Name
Department of Chemistry
SUNY/Oneonta
Chem 322 - Organic Chemistry II
Examination #3 - March 31, 2003 - ANSWERS
INSTRUCTIONS —
This examination has two parts. Part I is in multiple choice format and the
answers should be placed on the "Test Scoring Answer Sheet" which must be
turned in and will be machine graded.
Part II requires your responding to questions by writing answers into the
spaces provided in this booklet. This entire Exam Booklet must be handed in and
will be returned to you with a grade. Write your name in the space above NOW.
On the Test Scoring Answer Sheet, using a soft pencil, enter the following
data (in the appropriate places): your name, instructor's name, your OSC student
(or Social Security) number, course number (30032201) and the test number
(03); darken the appropriate bubbles under the entries, making dark black marks
which fill the bubbles (if you use your OSC number do not darken a bubble under
the A at the beginning of the number).
You may use a set of molecular models but no other aids during the exam.
Answer all questions. The questions on Part I are worth 2 points each.
You have 90 minutes. Good luck!
Explanations are in red type.
Answers are highlighted in red.
March 31, 2003
1.
Page #2 of 12
The structure for N-methylacetamide is
O
(a)
H 3C C
2.
Organic Chemistry II - Exam #3
CH2NH2
(b)
O
H 3C
C
NHCH3
(c)
O
H3CNH
C
NHCH3
(d)
O
H 3C
C
CH2NHCH3
Ph-CO-OCH2CH3 is
(a) phenyl ethanoate, (b) phenyl acetate, (c) phenoyl ethylate,
(d) ethyl benzoate. (e) ethyl phenoate
3.
CH3CH2O-CO-CH2-CO-OCH2CH3 is
(a) acetoacetic ester, (b) malonic ester, (c) Ester's ester
(d) acetoacetic anhydride, (e) malonic anhydride
4.
CH3-CO-O-CO-CH3 is
(a) acetic acetate, (b) ethyl acetate, (c) acetic anhydride, (d) formic anhydride,
(e) ethyl formate
5.
Rank the following compounds in order from most reactive to least reactive toward
nucleophilic acyl substitution:
I: (CH3)2CH-CO-Cl, II: (CH3)2CH-CO-OCH3, III: (CH3)2CH-CO-NH2 , IV: CH3-CO-NH2
(a) I > II > III > IV, (b) IV > III > II > I, (c) II > IV > I > III, (d) IV > III > I > II,
(e) I > II > IV > III
I is an acid chloride, II is an ester, and III and IV are amides. The order of reactivity is
most influenced by the functional group involved: I>II>III,IV. IV is more reactive than III
because the transition state of the rate limiting step is more sterically hindered in the
bulkier III than in IV.
6.
The behavior of esters and ketones in a reaction with a nucleophile is different because
(a) the carbonyl carbon of the ketone is more positive.
(b) the two oxygens of the ester sterically hinder nucleophilic attack.
(c) the ester cannot form a tetrahedral intermediate.
(d) there is a reasonable leaving group in the ester.
(e) the ketone is not easily oxidized.
March 31, 2003
Organic Chemistry II - Exam #3
Page #3 of 12
7.
O
CH3CH2
C
Cl
major product(s)
OH
+
O
(a) CH3CH2
C
O
OH
OH
(c) CH3CH2
C
(b) CH3CH2
Cl
C OH +
O
O
Cl (d) CH3CH2
C
(e) CH3CH2
C
OH +
Cl
O
OH
The following three questions consist of a statement followed by the connecting word
because followed by a reason. For each question choose the correct description of the
statement and reason from the list below -(a)
(b)
(c)
(d)
(e)
8.
The statement and the reason are both factually true, and the reason is the
correct explanation of the statement.
The statement and the reason are both factually true, but the reason is not the
correct explanation of the statement.
The statement is true and the reason is false.
The statement is false and the reason is true.
Both statement and reason are false.
Under basic conditions (H2O/NaOH) compound I on the right
will hydrolyze more quickly than II because the transition state
of the rate limiting step is more crowded for I than for II.
