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Which reaction product is formed from benzaldehyde and methyl amine? (a) \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{2}\) (b) \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CH}=\mathrm{NCH}_{3}\) (c) \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CH}_{2} \mathrm{NH}_{2}\) (d) \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CONH}_{2}\)

Short Answer

Expert verified
The product is 83C_6H_5CH=NCH_3 (option b).

Step by step solution

01

Understanding the Reactants

Benzaldehyde (83C_6H_5CHO) is an aromatic aldehyde, while methyl amine (83CH_3NH_2) is a primary amine. When an aldehyde reacts with an amine, an imine or a similar compound is typically formed.
02

Recognize the Reaction Type

In a reaction between an aldehyde and a primary amine, a nucleophilic addition typically occurs, resulting in the formation of an imine. This process involves the replacement of the carbon-oxygen double bond with a carbon-nitrogen double bond.
03

Form the Imine Product

In this reaction, methyl amine reacts with benzaldehyde to form an imine by eliminating a water molecule from the newly formed compound. The expected imine structure would be 83C_6H_5CH=NCH_3.
04

Identify the Correct Option

From the provided options, 83C_6H_5CH=NCH_3 matches the predicted imine structure formed from the reaction of benzaldehyde and methyl amine.

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Key Concepts

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

Aldehyde and amine reactions
When aldehydes interact with amines, intriguing transformations occur. Both benzaldehyde and primary amines, like methyl amine, are essential players in creating versatile organic compounds. The key starting materials for aldehyde and amine reactions are:
  • Aldehydes: Compounds with a carbonyl group, where the carbon atom is bonded to a hydrogen atom and is frequently reactive due to the electronegativity of oxygen.
  • Amines: Organic derivatives of ammonia, categorized by their structure as primary (one alkyl or aryl group), secondary, or tertiary.
When an aldehyde like benzaldehyde reacts with a primary amine such as methyl amine, the primary reaction pathway is the formation of a new type of compound called imines. In the succeeding sections, we will explore how this fascinating transformation occurs.
Imine formation
The formation of imines is an exciting aspect of aldehyde and amine chemistry. Imine formation involves a multi-step mechanism carried out under mild conditions. This transformation begins with:
  • Nucleophilic Attack: The lone pair of electrons on the nitrogen atom in the amine attacks the electrophilic carbon atom in the aldehyde's carbonyl group, forming a nucleophilic addition product.
  • Proton Transfer: A proton is transferred, facilitating the departure of a water molecule, crucial for finalizing the imine structure.
The reaction between benzaldehyde and methyl amine, thus, results in the imine molecule formed by replacing the C=O double bond of the aldehyde with a C=N double bond. The structure of the product, C_6H_5CH=NCH_3, highlights the newly formed connection between the amine's nitrogen and the aldehyde's former carbonyl carbon.
Nucleophilic addition reactions
Nucleophilic addition reactions are foundational in organic chemistry, where nucleophiles and electrophiles interact to form more complex structures. Benzaldehyde and amine reactions are classic examples showcasing this versatile chemistry.
The mechanism includes:
  • Electrophilic Site: The carbon atom in the carbonyl group of benzaldehyde serves as the electrophilic center.
  • Nucleophile: Methyl amine acts as the nucleophile, utilizing its lone pair of electrons to attack the electrophilic carbon.
  • Intermediate Formation: This attack forms a tetrahedral intermediate, which quickly rearranges as a result of a subsequent proton transfer.
These steps culminate in the formation of imines, showcasing nucleophilic additions as dynamic pathways for synthesizing tailored molecules in organic synthesis.

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Most popular questions from this chapter

A substance \(\mathrm{C}_{4} \mathrm{H}_{10} \mathrm{O}\) yields on oxidation a compound, \(\mathrm{C}_{4} \mathrm{H}_{8} \mathrm{O}\) which gives an oxime and a positive iodoform test. The original substance on treatment with conc. \(\mathrm{H}_{2} \mathrm{SO}_{4}\) gives \(\mathrm{C}_{4} \mathrm{H}_{\mathrm{g}}\). The structure of the compound is (a) \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{COH}\) (b) \(\mathrm{CH}_{3} \mathrm{CHOHCH}_{2} \mathrm{CH}_{3}\) (c) \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{OH}\) (d) \(\mathrm{CH}_{3} \mathrm{CH}_{2}-\mathrm{O}-\mathrm{CH}_{2} \mathrm{CH}_{3}\)

On mixing ethyl acetate with aqueous sodium chloride, the composition of the resultant solution is (a) \(\mathrm{CH}_{3} \mathrm{COOC}_{2} \mathrm{H}_{5}+\mathrm{NaCl}\) (b) \(\mathrm{CH}_{3} \mathrm{COONa}+\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\) (c) \(\mathrm{CH}_{3} \mathrm{COCl}+\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}+\mathrm{NaOH}\) (d) \(\mathrm{CH}_{3} \mathrm{Cl}+\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{COONa}\)

Sodium ethoxide has reacted with ethanoyl chloride. The compound that is produced in the above reaction is: (a) 2-Butanone (b) Ethyl chloride (c) Ethyl ethanoate (d) Diethyl ether

The end product (c) in this reaction, \(\mathrm{CH}_{3} \mathrm{COOH} \stackrel{\mathrm{CaCO}_{3}}{\longrightarrow} \mathrm{A} \stackrel{\text { Heat }}{\longrightarrow} \mathrm{B}\) \(\stackrel{\mathrm{NH}_{2} \mathrm{OH}}{\longrightarrow} \mathrm{C}\), is (a) Acetaldehyde oxime (b) Aceto oxime (c) Acetamide (d) Ethane nitrile

The correct order of increasing acid strength of the compound is (a) \(\mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{H}\) (b) \(\mathrm{MeOCH}_{2} \mathrm{CO}_{2} \mathrm{H}\) (c) \(\mathrm{CF}_{3} \mathrm{CO}_{2} \mathrm{H}\) (d) \((\mathrm{Me})_{2} \mathrm{CH} \mathrm{CO}_{2} \mathrm{H}\) (a) \(\mathrm{b}<\mathrm{d}<\mathrm{a}<\mathrm{c}\) (b) \(\mathrm{d}<\mathrm{a}<\mathrm{c}<\mathrm{b}\) (c) \(d

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