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Chemistry 251/253 |
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Homework 7 |
1. In the following reaction, the ethyl propionate was added to a mixture of benzaldehye and sodium ethoxide in ethanol. Draw the structure of the major product.CCOC(=O)C(C)C(O)c1ccccc1
This reaction is a combination of an aldol condensation and a Claisen reaction. The benzaldehyde, which has no a-hydrogens, acts as the electrophilic component, while the ethyl acetate reacts with the NaOEt to produce the nucleophilic component.
2. In the following reaction, the acetaldehyde was added to a mixture of ethyl benzoate and sodium ethoxide in ethanol. Draw the structure of the major product.O=CCC(=O)c1ccccc1
This reaction is a also combination of an aldol condensation and a Claisen reaction. The ethyl benzoate, which has no a-hydrogens, acts as the electrophilic component, while the acetaldehyde reacts with the NaOEt to produce the nucleophilic component.
3. The reaction shown below produced a product with the following 1H-NMR data: d, ppm 1.26, t, J= 6.8, 3H; 3.21, d, J = 2.6, 2H; 4.18, q, J = 6.8, 2H; 6.55, m, J = 2.6, 5.6, 1H; 6.68, d, J = 5.6, 1H; 7.18-7.58, m, 5H. Draw the structure of the product.CCOC(=O)C1=C(C=CC1)c2ccccc2
This reaction involves an intramolecular nucleophilic addition of an enolate ion to the carbonyl carbon of a ketone. The resulting alcohol undergoes dehydration during the work-up.
Draw the structure of the major product in each of the
following reactions. If you do not expect a reaction to occur, enter
NR in the appropriate text field.
4.
O=CCC(=O)c1ccccc1 In this reaction the ethyl
formate does not have any a-hydrogens,
so it can only act as an electrophile. The LDA deprotonates the
methyl group of the acetophenone to produce a high concentration of
enolate ion, which reacts with the ester in a nucleophilic acyl
substitution reaction. Compare this question to question
2.
5.CCC#CC(=O)c1ccccc1
In this reaction the LDA
deprotonates the terminal alkyne. The resulting carbanion then acts
as a nucelophile in an Sn2 reaction with the ethyl
bromide.
6.
CC(=O)C(Cc1ccccc1)C(=O)C
7.
CC(C)CC(C#N)C(=O)C It is
useful to think of the cyano group as a "pseudo-carbonyl group". It
stabilizes the conjugate base that is produced by deprotonation of
the a-carbon.
The pKa of the methylene protons in the starting ketone should be
higher than, but close to, 9.
8.
No reaction. The NaOEt will deprotonate the b-ketoester
completely (Keq = 106), but the resulting
enolate ion will not react with the 1-chloropropene because vinyl
halides do not undergo Sn2 reactions. The Sn2 reaction occurs at
sp3 hybridized carbon atoms.
9.
CCC(=O)c1ccccc1 The first
part of this transformation involves saponification of the ester.
Acidification of the basic reaction mixture converts the carboxylate
salt to the carboxylic acid, which, when heated, undergoes
decarboxylation to produce the ketone.
10.
CC(C)CCCC=O This is a
straightforward alkylation of an aldehyde.
11.CC1CC2C3CCCCC3(C)C=CC2=CC1=O
This is a straightforward
alkylation of a ketone.
12.CCC(C(=O)CC)C(=O)OC
13.
CSC(SC)c1ccccc1 This
is an example of the formation of a thioacetal. The reaction is
analogous to the nucleophilic addition of alcohols to aldehydes and
ketones.
14.
CCN(CC)CC1CC2(C)CCCC(C)(C)C2=CC1=O
This is an example of a
Mannich reaction.
15.CCC=CC(=O)c1ccccc1
In this reaction the methyl
protons of the a,b-unsaturated
ketone are the most acidic protons. They have a pK of approximately
19. The LDA deprotonates the methyl group and the resulting carbanion
attacks the methyl iodide.
16.No
reaction.
