Animated Solution for Chemistry - Organic Compounds Containing Nitrogen: In the following reaction sequence, the major product $B$ is:

Protecting the Amine

When dealing with highly reactive aromatic compounds like anilines, chemists often face a problem: the molecule is too reactive. The amino group ($-N{H}_2$) is a powerful activating group due to the strong $+R$ (resonance) effect of the nitrogen's lone pair. If we were to react $p$-toluidine directly with bromine, we would likely end up with a poly-brominated mess.

To control this beast, we employ a classic synthetic strategy: protection. By reacting $p$-toluidine with acetic anhydride ($A{c}_{2}O$), we convert the amino group into an acetamido group ($-NHCOC{H}_3$). This forms our intermediate product A, $p$-methylacetanilide.

Why does this help? The lone pair on the nitrogen atom is now torn between two lovers. It can either delocalize into the benzene ring or into the adjacent carbonyl group. Because the carbonyl oxygen is highly electronegative, it pulls a significant portion of that electron density away from the ring. This makes the $-NHCOC{H}_3$ group only moderately activating, allowing us to perform a controlled, mono-substitution in the next step.

Resonance vs Hyperconjugation

Now we introduce our electrophile: bromine in acetic acid ($B{r}_2/AcOH$). The stage is set for an electrophilic aromatic substitution. But we have a dilemma. There are two directing groups on the benzene ring, and they are fighting for control over where the incoming bromine will land.

Let's analyze the competitors: 1. The Acetamido Group ($-NHCOC{H}_3$): This is an ortho/para director. Since the para position is already occupied by the methyl group, it directs the bromine to its ortho positions. 2. The Methyl Group ($-C{H}_3$): This is also an ortho/para director. Its para position is blocked by the acetamido group, so it also directs the bromine to its ortho positions.

We have a direct conflict. The bromine can either go ortho to the acetamido group or ortho to the methyl group. Who wins?

Predicting the Major Product

In the world of electrophilic aromatic substitution, the rule is simple: the stronger activating group dictates the position of the incoming electrophile.

The acetamido group activates the ring through the $+R$ resonance effect. Even though it's weakened by the carbonyl group, resonance is still a very powerful force. On the other hand, the methyl group activates the ring through $+H$ (hyperconjugation) and $+I$ (inductive) effects. These are relatively weak forces compared to resonance.

Therefore, the $-NHCOC{H}_3$ group overpowers the $-C{H}_3$ group. The bromine atom will be directed to the position ortho to the acetamido group.

This leads us to our final major product B: 2-bromo-4-methylacetanilide. If our ultimate goal was to synthesize 2-bromo-4-methylaniline, we would simply add one final step: hydrolysis to remove the acetyl protecting group. This elegant dance of protection, directed substitution, and deprotection is a cornerstone of organic synthesis.