1A
ANSWER KEY
2.222 In-class Quiz #1 Friday January 18, 2002.
CIRCLE the correct response for each of the following questions.
1. Which of the following alkenes would react most rapidly with HBr in CH2Cl2?
O
Br
A
B
C
E
D
Since protonation of the alkene is rate limiting, the alkene that will react fastest
will be the one that forms the most stable intermediate cation. As the textbook
points out, the activation energy for this step will correlate with cation stability.
Note that delocalization of the positive charge onto the aromatic ring is a highly
stabilizing factor.
2. Which one of compounds A-E below would be the major product of the following
reaction?
?
H2SO4, H2O
OH
OH
OH
OH
OH
A
B
C
D
E
This reaction involves a 1,2-migration of a methyl group to convert the initially
formed secondary cation into the more stable tertiary cation. Note that structure
A also arises from a 1,2-shift, but that it is disfavoured by the ring strain in the
cyclobutane ring. Structures D and E are less favourable since they arise from
secondary cations.
1A
ANSWER KEY
3. Which of the following is NOT a feasible reaction?
Br
H
Br2, CH2Cl2
A
H
Br
OH
HCl, H2O
B
C
H3C
C
C
CH3
HBr, H2O
Br
H3C
Br
C
CH2 CH3
Cl
D
NaOH (aq.), Bu4N OH
CHCl3
Cl
H
E
1) Hg(OAc)2, H2O
OH
2) NaBH4
As we discussed in class, the preparation of a geminal dibromide from an alkyne
is not really possible in water because the dibromide will react further to form a
ketone under those conditions. Your textbook doesn’t actually say that the
reaction can be done in water, but in Figure 10.8 it shows the participation of
H3O+, which implies the presence of water. I disagree with this.
All the other reactions are ones we have discussed – notice that D is an addition
of singlet dichlorocarbene, generated in situ by base treatment of chloroform.
1B
ANSWERS
2.222 In-class Quiz #1 Friday January 18, 2002.
CIRCLE the correct response for each of the following questions.
1. Which of the following alkenes would react SLOWEST with HBr in CH2Cl2?
O
A
B
C
E
D
The initial protonation step of the reaction is rate limiting, and its activation
barrier is correlated with the stability of the resulting cation. The leaststable
cation will be the one substituted with an electron-withdrawing group, and a
ketone is a very good EWG. All the other alkenes are substituted with H and/or
donor groups.
2. Which one of compounds A-E below would be the major product of the following
reaction?
?
H2SO4, H2O
OH
OH
OSO3H
HO
OH
A
B
C
D
E
This question is basically the same as Exercise 10.7a in the textbook. Protonation
of the alkene forms a secondary cation, which rearranges to a more-stable
tertiary cation by a 1,2-shift of a methyl group. Note that structures D and E are
impossible to obtain under these conditions, while structure B is extremely
unlikely since the conjugate base of sulfuric acid is a poor nucleophile.
1B
ANSWERS
3. Only ONE of the following IS a feasible reaction. Which one is it?
I
H
I2, CH2Cl2
A
H
I
OH
HCl, H2O
B
C
H3CH2C
C
CH
O
H2SO4, H2O,
heat
H3CH2C
Cl
D
E
NaOH (aq.), Bu4N OH
CHCl3
1) Hg(OAc)2, H2O
2) NaBH4
H
C
CH3
Cl
H
HO
Acid-catalyzed hydrolysis of a terminal alkyne to a ketone proceeds with heating
as shown here, or at room temperature in the presence of mercuric acetate.
Notice that reaction B looks a bit like examples we have seen, but that there is no
driving force for a 1,2-shift when the ring is NOT aromatic! Reaction D would
require a two-step mechanism consistent with a triplet carbene, but these
conditions form singlet dichlorocarbene.
1C
ANSWERS
2.222 In-class Quiz #1 Friday January 18, 2002.
CIRCLE the correct response for each of the following questions.
1. Which of the following alkenes would react with HBr in CH2Cl2 to give a CHIRAL
product?
Br
A
B
C
E
D
In all cases (except A) the bromide ion will attack the more-substituted position and the
proton will end up on the other carbon of the alkene. The resulting bromides from A, B, D
and E all have a plane of symmetry passing through the Br-substituted position, but in the
case of C the molecule is not symmetrical. Of course, the chiral bromides obtained from C
are racemic.
Br
if addition is syn
HBr
H
Br
if addition is anti
H
2. Which one of compounds A-E below would be the major product of the following reaction?
H2SO4, H2O
OH
?
