃CO– (methyl ketone) group, so it gives iodoform test → yellow precipitate.

(B) Sec-Butyl alcohol (2-butanol)
Structure: CH₃–CHOH–CH₂–CH₃
It contains the CH₃–CHOH– group, which on oxidation forms 2-butanone (a methyl ketone) → gives iodoform test.

(C) Phenylethanal
Structure: C₆H₅–CH₂–CHO
It does not contain the CH₃CO– group and cannot form a methyl ketone → does not give iodoform test.

(D) Methyl n-propyl ketone (2-pentanone)
Structure: CH₃COCH₂CH₂CH₃
It contains the CH₃CO– group, so it gives iodoform test.

Therefore, (A), (B), and (D) give the yellow precipitate of iodoform.