1. The steroid shown below has two separate absorptions
in the UV. One has a lmax
of 235 nm and an e of 19,000 M-1
cm-1. The other has a lmax
of 300 nm and an e of 20 M-1
cm-1. Identify the portions of the molecule responsible
for each of these absorptions (the chromophores). Briefly explain
the difference in e. (10 points)
ANSWER: There are two chromophores in this molecule, the conjugated
diene which is found in the two six membered rings (the A and
B rings) and the carbonyl found on the five membered ring (the
D ring). Conjugated dienes have p
to p* absorptions which are found between
210 and 250 nm and are very intense (e
is large). Carbonyls have n to p*
absorptions which are found around 300 nm and are weak (e
is small). Therefore the conjugated diene is responsible for
the absorption with lmax
of 235 nm and an e of 19,000 M-1
cm-1, while the carbonyl is responsible for the absorption
with lmax of 300 nm and
an e of 20 M-1 cm-1.
2. From the first order spin-spin coupling concepts discussed
in class, predict the patterns of signals expected for each type
of proton in the NMR spectra of the following compounds (i.e,
singlet, doublet, triplet, etc. or doublet of doublets, triplet
of quartets, etc.). (10 points)
a)
ANSWER: Working from left to right, this compound will have: a
doublet from the two methyl groups coupled to the methine (CH);
a septuplet from the methine coupled to the two equivalent methyl
groups; a quartet from the methylene (CH2) coupled
to the single methyl at the far right; and a triplet from the
methyl group coupled to the methylene.
b)
ANSWER: All protons are equivalent and this compound will give
a singlet.
c)
ANSWER: From left to right: A doublet from the proton of the
CHCl2 group; a doublet of triplets (or triplet of doublets)
from the CHCl group coupled to both terminal groups; a doublet
from the CH2 group; and a singlet from the OH which
is decoupled by chemical exchange.
3. You have a solution of the compounds shown below in
dichloromethane. Each is soluble in aqueous solution when deprotonated
to the corresponding anion. You have the following reagents:
0.1 M aqueous NaOH (pH = 13), saturated aqueous CaCO3
(pH = 9.4), 0.1 M aqueous NaHCO3 (pH = 8.4), water
(pH = 7), and 0.1 M aqueous HCl (pH = 1). Demonstrate how you
might effect a separation of these compounds. (10 points)
ANSWER: Each compound will be soluble when deprotonated and will
deprotonate in any solution where pH is higher than the pKa.
So begin by removing the most acidic compound, benzoic acid pKa
4.2, by extraction with aqueous sodium bicarbonate pH = 8.4.
Then remove the next most acidic compound, p-chlorophenol pKa
9.1, with aqueous calcium carbonate pH = 9.4. Finally remove
the next most acidic compound, p-methoxyphenol pKa 10.2, with
aqueoue sodium hydroxide pH = 13. At this time only the least
acidic compound, cyclopentanol pKa 18, remains in the organic
solution.
4. Beginning with benzene and using any other necessary
organic or inorganic reagents, propose a synthetic route for the
following compounds. (21 points)
a)
ANSWER: Note that the two groups are meta to each other yet both
are o,p directors. Yet both derive from meta directors. The
best synthesis for this compound is as follows: Friedel-Crafts
acylation of benzene to make propanoyl benzene. Nitration of
propanoylbenzene, this occurs at the meta position (3-nitropropanoylbenzene).
Reduction of the carbonyl to make the alkyl group (3-nitropropylbenzene).
Reduction of the nitro group to the amine (3-propylaniline).
Diazotization of the amine followed by metathesis with fluoroboric
acid and thermolysis to give the 3-propylfluorobenzene.
b)
ANSWER: There are several ways to make this compound. A good
synthesis is as follows: Nitration of benzene to nitrobenzene.
Reduction to aniline. Diazotization and hydrolysis to phenol.