17.O=C(Cc1ccccc1)Cc2ccccc2
The first part of this
sequence involves a Claisen condensation to produce compound A.
Saponification of compound A produces the corresponding
b-ketocarboxylic
acid, which decarboxylates when heated in aqueous acid.
18. Select the compound that you could not
prepare fromusing
a Mannich reaction:
C The Mannich reaction involves the generation of a nucleophilic carbon atom at a position a to a carbonyl group.
19.The Mannich reaction shown below produces a mixture of isomeric substituted phenols. Select the phrase that best describes the composition of that mixture.
You need to draw all the resonance structures that are involved in this reaction to predict the product mixture.
20. How many of these compounds will produce a ketone when subjected
to saponification followed by vigorous heating in aqueous
acid? 3 The first compound
is not a b-ketoester,
so it will not decarboxylate. The last compound will decarboxylate,
but the product will be a carboxylic acid, not a ketone.
21. How many of these compounds will produce
the same product when subjected to saponification followed by
vigorous heating in aqueous acid? 2
The third and fourth compounds would produce the same ketone.
Questions 22-27 are related to the
structures of 2-bromo-2,3-dichloropropionitrile, ,
2,2,3-trichloropropionitrile,
,
and 2,3-dichloropropionitrile
.
22. The methylene protons in 2-bromo-2,3-dichloropropionitrile
are
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23. In the 1H-NMR spectrum of 2-bromo-2,3-dichloropropionitrile the methylene protons would appear as
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24. The methylene protons in 2,2,3-trichloropropionitrile are
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25. In the 1H-NMR spectrum of 2,2,3-trichloropropionitrile the methylene protons would appear as
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26. In the 1H-NMR spectrum of 2,3-dichloropropionitrile the methylene protons would appear as
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A sawhorse projection of
2,3-dichloropropionitrile, ,
reveals that Ha and Hb are diastereotopic,
which means that they are not identical and that they have different
chemical shifts in the 1H-NMR; Ha splits
Hb into a doublet and vice versa. Likewise Hc
splits each branch of these two doublets into doublets, so that the
methylene protons appear as four doublets. Diagramatically the
spin-spin splitting interactions look like this:
27. In 2,3-dichloropropionitrile, when the dihedral angle between the two C-Cl bonds is 180o, the dihedral angles between H-1 and the two H-2 protons are
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The sawhorse projection also shows that the dihedral angle between Ha and Hc is 60o, while that between Hc and Hb is 180o. That is why the coupling constants JBC and JAC do not have the same values.
28. Which of the following statements best
describes the methylene protons in 3-chloro-3-methylpentane,
?
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The molecule has a plane of symmetry that passes through the carbon atom of the methyl group attached to C-3, C-3, and the Cl atom attached to C-3. Therefore the two ethyl groups, and therefore the two methylene groups, are identical. However, the protons of the two methylene groups are not identical, they are diastereotopic as the diagram below indicates:
While C-3 of 3-chloro-3-methylpentane is not chiral, replacing HB of C-2 with a D produces a new structure in which both C-2 and C-3 are chiral. The stereochemistries of the two chiral centers are R and S. respectively. Replacing HA of C-2 generates a diastereomer of the compound produced by replacing HB. Its stereochemistry is S,S. Exactly the same situation applies when you consider the two protons attached to C-4.
29. Enter any letters which correspond to
reaction conditions that would achieve the following
transformation:.ABThe
first two steps of sequence D will generate an ester that will react
with the ethyl lithium in the third stepto produce the desired
product initially. However, that product is more reactive towards
ethyl lithium than the ester is, so the actual product that sequence
D would produce is
.
30. To an organic chemist, the reaction
is
similar to each of the four reactions shown below. However, one of
the four reactions shares a similarity that the others do not. Which
one? A In both cases, the
acidity of the proton that is abstracted by the base depends upon the
ability of the adjacent heteroatom, S or P, to accomodate more than 8
electrons in its valence shell.