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 \(\mathrm{C}_{4} \mathrm{H}_{\mathrm{g}}\). 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 of the compound is (b) \( \mathrm{CH}_{3} \mathrm{CHOHCH}_{2} \mathrm{CH}_{3} \).

Step 1: Analyzing Oxidation Product

The compound with formula \( \mathrm{C}_{4} \mathrm{H}_{10} \mathrm{O} \) is oxidized to form \( \mathrm{C}_{4} \mathrm{H}_{8} \mathrm{O} \). This suggests the conversion of an alcohol group (\(-\mathrm{OH}\)) to a carbonyl group (\(\mathrm{C}=\mathrm{O}\)). We need to consider alcohols that can be oxidized to ketones or aldehydes with four carbon atoms.

Step 2: Considering the Iodoform Test Result

The resulting compound \( \mathrm{C}_{4} \mathrm{H}_{8} \mathrm{O} \) gives a positive iodoform test. The iodoform test is positive for methyl ketones or secondary alcohols that retain a methyl group after oxidation. This eliminates the primary alcohol option \("\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{OH}"\).

Step 3: Matching with Oxime Formation

The compound \( \mathrm{C}_{4} \mathrm{H}_{8} \mathrm{O} \) also forms an oxime. Both aldehydes and ketones can form oximes, suggesting the compound is a carbonyl compound that matches the results from previous steps.

Step 4: Analyzing Dehydration Reaction

The original substance, on reaction with concentrated \( \mathrm{H}_{2} \mathrm{SO}_{4} \), forms \( \mathrm{C}_{4} \mathrm{H}_{8} \), indicating a dehydration reaction to form an alkene. This is consistent with secondary alcohols, as they readily dehydrate to form alkenes.

Step 5: Eliminating Options and Finding the Compound

Examine the options: (a), (b), (c), (d). The structure \( \mathrm{CH}_{3} \mathrm{CHOHCH}_{2} \mathrm{CH}_{3} \) matches the criteria: oxidation to a methyl ketone (\( \mathrm{CH}_{3} \mathrm{COCH}_{2} \mathrm{CH}_{3} \)), forms an oxime, and gives a positive iodoform test. It also dehydrates to form butene \( \mathrm{C}_{4} \mathrm{H}_{8} \). Therefore, the correct structure of the compound is (b).

Key Concepts

Iodoform Test

The iodoform test is a chemical reaction that can help identify certain types of organic compounds, specifically those containing either methyl ketones or secondary alcohols with at least one methyl group. This test involves the addition of iodine and a base, often sodium hydroxide, to the substance being tested. If the test is positive, a yellow precipitate of iodoform (CHCl₃) forms. This distinct color change makes it easier to identify the presence of these particular groups.

In the context of our exercise, the compound \( \mathrm{C}_{4} \mathrm{H}_{8} \mathrm{O} \) gives a positive iodoform test, suggesting it is a methyl ketone or has a methyl group on a secondary alcohol. Since the test is negative for primary alcohols, we can deduce that the original compound must fall under the categories that show a positive result.

Oxidation Reactions

Oxidation reactions are central to organic chemistry and involve the increase in oxidation state of a molecule, typically by the loss of electrons. In simpler terms, oxidation in organic molecules is often seen as the addition of oxygen or the removal of hydrogen.

For the substance\( \mathrm{C}_{4} \mathrm{H}_{10} \mathrm{O} \), oxidation converts an alcohol group (\(-\mathrm{OH}\)) into a carbonyl group (\(\mathrm{C} = \mathrm{O}\)). An alcohol becomes either an aldehyde or ketone, depending on its structure. Secondary alcohols generally oxidize into ketones.

In the exercise above, the oxidation of the alcohol leads to a compound\( \mathrm{C}_{4} \mathrm{H}_{8} \mathrm{O} \) that has a carbonyl group, because it forms both an oxime and passes the iodoform test. This process helps identify the type of alcohol at the starting point: it cannot be a primary alcohol, leaving secondary alcohols as the probable candidates.

Dehydration Reactions

Dehydration reactions are a type of elimination reaction where water is removed from a molecule. In organic chemistry, this commonly occurs with alcohols using a strong acid like sulfuric acid (\(\mathrm{H}_{2} \mathrm{SO}_{4}\)) to form an alkene and water.

The exercise describes the conversion of\( \mathrm{C}_{4} \mathrm{H}_{10} \mathrm{O} \) to\( \mathrm{C}_{4} \mathrm{H}_{8} \) using concentrated \( \mathrm{H}_{2} \mathrm{SO}_{4} \). This reaction signals that the structure of the original molecule likely includes a secondary alcohol that can dehydrate to form an alkene. Secondary alcohols are prone to such reactions as they generally form stable carbocations, which facilitates the removal of a water molecule.

Overall, understanding dehydration reactions helps deduce the original structure of organic compounds, as secondary alcohols fit the criteria of forming alkenes after dehydration, further assisting in verifying the exercise solution.