July 7, 1998
1. Draw the following compounds (where relevant, assume
all double bonds are trans), then indicate if they are
conjugated, cumulated, or isolated polyenes. Finally, predict
how their UV/Vis absorptions should order in terms of the energy
or wavelength of the absorption. (16 points)
a cumulated diene
absorption at second highest energy=second shortest wavelength,
second highest frequency
a conjugated diene
absorption at second lowest energy = second longest wavelength,
second lowest frequency
an isolated diene
absorption at highest energy = shortest wavelength, highest frequency
a conjugated triene
absorption at lowest energy = longest wavelength, lowest frequency
2. The attached spectral data corresponds to one of the
structures shown below. Select the appropriate structure and
defend your choice. (12 points)
Answer: Structure (c). It is not (b) or (e) since these are
tertiary amides with no N-H bonds, but there is an N-H stretch
in the IR spectrum. Compound (a) can be eliminated by the NMR
integration pattern: the NMR has ratios of 3:3:2:1 whereas compound
(a) requires 3:2:2:2. Compound (d) can be eliminated based on
the splitting pattern: the isopropyl group of (b) requires a septet
and doublet, which is not observed. Compound (c) should have
the following spectrum: a triplet at about 1.0 ppm with an integration
of 3 for the methyl of the ethyl group; a quartet at about 2.8
ppm with an inegration of 2 for the methylene of the ethyl group;
a singlet at variable ppm with an integration of 1 for the NH;
and a singlet at about 2.0 ppm with an integration of 3 for the
methyl attached to the carbonyl. The IR should show the NH stretch
(about 3300 cm-1) and the carbonyl (about 1600 cm-1).
3. When 3-methyl-1-butene reacts with water under acidic
conditions (acid catalyzed hydration), several isomeric products
are formed. The spectral data for one of those products are provided
on the following page. Identify this compound, and support your
identification. (12 points)
Answer: The product is 2-methyl-2-butanol. This occurs by rearrangement
of the carbocation to a tertiary carbocation prior to reaction
with water. The alcohol is demonstrated in the IR by the presence
of the OH stretch at about 3500 cm-1. The alkane nature
of the compound is seen in the CH stretches: all below 3000 cm-1.
The NMR is consistent: a singlet at about 1.0 ppm with an integration
of 6 for the two methyl groups, a triplet at 0.9 integrating to
3H and a quartet at 1.4 integrating to 2H for the ethyl group,
and the singlet with integration of 1 for the hydroxyl.
4. Predict spectral data as indicated for the following
compounds. (16 points)
a) NMR data:approximate d values,
multiplicity (i.e., singlet, doublet, etc.), and integration.
Answer: Propyloxy group (O-CH2-CH2-CH3): The terminal methyl
will lie at about 0.9 ppm and be a triplet with an integration
of 3H; the central methylene will lie at about 1.2 or 1.3 ppm
and will be a quartet of triplets (or triplet of quartets) with
an integration of 2H; and the methylene bound to oxygen will lie
at about 3.5 ppm and will be a triplet with an integration of
2H. Isopropyl group: The two methyl groups are equivalent and
will lie at about 0.9 ppm and will be a doublet with an integration
of 6H; the methine will lie at about 2.4 ppm and will be a septuplet
with an integration of 1H.
b) IR data: approximate diagnostic vibrational frequencies and
Answer: This compound will have C-H stretches corrsponding to
both sp2 (above 3000 cm-1) and sp3 (below 3000 cm-1)
carbons (weak to moderate intensities); an O-H stretch at about
3500 cm-1 (strong and broad); a carbonyl C=O stretch
at about 1700 cm-1 (strong); a C=C stretch at about
1600 cm-1 (moderate); and a C-Cl stretch at about 800
cm-1 (moderate). There will also be a C-O stretch
at about 1150 cm-1 and CH wag at about 1450 cm-1.
5. For the allyl radical, shown below: (16 points)
a) Give one property that is not explained by the simple Lewis structure given below.
b) Draw all resonance structures, and show how they relate to the property in (a).
c) Sketch a molecular orbital diagram (this should include the
energy of the orbitals and their phase relationships), and show
how this relates to the property in (a).
a) Two properties could answer this question. One is the unusual
stability of the allyl radical compared to other primary structures;
the other is the equal reactivity of both ends of an allyl radical.
b) Two resonance structures should have been drawn, one with
the radical on the right, one with the radical on the left. The
presence of two equivalent structures means a large resonance
stabilization in resonance theory. Since the radical center is
half on the left and half on the right, this explains the equal
reactivity of the two ends of the radical.
c) There are three pi orbitals: a low energy bonding orbital,
a non-bonding orbital at the same energy as a p-orbital not interacting
with other orbitals, and a high energy antibonding orbital. The
bonding orbital has zero nodes, the non-bonding orbital has one
node at the central carbon, the antibonding orbital has two nodes,
between each pair of carbons (1 and 2; 2 and 3). The bonding
orbital is lower in energy than a simple ethylene bonding pi orbital:
this explains the lower energy (extra stability) of the radical.
There are two electrons in the bonding orbital. The non-bonding
orbital has electron density at carbons 1 and 3 and is singly
occupied. This explains the presence of radical density at carbons
1 and 3.
6. The 2,4-pentadienyl cation is shown below. (18 points)
a) Draw the three resonance structures that make the largest contributions to the resonance hybrid.
b) Reaction of this cation with nucleophilic bromide ion gives two products. Draw the structures of these two products.
c) Predict which product should be the kinetic product (forms
fastest) and which should be the equilibrium product (most stable,
i.e. lowest energy). (They could be the same!)
a) The resonance structures have the positive charge on the left (C1), in the center (C3), and on the right (C5, shown in the test).
b) The two products are 5-bromo-1,3-pentadiene (from reaction at either end!!) and 3-bromo-1,4-pentadiene from reaction in the middle.
c) The kinetic product comes from reaction at the center with
the most positive charge: this would be the secondary carbon C3
and would give 3-bromo-1,4-pentadiene. The equilibrium product
is the most stable, and this is the conjugated diene (not the
isolated diene!), that is, 5-bromo-1,3-pentadiene.
7. For each of the following types of spectroscopies:
a) Identify the type of radiation involved and the approximate wavelengths or frequencies.
b) briefly explain how this radiation interacts with matter.
a) Ultraviolet/Visible radiation, 200-800 nm in wavelength.
b) Promotes electrons from filled bonding orbitals to empty antibonding
a) Infrared radiation, 4000-400 cm-1 in frequency; 2.5-25 micrometers in wavelength.
b) Promotes molecules to higher vibrational states.
Nuclear Magnetic Resonance Spectroscopy
a) Radio waves, 60-600 Mhz in frequency
b) Promotes nuclei to higher magnetic spin states.