CUET Preparation Today
Which of the following will not give yellow precipitate when treated with $I_2$ in presence of $NaOH$? |
|
|
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
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.
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.
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. |
