Question:

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:

Updated On: Mar 26, 2025
  • A, B, and D only
  • A, B, and C only
  • A, B, C, and D
  • B, C, and D only
Hide Solution
collegedunia
Verified By Collegedunia

The Correct Option is B

Approach Solution - 1

Sodium iodide in the presence of iodine (NaOI) reacts with certain carbonyl compounds to give yellow precipitates. Phenylethanone, sec-butyl alcohol (secondary alcohol), and phenylethanal (aldehyde) all react with NaOI to form yellow precipitates. Methyl n-propylketone is less likely to form a yellow precipitate under these conditions.
Was this answer helpful?
0
0
Hide Solution
collegedunia
Verified By Collegedunia

Approach Solution -2

Sodium iodide in the presence of iodine (NaOI) is commonly used to react with certain carbonyl compounds, resulting in the formation of yellow precipitates. This reaction occurs due to the formation of an iodoform (CHI3) as a product, which is characteristic of carbonyl compounds with a structure capable of undergoing this transformation.

Phenylethanone (acetophenone), a ketone, reacts with NaOI to form a yellow precipitate due to the presence of the methyl group adjacent to the carbonyl group. This structure allows the formation of iodoform when treated with NaOI.

Ssec-butyl alcohol (a secondary alcohol) also reacts with NaOI to form a yellow precipitate. Secondary alcohols with adjacent methyl groups undergo oxidation and subsequent halogenation to give iodoform under these conditions.

Phenylethanol (phenylethanal), an aldehyde, reacts with NaOI to form a yellow precipitate as well. Aldehydes, like phenylethanol, are more readily oxidized and undergo the iodoform reaction, forming the yellow precipitate.

Methyl n-propylketone, although a ketone, is less likely to form a yellow precipitate under these conditions. This is due to its structure lacking the necessary adjacent groups (like a methyl group attached to the carbonyl carbon) that are essential for the iodoform reaction.


In conclusion, NaOI in the presence of iodine gives yellow precipitates with certain carbonyl compounds, particularly those with methyl or secondary alcohol groups adjacent to the carbonyl. However, compounds like methyl n-propylketone are less likely to react under these conditions.

Was this answer helpful?
0
0