Organic Chemistry 1

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Organic Chemistry 1

• Autumn 2004
• Review Problem Set Solutions

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• Give structures for
• 2-Methyl-N,N-diethylbutanamide
• Start by drawing butanamide

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• Next put in the 2-methyl group

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• Finally we put in the two ethyl groups on
  nitrogen

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• 4-(2-bromopropyl)3-oxooctanal
• First draw the structure of octanal

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• Next, put in the oxo group, i.e. a ketone, at
  carbon-3

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• Finally, at carbon-4 we need a propyl
  substituent, with a bromine attached to its
  second carbon

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• 2,4-dimethyl-1,3-heptadiene
• First we write down 1,3-heptadiene

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• Then add in the methyl groups at the 2- and
  4-positions

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• Give a systematic name for each of the following
  molecules

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• The functional group that takes priority here is
  an ester so we will number the longest chain
  starting with that carbonyl group as carbon-1

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• There are methyl substituents at carbons 2 and 3,
  and a ketone at carbon-5.
• We also have to remember the alcohol part of the
  molecule - its an ethyl ester
• So our name is
• ethyl 2,3-dimethyl-5-oxohexanoate

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• Which of the following pairs of compounds would
  you expect to me more soluble in water? Why?
• Bu4NBr (C16H33)4NBr
• Bu4NBr.
• The long hydrocarbon chain of the other salt is
  hydrophobic

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• CH3(CH2)3NH2 CH3(CH2)3NMe2
• CH3(CH2)3NH2 - hydrogen bonding

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• Explain the following observations
• Hexanoic acid is soluble in an aqueous solution
  of NaHCO3, but hexanol is not.
• pKa of hexanoic acid is approx. 5.
• pKa of hexanol is approx. 15.
• Therefore NaHCO3 converts hexanoic acid to its
  sodium salt, which is soluble in water.
• However, it cant convert the alcohol to an
  anion, and the length of the carbon chain means
  that the alcohol is too hydrophobic to be water
  soluble.

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• In the reaction
• at 25 0C in 0.1 M H2SO4, the compounds A, B and C
  gave respectively 10-5 , 10-15  and 50  of
  Ar3C at equilibrium.

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• This is about carbocation stability.
• The data tell you that (4-MeOC6H4)3C is more
  stable than Ph3C is more stable than
  (4-O2NC6H4)3C.
• This is exactly what you would expect on the
  basis that MeO is a resonance donor of electrons
  and NO2 is a resonance acceptor of electrons

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• Attack of PhS- on D is 58 times faster than the
  reaction with E

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• The first thing we need to do is to draw out
  chair forms of each of the compounds, bearing in
  mind that the tert-butyl group is very large, and
  has a strong preference for the equatorial
  position.

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• The t-Bu group will always be equatorial because
  of its size, and the conformations of D and E
  will be locked.
• In D the nucleophile (which is quite bulky) must
  approach from the equatorial direction.
• In E it must approach approximately axially.
  Since axial approach is more sterically hindered,
  the reaction is slower.

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• 50  of each implies a symmetric intermediate.
• Therefore

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• This first substitution does indeed go with
  inversion

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• But if we now redraw the molecule, we can see
  that there is a possibility for neighboring group
  participation

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• 7-Bromo-1,3,5-cycloheptatriene is water soluble

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• The first step is to protonate the tosylate to
  make it into a leaving group

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• In fact the truth is a little more complicated -
  this really goes by a pseudo-SN1 mechanism to
  give a cation with a non-classical delocalized
  structure.
• This can then be attacked at any of the three
  carbons involved in the delocalization to give
  the observed products.

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• Compound F exchanges H for deuterium in slightly
  basic D2O, but compound G does not.

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• The bridges mean that these molecules are not
  quite planar, but there is still reasonable
  overlap in the ?-systems

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• What stereoisomers would you expect for the
  following compounds?
• Label double bonds E or Z and chiral centers R or
  S.
• Distinguish enantiomers and diastereoisomers.
• Which isomers would you expect to be optically
  active?

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• Draw each of the chair forms of the following
  molecules.
• Predict which will be more stable, in each case
  and give your reasons.

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• Study the molecule below
• Order the substituents according the
  Cahn-Ingold-Prelog system
• State the absolute configuration
• Give a systematic name for the molecule

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• OCH3gtCH2CH2CH2COOHgtCH2CH3gtH

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• In this orientation we are not looking at the
  molecule from the right direction so we need to
  do a switch so that the hydrogen points away from
  us

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We now need to name the molecule

• S-5-methoxyheptanoic acid

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• For each of the following pairs of molecules,
  state which you expect to be the more acidic, and
  give your reasons.

