Problem 105 - A Complete Resource Book in Chemistry for JEE Main

Question

A substance \(\mathrm{C}_{4} \mathrm{H}_{10} \mathrm{O}\) yields on oxidation a compound, \(\mathrm{C}_{4} \mathrm{H}_{8} \mathrm{O}\) which gives an oxime and a positive iodoform test. The original substance on treatment with conc. \(\mathrm{H}_{2} \mathrm{SO}_{4}\) gives \(\mathrm{C}_{4} \mathrm{H}_{8} .\) The structure of the compound is (a) \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{COH}\) (b) \(\mathrm{CH}_{3} \mathrm{CHOHCH}_{2} \mathrm{CH}_{3}\) (c) \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{OH}\) (d) \(\mathrm{CH}_{3} \mathrm{CH}_{2}-\mathrm{O}-\mathrm{CH}_{2} \mathrm{CH}_{3}\)

Step-by-Step Solution

Answer

The correct structure is (b) \( \mathrm{CH}_{3} \mathrm{CHOHCH}_{2} \mathrm{CH}_{3} \).

Step 1: Analyze Oxidation Product

The compound \( \mathrm{C}_{4} \mathrm{H}_{10} \mathrm{O} \) oxidizes to form \( \mathrm{C}_{4} \mathrm{H}_{8} \mathrm{O} \), which is known to give an oxime and a positive iodoform test. These observations indicate that the oxidation product is a methyl ketone. The positive iodoform test specifically confirms a \( \mathrm{CH_{3}C=O} \) group presence. Thus, the oxidation product is most likely butanone (\( \mathrm{CH_{3}COCH_{2}CH_{3}} \)).

Step 2: Identify Parent Alcohol

Since the oxidation product is butanone, the original alcohol (\( \mathrm{C}_{4} \mathrm{H}_{10} \mathrm{O} \)) must be consistent with converting into butanone. This implies that upon oxidation, a secondary alcohol group is converted into the ketone. Given this condition, the original alcohol structure should be \( \mathrm{CH}_{3} \mathrm{CHOHCH}_{2} \mathrm{CH}_{3} \).

Step 3: Consider Reaction with Conc. \( \mathrm{H}_{2} \mathrm{SO}_{4} \)

The original compound \( \mathrm{C}_{4} \mathrm{H}_{10} \mathrm{O} \) when treated with conc. \( \mathrm{H}_{2} \mathrm{SO}_{4} \) forms \( \mathrm{C}_{4} \mathrm{H}_{8} \). This dehydration reaction confirms the presence of a secondary alcohol group, which upon losing water, forms a double-bonded compound. In the case of \( \mathrm{CH}_{3} \mathrm{CHOHCH}_{2} \mathrm{CH}_{3} \), the dehydration yields \( \mathrm{C}_{4} \mathrm{H}_{8} \) (2-butene).

Step 4: Choose the Correct Structure

Analyze the choices with our findings: The structure \( \mathrm{CH}_{3} \mathrm{CHOHCH}_{2} \mathrm{CH}_{3} \) is consistent with forming butanone upon oxidation, undergoing a positive iodoform test, and producing \( \mathrm{C}_{4} \mathrm{H}_{8} \) upon dehydration. Thus, the correct structure of the original compound is option (b).

Key Concepts

Oxidation Reaction

Oxidation reactions in organic chemistry often involve the gain of oxygen or the loss of hydrogen from a molecule. In this context, we have an alcohol with the formula \(\mathrm{C}_{4} \mathrm{H}_{10} \mathrm{O}\). Upon oxidation, this compound forms \(\mathrm{C}_{4} \mathrm{H}_{8} \mathrm{O}\), which indicates that the alcohol likely loses hydrogen to form a carbonyl group.
This specific reaction transforms a secondary alcohol into a ketone. Ketones are typically identified by the carbonyl group \((\mathrm{C=O})\) bonded to two alkyl groups. Oxidation is crucial for converting secondary alcohols to ketones since it transforms the hydroxyl (--OH) group into a carbonyl functional group.

Iodoform Test

The iodoform test is a qualitative analysis method used to detect the presence of a methyl ketone. When a compound reacts with iodine in the presence of a base, a yellow precipitate of iodoform (\(\mathrm{CHI}_{3}\)) forms if a \(\mathrm{CH}_{3}\mathrm{C=O}\) group is present.
In our exercise, the oxidation product \(\mathrm{C}_{4} \mathrm{H}_{8} \mathrm{O}\) gave a positive iodoform test, indicating it contains a methyl ketone structure, most likely butanone (2-butanone).

• This test helps confirm the functional groups present.
• It's a simple but effective method for identifying methyl ketones in organic compounds.

Dehydration Reaction

Dehydration reactions in chemistry often involve removing a molecule of water from a compound. In this case, our original alcohol \( \mathrm{C}_{4} \mathrm{H}_{10} \mathrm{O} \) is treated with concentrated sulfuric acid (\( \mathrm{H}_{2} \mathrm{SO}_{4} \)). This treatment typically leads to the elimination of water, converting the alcohol group into a double bond to form an alkene.
Specifically, the secondary alcohol \( \mathrm{CH}_{3} \mathrm{CHOHCH}_{2} \mathrm{CH}_{3} \) undergoes dehydration to give \( \mathrm{C}_{4} \mathrm{H}_{8} \), known as 2-butene in this context.

• Dehydration is a vital reaction for forming alkenes from alcohols.
• It reflects the versatile reactivity pathways of alcohols.

Secondary Alcohol

Secondary alcohols are characterized by their functional hydroxyl group (\(--OH\)) bonded to a carbon atom that is also attached to two other carbon atoms. In our exercise, \(\mathrm{CH}_{3} \mathrm{CHOHCH}_{2} \mathrm{CH}_{3}\) is the secondary alcohol.
Upon oxidation with an appropriate reagent, secondary alcohols typically form ketones. This context explains why the substance undergoes a transformation to \(\mathrm{C}_{4} \mathrm{H}_{8} \mathrm{O}\), a ketone, upon oxidation.

• Secondary alcohols are notable for their ability to form ketones.
• They typically do not oxidize further, which differentiates them from primary alcohols that can form aldehydes and then acids.

Methyl Ketone

Methyl ketones are an important class of ketones in organic chemistry where the carbonyl group is bonded to at least one methyl group. The presence of this structure is crucial for the positive iodoform test. Butanone, or 2-butanone, is a common methyl ketone that arises from the oxidation of secondary alcohols.
The formula \(\mathrm{C}_{4} \mathrm{H}_{8} \mathrm{O}\) confirms its ketonic nature by both the ability to form an oxime and produce a yellow precipitate in the iodoform test.

• Methyl ketones are useful in both synthetic and applied chemistry due to their reactivity.
• They serve as intermediates in various chemical synthesis processes.