(d)
(I) O
(CH3)3C C OCH3
(II)
CH3
O
C OCH3
March 31, 2003
9.
Organic Chemistry II - Exam #3
Branched alkylbenzenesulfonates, eg the
compound to the right, are widely
employed as detergents because they are
inexpensive to make.
SO3- Na+
A Branched Alkylbenzenesulfonate.
(d)
10.
Page #4 of 12
A Dean-Stark apparatus is sometimes used in Fisher esterification syntheses because
equilibrium constants in these reactions are typically about 1 and the Dean-Stark tube
removes water thereby driving the reaction in the forward direction.
(a)
11.
Which of the following procedures give(s) benzyl alcohol, Ph-CH2OH, starting with
methyl benzoate, Ph-CO-OCH3?
(a) 1. DIBAH (diisobutylaluminum hydride) in toluene, -78oC; 2. H3O+,
(b) H2, Pd/BaSO4, (c) 1. LiAlH4, ether; 2. H3O+,
(d) a & b, (e) a & c
12.
CH3CH2CH2C/N + H2O
H SO , heat
> major organic product
2
4
_________________
(a) CH3CH2CH2CH2-NH2, (b) CH3CH2CH2CH2-N=N-CH2CH2CH2CH3
(c) CH3CH2CH2CO-NH2, (d) CH3CH2CH2CH2-NO2, (e) CH3CH2CH2COOH
13.
(CH3)3CBr + Na+CN-
____________
> major product
(a) (CH3)3CCN, (b) (CH3)2C=CHCN, (c) (CH3)2C=CH2, (d) (CH3)3CCOOH
(e) none of the above
14.
Rank the protons in 3,5-heptanedione in order of
decreasing acidity (most acidic first).
(a) I > II > III, (b) III > II > I,
(c) II > III > I, (d) II > I > III
H3C CH2
I
II
O
O
C
CH2 C
III
CH2 CH3
The I protons have no adjacent carbonyl group, the II protons have one, and the III
protons have two. The carbonyl groups stabilize the negative charge on the anion
through resonance: 2 is better than 1 and 1 is better than none.
March 31, 2003
15.
Organic Chemistry II - Exam #3
Page #5 of 12
Select the major product of the following reaction.
OH
COOH
1. BH3, THF
2. H3O
(b)
(a)
COOH
+
O
OH
(c)
COOH
(d)
CH
CH2OH
OH
(e)
16.
CH2OH
The structures shown to the right are
H3 C
(a) tautomers, (b) resonance structures,
(c) enantiomers, (d) diastereomers,
(e) none of the above.
17.
N
The structures shown to the right are
H3 C
(a) tautomers, (b) resonance structures,
(c) enantiomers, (d) diastereomers,
(e) none of the above.
N
O
H3 C
O
O
H2 C
O
N
N
O
O
OH
O
When drawing resonance structures we move (usually unshared and π) electrons to draw
different Lewis structures of the same compound. Tautomers are isomers that are in
(usually fairly rapid)
equilibrium with each other.
O
For questions 18 and 19
consider the mechanism for
the acid-catalyzed bromination
of acetone shown to the right.
[Note: not all resonance
structures are shown.]
O H
(a) CH3CCH3 + H3O
O H
(b) CH3CCH3
H 2O
+
OH
+ H 2O
CH3C
CH2
+
H 3O
CH3CCH2Br
+
Br
OH
OH
(c) CH3C
CH3CCH3
CH2 + Br2
OH
(d) CH3CCH2Br + H2O
O
CH3CCH2Br
+
H 3O
March 31, 2003
18.
Organic Chemistry II - Exam #3
Page #6 of 12
Which of the steps shown is incorrect?
(a), (b), (c), (d), (e) All of the steps are essentially correct.
19.
Which of the steps is the slow step in the mechanism?
(a), (b), (c), (d), (e) There is no slow step.
Transfer of a proton off a carbon is frequently a slow process while transfer of a proton
off of oxygen or nitrogen is usually fairly fast.
20.
Which of the following compounds will react with iodine in aqueous sodium hydroxide
solution to give a precipitate of iodoform, CHI3?
