Question

Consider the following sequence of reactions. The final product (Q) is : $${\mathrm{C}}_6{\mathrm{H}}_5\mathrm{CHO}+{\mathrm{CH}}_3\mathrm{CHO}\frac{\mathrm{NaOH}}{\mathrm{Nent}}(\mathrm{P})\frac{(\mathrm{i})\mathrm{L}\mathrm{L}{\mathrm{HH}}_4}{\text{ (ii) }{\mathrm{H}}_2\mathrm{O}}\text{ (Q) }$$ (A) ${\mathrm{C}}_6{\mathrm{H}}_5{\mathrm{CHCHOHCH}}_2{\mathrm{CH}}_2\mathrm{OH}$ (B) ${\mathrm{C}}_6{\mathrm{H}}_5{\mathrm{CH}}_2{\mathrm{CH}}_2{\mathrm{CH}}_2\mathrm{OH}$ (C) $\mathrm{C}\cdot {\mathrm{H}}_2\mathrm{CH}={\mathrm{CHCH}}_2\mathrm{OH}$ (D) ${\mathrm{C}}_6{\mathrm{H}}_5{\mathrm{CH}}_2{\mathrm{CH}}_2\mathrm{CHO}$

Short answer

The final product (Q) is: (A) $C_6{H}_{5}CHCH(OH)C{H}_{2}C{H}_{2}OH$.

Step-by-step solution

Identify the reactants and steps in the reactions

The reactants given are: 1. C₆H₅CHO (which represents an aldehyde group attached to a benzene ring) 2. CH₃CHO (which represents acetaldehyde) The reaction has two steps: Step 1: $C_6{H}_{5}CHO+C{H}_{3}CHO\stackrel{NaOH}{\to }(P)$ Step 2: $(P)\stackrel{\genfrac{}{}{0ex}{}{(i)LiAl{H}_4}{(ii)H_{2}O}}{\to }(Q)$ We will now analyze each step of the reaction to determine the product formed.

Reaction with NaOH (Formation of P)

In the first step, both C₆H₅CHO and CH₃CHO react with NaOH. From the given reagent (NaOH), we can deduce that this reaction is likely a nucleophilic addition reaction. Based on the reaction conditions, benzaldehyde (C₆H₅CHO) reacts with acetaldehyde (CH₃CHO) in the presence of aqueous NaOH. In this process, a nucleophilic addition of acetaldehyde to benzaldehyde occurs, leading to the formation of an intermediate product P.

Reaction with LiAlH₄ and H₂O (Formation of Q)

The intermediate product P from step 1 reaction then undergoes reaction with LiAlH₄ (Step (i)), followed by the addition of water (H₂O, Step (ii)). LiAlH₄ is a strong and selective reducing agent, typically used to reduce carbonyl compounds to alcohols. In the presence of LiAlH₄, the intermediate product P will undergo reduction to form an alcohol. Subsequently, by the addition of water, we will obtain the final product Q.

Identify the correct product Q from the given options

By analyzing the reactions and considering the reagents and sequence of reactions given, we can conclude that the correct final product Q should be an alcohol, having the benzene ring connected to an aliphatic chain. Comparing the given options, we can identify that option (A): $C₆H₅CHCH(OH)CH₂CH₂OH$ is the correct product Q. This is because option (A) has the expected alcohol product formed due to the nucleophilic addition and reduction reactions that have taken place in the sequence of reactions. So the correct answer to the given problem is: Q = (A) $C_6{H}_{5}CHCH(OH)C{H}_{2}C{H}_{2}OH$

Key concepts

Nucleophilic Addition

Nucleophilic addition is a fundamental step in many organic reactions that involve carbonyl compounds, like aldehydes or ketones. In this type of reaction, a nucleophile attacks the electrophilic carbon atom of the carbonyl group (a carbon double-bonded to oxygen, C=O). The pi bond breaks and forms a new sigma bond, resulting in the addition of the nucleophile to the carbon atom.

In our specific problem, this process occurs when acetaldehyde ( CH₃CHO ) and benzaldehyde ( C₆H₅CHO ) are reacted in the presence of a base such as NaOH . Here are the key aspects to understand this mechanism:

• The carbonyl carbon in aldehydes is positively polarized, making it a perfect site for nucleophiles, which are electron-rich species, to attack.
• The presence of a base facilitates the reaction, possibly by mildly deprotonating the nucleophile, enhancing its reactivity.
• The result is typically an alcohol, resulting from the addition of the nucleophile and rearrangement of electrons.

Understanding nucleophilic addition helps explain the transformation from reactants to intermediate products, setting up the stage for further reactions.

Reduction Reaction

Reduction reactions in organic chemistry often involve the gain of electrons, hydrogen, or the loss of oxygen in a molecule. The reduction of carbonyl compounds, such as aldehydes or ketones, to alcohols is a common and useful reaction.

In the exercise, the reduced product (Q) comes from the transformation of intermediate (P) through reduction. Here's a deeper look into what happens during this reaction:

• Aldehydes like benzaldehyde can be reduced to primary alcohols, while ketones would reduce to secondary alcohols.
• The reducing agent, LiAlH₄, donates hydride ions (H⁻), which act as nucleophiles attacking the positively charged carbon in carbonyl groups.
• The carbonyl's oxygen is protonated, either through the reagent itself or subsequent water addition, finalizing the conversion to an alcohol.

Reduction reactions are crucial as they often convert highly reactive carbonyl groups to more stable alcohols, allowing further functionalization or application of the organic molecule.

Reagent LiAlH₄

LiAlH₄, or lithium aluminium hydride, is an impressively potent reducing agent used primarily in organic synthesis to reduce esters, aldehydes, ketones, and other carbonyl containing substances to alcohols.

Some important characteristics of LiAlH₄ to understand its function in our context are:

• LiAlH₄ is powerful and selective, providing controlled reduction, especially effective for carbonyl compounds.
• It operates by delivering hydride ions, the active reducing agent, to the electrophilic carbon atom of the carbonyl group.
• Water or an acid is often added after the reaction, to hydrolyze the alkoxide intermediate and generate the alcohol product.

In the reaction sequence given, LiAlH₄ is employed to reduce the intermediate compound (P) formed after nucleophilic addition, leading to the alcohol product (Q). Understanding the role and behavior of LiAlH₄ in reduction reactions empowers chemists to predict outcomes and further manipulate organic compounds.