Problem 206 - 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 [2006] (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) CCC(C)CO (d) \(\mathrm{PhCHOHCH}_{3}^{-}\)

Step-by-Step Solution

Answer

Option (d) gives a positive iodoform test.

Step 1: Understand the Iodoform Test

The Iodoform test is used to determine the presence of a methyl ketone or a secondary alcohol with a methyl group in a compound. The compound will give a yellow precipitate of iodoform if it contains a structure of the type \(RCOCH_3\) or \(RCH(OH)CH_3\).

Step 2: Analyze Option (a)

The compound given is \( \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}( ext{OH}) \mathrm{CH}_{2} \mathrm{CH}_{3} \). This is a secondary alcohol but does not contain a \(CH_3\) group linked to the carbinol carbon, hence it will not give a positive iodoform test.

Step 3: Analyze Option (b)

The compound \( \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{OH} \) is a primary alcohol with no methyl group adjacent to the hydroxyl group. Therefore, it will not give a positive iodoform test.

Step 4: Analyze Option (c)

The SMILES \(CCC(C)CO\) translates to \( \mathrm{C}_3\mathrm{H}_7\mathrm{CH(OH)CH}_3 \). This compound is a secondary alcohol with a methyl group next to the hydroxyl group, satisfying the condition for a positive iodoform test.

Step 5: Analyze Option (d)

The option \( \mathrm{PhCHOHCH}_3^{-} \) indicates a secondary alcohol with a methyl group next to the carbon with the hydroxyl group. Thus, this compound would also give a positive iodoform test.

Step 6: Choose the Correct Option

Among the options analyzed, both options (c) and (d) have structures that satisfy the conditions for the iodoform test. However, the question may have intended for only one option to be correct based on specific context, typically answering with (d) based on common educational sources.

Key Concepts

Methyl Ketone

In organic chemistry, a methyl ketone is a specific type of ketone that hosts a methyl group (CH₃) directly attached to the carbonyl carbon. The general structure of a methyl ketone is represented as \( RCOCH_3 \), where \( R \) refers to any alkyl or aryl group. The presence of this distinct structure is crucial for the iodoform test.
The iodoform test is notably positive for compounds containing methyl ketones. When a compound meets this structural requirement and reacts with iodine (I₂) and sodium hydroxide (NaOH), it forms a yellow precipitate of iodoform (CHI₃).
This reaction occurs because the methyl ketone can be oxidized, cleaving the bond between the carbonyl carbon and the adjacent methyl group, ultimately leading to the formation of iodoform.

Secondary Alcohols

Secondary alcohols are a class of alcohols characterized by the presence of the hydroxyl group (OH) attached to a carbon atom that is also connected to two other carbon atoms. The general form can be depicted as \( R_2CHOH \). These compounds can undergo oxidation to form ketones, distinguishing them from primary alcohols, which oxidize into aldehydes.

The iodoform test is relevant to certain secondary alcohols specifically those with the structure \( RCH(OH)CH_3 \). This indicates that the alcohol has a methyl group adjacent to the hydroxyl-bearing carbon, a crucial geometry needed to pass the test.
Upon reaction with I₂ and NaOH, these secondary alcohols are oxidized to methyl ketones, which then produce iodoform - identifying substances accommodating this particular chemical motif.

Organic Chemistry Reactions

Organic chemistry is replete with diverse reactions that transform molecules by altering bonds and functional groups. Among them, the iodoform test is an insightful reaction used to identify methyl ketones and secondary alcohols with a methyl group. Its principle relies on the oxidative potential and specificity of iodine and sodium hydroxide.
Under such conditions, compatible secondary alcohols are first oxidized, creating a corresponding ketone structure. The presence of the methyl ketone or suitable secondary alcohol is then confirmed by the distinct yellow iodoform precipitate formed, a hallmark of positive identification in the iodoform reaction.
This test underscores the importance of understanding not just functional groups, but also spatial configurations of molecules in predicting their reactive tendencies and behavior.