In the following reactions, P, Q, R, and S are the major products. [JEE(Advanced) 2023]

(1) CH3CH2C

Q: Why doesn't the organocadmium reagent in reaction R react further with the ketone?

Organocadmium reagents are significantly less nucleophilic than Grignard or organolithium reagents. They are reactive enough to attack the highly electrophilic carbonyl carbon of an acid chloride, but not reactive enough to attack the less electrophilic carbonyl carbon of a ketone. This makes them excellent reagents for synthesizing ketones from acid chlorides!

Q: In reaction S, why does dehydration occur so readily in the final step?

The dehydration of the cyanohydrin (after hydrolysis to a carboxylic acid) forms an alpha-beta unsaturated carbonyl compound. This newly formed double bond is in conjugation with both the phenyl ring and the carboxylic acid group, providing immense thermodynamic stability through extended resonance. Heat ($\Delta$) provides the necessary activation energy for this elimination.

Q: How do I quickly spot an asymmetric carbon?

Look for an $sp^3$ hybridized carbon (a carbon with four single bonds) that is attached to four completely different groups. If any two groups are identical (like two hydrogen atoms, two methyl groups, or two phenyl rings), the carbon is achiral.

Q: Why does reaction P yield a chiral center if the carbon attacked by the Grignard ends up with two phenyl groups?

You are absolutely right that the carbon attacked by the Grignard reagent becomes achiral because it has two identical phenyl groups. However, the original carbon chain already contained a chiral center at the $CH(CH_3)$ position! This center remains untouched during the reaction, so the final molecule as a whole still possesses an asymmetric carbon.

Q: What is the role of $CrO_3$ and dilute $H_2SO_4$ in reaction S?

This combination is known as the Jones reagent. It is a strong oxidizing agent. In this specific reaction, it oxidizes the secondary alcohol (formed from the initial Grignard addition to the aldehyde) into a ketone.

Question

In the following reactions, P, Q, R, and S are the major products.

Options:

(A) Both P and Q have asymmetric carbon(s).

(B) Both Q and R have asymmetric carbon(s).

(C) Both P and R have asymmetric carbon(s).

(D) P has asymmetric carbon(s), S does not have any asymmetric carbon.

Sigma Percentile · Accepted Answer

The beauty of organic chemistry lies in its logical sequences. In this problem, we are tasked with predicting the major products of four distinct reaction pathways and then analyzing them for chirality. Let's break down each transformation step-by-step.

### Decoding the Nitrile: Formation of Product P

Our first journey begins with a substituted nitrile: $CH_3CH_2CH(CH_3)CH_2CN$.

When this nitrile is treated with phenyl magnesium bromide ($PhMgBr$), the nucleophilic phenyl group attacks the electrophilic carbon of the cyano group. Following acidic hydrolysis ($H_3O^+$), the intermediate imine salt is converted into a ketone: $CH_3CH_2CH(CH_3)CH_2COPh$.

But the reaction doesn't stop there! A second equivalent of the Grignard reagent attacks this newly formed ketone. The carbonyl carbon is transformed into a tertiary alcohol, yielding our final product P: $CH_3CH_2CH(CH_3)CH_2C(OH)Ph_2$.

Now, let's check for chirality. The carbon attached to the $-OH$ group is bonded to two identical phenyl rings, making it achiral. However, if we look back at the carbon chain, the $-CH(CH_3)-$ carbon is bonded to four different groups: a hydrogen atom, a methyl group, an ethyl group, and the bulky alcohol-containing chain.

**Therefore, product P possesses an asymmetric carbon.**

### Friedel-Crafts and Grignard: Unveiling Product Q

Reaction two starts with benzene ($Ph-H$) and acetyl chloride ($CH_3COCl$) in the presence of anhydrous $AlCl_3$.

This is a classic **Friedel-Crafts acylation**. The electrophilic acylium ion attacks the benzene ring, producing acetophenone ($PhCOCH_3$).

Next, we introduce $PhMgBr$ followed by aqueous workup. The Grignard reagent attacks the carbonyl carbon of acetophenone, forming a tertiary alcohol: $PhC(OH)(CH_3)Ph$.