(a) PhCHO, (b) PhCOCH3, (c) CH3CH2CHOHCH2CH3, (d) CH3CH2CHO
21.
One of the principal products of the following reaction would be:
O
1. excess LiAlH4, ether
+
2. H3O
CO2C2H5
(a)
(b)
CO2C2H5
22.
O
OH
O
(d)
(c)
CO2C2H5
OH
OH
CH2OH
(e)
CH2OH
CH2OH
Some nucleophiles are said to be ambident. Which of the following nucleophiles are in
this category?
(a) I&II, (b) II&III, (c) I, II&III, (d) IV,
(e) I&III, (f) II&IV
O
CH3CH2O
Ambident (literally, two teeth) means
that there are two atoms that can use a
pair of electrons to make a nucleophilic attack.
I
C
N
II
H 2C
C
III
CH 3
CH 3NH2
IV
March 31, 2003
23.
Organic Chemistry II - Exam #3
Page #7 of 12
Select the compound that would be formed in greatest amount.
O
H3 C
BrCH2Ph
ether
LiN[CH(CH3)2]2
THF
O
H3 C
H3 C
CH2Ph
O
O
H3 C
(c)
PhCH2
(b)
(a)
O
major product
O
CH2Ph
H3 C
H3 C
(e)
(d)
CH2Ph
PhCH2
24.
O
O
1. NaOC2H5, C2H5OH
2. CH3I
O
O
O
O
H3 C
CH3
H3 C
(a)
major product
(b)
O H3 C
(d)
(c)
O
O
O
CH3
March 31, 2003
25.
Organic Chemistry II - Exam #3
Select the product that would be formed in greatest amount. Hell-Volhard-Zelinskii
reaction.
1. Br2, PBr3
OH
2. H2O
O
Br
(a)
O
Br
OH
CHBr2 (c)
(b)
O
Br
OH
(d)
O
26.
Page #8 of 12
C OH
Br Br
(e)
Acetoacetic ester synthesis.
O
O
NaOC2H5
C2H5OH
CH3CCH2 COC2 H5
+
CH3 CH2 CH2CH2Br
H3O
heat
O
O
(a) CH3CCH2(CH2 )3CH3
O
(b) CH3(CH2)3CH2COC2 H5 (c) CH3 CCH3
O
(d) CH3 (CH2)3CH2CCH2 (CH2)3CH3
27.
O
The reaction shown to the right is an
example of a(n)
2 CH3CH
base
OH
O
CH3CH CH2CH
(a) Claisen condensation. (b) aldol reaction.
(c) demerol reaction. (d) Knovenagel reaction. (e) Diels-Alder reaction.
In fact the common name for the product here is aldol.
28.
The reaction shown to the right does not
give as good a yield as the similar reaction
shown in question #27. A reasonable
explanation for this would be that
O
2
CH3CCH3
base
OH
O
CH3 C CH2CCH3
CH3
(a) the aldehydic proton is more acidic than
the methyl protons.
(b) the product formed from acetone undergoes a subsequent reaction, thus reducing the
yield.
March 31, 2003
Organic Chemistry II - Exam #3
Page #9 of 12
(c) a steric factor is involved as a result of the intermediate (and product) being more
crowded in these reactions and the ketone case is worse than the aldehyde in this
respect.
(d) the ketone has 6 acidic hydrogens while the aldehyde has only 3.
(e) Trick question, dude! Like, this reaction gives a better yield than the one in question
#27.
29.
Which of the following describe common situations encountered with aldol reactions?
(a) Product reverts to reactant. (b) Product undergoes a spontaneous dehydration,
producing a conjugated enone. (c) Product undergoes a spontaneous dehydration,
producing a non-conjugated enone. (d) Product undergoes reduction, producing a diol.
(e) a & b, (f) a & c, (g) b & c, (h) a, b & c.
30.
Which of the following pairs of compounds would be the most reasonable choice for an
attempt at a "mixed" or "crossed" aldol condensation?
