For the reaction sequence given below, the correct statement(s) is(are)
(A) P is optically active.
(

Q: Why is product P optically inactive even though the reactant has a wedge bond?

The wedge bond in the reactant indicates a specific stereocenter. However, when the lactone ring opens during reduction, the resulting molecule (P) has two identical $-\text{CH}_2\text{CH}_2\text{OH}$ groups attached to the central carbon. This symmetry makes the molecule achiral, so it loses its optical activity regardless of the starting stereochemistry.

Q: What is the role of $\text{CrO}_3-\text{H}_2\text{SO}_4$ in forming Q?

This mixture is known as the Jones reagent, a strong oxidizing agent. It oxidizes the primary alcohol groups at both ends of product P all the way to carboxylic acid groups, forming the dicarboxylic acid Q.

Q: How does soda lime ($\text{NaOH}/\text{CaO}$) convert Q to R?

Soda lime is used for decarboxylation. When heated with a carboxylic acid, it removes the $-\text{COOH}$ group as carbon dioxide (forming sodium carbonate) and replaces it with a hydrogen atom. Since Q has two $-\text{COOH}$ groups, both are removed.

Q: Why does dehydration of P form a diene instead of a cyclic ether?

While 1,4-diols can form cyclic ethers under mild acidic conditions, the high temperature (443 K) and concentrated $\text{H}_2\text{SO}_4$ strongly favor elimination (dehydration) over substitution, leading to the formation of the diene S.

Q: What is Bayer's test and why does S give a positive result?

Bayer's test uses cold, dilute, alkaline $\text{KMnO}_4$ to detect unsaturation (carbon-carbon double or triple bonds). Product S is a diene, so its double bonds react with the reagent, decolorizing the purple $\text{KMnO}_4$ and forming a brown $\text{MnO}_2$ precipitate.

Question

For the reaction sequence given below, the correct statement(s) is(are)

Options:

(A) P is optically active.

(B) S gives Bayer's test.

(C) Q gives effervescence with aq. NaHCO$_3$.

(D) R is an alkyne.

Sigma Percentile · Accepted Answer

Imagine you are a molecular architect, and your job is to dismantle a complex structure and rebuild it into entirely new forms using a specific set of chemical tools. This problem takes us on exactly that kind of thrilling journey through a multi-step organic synthesis. We start with a cyclic ester, break it open, oxidize it, strip away its acid groups, and even dehydrate it to see what it becomes. Let's break down this beautiful sequence step by step.

### Analyzing the Starting Material
Our journey begins with a six-membered ring containing an ester linkage. This makes it a **lactone**, specifically a $\delta$-lactone. If you look closely at the fourth position of the ring, there is a phenyl group attached via a solid wedge bond. This wedge is crucial—it tells us that we are starting with a specific stereoisomer, a single enantiomer. The spatial arrangement is fixed, and we must keep an eye on this chiral center as we proceed.

### The Power of Lithium Aluminum Hydride
Our first chemical tool is $	ext{LiAlH}_4$, a heavy-duty reducing agent. Think of it as a pair of molecular scissors that specifically targets the carbon-oxygen bonds of the ester. It cleaves the ring open, reducing both the carbonyl carbon and the ring oxygen into primary alcohols.

The result is an open-chain diol, **Product P**. Its chemical name is 3-phenylpentane-1,5-diol.

Now, let's address the stereochemistry. The original molecule had a chiral center, so is P optically active? Let's inspect the central carbon of P. It is attached to a hydrogen atom, a phenyl group, and... two identical $-	ext{CH}_2	ext{CH}_2	ext{OH}$ groups! Because it possesses two identical substituents, this carbon is no longer a chiral center. The molecule has become **achiral** and is therefore optically inactive. This elegantly disproves Option (A).

