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 \(\mathrm{C}_{2} \mathrm{H}_{6} \mathrm{O}_{2} . \mathrm{R}\), 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) \(\mathrm{CH}_{3}^{2} \mathrm{CCl}_{2} \mathrm{CH}_{3}^{2}\) (c) \(\mathrm{CH}_{1} \mathrm{CHClCH}_{2} \mathrm{Cl}\) (d) \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CHCl}_{2}\)

Step-by-Step Solution

Answer

The compound \( P \) is option (d): \( \mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CHCl}_2 \).

Step 1: Analyze the Reaction from P to Q or R

The compound \( P \) with formula \( \mathrm{C}_3 \mathrm{H}_6 \mathrm{Cl}_2 \) reacts with an alkali which results in two pathways: forming either \( Q \) (\( \mathrm{C}_3 \mathrm{H}_6 \mathrm{O} \)) or \( R \) (\( \mathrm{C}_3 \mathrm{H}_4 \)). The presence of two chlorines suggests it can undergo elimination with alkali to form an alkene \( R \), or substitution to form an alcohol \( Q \).

Step 2: Verify the Oxidation Pathway for Q

The compound \( Q \), when oxidized, produces \( \mathrm{C}_2 \mathrm{H}_6 \mathrm{O}_2 \). This implies that \( Q \) is likely a primary alcohol, which can oxidize to carboxylic acid or a related dicarbonyl compound. The likely compound for this oxidation is propan-1-ol, leading to acetic acid/ethanediol type product.

Step 3: Analyze Reaction Pathway for R

The compound \( R \) becomes \( S \), camphthing through hydration via \( \mathrm{H}_2\mathrm{SO}_4 \) with \( \mathrm{Hg}^{2+} \), suggesting it is a terminal alkyne forming acetone (a ketone) \( S \). \( S \) reacting with bromine and alkali yielding \( \mathrm{C}_2 \mathrm{H}_4 \mathrm{O}_2 \) (acetic acid), reaffirms \( S \) is acetone.

Step 4: Match Patterns with Possible Compounds for P

Explore the given options to determine which can yield both \( Q \) and \( R \) under the reactions described.- \( \mathrm{CH}_2 \mathrm{ClCH}_2 \mathrm{CH}_2 \mathrm{Cl} \) is unlikely due to its structure not supporting transformation to \( C_3H_6O \) or \( C_3H_4 \).- \( \mathrm{CH}_3 \mathrm{CCl}_2 \mathrm{CH}_3 \) does not support the formation of primary alcohol for \( Q \).- \( \mathrm{CH} \equiv \mathrm{CHClCH}_2 \mathrm{Cl} \) and \( \mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CHCl}_2 \) support elimination/substitution to provide alkenes and alcohols.\(\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CHCl}_2\) is selected for secondary to primary reactions.

Step 5: Conclusion of Compound Selection

Based on oxidation and reaction analysis, option (d) \( \mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CHCl}_2 \) upon base gives propene and propanol fitting product paths for \( Q \) and \( R \). This supports pathway products, consistent with all transformations shown within given data.

Key Concepts

Elimination Reaction

The elimination reaction is a fundamental concept in organic chemistry where elements of a molecule, such as water or hydrogen halide, are removed, resulting in the formation of a double bond. In the exercise given, compound P undergoes an elimination reaction when treated with an alkali, resulting in the formation of compound R, an alkene with the formula \( \mathrm{C}_3 \mathrm{H}_4 \). This transformation typically occurs through an E2 mechanism.

• Mechanism: In an E2 reaction, a base abstracts a proton while leaving a good leaving group, like \( \mathrm{Cl^-} \), thus forming a double bond.
• Result: The result is the conversion of a haloalkane to an alkene.

Understanding this process helps explain the reactivity patterns observed in the transformation from compound P to R.

Oxidation of Alcohols

Oxidation reactions involve the increase of an element's oxidation state through the loss of electrons. In organic chemistry, alcohol oxidation serves to convert alcohols into carbonyl compounds.

• Primary Alcohol Oxidation: This is relevant to compound Q, which on oxidation produces \(\mathrm{C}_2 \mathrm{H}_6 \mathrm{O}_2\). Primary alcohols like propan-1-ol are typically oxidized to form either aldehydes or subsequently carboxylic acids, depending on the reaction conditions.
• Outcome: In this specific exercise, Q leads to a dicarbonyl compound, likely due to the mild to intermediate oxidation steps involved. Understanding this helps in deducing the identity of initially formed Q.

Alkyne Hydration

Alkyne hydration is a chemical reaction that adds water across the triple bond of an alkyne, resulting in an enol. This enol tautomerizes to a more stable carbonyl compound.

• Reaction Details: In the exercise, compound R, which is a terminal alkyne, reacts with \(\mathrm{H}_2\mathrm{SO}_4\) and \(\mathrm{Hg}^{2+}\) to form compound S. This reaction allows the transformation of an alkyne to a ketone.
• Tautomerization Process: Following hydration, the enol intermediate rapidly rearranges, leading to the formation of a keto product like acetone in this reactivity sequence.

Comprehending this reaction helps explain the formation and transformation mechanisms leading to the product S.

Alkenes and Alkynes

Alkenes and alkynes are unsaturated hydrocarbons known for containing double and triple carbon-carbon bonds, respectively. These functional groups are crucial in many organic reactions because their bonds can undergo transformations.

• Alkenes: As seen with compound R, alkenes are usually formed via elimination reactions and serve as intermediates in further chemical transformations.
• Alkynes: In the given exercise, understanding alkynes' reactivity guides how compound P transitions to different products via its reaction pathways. These include hydration, leading to further transformations.

Mastering the properties of these hydrocarbons helps dissect their roles in reactions and subsequent product formations.

Substitution Reaction

Substitution reactions are hallmark processes in organic chemistry where an atom or group in a molecule is replaced with a different atom or group. These reactions are characteristic of haloalkanes.

• Involved Mechanism: Typically, substitution can occur through either an \(\mathrm{S}_N1\) or \(\mathrm{S}_N2\) pathway.
• In Context: For compound P, substitution with an alkali leads to alcohol formation (compound Q), showcasing a classic example where a chlorine atom is substituted by a hydroxyl group due to the reaction with a base.

Understanding substitution helps trace back the steps leading from P to Q, illuminating reaction sequences.

Reaction Mechanisms

Reaction mechanisms in organic chemistry describe the step-by-step process through which a chemical reaction proceeds. These are crucial for predicting the products of a given set of reactants.

• Detailed Analysis: They involve breaking and forming chemical bonds through a series of elementary steps, as seen in transformations from compound P to Q and R.
• Pathway Clarity: Detailed mechanisms elucidate how particular conditions and reagents steer diverse outcomes, underlining the importance of understanding every elementary step involved.

These insights help interpret the complex pathways seen in the exercise, as all transformations from compound P hinge on these fundamental processes.