Problem 115 - A Complete Resource Book in Chemistry for JEE Main

Question

A compound \(\mathrm{P}\left(\mathrm{C}_{3} \mathrm{H}_{6} \mathrm{Cl}_{2}\right)\) on reaction with an alkali either gives a compound \(\mathrm{Q}\left(\mathrm{C}_{3} \mathrm{H}_{6} \mathrm{O}\right)\) or \(\mathrm{R}\left(\mathrm{C}_{3} \mathrm{H}_{4}\right) .\) On oxidation, Q gives a compound \(\mathrm{C}_{3} \mathrm{H}_{6} \mathrm{O}_{2} . \mathrm{R}\), on reacting with dilute \(\mathrm{H}_{2} \mathrm{SO}_{4}\) containing \(\mathrm{Hg}^{2+}\) ion, gives a compound \(\mathrm{S}\left(\mathrm{C}_{3} \mathrm{H}_{6} \mathrm{O}\right)\), which reacts with bromine and alkali to give sodium salt of \(\mathrm{C}_{2} \mathrm{H}_{4} \mathrm{O}_{2}\). Then \(\mathrm{P}\) is (a) \(\mathrm{CH}_{2} \mathrm{ClCH}_{2} \mathrm{CH}_{2} \mathrm{Cl}\) (b) \(\mathrm{CH}_{3} \mathrm{CCl}_{2} \mathrm{CH}_{3}^{2}\) (c) \(\mathrm{CH}_{3} \mathrm{CHClCH}_{3} \mathrm{Cl}\) (d) \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CHCl}_{2}\)

Step-by-Step Solution

Answer

P is (c) \( \mathrm{CH}_{3} \mathrm{CHClCH}_{3} \).

Step 1: Identify Possible Structures of P

Given P has the formula \( \mathrm{C}_{3} \mathrm{H}_{6} \mathrm{Cl}_{2} \), we have different structural isomers listed in the options:1. \( \mathrm{CH}_{2} \mathrm{ClCH}_{2} \mathrm{CH}_{2} \mathrm{Cl} \)2. \( \mathrm{CH}_{3} \mathrm{CCl}_{2} \mathrm{CH}_{3} \)3. \( \mathrm{CH}_{3} \mathrm{CHClCH}_{3} \)4. \( \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CHCl}_{2} \)We need to check which of these can form compounds Q and R under the given conditions.

Step 2: Analyze Path of Reaction for Q (C3H6O)

Q is formed when P reacts with an alkali. Structure (3) \( \mathrm{CH}_{3} \mathrm{CHClCH}_{3} \) can form \( \mathrm{CH}_3\mathrm{COCH}_3 \) (acetone), a ketone, while other options produce compounds that don't fit \( \mathrm{C}_{3} \mathrm{H}_{6} \mathrm{O} \) isomers effectively.

Step 3: Analyze Path for R (C3H4) and its Further Reactions

R is formed if P gives \( \mathrm{C}_{3} \mathrm{H}_{4} \) upon reaction with an alkali. Both 1-propyne and 2-propyne (RC triple bonds) are possible, but the reaction of R with dilute \( \mathrm{H}_{2} \mathrm{SO}_{4} \) and \( \mathrm{Hg}^{2+} \) forming \( \mathrm{S} \) suggests a rearrangement to a ketone or aldehyde. The structure \( \mathrm{CH}_{3} \mathrm{CHClCH}_{3} \) matches the given reactivity as on reacting with alkali it forms propyne, which gives acetone on above treatment with Hg²⁺.

Step 4: Verify the Answer with Bromine Test

S reacts with bromine and alkali to form a sodium salt of acetate \( \mathrm{C}_{2} \mathrm{H}_{4} \mathrm{O}_{2} \). Acetone reacts with bromine and alkali to form sodium acetate in the haloform reaction.Thus, the presence of \( \mathrm{CH}_{3} \mathrm{COCH}_{3} \) (acetone) supports option 3: \( \mathrm{CH}_{3} \mathrm{CHClCH}_{3} \).

Step 5: Conclude the Recognition of Compound P

After evaluating the reactions and confirming with the compounds formed, the correct structure of P is \( \mathrm{CH}_{3} \mathrm{CHClCH}_{3} \) supported through the formation of R and Q as required.

