Organic Chemistry 2

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Organic Chemistry 2

• Spring 2005
• Problem Set 5 - Solutions

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• 1
• Explain why the rate of formation of acetone
  oxime is a maximum at a pH of about 5, and is
  much lower at very high or very low pH. You will
  need to draw a detailed mechanism for the
  reaction.

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• Under neutral or basic conditions the rate
  controlling step is loss of water which depends
  on protonation.
• Therefore the rate increases with acidity.
• However, when conditions are much more acid, the
  hydroxylamine is protonated to give HONH3, which
  cannot attack the ketone in the first step.
• Thus, as the concentration of the free
  hydroxylamine falls, the first step becomes rate
  determining, and the rate is reduced as the
  acidity increases.
• The best rate is obtained around pH5.

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• Predict the product(s) for each of the following
  reactions, justifying your conclusion by drawing
  a mechanism.

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• How would you accomplish each of the following
  conversions?
• RCl ? RD
• (Use D2O as your source of deuterium)

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• PhCN ? PhCOCH3
• 1) MeMgI 2) H3O

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• Show how you would use Grignard reagents in the
  synthesis of each of the following alcohols.

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• Give a mechanism for the following reaction.

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• Compound C, C6H12O2, reacts with dilute H2SO4 to
  give D, C4H8O, which is readily distilled, and E,
  C2H6O2 which has a high boiling point. D reacts
  with MeMgI, followed by aqueous work-up to give
  F, C5H12O.
• The best strategy is to start with the simplest,
  smallest molecules these will be the easiest to
  identify

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• E IR spectrum 3360, 2945, 1085 cm-1
• 1H nmr spectrum
• ? 3.63 s 4H
• 4.36 s 2H, exchanges with D2O
• The formula, C2H6O2 implies no rings or double
  bonds
• The IR band at 3360 cm-1 implies that an OH group
  is present

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• The NMR signal at d 4.36 is for two protons
  which implies two OH groups
• The exchange with D2O tells us these protons
  should be assigned to OH groups.
• This leaves C2H4 to be accounted for, and as
  these four protons are equivalent, we can now
  assign the structure as HOCH2CH2OH for E

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• D IR spectrum 2985, 1720 cm-1
• 1H nmr spectrum ? 0.99 t 3H
• 2.07 s 3H
• 2.42 q 2H
• The band in the IR spectrum at 1720 cm-1
  indicates a carbonyl compound, probably a ketone.
• This accounts for the one dbe from the formula.

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• In the NMR spectrum the 3H triplet and the 2H
  quartet are clearly coupled, implying that we
  have a CH2CH3 grouping.
• This leaves us with just a CH3 group this gives
  rise to a singlet at d 2.07, typical of a CH3
  group next to a carbonyl.

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• Now we can assign the structure of D it has to
  be 2-butanone, CH3COCH2CH3.
• Lets now return to the original reaction to
  produced D and E from C

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• If we were to look at the reaction the other way
  round, and ask how 1,2-ethane diol and 2-butanone
  would react then the answer is straightforward.
• A cyclic diol would be formed, of structure

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• This has the correct formula, C6H12O2 for C. Do
  the spectra fit?
• C IR spectrum 2985, 1070 cm-1
• Mass spectrum m/z 116, 101, 87
• 1H nmr spectrum ? 0.88 t 3H
• 1.22 s 3H
• 1.61 q 2H
• 3.88 s 4H

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• IR spectrum shows neither carbonyl nor OH
  absorptions correct
• In the mass spectrum, the losses of 15 and 29
  mass units look reasonable for cleavage of methyl
  and ethyl groups
• In the NMR spectrum the 2H quartet and the 3H
  triplet can be assigned to the CH2 and CH3 of the
  ethyl group.

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• The CH3 singlet can be assigned to the methyl
  group.
• The 4H singlet at d 3.88 can be assigned to the
  OCH2 of the acetal ring this is a typical
  chemical shift for protons on a carbon bonded to
  oxygen.
• Since the reaction of ketones with Grignard
  reagents is well known, we can now complete the
  entire sequence.

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