O
O
O
O
(d) CH3CCH3 + CH3 CCH2CH3
O
(c) PhCH + HCH
(b) PhCH + CH3CCH3
(a) CH3CH + CH3 CCH3
O
O
O
O
O
(e) CH3 CH + CH3CCH2 CH3
Crossed aldol reactions typically work best when one of the components does not have
an α-hydrogen and, therefore, cannot form an enolate anion. This is true for (b), where
benzaldehyde does not have an α-hydrogen, while acetone does. So, acetone forms the
enolate anion which attacks benzaldehyde. The situation in (c) is that neither component
has an α-hydrogen so neither can form an enolate anion. In cases (a), (d), and (e) both
components have α-hydrogens.
31.
To prepare acetoacetic ester (ethyl acetoacetate), which of the following compounds
could you use in a Claisen condensation?
O
O
(a) CH3 COC2 H5
O
O
(c) HCCH2 COC2 H5
O
(d) CH3CCH2 CCH3
32.
(b) CH3CCH2OC2H5
O
(e) none of these answers is correct
Which of the following compounds would not give a good yield in a Claisen
condensation?
March 31, 2003
Organic Chemistry II - Exam #3
O
(a)
O
CH2 COC2H5
O
(b) CH3CH2 COC2H5 (c) (CH3 )2CH COC2 H5
O
O
CH2 CH2 COC2H5
(d)
Page #10 of 12
(e) CH3 CH2 COCH3
The issue here is that the ester that is used in a Claisen condensation needs two αhydrogens because equilibrium does not favor the β-ketoester product; however, the
deprotonated β-ketoester (formed by removal of the second α-hydrogen) is favored.
33.
Select the product that is formed in the following Michael addition.
CH2 (CO 2 C2 H5 )2
+
H2C
CHCOCH3
base
OH
(a) H2C
OH
CHC CH(CO2 C2 H5 )2
(b) H3C
CH2C CH(CO2C2H5)2
CH3
CH3
O
(c) H2 C
CH2C
O
CH3 (d) H3C
CH(CO2C2H5)2
34.
CHC
CH3
CH(CO2 C2H5 )2
(e) none of these
products is formed
A frequent aim of synthesis is to make larger molecules from smaller ones. Which of the
following syntheses does not accomplish this?
(a) aldol reaction, (b) Claisen condensation, (c) Grignard synthesis,
(d) Diels-Alder reaction, (e) saponification reaction.
35.
Which of the following bases would not convert acetone, essentially completely, to its
enolate anion?
(a) butyllithium, (b) sodium hydride, (c) lithium diisopropylamide, (d) sodium hydroxide
Sodium hydroxide is the weakest base of the bunch and is a weaker base than the
enolate anion formed from acetone. The others are stronger bases than the enolate anion
of acetone.
Do Not Detach The Following Sheets From The Rest Of The Exam
March 31, 2003
Organic Chemistry II - Exam #3
Page #11 of 12
Part II. Enter your answers in the space provided. If there is inadequate
room, continue on the back of the page and clearly indicate on the front of
the page that you have done this.
Hand in this entire exam booklet when you are finished; it will be returned
to you with your grade. Make sure your name is on the front sheet.
1.
Synthesis. Outline the following syntheses. Show all reagents and any important
conditions. Do not show mechanisms or balance equations.
(a) Make methyl benzoate, Ph-CO-OCH3, starting from benzoic acid. In addition to
benzoic acid, you may use any materials you wish.
O
The Fisher
O
O
CH3OH
SOCl2
esterification method
C OCH3
C OH
C Cl
was the one you used
in lab last semester.
One could also do a Fisher esterification, using benzoic acid, methanol,
and a catalytic amount of a strong acid like sulfuric acid.
(d) Outline a synthesis of 4-pentenoic acid, CH2=CHCH2CH2COOH, starting from malonic
ester (diethyl malonate), CH3CH2O-CO-CH2-CO-OCH2CH3. In addition to malonic ester
you may use any materials you wish.
O
O
H5C2OCCH2COC2H5
2H 5
CH
BrCH2
O
CH2
C2H5OH
O
H5C2OCCHCOC2H5
H2C HC
O
H3O+
CH2
H CHCOH
H2O
heat
2.
