What is the Meisenheimer complex?

• The strained triple bond formed by pulling HX off the ring instead
• A diazonium salt made at ice temperature
• The anion between nucleophile addition and leaving group departure

What does a ring need before a nucleophile will attack it by addition elimination?

• Strong withdrawing groups ortho or para to the leaving group
• No substituents at all on the ring
• An electron donating group meta to the leaving group helps most

Primary aromatic amines react with nitrous acid at ice temperature to give diazonium salts.

A carbon 14 label at the leaving carbon ends up split over two spots. What does that prove?

• The reaction ran through the symmetric benzyne intermediate
• The reaction ran through a diazonium salt at ice temperature
• The label proves nothing about any intermediate

How does the benzyne route differ from the addition elimination route?

• It needs withdrawing groups ortho or para to the leaving group first
• It runs through a diazonium salt at ice temperature
• It pulls HX off with very strong base to give a strained triple bond

Where do electrophiles attack pyrrole, and why?

• At C3, because the charge must avoid nitrogen entirely
• At C2, because that ion spreads the charge over more atoms including nitrogen
• At nitrogen itself, because nitrogen is the most exposed atom

Where will a nucleophile attack pyridine, and what is the reason?

• At C3, because that is where electrophiles attack too
• At any carbon with equal ease, since pyridine is symmetric
• At C2 or C4, where the negative charge can rest on nitrogen

Plain chlorobenzene meets an amine but nothing happens, while the ortho- or para-nitro substituted ring reacts smoothly. Why?

• The plain ring cannot form a diazonium salt at any temperature
• The plain ring lacks withdrawing groups to steady the Meisenheimer anion
• Amines only attack rings that carry donating groups