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• CHCl2COOH
• or
• CCl3COOH
• CCl3COOH - inductive effect of Cl

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• CH3COOH CF3COOH
• CF3COOH - inductive effect of F.

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• Circle the most acidic hydrogen in each of the
  following molecules

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• Write down a tautomer of each of the following
  molecules
• Remember the strategy
• Remove a proton to give an anion
• Write down a resonance from of the anion
• (If you cant do that go back to the start and
  choose another proton)
• Put the proton back on to the other resonance
  form

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• Similar strategies can also lead to

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• We use identical methodology for

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• Give resonance forms of the following molecules

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• How many p-electrons in each of the following?
  State whether each is aromatic, antiaromatic or
  non-aromatic.

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• Place the following compounds in order with
  respect to their reactivity in a typical SN2
  reaction, giving brief reasons.

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• Place the following compounds in order with
  respect to their reactivity in a typical SN1
  reaction, giving brief reasons.

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• Predict the product(s) of the following
  reactions, paying attention to stereochemistry
  where appropriate. Full credit may be obtained
  for the correct product, but partial credit may
  be obtained for a mechanism heading in the right
  direction if the product is wrong.

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SOCl2 - thionyl chloride

• In ether HCl is not dissociated, so there is no
  free Cl-

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• So
• This process gives an unusual retention of
  stereochemistry - the Cl- goes straight back in,
  before the cation has an opportunity to rotate.
  the mechanism is sometimes called SNi

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• Study the reactions shown below.
• Which would you expect to be the faster? Give
  your reasons and explain the observed
  stereochemistry.

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• The trans-isomer is correctly set up for an
  intramolecular SN2 process

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• In the cis-isomer, the sulfur is not correctly
  oriented to be able to do an intramolecular SN2
• So we get a slower, regular SN2 process with a
  single inversion

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• Rank the following three compounds in order of
  increasing ?max in their UV spectrum, briefly
  noting your reasons

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• For each of the following sets of spectroscopic
  data, identify the compound, and interpret as
  much of the data as you can.

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• C11H14O3
• Mass spectrum m/z 194 (M), 134, 121, 76
• IR spectrum 1730 cm-1
• 1H NMR spectrum
• ? 2.5 (t, J  8 Hz, 2H)
• 2.9 (t, J  8 Hz, 2H)
• 3.7 (s, 3H)
• 3.8 (s, 3H)
• 6.8 (d, J  7 Hz, 2H)
• 7.2 (d, J  7 Hz, 2H)

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• Dbe (2 x 11) 2 14/2 5
• The large number of dbe, together with the
  protons with d in the region of 7 strongly
  suggests an aromatic compound.
• The IR absorption at 1730 cm-1 implies a carbonyl
  group this will be our 5th dbe.

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• ? 2.5 (t, J  8 Hz, 2H)
• 2.9 (t, J  8 Hz, 2H)
• Taking these two together
• It is clear they are coupled to each other, and
  that neither has any other coupling.
• Each is a triplet because it is next to the other
  CH2
• So we have CH2CH2-

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• d 3.7 (s, 3H)
• 3.8 (s, 3H)
• The chemical shifts are typical of OCH3 groups
• So we have two non-equivalent OCH3 groups, and
  together with the carbonyl, this accounts for all
  the oxygen atoms

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• d 6.8 (d, J  7 Hz, 2H)
• 7.2 (d, J  7 Hz, 2H)
• From the chemical shifts, these most be four
  protons on an aromatic ring.
• So the aromatic ring is disubstituted
• We need to think about the pattern we would
  expect for the various orientations of the two
  substituents

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• This is the situation we have here two doublets
• So we have a 1,4-disubstituted benzene ring
• We also have CH2CH2-, a carbonyl group and two
  OCH3 groups
• This accounts for all the atoms of the molecular
  formula

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• 1730 cm-1 is a typical absorption for an ester
  that is not conjugated with an aromatic ring.
• This suggests that we have ArCH2CH2COOMe
• So we can now deduce the structure of the whole
  molecule

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• C6H10O2
• Mass spectrum m/z 114 (M), 99, 86, 69, 41
• IR spectrum 1720, 1660 cm-1
• 1H NMR spectrum
• ? 1.35 (t, J  7 Hz, 3H)
• 1.90 (d, J  6 Hz, 3H)
• 4.15 (q, J  7 Hz, 2H)
• 5.80 (d, J  16 Hz, 1H)
• 6.90 (m, 1H)

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• dbe (2 x 6) 2 10/2 2
• In the IR spectrum the absorption at 1720 cm-1
  must be a carbonyl group.
• The absorption at 1660 cm-1 might be another
  carbonyl if it was very conjugated, but given
  that the NMR spectrum has some protons that are
  in the right region for an alkene, CC is more
  likely

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• So we have found both double bonds
• In the NMR spectrum the signal at d 1.35, is a 3H
  triplet, so the CH3 is next to a CH2
• If we compare coupling constants, we can see that
  the CH3 is next to the CH2 that gives the signal
  at d 4.15. This signal is a quartet, confirming
  it is next to a CH3.
• Its chemical shift strongly suggests that it is
  attached to oxygen.