Na+ -OC
H2C HC
CH2
Mechanism.
(a) Show all steps in the mechanism of the following enolate methylation reaction
sequence leading to the major product (2,6-dimethylcyclohexanone). Be sure to show
relevant resonance structures of intermediates.
H
H
C
O
1) Li
H
C
N
H 3C
H
CH3
2) CH3I
C
O
H
C
56%
H
CH3
H
O
C
CH3
C
6%
CH3
March 31, 2003
n
H
O
Li
N
Ho
O
H
C
HH
H
O
H N
H
CH3
H
CH3
H
I
Organic Chemistry II - Exam #3
p
O
Page #12 of 12
H
CH3
H
O
CH3 H
H
CH3
H
CH3
+
I
(b) Based on the mechanisms for the formation of the major (56%) and minor (6%) products
above (which are the same except for the hydrogen replaced), explain why the amount of 2,6dimethylcyclohexanone produced is about 10 times greater than that of 2,2dimethylcyclohexanone.
The first step in this mechanism is the rate limiting step.
O
O
In this step enolate anion p is formed by removal of a
H
H
proton n from the ketone reactant to form the major
CH3
H
CH3
H
product. For the minor product to form, proton o
would be removed to form enolate anion q. Since the
p
ratio of major product to minor product is about 9 to 1,
O
O
the rate of formation of p to q is also about 9 to 1.
H
H
Why is this? Because the transition state leading to the
H
CH3
formation of p is lower in energy than that of q. By
H
CH3
q
the Hammond Postulate, the transition state energies
should be reflected in the energies of the enolate
anions themselves. Thus, p is more stable than q. This is reasonable in that q has some
negative charge on C-2, while p has some negative charge on C-6. The negative charge on
C-2 is intensified by the attached methyl group, raising the energy of q. There is no similar
alkyl group attached to C-6 in p. In other words, proton n is more acidic than protons o
owing to the electron donating methyl group.
March 31, 2003
3.
Organic Chemistry II - Exam #3
Page #13 of 12
Roadmap.
One mole of acetaldehyde reacts with 3 moles of Compound A. This constitutes a sequence
of three crossed aldol reactions that produce one mole of compound B. Concentrated
aqueous sodium hydroxide and one mole of Compound A are added to Compound B and the
mixture is heated, giving Compound C and sodium formate. A mole of compound C reacts
with 4 moles of nitric acid to produce a nitrate tetraester, pentaerythritol tetranitrate, which is
used in medicine as a vasodilator and to make Primacord fuse. Compounds A and B each
give a positive Tollen’s test; C does not. Draw the structures of A, B, and C.
O
H 3C
CH +
dilute
NaOH
H
3
C O
CH2OH
H 2O
HOCH2
H
CHO
C
CH2OH
Compound A
Compound B
conc. NaOH
H 2O
Compound A
O NO2
CH2
O H 2C
C
CH2 O NO2
NO2
CH2
HONO2
CH2OH
HOCH2
C
CH2OH
CH2OH
O NO2
pentaerythritol tetranitrate
Compound C
+ 4 H 2O
+ sodium formate
The chemically astute among you might have used the clue, “A mole of compound C reacts
with 4 moles of nitric acid to produce a nitrate tetraester, pentaerythritol tetranitrate,” to get at
the structure of compound C. You would need to recall that when esters form from alcohols
and acids, the acid loses an OH, the alcohol loses an H, and the ester bond forms. If you
were less astute, the structure of the final product should have told you that compound C has
five carbons and the 2,2-dimethylpropyl skeleton structure.
Since the final product has 5 carbons and the starting material,
acetaldehyde, has two, three carbons need to be added. The 3 molecules
of compound A will likely do this – one carbon per molecule. Since this
reaction is a crossed aldol, compound A must be an aldehyde or ketone.
Since it has one carbon, it must be formaldehyde. Another clue is the
sodium formate formed in the reaction that converts B to C, a Cannizzaro
reaction.
Part I (70)_________
Part II
1. (10)____________
2. (10)____________
3. (10)____________
Total(100)_________