Problem 91 - A Complete Resource Book in Chemistry for JEE Main

Question

Treatment of propionaldehyde with dil. \(\mathrm{NaOH}\) solution gives (a) \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{COCH}_{2} \mathrm{CH}_{2} \mathrm{CHO}\) (b) \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CHOHCH}_{2} \mathrm{CH}_{2} \mathrm{CHO}\) (c) \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CHOHCH}\left(\mathrm{CH}_{3}\right) \mathrm{CHO}\) (d) \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{COOCH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{3}\)

Step-by-Step Solution

Answer

The correct answer is (c),  3_33_23OHCH(CH_3)CHO.

Step 1: Identify the Reaction

The given reaction is an Aldol condensation. In this reaction, two molecules of propionaldehyde ( 3_33_23O) will react in the presence of a dilute  3aOH solution. This reaction will initially form a beta-hydroxy aldehyde and may further lead to dehydration to form an alpha, beta-unsaturated carbonyl compound.

Step 2: Structure of Propionaldehyde

Propionaldehyde has the molecular formula  3_33_23O. The structure consists of a three-carbon chain ending in an aldehyde group:  3_33_23H.

Step 3: Aldol Addition

In the presence of dilute  3aOH, one aldehyde molecule deprotonates to form an enolate ion, which then attacks the carbonyl carbon of the second aldehyde molecule to form a beta-hydroxy aldehyde intermediate:  3_33_23OHCH(CH_3)CHO.

Step 4: Identify the Product

Following the addition, further reaction conditions might lead to the dehydration of the beta-hydroxy aldehyde, but the options provided focus on the possible beta-hydroxy product without dehydration.

Step 5: Match with Provided Options

The likely beta-hydroxy aldehyde product matches option (c):  3_33_23OHCH(CH_3)CHO. This compound is formed from the aldol addition of propionaldehyde, reflecting the structure of option (c).

Key Concepts

Beta-Hydroxy Aldehyde

When propionaldehyde undergoes an Aldol condensation in the presence of dilute sodium hydroxide (\(\mathrm{NaOH} \)), the result is the formation of a beta-hydroxy aldehyde. To understand this process, let's break it down into simpler terms.

The term "beta-hydroxy aldehyde" refers to a molecule that has both an aldehyde and a hydroxyl group (OH group) attached to it, located one carbon apart. In this case, the product from the reaction is \(\mathrm{CH}_{3}\mathrm{CH}_{2}\mathrm{CHOHCH}(\mathrm{CH}_{3})\mathrm{CHO} \).

Here's how it is formed:

• Two propionaldehyde molecules interact during the initial Aldol addition step.
• One of these molecules becomes a nucleophilic enolate ion, which we'll discuss in the next section.
• The enolate ion attacks the carbonyl carbon of another propionaldehyde molecule.
• This reaction results in the formation of our beta-hydroxy aldehyde product.

This completed reaction reflects the structure of option (c) in the original exercise's multiple-choice question, highlighting a key product from this chemical reaction.

Enolate Ion

The formation of an enolate ion is a pivotal step in Aldol condensation. This process begins when one of the aldehyde molecules in the reaction mixture deprotonates.

Deprotonation occurs at the alpha carbon, which is the carbon adjacent to the carbonyl group. When this hydrogen is removed, the molecule gains a negative charge, forming what we call an "enolate ion."

It's essential to understand why the enolate ion is so reactive:

• The enolate ion's negative charge makes it a strong nucleophile.
• This nucleophilicity allows it to effectively attack another molecule's carbonyl carbon.
• This ability to attack leads to the formation of a new carbon-carbon bond, which is integral to the formation of the beta-hydroxy aldehyde.

The enolate ion thus acts as a bridge, connecting two molecules during Aldol addition. Its role as a key reactive species underscores its importance in organic synthesis and helps achieve the final product of beta-hydroxy aldehyde.

Alpha, Beta-Unsaturated Carbonyl Compound

In some cases, after the formation of a beta-hydroxy aldehyde, further reactions can occur. Specifically, dehydration might take place to yield an alpha, beta-unsaturated carbonyl compound.

Let's explore this transformation further:

• "Dehydration" refers to the removal of a water molecule (\(\mathrm{H}_2\mathrm{O}\) from the beta-hydroxy aldehyde).
• The result is the formation of a carbon-carbon double bond between the alpha and the beta carbons.
• This double bond configuration is typical of alpha, beta-unsaturated carbonyl compounds, adding a characteristic UV-absorbent bond to the molecule.

Alpha, beta-unsaturated carbonyl compounds are significant in organic chemistry because they can participate in a variety of reactions like Michael addition. While dehydration wasn't the focus of the multiple-choice question you encountered, recognizing this possibility provides deeper insights into Aldol condensation products.