A
The reaction is a free radical halogenation. The light causes
homolytic cleavage of the Br-Br bond. One of the resulting Br atoms
abstracts an H atom from the 3o, allylic carbon to produce
a 3o allylic free radical.|
Chemistry 251/253 |
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Homework 5 |
Select the picture that best represents the pi orbital array in the intermediate that is formed in each of the following reactions.
1.A
The reaction is a free radical halogenation. The light causes
homolytic cleavage of the Br-Br bond. One of the resulting Br atoms
abstracts an H atom from the 3o, allylic carbon to produce
a 3o allylic free radical.
2.
B.
This is an example of a Michael addition. The cyanide ion adds to the
b-carbon.
When it does, the hybridization of that carbon changes from
sp2 to sp3. Negative charge develops on the
a-carbon.
Choice D is not correct because there are not enough electrons
shown.
3.
B
4. Select the alkene that will add
HCl slowest:
B
The rate determining step in these reactions is the addition of the
proton to the double bond. The double bond with the lowest pi
electron density will react the slowest. The carbonyl group is an
electron withdrawing group that reduces the pi electron density of
the double bond the most.
5. Draw the structure of the major
product formed in the following reaction:
CCC1(O)CCC=C1
6. Treatment of 1-phenyl-4-methyl-1,3-pentadiene with aqueous acid produces 4 of the 6 alcohols shown in the scheme below. In alphabetical order, enter the letters corresponding to those compounds, e.g. enter ABCD if you think those four alcohols are formed. ADEF or CDEF This question was more difficult than I intended. Nonetheless, some answers are better than others. The products that are formed are determined by the relative stabilities of the intermediate carbocations, which can be assessed by the number of resonance contributors. The relative stabilities of the products is another factor that has to be considered. You can evaluate that by looking at the number of substituents on the double bond. Taking this approach, compound B is easily eliminated since the carbocation leading to its formation has no resonance stabilization. Then the question comes down to a choice between compounds A and C. Here the more stable compound is C. However, the tertiary resonance contributor makes the more important contribution to the hybrid structure, and you could argue that the nucleophile should attack that center preferentially, which would produce compound A.
7. The product distribution in a chemical reaction may be influenced by changing the reaction conditions. Which of the three possible reduction products shown in the reaction below would be the most difficult to form no matter what conditions you chose?B The kinetically preferred product should be compound A, which is formed by attack of hydride ion on the most electron deficient carbon. The thermodynamically preferred product should be C since the double bond is conjugated with the carbonyl group.
8. Draw the structure of the
1,4-addition product of the following reaction sequence:
COC1=CC(C)(C)CCC1
9. Draw the structure of the product
you would isolate from the following reaction sequence:
[2H]C1C(=O)CCCC1(C)C
Neutralization of
the enolate ion during the work-up of the reaction would produce the
enol, but that would tautomerize to yield the ketone, which is the
actual product you would isolate.
10. Al Kane used 
and
acetaldehyde as starting materials for the synthesis of 
.
Al Keene made the same compound using a different alkyl bromide and
acetaldehyde as his starting materials. Draw the structure of the
alkyl bromide Al Keene used.
CCC(Br)C(=C)C Al
Kane's synthesis involved generation of an organometallic reagent
from 1-bromo-2-methyl-2-pentene followed by addition to acetaldehyde.
The organometallic reagent is an allylic carbanion for which there
are two resonance contributors, A and B. Structure B implies the
alkyl bromide 3-bromo-2-methyl-1-pentene.
11. Draw the structure of the
thermodynamically controlled product of the following
reaction:
CC(C)C(CC(=O)C)C#N
12. Draw the tautomer of 
.
Oc1ccccc1
13.Would you expect the value of the equilibrium constant for the tautomerization of the compound shown in Question 12 to be
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In tautomerization
,
the formation of the enol is accompanied by the formation of an
aromatic ring. We have seen repeatedly that such a ring is expecially
stable. Hence, this is a case where the enol tautomer is more stable
than the keto form.
14. The transformation 
involves
a Michael addition. Draw the structure of compound X
.CC(=O)CCC(=O)C=C
15. Draw the structure of the product
of the following Robinson annulation:
CC1CC(=O)C(=C2CCCCC12)C
16. Draw the 1,6-addition product in
the following reaction:
CCC(C)(C)C=CC=C(C)O[Si](C)(C)C
17. Select the reaction conditions
that are best suited to effecting the following
transformation:
18. How many esters are there that
are isomers of butanoic acid,
CH3CH3CH3CO2H?
