Which of the following will not give yellow precipitate when treated…

The correct answer is Option (3) →

The name of the given compounds in the options are:

To determine which compound will not give a yellow precipitate when treated with iodine (\(I_2\)) in the presence of sodium hydroxide (\(NaOH\)), we need to understand the iodoform reactio*. This reaction typically occurs with alcohols that have at least one methyl group adjacent to the hydroxyl group and can be oxidized to a carbonyl compound.

Overview of the Iodoform Reaction

The iodoform reaction involves the following steps:

Oxidation: The alcohol is oxidized to form a ketone or aldehyde.

Iodination: The carbonyl compound then reacts with iodine in the presence of a base (like \(NaOH\)).

Formation of Yellow Precipitate: The yellow precipitate formed is iodoform (\(CHI_3\)), which is characteristic of compounds that meet the requirements for the reaction.

Analyzing Each Compound

1. Propan-2-ol (isopropanol):

This compound has a methyl group adjacent to the hydroxyl group, so it will produce iodoform and give a yellow precipitate.

2. Butan-2-ol:

Similar to propan-2-ol, butan-2-ol also has a methyl group adjacent to the hydroxyl group, allowing it to react to form iodoform. It will also give a yellow precipitate.

3. Methanol:

Methanol does not have a methyl group adjacent to the hydroxyl group; instead, it has only one carbon. Therefore, it will not undergo the iodoform reaction and will not produce a yellow precipitate.

4. 3-Methylpropan-2-ol:

This compound also has a methyl group adjacent to the hydroxyl group, allowing it to react to form iodoform and produce a yellow precipitate.

Conclusion

Based on the analysis above: Propan-2-ol, butan-2-ol, and 3-methylpropan-2-ol will give a yellow precipitate when treated with \(I_2\) in the presence of \(NaOH\). Methanol will not give a yellow precipitate.

Thus, the compound that will not give a yellow precipitate when treated with iodine in the presence of sodium hydroxide is Methanol.