Question

Major product (P) and (Q) in following reaction is: $$ \text { (Q) } \stackrel{\text { (i) } \mathrm{CH}_{3} \mathrm{MgBr}}{\text { (ii) } \mathrm{H}_{2} \mathrm{O}}=\mathrm{O} \frac{\text { (i) } \mathrm{NaBH}_{4}}{\text { (ii) } \mathrm{H}_{2} \mathrm{O}}(\mathrm{P}) $$ (A) CC1(O)C=CCC1 (B) CC1(O)C=CCC1 (C) CC1CCC(=O)C1 (D) CC1(C)CCCC1(C)O

Short answer

The major products of the reaction are: - Product (P): Compound (A) - \(CC1(O)C=CCC1\) - Product (Q): Compound (C) - \(CC1CCC(=O)C1\)

Step-by-step solution

React compound (Q) with CH3MgBr and water

First, we need to determine the product formed when the compound (Q) reacts with Grignard reagent (CH3MgBr) and then with water. To do this, we identify the functional groups on compound (Q) that can undergo reaction with CH3MgBr, which is a nucleophile. During this step, the nucleophilic addition to carbonyl will take place.

React compound (Q) with NaBH4 and water

Next, we need to determine the product formed when the compound (Q) reacts with NaBH4 (sodium borohydride) and then with water. To do this, we identify the functional groups on compound (Q) that can undergo reaction with NaBH4, which is a reducing agent. During this step, the reduction of carbonyl group will happen.

Identify the major products (P) and (Q)

Now, we will compare the products formed in Steps 1 and 2 with the given compounds (A), (B), (C), and (D) to identify the products (P) and (Q). Among the given choices, compound (C) with SMILES notation "CC1CCC(=O)C1" corresponds to the product obtained from the reaction of compound (Q) with CH3MgBr and water (Step 1), and compound (A) with SMILES notation "CC1(O)C=CCC1" corresponds to the product obtained from the reaction of compound (Q) with NaBH4 and water (Step 2). So, the major products of the reaction are: - Product (P): Compound (A) - CC1(O)C=CCC1 - Product (Q): Compound (C) - CC1CCC(=O)C1

Key concepts

Nucleophilic Addition

The concept of nucleophilic addition is key to understanding many organic reactions, including Grignard reactions. A Grignard reagent, such as \( \text{CH}_3\text{MgBr} \), is a powerful nucleophile, which means it has an electron pair to donate and typically attacks an electrophile—a positively charged or electron-deficient atom. In the context of a carbonyl compound, the carbon atom of the carbonyl group \( (>C=O) \) acts as the electrophile.

This reaction is a two-step process. First, the nucleophile attacks the carbonyl carbon, breaking the \( C=O \) double bond and leading to the formation of an alkoxide ion. The carbon of the Grignard reagent forms a new bond with the carbonyl carbon, effectively introducing a new carbon-carbon bond into the molecule. The acidic environment, like water in the second step, then protonates the alkoxide to form an alcohol.

This kind of reaction is pivotal because it provides a method for lengthening carbon chains and constructing complex molecules from simpler precursors, which is a frequent requirement in synthetic organic chemistry.

Carbonyl Compounds

Carbonyl compounds are integral in organic chemistry and are characterized by the carbonyl group \( (>C=O) \). This group appears in aldehydes, ketones, carboxylic acids, esters, and other derivatives. The carbon atom in the carbonyl group is electrophilic, making it susceptible to attack by nucleophiles, such as Grignard reagents or hydride ions.

In carbonyl-containing compounds, the polarity of the \( C=O \) bond, due to the oxygen's electronegativity, is particularly important. This polarity results in a partial positive charge on the carbon, which facilitates its role in nucleophilic addition reactions. Understanding this behavior is critical when predicting or explaining the course of chemical reactions involving these compounds.

Nucleophilic addition to carbonyl compounds is not limited to Grignard reagents. Other nucleophiles, including cyanides and amines, can also participate, leading to a broad array of possible transformations. This versatility makes carbonyl chemistry a fundamental building block in many synthetic strategies.

Reduction with Sodium Borohydride

Reduction is a key transformation in organic synthesis, allowing the conversion of carbonyl groups into alcohols. Sodium borohydride \( (\text{NaBH}_4) \) is a mild reducing agent used predominantly for reducing aldehydes and ketones. It does not typically reduce esters or carboxylic acids, which makes it more selective compared to others like lithium aluminum hydride.

The reduction mechanism involves the transfer of hydride ions (\( \text{H}^- \)) from sodium borohydride to the electrophilic carbonyl carbon, which results in the formation of an alkoxide intermediate. Subsequent protonation, usually by water or a mild acid, converts the intermediate into the corresponding alcohol.

This specificity and mild nature of sodium borohydride make it a favorable choice in multi-step syntheses where sensitivity to water or temperature could be an issue. Students exploring this concept should practice writing reaction schemes to solidify their understanding of this fundamental reduction process.