Let's evaluate product Q for chirality. The central carbon is bonded to a methyl group, a hydroxyl group, and **two identical phenyl groups**. Because of this symmetry, it cannot be a chiral center.

**Thus, product Q does not have any asymmetric carbon.**

### The Organocadmium Route: Synthesizing Product R

In the third sequence, we react an acid chloride ($CH_3CH_2COCl$) with dibenzylcadmium ($(PhCH_2)_2Cd$).

Organocadmium reagents are uniquely useful because they are less reactive than Grignard reagents. They successfully convert acid chlorides into ketones but do not react further with the resulting ketone. This gives us ethyl benzyl ketone: $CH_3CH_2COCH_2Ph$.

We then treat this ketone with $PhMgBr$ and water. The Grignard addition yields a tertiary alcohol: $CH_3CH_2C(OH)(Ph)(CH_2Ph)$.

Let's inspect the central carbon of product R. It is bonded to an ethyl group, a benzyl group, a phenyl group, and a hydroxyl group. All four substituents are entirely distinct!

**Consequently, product R clearly has an asymmetric carbon.**

### A Multi-Step Masterpiece: The Journey to Product S

The final reaction is a beautiful four-step sequence starting with phenylacetaldehyde ($PhCH_2CHO$).

First, Grignard addition ($PhMgBr, H_2O$) converts the aldehyde into a secondary alcohol: $PhCH_2CH(OH)Ph$.

Second, the Jones reagent ($CrO_3$, dil. $H_2SO_4$) oxidizes this secondary alcohol back into a ketone: $PhCH_2COPh$.

Third, the addition of hydrogen cyanide ($HCN$) forms a cyanohydrin intermediate: $PhCH_2C(OH)(CN)Ph$.

The final step is the grand finale: heating with sulfuric acid ($H_2SO_4, \Delta$). This harsh condition accomplishes two things simultaneously. It hydrolyzes the cyano group into a carboxylic acid, and it dehydrates the alcohol to form a double bond. The resulting product S is an $\alpha,\beta$-unsaturated acid: $PhCH=C(COOH)Ph$.

Because product S contains a double bond at the critical carbons and lacks any $sp^3$ hybridized carbon bonded to four different groups, it is achiral.

**Therefore, product S does not have any asymmetric carbon.**

### The Final Verdict: Chirality Check

Let's summarize our structural analysis:
- **Product P:** Has an asymmetric carbon.
- **Product Q:** No asymmetric carbon.
- **Product R:** Has an asymmetric carbon.
- **Product S:** No asymmetric carbon.

Evaluating the given options, statement (C) correctly identifies that both P and R have asymmetric carbons. Statement (D) correctly states that P has an asymmetric carbon while S does not.

**The correct options are (C) and (D).**

Animated Solution for Chemistry - Organic Chemistry: In the following reactions, P, Q, R, and S are the major products.

Select Answer:

Visualized Solution

$C H_{3} C H_{2} C H (C H_{3}) C H_{2} C N P h M g B r, H_{3} O^{+} C H_{3} C H_{2} C H (C H_{3}) C H_{2} C O P h$

$C H_{3} C H_{2} C H (C H_{3}) C H_{2} C O P h P h M g B r, H_{2} O C H_{3} C H_{2} C H (C H_{3}) C H_{2} C (O H) P h_{2}$

$P h H + C H_{3} C O C l A l C l_{3} P h C O C H_{3} P h M g B r, H_{2} O P h C (O H) (C H_{3}) P h$

$C H_{3} C H_{2} C O C l (P h C H_{2})_{2} C d C H_{3} C H_{2} C O C H_{2} P h P h M g B r, H_{2} O C H_{3} C H_{2} C (O H) (P h) (C H_{2} P h)$

$P h C H_{2} C H O P h M g B r P h C H_{2} C H (O H) P h C r O_{3} P h C H_{2} C O P h$

$P h C H_{2} C O P h H C N P h C H_{2} C (O H) (C N) P h H_{2} S O_{4}, Δ P h C H = C (C O O H) P h$

\text{Final Conclusion}

The Sigma Insight: Carbonyl Compounds

Decoding the Nitrile

Friedel-Crafts and Grignard

The Organocadmium Route

A Multi-Step Masterpiece

The Final Verdict

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