Animated Solution for Chemistry - Organic Chemistry: The number of chiral centres present in [B] is ............. .

A Journey of Two Grignards

Welcome to a fantastic organic chemistry puzzle! We are presented with a reaction sequence starting from a substituted cyclohexane derivative. Our ultimate goal is to determine the exact number of chiral centers present in the final product, [B].

To solve this, we must carefully trace the chemical transformations step by step, paying close attention to the stereochemistry and the structural changes induced by the reagents.

The First Grignard Attack

Our journey begins with the reactant, which features a nitrile group ($-C\equiv N$). The first reagent introduced is ethylmagnesium bromide ($C_2{H}_{5}MgBr$), a classic Grignard reagent.

Grignard reagents are incredibly useful because they act as a source of strong nucleophilic carbanions. In this case, it provides an ethyl carbanion ($C_2{H}_5^{-}$). This nucleophile aggressively attacks the electrophilic carbon of the nitrile group, pushing the pi electrons onto the nitrogen to form an imine salt intermediate.

Following this attack, the reaction mixture is subjected to acidic hydrolysis ($H_3{O}^{+}$). This hydrolysis step is crucial; it converts the intermediate imine salt directly into a ketone. Thus, our intermediate product [A] is formed, featuring a newly established carbonyl group attached to the ethyl chain.

The Second Grignard Attack

With product [A] in hand, we proceed to the second phase of the sequence. We introduce another Grignard reagent, methylmagnesium bromide ($C{H}_{3}MgBr$).

This time, the methyl carbanion ($C{H}_3^{-}$) acts as our nucleophile. It targets the highly electrophilic carbonyl carbon of the ketone we just synthesized. The nucleophilic attack breaks the carbon-oxygen pi bond, forming a tetrahedral alkoxide intermediate.

To complete the reaction, a final hydrolysis step with water ($H_{2}O$) protonates the alkoxide oxygen. This transforms the carbonyl group into a tertiary alcohol. We have now successfully synthesized our final product, [B].

Hunting for Chiral Centers

The core of the question asks us to identify the number of chiral centers in product [B]. Remember the golden rule: a chiral center must be an $s{p}^3$ hybridized carbon atom bonded to four completely different groups.

Let's systematically evaluate the potential candidates in our molecule:

1. The Ring Carbon: Let's examine the carbon on the cyclohexane ring that connects to the side chain. Is it chiral? If we trace the path around the ring in both clockwise and counterclockwise directions, we find that the two paths are perfectly identical. Because it is attached to two identical paths, it does not possess four unique groups. Therefore, this carbon is achiral.

2. The First Side-Chain Carbon: Moving to the side chain, look at the carbon atom immediately adjacent to the ring. It is bonded to a hydrogen atom, a methyl group, the entire cyclohexane ring, and the rest of the complex side chain. All four of these groups are structurally distinct! Thus, this is our first chiral center.

3. The Alcohol Carbon: Finally, let's inspect the carbon bearing the hydroxyl ($-OH$) group. It is bonded to the hydroxyl group, a methyl group, an ethyl group, and the rest of the molecule leading back to the ring. Once again, we have four entirely different groups attached. This is our second chiral center.

Having thoroughly checked the molecule, we conclude that there are exactly two chiral centers. The correct integer answer to this question is 2.