4 
19. Draw the structure of the mixed
anhydride that would be formed from formic acid, HCO2H,
and acetic acid, CH3CO2H.
CC(=O)OC=O
20. Draw the structure of the major
product of the following reaction sequence.
OC(=O)C1CCC(=O)C1
21. Draw the structure of the
following acid catalysed reaction.
O=C1CCN2CCCCC2C1
22.* 
is
prepared by treatment of a dicarboxylic acid with concentrated
sulfuric acid. Draw the structure of the dicarboxylic acid.
OC(=O)/C=C\C(=O)O
23. Draw the structure of the alcohol
that was used in the preparation of 
.
OC1CCCCC1
24. Cyclic esters are called
lactones. Draw the structure of the compound from which the following
lactone was prepared.
CC(O)CCCC(C)C(=O)O
25. Trying to think like an organic
chemist, a student attempted to brominate the a-carbon
of acetic acid in the same way he had read that it was possible to
brominate the a-carbon
of acetone:
.
Unfortunately this eminently logical extension of what he had learned
did not yield any a-bromoacetic
acid. Draw the structure of the compound the student actually
obtained.
CC(=O)O
The NaOH deprotonated the carboxylic acid group rather than the a-carbon. The carboxylate anion does not react with the Br2, so when the solution was acidified during work-up the starting material was regenerated.
26. Draw the structure of the
carboxylic acid from which you would prepare the insect repellant
DEET.
Cc1cccc(c1)C(=O)O
27. Select the compound that would
react with a solution of Br2 in CCL4 most
rapidly.
D
Br2 is electrophilic. Therefore it will react fastest with
the double bond that has the highest electron density. The carbonyl
groups in compounds A-C reduce the electron density in the double
bond.
28. When Al Keene reacted 1 mmol of 2-methyl-1,3-butadiene with 1 mmol of HCl(g) , GC analysis indicated that four new compounds were formed. The 1H-NMR spectrum of one of them is attached. Draw the structure of this compound.
CC(=CCCl)C Protonation at C-1 produces a resonance stabilized allylic carbocation. The electron density in this ion is lowest at C-2 and C-4 as indicated by the two resonance contributors shown. Attack of chloride ion at these two carbons produces two isomeric alkyl halides. A similar situation arises from protonation at C-4.
The NMR spectrum of this compound is interesting on several accounts. First, the two methyl groups are not identical since one is "cis" to an H atom while the other is "cis" to a CH2Cl group. Hence, these two methyls appear as two singlets. Second, the vinyl proton appears as an overlapping doublet of doublets rather than as a triplet. This is because the two methylene protons are not identical. Rather, they are diastereotopic. Hence they are magnetically non-equivalent. They appear as two doublets, although their chemical shift differences are so small that it is very difficult to observe this splitting.
29. One day Polly Ester received a
package in the mail with a note saying "I suspect the sample in the
vial is 
.
I would appreciate it if you would analyse the material to confirm my
suspicion. Please call me at 225-3778. Sincerely yours, S. Holmes."
Having nothing better to do, Polly decided to give it a go. When she
treated the sample with O3, she obtained two new compounds
which she labeled S and H. She tested both of these compounds with
aqueous chromic acid; compound S did not react, but compound H turned
the chromic acid solution green immediately. Polly labeled the
compound that was produced by treatment of compound H with chromic
acid compound W. After recording the 1H-NMR
spectrum of compound W, Polly called S. Holmes and reported that his
suspicions were incorrect. The NMR spectrum was not consistent with
the structure she had expected to obtain. "My suspicions are seldom
wrong young lady", Mr. Holmes replied sternly. "You had better take
another look at that NMR spectrum." When she did, Polly realized that
Mr. Holmes had been correct and that the spectrum of compound W was,
indeed, consistent with Mr. Holmes' suspicion. Draw the structure of
compound W.
CC1CC(C)(OC1=O)c2ccccc2
Oxidation of compound H produced the carboxylic acid that Polly expected, but this material cyclized spontaneously under the acidic reaction conditions. The OH group reacted with the carboxylic acid group to form a cyclic ester, a lactone. The NMR spectrum of compound W does not contain any exchangeable protons, which indicates that the structure does not contain either a CO2H or an OH group.
30. How many isomeric amides are there with the molecular formula C5H11NO?18
