CHEM 2312



Answer Key

July 7, 1998

Dr. Hanson

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) 2,3-pentadiene

a cumulated diene

absorption at second highest energy=second shortest wavelength, second highest frequency

b) 1,3-pentadiene

a conjugated diene

absorption at second lowest energy = second longest wavelength, second lowest frequency

c) 1,5-hexadiene

an isolated diene

absorption at highest energy = shortest wavelength, highest frequency

d) 2,4,6-octatriene

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 approximate intensities

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.

Ultraviolet/Visible spectroscopy


a) Ultraviolet/Visible radiation, 200-800 nm in wavelength.

b) Promotes electrons from filled bonding orbitals to empty antibonding orbitals.

Infrared Spectroscopy


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.