ORGANIC CHEMISTRY 2

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ORGANIC CHEMISTRY 2

• SPRING 2005
• PROBLEM SET 3 - SOLUTIONS

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• Give a mechanism for each of the following
  transformations.

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• Predict the predominant site of electrophilic
  substitution on each of the following molecules.
  State whether each is activated or deactivated
  relative to benzene.

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• Arrange in order of decreasing reactivity towards
  electrophiles
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• PhCl PhMe PhOMe PhNMe2 PhNMe3 PhNO2
  PhCOOEt PhOCOMe
• Resonance donors will be more activating than
  inductive donors and we have three here

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• PhOMe PhNMe2 PhOCOMe
• Because it is less electronegative than oxygen,
  nitrogen is more willing to share/donate its lone
  pair of electrons in the positively charged
  intermediate.
• So PhNMe2 is the most reactive

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• PhOMe and PhOCOMe are both sharing lone pairs
  at oxygen
• In PhOCOMe, the lone pair is less available, as
  it is also shared with the carbonyl group.
• So we have
• PhNMe2 gt PhOMe gt PhOCOMe

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• Next we look for inductive donors there is just
  one in this collection, the methyl group of PhMe.
• So, we have
• PhNMe2 gt PhOMe gt PhOCOMe gt PhMe

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• Next well look at the other end of the scale
  the worst withdrawers.
• We know that -NO2 is very electron withdrawing
• But here we have something even worse the
  NMe3 group which bears a full positive charge

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• So now we have
• -NO2 gt -NMe3
• This leave us with PhCl and PhCOOEt in the
  middle.
• Cl is an inductive withdrawer (but a resonance
  donor)
• COOEt is a resonance withdrawer

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• As usual, resonance beats inductive so we can now
  put the whole set in order
• PhNMe2 gt PhOMe gt PhOCOMe gt PhMe gt PhCl gt PhCOOEt
  gt PhNO2 gt PhNMe3

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• Predict the site of mononitration of each of the
  following molecules.
• We need to choose the more activated ring, and to
  use the directing effect of the substituents on it

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• Predict the product of each of the following
  reactions

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• Aniline, PhNH2, reacts with Br2 to give o- and
  p-bromoaniline. However, with HNO3/H2SO4 it gives
  m-nitroaniline, in a slow reaction. Explain.
• The NH2 group in aniline is o,p-directing and
  activating.
• However, in strong acid it is converted to -NH3,
  which is m-directing and strongly deactivating.

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Electrophilic substitution of naphthalene
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• Hence electrophilic addition takes place at the
  1-position, when under conditions of kinetic
  control.
• Naphthalene is less aromatic than benzene
• Therefore substitution is faster - less resonance
  energy is lost in forming the intermediate.