Nucleophilic Aromatic Substitution, Diazonium Salts and Heterocycles · seed 1 · A4, ink-friendly. The answer key prints on its own page for grown-ups.

When rings welcome attackers and swap groups

Science · Chemistry · ages 20-22
Name ______________________   Date ____________

  1. What is the Meisenheimer complex?

    • The strained triple bond formed by pulling HX off the ring
    • A diazonium salt made at ice temperature
    • The settled anion formed when the nucleophile adds before the leaving group departs

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

    • Strong electron 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

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

    Circle one:   True   False

  4. 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
    • The label proves nothing about any intermediate

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

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

  6. 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

  7. 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

  8. Plain chlorobenzene meets an amine but nothing happens, while the 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
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Answer key

For grown-ups. Fold this page away before handing over the rest.

When rings welcome attackers and swap groups W1-mt_I4EtpYcrEq-s1

  1. The settled anion formed when the nucleophile adds before the leaving group departs · It is the anion sitting between addition of the nucleophile and departure of the leaving group.
  2. Strong electron withdrawing groups ortho or para to the leaving group · Those groups steady the Meisenheimer anion, which is what lets the attack happen.
  3. True · That cold reaction is the standard entry into diazonium chemistry.
  4. The reaction ran through the symmetric benzyne intermediate · Only the symmetric triple bond can share the label evenly between two spots.
  5. It pulls HX off with very strong base to give a strained triple bond the nucleophile can join at either end · No withdrawing groups are needed, and the symmetric triple bond explains attack at either end.
  6. At C2, because that ion spreads the charge over more atoms including nitrogen · The C2 intermediate shares the charge most widely, which makes that route the steadiest.
  7. At C2 or C4, where the negative charge can rest on nitrogen · Drawing each anion shows the charge landing on nitrogen only for the C2 and C4 routes.
  8. The plain ring lacks withdrawing groups to steady the Meisenheimer anion · Without withdrawing groups beside the leaving group, the anion has nowhere to rest its charge.
Worksheet · LightMySky