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Reaction of aniline with benzaldehyde is a \ldots...... reaction? (a) addition (b) condensation (c) polymerization (d) substitution

Short Answer

Expert verified
The reaction is a condensation reaction.

Step by step solution

01

Understand the Reactants

Aniline is an aromatic amine with the formula \( C_6H_5NH_2 \), and benzaldehyde is an aromatic aldehyde with the formula \( C_6H_5CHO \). We need to determine the type of reaction they undergo when combined.
02

Identify Possible Reaction Types

The options given are addition, condensation, polymerization, and substitution. Condensation reactions are characterized by the combination of two molecules with the loss of a small molecule, often water.
03

Analyze Reaction Characteristics

When aniline reacts with benzaldehyde, they commonly undergo a condensation reaction to form a Schiff base by losing a molecule of water. This is noted by the formation of an imine (\( C=N \)).
04

Conclusion on Reaction Type

Based on the analysis, the reaction between aniline and benzaldehyde, involving the loss of water, fits the criteria of a condensation reaction.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Aniline and Benzaldehyde Reaction
Aniline and benzaldehyde are two important organic compounds. Aniline is an aromatic amine with the formula \( C_6H_5NH_2 \), while benzaldehyde is an aromatic aldehyde with the formula \( C_6H_5CHO \). When these two compounds come together, they participate in a particular type of organic chemical reaction.
This reaction is known as a condensation reaction. In a condensation reaction, two molecules join together, typically resulting in the elimination of a small molecule such as water. The specific reaction between aniline and benzaldehyde is often discussed because it leads to the formation of a Schiff base.
This process is vital in synthetic organic chemistry as it facilitates the creation of new compounds with varying degrees of complexity. Understanding how aniline and benzaldehyde react not only deepens comprehension of basic organic reactions but also lays the groundwork for advanced chemical synthesis.
Formation of Schiff Base
A Schiff base is a compound with a functional group that features a carbon-nitrogen double bond, also known as an imine, denoted by \( C=N \). This structure forms when a primary amine (such as aniline) reacts with an aldehyde (like benzaldehyde).
The formation of a Schiff base is not just a simple union of two molecules. It's a fascinating process involving a series of intermediate steps:
  • The nucleophilic nitrogen in the aniline attacks the electrophilic carbon in the benzaldehyde.
  • Initially, this forms an unstable intermediary known as a carbinolamine.
  • Eventually, the carbinolamine loses a molecule of water to form the more stable imine structure, which is the Schiff base.
This reaction is highly advantageous in synthetic organic chemistry for creating a variety of complex molecules. Schiff bases are pivotal not only in laboratory settings but also in biological systems, playing roles in various enzymatic and biochemical processes.
Loss of Water in Reactions
One noteworthy characteristic of many condensation reactions, including that between aniline and benzaldehyde, is the loss of a small molecule like water. This loss of water is pivotal in defining a reaction as a condensation reaction.
In the reaction between aniline and benzaldehyde, once the initial bond forms between the nitrogen of the amine and the carbonyl carbon of the aldehyde, a water molecule is eliminated. This elimination helps drive the reaction forward, resulting in the formation of the stable Schiff base with the imine functional group.
Losing a water molecule is common in various organic reactions, helping stabilize the formed compound while also occasionally altering its reactivity. This step is crucial because it represents the transformation from the initial carbinolamine to the final imine. Understanding the thermodynamic and kinetic aspects of water loss provides insight into why certain reactions proceed the way they do, offering a deeper appreciation of reaction mechanisms and outcomes.

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Most popular questions from this chapter

Nitrobenzene gives n-phenylhydroxyl-amine by (a) \(\mathrm{H}_{2} / \mathrm{Pd}-\mathrm{C}\) (b) \(\mathrm{Sn} / \mathrm{HCl}\) (c) \(\mathrm{Zn} / \mathrm{NH}_{4} \mathrm{Cl}\) (d) \(\mathrm{Zn} / \mathrm{NaOH}\)

How many of the following amines can be used for the symmetrical cleavage of diborane? \(\mathrm{CH}_{3} \mathrm{NH}_{2},\left(\mathrm{CH}_{3}\right)_{2} \mathrm{NH},\left(\mathrm{C}_{2} \mathrm{H}_{3}\right)_{3} \mathrm{~N},\left(\mathrm{CH}_{3}\right)_{3} \mathrm{~N},\left(\mathrm{C}_{2} \mathrm{H}_{3}\right)_{2} \mathrm{NH}\), \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{NH}_{2}, \mathrm{C}_{5} \mathrm{H}_{5} \mathrm{~N}\), (i \(\left.-\mathrm{Pr}\right)_{3} \mathrm{~N}\), quinuclidine

Among the following compounds, which will react with acetone to give a product containing \(>\mathrm{C}=\mathrm{N}-\) ? 1\. \(\mathrm{C}_{6} \mathrm{H}_{3} \mathrm{NH}_{2}\) 2\. \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{~N}\) 3\. \(\mathrm{C}_{6} \mathrm{H}_{3} \mathrm{NHC}_{6} \mathrm{H}_{5}\) 4\. \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NHNH}_{2}\) (a) 3,4 (b) \(1,3,4\) (c) 1,3 (d) 1,4

Which of the following compounds cannot by identified by carbylamine test? (a) \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{NH}_{2}\) (b) \(\mathrm{CH}_{3}-\mathrm{CH}_{2}-\mathrm{CH}_{2}-\mathrm{NH}_{2}\) (c) \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{2}\) (d) \(\mathrm{C}_{2} \mathrm{H}_{5}-\mathrm{NH}-\mathrm{C}_{2} \mathrm{H}_{5}\)

The correct order of increasing basic nature for the bases \(\mathrm{NH}_{3}, \mathrm{CH}_{3} \mathrm{NH}_{2}\), and \(\left(\mathrm{CH}_{3}\right)_{2} \mathrm{NH}\) is \([\mathbf{2 0 0 3}]\) (a) \(\mathrm{NH}_{3}<\mathrm{CH}_{3} \mathrm{NH}_{2}<\left(\mathrm{CH}_{3}\right)_{2} \mathrm{NH}\) (b) \(\left(\mathrm{CH}_{3}\right)_{2} \mathrm{NH}<\mathrm{NH}_{3}<\mathrm{CH}_{3} \mathrm{NH}_{2}\) (c) \(\mathrm{CH}_{3} \mathrm{NH}_{2}<\left(\mathrm{CH}_{3}\right)_{2} \mathrm{NH}<\mathrm{NH}_{3}\) (d) \(\mathrm{CH}_{3} \mathrm{NH}_{2}<\mathrm{NH}_{3}<\left(\mathrm{CH}_{3}\right)_{2} \mathrm{NH}\)

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