### The Heavy Artillery: Jones Oxidation
Next, we take our diol P and treat it with $	ext{CrO}_3$ in aqueous $	ext{H}_2	ext{SO}_4$, commonly known as Jones reagent. This is a strong oxidizing agent that doesn't stop halfway. It oxidizes both primary alcohol groups at the ends of the chain all the way up to carboxylic acids.

This transformation yields **Product Q**, which is 3-phenylglutaric acid. Because Q is a dicarboxylic acid, it exhibits classic acidic properties. If you drop it into aqueous sodium bicarbonate ($	ext{NaHCO}_3$), it will vigorously react to release carbon dioxide gas, creating a visible fizz or effervescence. This confirms that Option (C) is absolutely correct.

### The Molecular Weight-Loss Program
What happens when we heat Q with soda lime (a mixture of $	ext{NaOH}$ and $	ext{CaO}$)? Soda lime is famous for its ability to perform **decarboxylation**. It effectively rips the $-	ext{COOH}$ groups off the molecule, releasing them as carbon dioxide and replacing them with simple hydrogen atoms.

Since Q has two carboxylic acid groups, both are removed. The resulting molecule, **Product R**, is isopropylbenzene, widely known as cumene. Looking at its structure, it is an aromatic hydrocarbon with an alkyl side chain. There are no triple bonds to be found, meaning R is definitely not an alkyne. Thus, Option (D) is incorrect.

### Turning Up the Heat: Dehydration
Finally, let's rewind back to our diol, Product P, and subject it to a different extreme: concentrated sulfuric acid at a scorching 443 K ($170^\circ	ext{C}$). These are classic, harsh conditions for the dehydration of alcohols.

The high heat forces a water molecule to be eliminated from each side of the chain, creating carbon-carbon double bonds. The result is **Product S**, 3-phenylpenta-1,4-diene. This is a conjugated diene, rich in $\pi$-electrons.

Because S contains these carbon-carbon double bonds, it is highly unsaturated. If we introduce it to Bayer's reagent (cold, dilute, alkaline $	ext{KMnO}_4$), the double bonds will readily react, decolorizing the vibrant purple solution and leaving behind a brown precipitate of $	ext{MnO}_2$. This positive Bayer's test confirms that Option (B) is correct.

### The Final Verdict
By carefully tracing the mechanistic pathway of each reagent, we've unraveled the entire sequence. We discovered that the initial chirality is lost upon ring opening, making P optically inactive. We confirmed the acidic nature of Q and the unsaturated nature of S, while proving R is merely an aromatic alkane. The correct statements are indeed (B) and (C).

Animated Solution for Chemistry - Organic Chemistry: For the reaction sequence given below, the correct statement(s) is(are)

Select Answer:

Visualized Solution

$Analyzing the Reactant$

$Reduction to Product P$

$Stereochemistry of P$

$Oxidation to Product Q$

$Properties of Q$

$Decarboxylation to Product R$

$Properties of R$

$Dehydration to Product S$

$Properties of S$

$Conclusion$

The Sigma Insight: Carbonyl Compounds

Analyzing the Starting Material Our journey begins with a six-membered ring containing an ester linkage

The Power of Lithium Aluminum Hydride Our first chemical tool is $LiAlH_{4}$ , a heavy-duty reducing agent

The Heavy Artillery

The Molecular Weight-Loss Program What happens when we heat Q with soda lime (a mixture of $NaOH$ and $CaO$ )? Soda lime is famous for its ability to perform decarboxylation

Turning Up the Heat

The Final Verdict By carefully tracing the mechanistic pathway of each reagent, we've unraveled the entire sequence

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Animated Solution for Chemistry - Organic Chemistry: For the reaction sequence given below, the correct statement(s) is(are)

Select Answer:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Analyzing the Reactant

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Reduction to Product P

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Stereochemistry of P

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Oxidation to Product Q

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Properties of Q

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Decarboxylation to Product R

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Properties of R

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Dehydration to Product S

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Properties of S

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Conclusion