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:

• A. A, B, and D only
• B. A, B, and C only
• C. A, B, C, and D
• D. B, C, and D only

Answer 1

The Correct Option is B

To determine which compounds will yield yellow precipitates when reacted with NaOI, we need to consider the haloform reaction. This reaction typically occurs with methyl ketones, alcohols with adjacent methyl groups, and certain aldehydes.

A. Phenylethanone (Acetophenone): Phenylethanone contains a methyl group adjacent to a carbonyl, which undergoes haloform reaction producing iodoform (yellow precipitate). Thus, it will give yellow precipitates.

B. Sec-Butyl alcohol: This alcohol can be oxidized to form methyl ketones containing an adjacent methyl group, which reacts to form iodoform. Hence, it will also produce yellow precipitates.

C. Phenylethanal: This compound is an aldehyde with a methyl group adjacent. It can participate in the haloform reaction, leading to the formation of yellow precipitates.

D. Methyl n-propylketone: This compound does not have a methyl group adjacent to the carbonyl group necessary for the haloform reaction, so it will not give yellow precipitates.

Thus, the compounds that will yield yellow precipitates with NaOI are A, B, and C. Therefore, the correct answer is: A, B, and C only

Answer 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.