Key Concepts

Structural Isomerism

Structural isomerism occurs when compounds have the same molecular formula but different structural arrangements. In the context of organic chemistry, this is a fascinating concept. For instance, compounds with the formula \( \mathrm{C}_{3} \mathrm{H}_{6} \mathrm{Cl}_{2} \) can form different structures. These isomers include:

• 1,1-dichloropropane \( (\mathrm{CH}_{3}\mathrm{CHClCH}_{3}) \)
• 1,2-dichloropropane \( (\mathrm{CH}_{2}\mathrm{ClCH}_{2}\mathrm{CH}_{2}\mathrm{Cl}) \)
• 1,3-dichloropropane \( (\mathrm{CH}_{3}\mathrm{CCl}_{2}\mathrm{CH}_{3}) \)
• Others like the ones mentioned in the original exercise

Each of these structures is a structural isomer of the others, having the same molecular formula but differing in the arrangement of atoms.
This variation in structure can drastically change the chemical properties and reactivities of the compounds, which is crucial when determining chemical behaviors and reaction pathways.

Reaction Mechanisms

Understanding reaction mechanisms in organic chemistry is essential for predicting the outcomes of chemical reactions. Let's break down the exercises mentioned:In the case of compound \( \mathrm{P}\), which reacts with an alkali, it's important to consider the mechanism that leads from \( \mathrm{P} \) to \( \mathrm{Q} \) and \( \mathrm{R} \).

• For \( \mathrm{Q} \): The alkali facilitates the transformation of \( \mathrm{CH}_{3}\mathrm{CHClCH}_{3} \) into acetone (\( \mathrm{CH}_{3}\mathrm{COCH}_{3} \)).
  This is known as a dehydrohalogenation reaction, where a chlorine atom and a hydrogen atom are lost, forming a double bond that rearranges to produce the ketone.
• For \( \mathrm{R} \): The same structural isomer could, under different conditions, eliminate two atoms to produce propyne (\( \mathrm{C}_{3}\mathrm{H}_{4} \)).
  These pathways highlight the multi-step nature of reaction mechanisms, where intermediates may form and rearrange into final products.

Understanding these reaction mechanisms helps identify possible transformation paths when analyzing organic compounds.

Alkali Reactions

Reactions involving alkalis are common in organic chemistry and often involve elimination or substitution processes. Here, the role of an alkali can be understood through its ability to deprotonate or remove halogens.Consider the reaction where compound \( \mathrm{P} \) (\( \mathrm{CH}_{3}\mathrm{CHClCH}_{3} \)) reacts with an alkali:

• The alkali can remove the chloride groups and hydrogen atoms adjacent to chloride atoms.
  This typically results in elimination reactions leading to the formation of unsaturated hydrocarbons like \( \mathrm{C}_{3}\mathrm{H}_{4} \) (propyne).
• Additionally, the alkali may facilitate hydroxide substitution, leading to the formation of an alcohol or a ketone, \( \mathrm{Q} \) (acetone).

These reactions underscore how alkalis can drive significant changes in compound structures leading to a range of products due to different mechanistic pathways.

Oxidation Reactions

Oxidation reactions involve the increase in the oxidation state of a molecule. This can often mean the addition of oxygen or removal of hydrogen in the context of organic compounds. In the exercise, oxidation plays a vital role.

• When \( \mathrm{Q} \) (acetone) undergoes oxidation, it forms a product with the formula \( \mathrm{C}_{3}\mathrm{H}_{6}\mathrm{O}_{2} \) (likely a carboxylic acid or related compound).
  This transformation indicates typical oxidative processes, possibly involving the increase in carbon's oxidation state from a ketone to a carboxylic acid.
• In another reaction pathway, propyne formed from \( \mathrm{R} \) undergoes rearrangement and conversion in the presence of \( \mathrm{Hg}^{2+} \) and sulfuric acid that involves oxidative processes to ultimately produce acetone \( \mathrm{S} \).

Understanding oxidation reactions allows organic chemists to predict how compounds' functional groups transform, providing insights into possible synthetic routes for desired chemical products. This understanding is crucial in synthesizing complex molecules and developing new reactions.