Problem 117
Question
Cinnamic acid is formed when \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CHO}\) condensation with \(\left(\mathrm{CH}_{3} \mathrm{CO}\right)_{2} \mathrm{O}\) in presence of (a) sodium metal (b) sodium acetate (c) anhydrous \(\mathrm{ZnCl}_{2}\) (d) conc. \(\mathrm{H}_{2} \mathrm{SO}_{4}\)
Step-by-Step Solution
Verified Answer
Sodium acetate is used in the Perkin reaction to synthesize cinnamic acid from benzaldehyde and acetic anhydride.
1Step 1: Identify the Reaction Type
The formation of cinnamic acid in the given context is likely to involve a type of condensation reaction. It requires the combination of an aldehyde, such as benzaldehyde (\(\text{C}_6 \text{H}_5 \text{CHO}\)), with acetic anhydride (\((\text{CH}_3 \text{CO})_2 \text{O}\)). This process is related to the Perkin reaction.
2Step 2: Explore the Role of Different Reagents
In the Perkin reaction context, the reagents facilitate the acylation of the aldehyde. Sodium metal and sodium acetate are potential catalysts for such reactions, where sodium acetate helps in generating an enolate ion from acetic anhydride needed for the condensation.
3Step 3: Determine the Correct Reagent for Perkin Reaction
Among the options given, sodium acetate is widely known to be used in the Perkin reaction to facilitate the condensation of benzaldehyde with acetic anhydride to form cinnamic acid.
Key Concepts
Cinnamic Acid FormationCondensation ReactionRole of Sodium Acetate
Cinnamic Acid Formation
Cinnamic acid is a key organic compound in the realm of synthetic chemistry. It is an aromatic acid with a sweet smell and is often used in flavorings and fragrances. The formation of cinnamic acid from benzaldehyde (\(\text{C}_6\text{H}_5\text{CHO}\)) and acetic anhydride (\((\text{CH}_3\text{CO})_2\text{O}\)) involves a well-known reaction called the Perkin reaction.
- The starting material benzaldehyde provides the aromatic ring necessary for cinnamic acid's structure.
- Acetic anhydride acts as an acylating agent, contributing its acetyl group to the reaction.
Condensation Reaction
Condensation reactions are a fundamental class of chemical reactions where two molecules combine to form a larger molecule, typically with the loss of a small molecule such as water. In the context of cinnamic acid synthesis, the condensation is between benzaldehyde and acetic anhydride. This is orchestrated as a cross aldol condensation, which is part of a broader family known as carbon-carbon bond-forming reactions.
Cross aldol condensation involves nucleophilic attack from an enolate derived from acetic anhydride onto the carbonyl carbon of benzaldehyde. The result is the formation of a beta-hydroxy carbonyl that subsequently dehydrates to form the alpha, beta-unsaturated cinnamic acid.
Cross aldol condensation involves nucleophilic attack from an enolate derived from acetic anhydride onto the carbonyl carbon of benzaldehyde. The result is the formation of a beta-hydroxy carbonyl that subsequently dehydrates to form the alpha, beta-unsaturated cinnamic acid.
- This type of reaction can form a new carbon-carbon bond.
- It is crucial for constructing complex organic molecules in a step-by-step, reliable manner.
- Loss of simpler molecules like water or acetic acid is common, streamlining the formation of the desired product.
Role of Sodium Acetate
Sodium acetate plays a pivotal role in the Perkin reaction, facilitating the efficient formation of cinnamic acid from benzaldehyde and acetic anhydride. It is specifically noted for its function as a catalyst in this process.
Here's how its role unfolds:
Here's how its role unfolds:
- Sodium acetate aids in generating the enolate ion from acetic anhydride. This enolate ion is the active nucleophile necessary for the subsequent attack on benzaldehyde.
- The presence of sodium ions stabilizes the enolate, enhancing the reaction's progress.
- Using a weak base such as sodium acetate minimizes unwanted side reactions and helps maintain the reaction conditions conducive to forming cinnamic acid.
Other exercises in this chapter
Problem 113
Predict the product 'B' in the sequence of reaction $$ \mathrm{HC} \equiv \mathrm{CH} \frac{30 \% \mathrm{H}_{2} \mathrm{SO}_{4}}{\mathrm{HgSO}_{4}}-\mathrm{A}
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An organic compound 'A' has the molecular formula \(\mathrm{C}_{3} \mathrm{H}_{6} \mathrm{O} .\) It undergoes iodoform test. When saturated with HCl it gives 'B
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A compound gives a yellow ppt. on warming with the aqueous solution of \(\mathrm{NaOH}\). Its vapour density is 29 . The compound is (a) \(\mathrm{CH}_{3} \math
View solution Problem 119
In the reaction \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{C} \equiv \mathrm{CH} \stackrel{\mathrm{H}_{2} \mathrm{SO}_{4}, \mathrm{HgSO}_{4}}{\longrightarrow}[\m
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