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• So we have OCH2CH3
• The signal at d 1.9 is for another methyl group,
  this time coupled to a CH (since it is a doublet)
• The signals at d 5.8 and 6.9 must come from
  protons attached to a carbon-carbon double bond.
• The signal at d 5.8 has one coupling of 16Hz

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• This tells us that it is coupled only to the
  other proton on the double bond, and that it is
  trans to it (trans couplings are in the range
  12-18 Hz, compared with 7-10 Hz for
  cis-couplings)
• So we have

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• We now have all the pieces of the molecule, and
  all we have to do is put them together

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• Given that our carbonyl stretch is 1720 cm-1, we
  can deduce that we have an unsaturated ester, so
  overall the molecule must be

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• C9H10O2
• Mass spectrum m/z 150 (M), 135, 107, 77
• IR spectrum 1670 cm-1
• 1H NMR spectrum
• ? 2.6 (s, 3H)
• 3.9 (s, 3H)
• 7.0 (d, J  7 Hz, 2H)
• 7.9 (d, J  7 Hz, 2H)

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• dbe (2 x 9) 2 10/5
• With the signals in the NMR spectrum at ? 7.0 and
  7.9, this strongly suggests an aromatic ring
• In the IR spectrum, 1670 cm-1 suggests a
  conjugated carbonyl typical for an aromatic
  ketone

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• The singlet at ? 2.6 in the NMR spectrum is a CH3
  group the chemical shift would be reasonable
  either for a methyl group attached to an aromatic
  ring or to a carbonyl
• The singlet at ? 3.9 in the NMR spectrum is a CH3
  group the chemical shift would be reasonable
  for OCH3

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• The two doublets in the NMR spectrum at d 7.0 and
  7.9 are again typical of a 1,4-disubstituted
  aromatic ring.
• So we have found all the bits of our molecule,
  and there are just two possible ways of putting
  them together

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• We choose between these on the basis of the IR
  spectrum. 1670 cm-1 is reasonable for an aromatic
  ketone, but we would expect an aromatic ester to
  be nearer to 1710-1720 cm-1. Hence we have

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• C11H20O
• Mass spectrum m/z 168 (M), 126, 111, 83, 57
• IR spectrum 3020, 2990, 1680, 1620 cm-1
• 1H NMR spectrum
• ? 0.9 (d, J  8 Hz, 6H)
• 1.1 (s, 9H)
• 1.5 (nonet, J  8 Hz, 1H)
• 2.3 (d, J  8 Hz, 2H)
• 6.1 (d, J  18 Hz, 1H)
• 6.9 (d, J  18 Hz, 1H)

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• dbe (2 x 11) 2 20/2 2
• In the IR spectrum there is an absorption at 3020
  cm-1, which indicates sp2 CH
• Absorptions at 1680 and 1620 cm-1 suggest
  respectively CO and CC, almost certainly
  conjugated. Since there is only one oxygen, and
  no sign of an aldehyde proton in the NMR
  spectrum, this is probably an a,b-unsaturated
  ketone

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• ? 0.9 (d, J  8 Hz, 6H)
• This suggests two CH3 groups attached to one
  carbon, that bears a single hydrogen

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• 1.1 (s, 9H)
• This indicates a tert butyl group

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• d 6.1 (d, J  18 Hz, 1H)
• 6.9 (d, J  18 Hz, 1H)
• The chemical shift suggests that these
• protons are attached to our double bond
• The coupling constant of 18 Hz means
• that they must be trans.

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• The fact that neither of these protons shows any
  other coupling means that the carbons attached to
  the double bond dont have any attached protons.
• We already know one is a carbonyl, since we know
  the molecule is conjugated.
• The other must be the tert-butyl group
• So we have

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• This part of the structure accounts for C7H11O of
  our structure, leaving us with 4 carbons and nine
  hydrogens to find.
• The NMR signal at d 2.3 (d, J  8 Hz, 2H) implies
  a CH2 group, and we already know that we have a
  CH(CH3)2 unit.
• The CH proton is a nonet because it is next to
  two methyl groups and one CH2

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• So we must have