|
Chemistry 251/253 |
Homework 8 Answer Key |
1.
CC(C)(C)c1ccc(Cl)cc1
2. 
COc1ccc(cc1[N+](=O)[O-])S(=O)(=O)O or COc1c(cccc1S(=O)(=O)O)[N+](=O)[O-]
3.
CN(C)c1ccc(Br)c(C#N)c1 The dimethylamino group activates the positions ortho
and para to it. The para position is preferred since it is less
hindered sterically; the cyano group is smaller than the
dimethylamino group.
4.
[O-][N+](=O)c1ccc(c2ccccc2)c(c1)[N+](=O)[O-] This is a tricky question. It deals with the effects of substitutent groups on both the rate and the orientation in electrophilic aromatic substitution reactions. First let's deal with the rate. The right hand ring contains two substituents, a strongly deactivating nitro group and a weakly activating phenyl ring. It should react more slowly than benzene. The left hand ring contains one substituent, the nitro phenyl ring. This is a deactivting group, and this ring should also react more slowly than benzene. Since there's no direct way to compare these effects, either of the two structures shown above is acceptable. In the first structure the orientation of the incoming nitro group is ortho to the phenyl ring and meta to the nitro group as expected. In the second structure the nitro group is meta to the deactivating, meta-directing, nitrophenyl substituent.
5.
CC(=CC(=O)c1ccccc1)c2ccccc2 The first reaction is a Friedel-Crafts acylation. The
product is acetophenone. The second reaction is an aldol condensation
of two molecules of acetophenone. The initially formed aldol
spontaneously dehydrates under the reaction conditions.
6.
O=C(c1cc2ccccc2o1)c3ccccc3 If you compare the structure of the intermediate
carbocation that is produced by reaction at C-2 with the one formed
by reaction at C-3, you are forced to conclude that reaction at C-3
is preferred because there is one resonance contributor in which
every atom has an octet of electrons for the latter case. In other
words, you should expect preferential substitution at C-3. However, I
said in class that reaction occurred preferentially at C-2, and in
fact, that is correct, but it appears to be an anomoly, i.e. it is an
example where resonance theory fails to predict the actual outcome of
a reaction.
7.
CC(=O)CCc1c[nH]c2ccccc12 or CC(=O)CCc1cc2ccccc2[nH]1
If you draw resonance structures for the intermediate ions, it
becomes apparent that electrophilic attack at the 3 position is
preferred. This should also be the case in Question 6, but it is not.
8.
OC(=O)C1C=Nc2ccccc12 The methyl lithium deprotonates
the amine. The resonance-stabilized conjugate base then acts as a
nucleophile towards the carbon dioxide. Note that the NH group is
part of an enamine, which means that the beta carbon is electron
rich. This is where the electrophile reacts.
9.
Cc1ccc(cn1)S(=O)(=O)O The methyl group directs the electrophile to the para
position.
The following reaction sequence relates to questions 10 and 11: A = ClC(=O)CNc1ccccc1; B = O=C1CNc2ccccc12
The signal at 1715 cm-1 indicates a ketone rather than an amide. The NMR show 7 signals, each integrating to 1H. The doublets at 3.6 and 4.6 ppm correspond to the two protons of the methylene group.
12. Select the compound that is most reactive in electrophilic aromatic substitution reactions. A The two 5-membered rings are both electron rich, while the 6-membered rings are electron deficient. The electron/proton ratio in imidazole A is higher than it is in oxazole B.
13. and 14. C= Nc1ccccc1O: D = O=c1[nH]c2ccccc2o1 or Oc1nc2ccccc2o1.
Nitration
occurs preferentially at the position ortho to the OH group because
the nitro group can form an intramolecular H-bond to the H atom of
the OH group. Reduction of the nitro group then yields the
aminophenol C which reacts with the difunctional acid chloride to
form a ring. I accepted either tautomer
of compound D. Note the similarity between the structure of the
"enol" tautomer and that of a carboxylic acid; the OH proton gives a
signal at 8.7 ppm.
15.
CCCNC(=O)c1ccccc1 This
is a simple example of a nucleophilic acyl substitution
reaction.
16-18. E = CC=CCc1cccnc1NC(=O)C, F = CC(=O)Nc1ncccc1CC=O, G = CC(=O)Nc1ncccc1CC=O.
19. Redraw the structure of 
so that its
aromatic character is apparent. Oc1cc(O)ncn1
20.
[I-].C[N+]1(C)CCCC1 or CC1CCCN1C
Apparently I misstated the logic involved in controlling methylation
of amines. If you use an excess of methylating agent, as stated in
this problem, then you will completely methylate the amine and form
an ammonium ion. If you want to limit the methylation to
monomethylation, then you use a large excess of the starting
material. Under these conditions, i.e. when the methyl iodide is the
limiting reagent, you will have a lot of starting material left over,
but that proportion of it that has reacted will be methylated just
once. Because it was my error, I have given credit for both
answers.
21. Select the compound that would be most basic: I intended to say least basic. In that case the answer is clearly compound A. Since it is more difficult to compare compounds B, C, and D, I accepted any of those three choices.
22.
CC=Cc1ccccc1
23.
CC1C(O)Cc2ccccc12 This is the product of
anti-Markownikov addition.
24.and 25. H = CC1CC2CCCCC2(CC31OCCO3)C#N, I
= CC1CC2CCCCC2(CN)CC1=O 
26. Draw the structure of the major product of the
following reaction:
CCC=C(C)C The starting
material is a tertiary alkyl bromide. It undergoes dehydrobromination
upon exposure to ethylamine. The more highly substituted alkene is
the preferred product.
27. Calculate the electron/proton ratio in purine. Then convert that to its equivalent ratio of electrons per nucleus for an "imaginary" benzene molecule. Express your answer to two decimal places. (10 e/58 p) x 6 = 1.03.
28. Specify the R and S designations of C-2 and C-3, respectively
in
.C2 = R, C-3
= S
29. Consider the following possibilities:
It is possible to predict the experimental outcome by comparing
the resonance contributors involved in both reactions. Draw the
structure of a resonance contributor that convinces me you will make
the correct prediction.
ClC1[N+]C=CC=C1 The nitrogen has a positive charge,
but it does not have a filled shell in the structure shown in the
box.
30.
CC1=NCNC(=O)C1 This
reaction involves both nucleophilic addition of an amine to a ketone
and nucleophilic acyl substitution of an amine on an ester.