Alkylation with iodoethane to give ethoxybenzene. Double nitration
to give 2,4-dinitroethoxybenzene. Controlled reduction to give
2-ethoxy-5-nitroaniline (this would need to be separated from
the 3-nitro-4-ethoxyaniline isomer formed by reduction of the
other nitro group). Diazotization and reaction with potassium
iodide to give 2-iodo-4-nitroethoxybenzene. In an alternative
synthesis, the intermediate compound 2,4-dinitroethoxybenzene
can also be made by nucleophilic aromatic substitution of 2,4-dinitrochlorobenzene
with ethoxide ion.
c)
ANSWER: The key to this synthesis is recognizing that the m-directing
carboxylic acid comes from an o,p-directing alkyl group. Friedel-Crafts
alkylation of benzene gives toluene. Electrophilic bromination
of toluene gives 4-bromotoluene. Oxidation with potasium permanganate
gives 4-bromobenzoic acid. Sulfonation with fuming sulfuric acid
gives 2-bromo-5-carboxybenzenesulfonic acid.
5. Complete the following reactions, indicating reactants,
products, or reagents as required. (20 points)
a)
ANSWER: the products are pentafluoroehtoxybenzene and sodium
bromide.
b)
ANSWER: The products are phenol and bromoethane.
c)
ANSWER: The product is 3-methylbenzonitrile.
d)
ANSWER: The reagents are 1. NaOH 2. CO2.
e)
ANSWER: The product is an azobenzene: C6H5-N=N-C6H3CH3NH2,
where the 3-methylaniline has coupled at the position para to
the amine.
6. Compound X has a MW of 114 and the spectral information
on the following page. Which of the four structures provided
corresponds to Compound X. (10 points)
a) b)
c) d)
ANSWER: All of the compounds are isomers, so the MW information
can not make the distinction. All are esters (carbonyl stretch
at 1730) and all have both alkane and alkene portions so the IR
is also not able to distinguish these compounds. The NMR can,
however. compounds a and b have the ethyl group adjacent to the
carbonyl while compounds c and d have the ethyl group adjacent
to the oxygen. The quartet from the ethyl CH2 is found
at 4.2 ppm, consistent with the ethyl adjacent to the oxygen,
so the compound is either c or d. Compound c has both alkene
protons on one carbon (although they are NOT equivalent: one is
cis to the methyl and one is trans!) while compound d has them
on two different carbons. One would expect compound c to have
similar chemical shifts for these alkene protons while they would
be fairly different for d, since one is adjacent to an electron
donor (CH3) and one is adjacent to an electron withdrawing
group (carbonyl). The signals are at quite different chemical
shifts, with one doublet of quartets centerd at about 5.75 ppm
and the other at about 6.9 ppm. Therefore compound d is the correct
choice.
7. When toluene is dissolved in 2-propanol and treated
with a strong acid, a compound known as cymene is isolated as
a product. Cymene has a MW of 134 and gives the IR and NMR spectra
shown on the following page. Propose a structure for cymene.
(12 points)
ANSWER: Cymene has an isopropyl group and a methyl group on the
benzene ring. (It results from Friedel-Crafts alkylation of toluene
by the cation derived from 2-propanol.) The major isomer formed
is the para isomer, and the spectra here are from that compound,
p-cymene. Note the following: the IR has no bands for a carbonyl
C=O stretch or an ether C-O stretch. There are the expected bands
for the alkyl (~2960) and aromatic (~3050) C-H stretches. There
is also an aromatic C=C stretch at 1620. The NMR has a doublet
at 1.3 ppm and a septuplet at 2.8 ppm arising from the isopropyl
group and a singlet at 2.3 from the methyl group. The para orientation
of the substituents can be inferred from the singlet for the aromatic
protons at 7.1 ppm. For ortho or meta orientation a more complex
set of peaks would be expected.
8. Provide a step-by -step mechanism for the following
transformation. Show all steps and all intermediates. (8 points)
ANSWER: This is a nucleophilic aromatic substitution reaction. The first step is attack of the hydroxide nucleophile at the site of chloro substitution on the benzene ring to generate a delocalized anion. This anion is stabilized by the electron withdrawing nitro groups. The second step is loss of chloride from this delocalized anion to give the phenol and chloride anion. The sodium cation does not participate in the reaction.