Problem 159 - A Complete Resource Book in Chemistry for JEE Main

Question

Among the following the one that gives positive iodoform test upon reaction with \(\mathrm{I}_{2}\) and \(\mathrm{NaOH}\) is (a) \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}(\mathrm{OH}) \mathrm{CH}_{2} \mathrm{CH}_{3}\) (b) \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{OH}\) (c) \(\mathrm{CH}_{3}-\mathrm{CHCH}_{2} \mathrm{OH}\) (d) \(\mathrm{PhCHOHCH}_{3}\)

Step-by-Step Solution

Answer

Option (d) \( \text{PhCHOHCH}_3 \) gives a positive iodoform test.

Step 1: Understand the Iodoform Test

The iodoform test is used to identify methyl ketones or secondary alcohols that can be oxidized to methyl ketones. The compound should contain a \( \text{CH}_3\text{C}=O\) group or a \( \text{CH}(OH)\text{CH}_3\) group for the test to be positive.

Step 2: Analyze Option (a)

The compound \( \text{CH}_3 \text{CH}_2 \text{CH}(\text{OH}) \text{CH}_2 \text{CH}_3 \) contains a secondary alcohol but it is not directly adjacent to a methyl group. Therefore, it will not give a positive iodoform test.

Step 3: Analyze Option (b)

The compound \( \text{C}_6 \text{H}_5 \text{CH}_2 \text{CH}_2 \text{OH} \) is a primary alcohol and does not contain the necessary methyl group required for a positive iodoform test.

Step 4: Analyze Option (c)

The compound \( \text{CH}_3-\text{CHCH}_2 \text{OH} \) is a primary alcohol. It does not have the \( \text{CH}(OH)\text{CH}_3 \) structure needed for the iodoform test to be positive.

Step 5: Analyze Option (d)

The compound \( \text{PhCHOHCH}_3 \) contains a \( \text{CH}(OH)\text{CH}_3 \) group, which upon oxidation forms a methyl ketone. Therefore, it will give a positive iodoform test.

Key Concepts

Methyl Ketones

Methyl ketones are specific types of ketones that contain a methyl group (CH extsubscript{3}) attached to the carbonyl carbon (C=O). This structural feature is crucial for them to participate in the iodoform test, which identifies compounds with a CH extsubscript{3}C=O group as capable of producing a yellow precipitate known as iodoform (CHI extsubscript{3}).

The iodoform test is particularly useful in organic chemistry because it helps distinguish methyl ketones from other ketones. This test occurs in alkaline conditions using iodine (I extsubscript{2}) and a hydroxide, commonly sodium hydroxide (NaOH).
Such ketones react with iodine and base to undergo haloform reaction, resulting in the removal of the methyl group as iodoform:

• The carbonyl carbon is oxidized to form a carboxylate and iodoform precipitates out.
• This reaction is selective for methyl ketones because they have the specific chemical structure needed for the reaction pathway.

Understanding the presence and role of methyl ketones in the iodoform test allows chemists to identify key functional groups within a compound.

Secondary Alcohols

Secondary alcohols are a type of alcohol where the hydroxyl group (-OH) is connected to a secondary carbon atom, which is a carbon atom bonded to two other carbon atoms. These alcohols can sometimes participate in the iodoform test if they can be oxidized to form methyl ketones.

For a secondary alcohol to give a positive iodoform test, it must include a structure that can oxidize to form a specific type of ketone:

• The alcohol must have a configuration that includes a CH(OH)CH extsubscript{3} group.
• Upon oxidation, this group will transform into a methyl ketone, thus engaging in the iodoform reaction and producing the yellow iodoform precipitate.

Secondary alcohols without the methyl-adjacent hydroxyl setup will not participate positively in this test. It's important to note that not every secondary alcohol falls into this category; the molecule must specifically be able to convert to a structure susceptible to the haloform reaction.

Chemical Reactions

Chemical reactions lie at the heart of transforming substances into different chemicals, involving breaking and forming bonds to shape new molecular structures. The iodoform test is a specific type of chemical reaction tailored to identify particular compounds, namely methyl ketones and some secondary alcohols.

This reaction comprises several steps:

• The hydrogen atoms in the methyl group adjacent to a carbonyl or oxidized alcohol are substituted with iodine atoms.
• A base, such as NaOH, arranges for the removal of the substituted iodo group as a form of iodoform, given yellow precipitate form as the test result.

Whether dealing with organic synthesis or analytical assessment, understanding the basics of chemical reactions like the iodoform test helps clarify the molecular changes and pathways that a compound might undergo. In practice, successful applications involve balancing the conditions applied (such as pH and temperature) to ensure the reaction proceeds correctly.