Question

Give the structures of the major organic products you would expect from reaction of \(m\) -toluidine (m-methylaniline) with the following reagents: (a) \(\mathrm{Br}_{2}\) ( 1 equivalent) (b) \(\mathrm{CH}_{3} \mathrm{I}\) (excess) (c) \(\mathrm{CH}_{3} \mathrm{COCl}\) in pyridine (d) The product of (c), then \(\mathrm{HSO}_{3} \mathrm{Cl}\)

Short answer

(a) 4-bromo-3-methylaniline, (b) N,N,N-trimethyl-3-methylaniline, (c) 3-methylacetanilide, (d) para-acetamido-methylsulfonyl chloride.

Step-by-step solution

Reaction with Bromine

When m-toluidine reacts with one equivalent of bromine (\( \mathrm{Br}_{2} \)), the bromine molecule will substitute primarily at the ortho and para positions to the amino group, since the NH2 group is an activating and ortho/para-directing group. However, given that there is steric hindrance at the ortho position due to the methyl group, as major product, you will expect para substitution to the amino group, given that one equivalent of bromine is used. Thus, the major product is 4-bromo-3-methylaniline.

Reaction with Excess Methyl Iodide

When m-toluidine undergoes reaction with excess methyl iodide (\( \mathrm{CH}_{3} \mathrm{I} \)), the nitrogen of the amino group gets methylated. Each hydrogen atom can be replaced by a methyl group, leading to the formation of N,N,N-trimethyl-3-methylaniline, where the nitrogen is fully methylated.

Acylation with Acetyl Chloride

In the presence of pyridine, m-toluidine reacts with acetyl chloride (\( \mathrm{CH}_{3} \mathrm{COCl} \)) to form an amide. The acetyl group (\( \mathrm{CH}_{3} \mathrm{CO} \)) will substitute the hydrogen of the amino group, yielding 3-methylacetanilide, where the NH2 group becomes NHCOCH3.

Reaction of Acylation Product with Chlorosulfuric Acid

The product from Step 3 (3-methylacetanilide) will further react with chlorosulfonic acid (\( \mathrm{HSO}_{3} \mathrm{Cl} \)). This reagent introduces a sulfonyl chloride group to the para or ortho position relative to the NHCOCH3 group, preferably para due to steric hindrance by the acetyl group. Thus, the major product is para-acetamido-methylsulfonyl chloride.

Key concepts

Understanding m-Toluidine

M-toluidine, also known as m-methylaniline, is an aromatic compound that is typically characterized by a benzene ring with two distinct functional groups attached. It features a methyl group (-CH₃) and an amine group (-NH₂) positioned at a meta orientation to each other on the benzene ring. M-toluidine is significant in organic chemistry due to its reactivity, largely influenced by these groups.

The amine group is particularly important, as it acts as an activator, making the benzene ring more reactive. This group directs substitutions to the ortho and para positions, but steric factors often influence the final product. The presence of the methyl group slightly affects reactivity due to its electron-donating properties, though it mainly contributes to steric effects more than electronic ones.

Bromine Substitution Explained

Bromine substitution is a common reaction in organic chemistry where bromine atoms replace hydrogen atoms on an aromatic ring. For m-toluidine, this involves reaction with one equivalent of bromine (\(\mathrm{Br}_{2}\)). The amino group directs the substitution to the ortho and para positions.

However, the methyl group at the meta position can cause steric hindrance, making ortho substitution less favorable. Consequently, the major product formed is 4-bromo-3-methylaniline, where bromine substitutes at the para position relative to the amino group. This reaction showcases how even small groups can significantly influence the site of substitution on an aromatic ring.

Methylation Process

The methylation process involves the introduction of methyl groups to a compound. In the context of m-toluidine, the reaction with excess methyl iodide (\(\mathrm{CH}_{3} \mathrm{I}\)) targets the nitrogen atom of the amino group. Under these conditions, each hydrogen atom bonded to the nitrogen can be replaced by a methyl group.

This results in N,N,N-trimethyl-3-methylaniline, where the nitrogen becomes fully methylated. Methylation affects properties like solubility and reactivity, and in this case, it enhances the steric bulk around the nitrogen, impacting further reactivity and interactions within the molecule.

The Acylation Reaction

Acylation involves the introduction of an acyl group into a compound, often using an acyl chloride. For m-toluidine, reaction with acetyl chloride (\(\mathrm{CH}_{3} \mathrm{COCl}\)) in pyridine leads to the formation of an amide bond.

The acetyl group (\(\mathrm{CH}_{3} \mathrm{CO}\)) displaces a hydrogen atom from the amino group, converting it into an amide, specifically, 3-methylacetanilide (NHCOCH₃). This transformation decreases the electron-donating capacity of the nitrogen, altering the overall reactivity and properties of the molecule.

Understanding Sulfonation

Sulfonation introduces a sulfonyl chloride group into an aromatic compound. In the case of the acylation product (3-methylacetanilide), chlorosulfonic acid (\(\mathrm{HSO}_{3} \mathrm{Cl}\)) reacts with the arenes.

The orientation of the sulfonyl chloride group is influenced by existing groups on the molecule. Typically, para substitution to the acetamido group is favored due to steric hindrance from the methyl group. The final product, para-acetamido-methylsulfonyl chloride, highlights the impacts of various substituents on a benzene ring that determine both the position and type of substitution.