Textbook: Loudon 5th Ed. (2009)

Chapter 25: The Chemistry of The Aromatic Heterocycles

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Individual Problems

Individual Problems

Problem # 584

Imidazole (shown below) has two nitrogen atoms, N-1 and N-3. Which nitrogen is more basic?

To answer this problem, draw the product after each nitrogen protonates, and compare their stabilities. Explain your reasoning.

Solution

Imidazole is aromatic. When N-3 protonates, the product is still aromatic.

But when N-1 protonates, the product is no longer aromatic (and therefore significantly less stable).

Because of this, N-3 is much more basic than N-1. Another way of thinking of this is that the lone pair on N-1 is involved in the aromatic circuit, and so is not available to pick up a proton.

Problem Details

MendelSet practice problem # 584 submitted by Matt on July 9, 2011.

Subject: Organic Chemistry

Subject Topic(s): Benzene and Aromaticity, Amines (Basicity, Hoffman Elimination, Gabriel Amine Synthesis), Pyridine and Heteroaromatic EAS

Question Type: Concept (explain why this is so..)

Keyword(s): base strength, aromaticity, amine bases

Problem # 591

Pyrrole undergoes eletrophilic aromatic substitution at C-2. Let's compare the resonance forms of EAS carbocation intermediates to see why this is the case. What do you think? Why C-2 and not C-3?

Solution

Electrophilic substitution at C-2 leads to a carbocation intermediate with three resonance forms, while substitution at C-3 leads to a carbocation intermediate with only two resonance forms.

The C-2 intermediate has more resonance forms than the C-3 intermediate, and so is more stable. Therefore, EAS occurs at C-2.

Problem Details

MendelSet practice problem # 591 submitted by Matt on July 9, 2011.

Subject: Organic Chemistry

Subject Topic(s): Pyridine and Heteroaromatic EAS

Question Type: Concept (explain why this is so..)

Keyword(s): heteroaromatic compounds, EAS mechanism

Problem # 593

a) Rationalize the relative stabilities of the cation species below.

b) Pyridine undergoes eletrophilic substitution at C-3. Let's compare the resonance forms of EAS carbocation intermediates to see why this is the case. Consider part a) in your explanation.

Solution

a) The species on the left are much more stable than the species on the right because they have complete octets.

Nitrogen is less electronegative than oxygen, so a positive nitrogen (with a full octect) is more stable than a positive oxygen; oxygen is electronegative and so doesn't "like" being positive (have low electron density).

The carbocation and the "nitrocation" are both very unstable because they only have 6 valence electrons. Because nitrogen is more electronegative than carbon, the nitrocation is even less stable than the carbocation, because its worse at handling a positive charge.

b) The cation intermediates that result from substitution at C-3 only contain carbocations, while the intermdiates from C-2 or C-4 substitution contain a "nitrocation," which is less stable than a regular carbocation, and so are not favored. So pyridine undergoes EAS reactions at C-3.

One more note: For substitution at C-2 and C-4, it's tempting to draw a resonance form where the positive charge is on the nitrogen, which is flanked by two double bonds. But this cyclic compound is too strained to exist, and so doesn't contribute to this analysis.