HO
HO
A
B
C
OH
D
HO
E
Reaction will occur fastest at the most-substituted alkene group, and of course will give the
Markovnikov product. Note that if the reaction were left long enough, both alkenes would
react to give a diol, but I didn’t include that option in this question.
1C
ANSWERS
3. Which ONE of the following DOES NOT involve electrophilic addition to an alkene?
Cl
ICl, CH2Cl2
A
I
OH
HCl, H2O
B
C
H3CH2C
OH
D
E
C
CH
O
H2SO4, H2O,
heat
H3CH2C
C
CH3
H2SO4, H2O,
heat
OH
1) Hg(OAc)2, H2O
OH
2) NaBH4
The reaction in D proceeds by protonation of oxygen, loss of water, and isomerization of the
double bond to the internal position. Although it is disfavoured by formation of a primary
cation, the product is more stable than the reactant because the alkene group is moresubstituted.
2A
ANSWER KEY
2.222 In-class Quiz #2 Friday February 1, 2002.
CIRCLE the correct response for each of the following questions.
1. What is the Major Product of the following reaction?
A
B
?
AlCl3
+
OH
Cl
C
D
E
This Freidel-Crafts Alkylation proceeds via the more-stable secondary cation
formed from the alkene on treatment with the Lewis Acid AlCl3. Notice that this is
the same Markovnikov selectivity that you would expect in an acid-catalyzed
addition to the alkene – in fact this reaction actually IS an acid-catalyzed addition
from the alkene’s point of view!
2. Which set of reagents will perform the following transformation?
O
O
H
?
H
A. Zn(Hg)/HCl (aq.)
B. Na (s), NH3 (l), EtOH
C. H2 (g), Pd (cat.), AcOH
D. 1) Hg(OAc)2, H2O; 2) NaBH4
E. 1) B2H6, THF; 2) H2O2, NaOH (aq.)
Selective reduction of the alkyne by dissolving metal always gives the trans alkene
product. Option A is a Clemmensen Reduction, which would remove the
carbonyl, while option C would hydrogenate both the carbonyl and the alkyne.
The alkyne would be reduced all the way to an alkane. Options D and E would
result in the formation of ketone products via intermediate enols.
2A
ANSWER KEY
3. Which set of compounds could be used to prepare the following molecule using
the conditions specified?
?
heat
CO2CH3
+ Et2Zn + CH2I2
A
+ H2C
B
CO2CH3
CH2
CO2CH3
+
CO2CH3
CO2CH3
H3CO2C
CO2CH3
C
D
+
+
H3CO2C
H3CO2C
E
+ KMnO4
HO2C
HO2C
CO2H
HO2C
CO2H
HO2C
This is a Diels-Alder cycloaddition. Recall that this type of reaction always
occurs between a diene and a dienophile (alkene), and that the dienophile should
be substituted with an electron-withdrawing group. Also, recall that the product
will be a cyclohexene ring, in which the double bond is located between what
were the C2 and C3 positions of the diene.
Option A shows conditions for a Simmons-Smith cyclopropanation, but the
product is NOT a cyclopropane – note that the bridging CH2 group is attached
across the ring, not to two adjacent carbons from one of the alkenes. Option E
describes an oxidative cleavage of the alkenes, which would form carboxylic
acids, but it would cleave BOTH alkene groups, not just the exocyclic one.
2A
ANSWER KEY
4. What product is obtained from the following 2-step sequence of reactions?
O
N
HNO3, H2SO4
Cl
AlCl3
N
N
N
N
NO2
NO2
O
N
NO2
O
O
O
NO2
NO2
O
A
?
B
C
D
E
The N,N-dimethylamino group is a strongly activating ortho/para director. Thus,
Freidel-Crafts acylation will occur primarily para. The second reaction will be
influenced by both the dimethylamino and the acyl groups, but both these groups
favor nitration at the same position – ortho to the amine and meta to the ketone.
5. Which set of reaction conditions would perform the following transformation?
OH
?
A. 1) Hg(OAc)2, H2O; 2) NaBH4
B. H2SO4, H2O
C. OsO4
D. 1) O3, CH2Cl2; 2) Zn, AcOH
E. 1) B2H6, THF; 2) H2O2, NaOH (aq.)
Only Hydroboration/Oxidation can add a hydroxyl group to an alkene in an antiMarkovnikov fashion. Option A and Option B would both form the tertiary
alcohol, while Option C would form a DIOL. Option D is an oxidative cleavage,
which would chop off the terminal CH2 group and form a ketone in its place.
2B
ANSWER KEY
2.222 In-class Quiz #2 Friday February 1, 2002.
CIRCLE the correct response for each of the following questions.
1. What is the Major Product of the following reaction?
Br
+
Br
OH
A
B
C
?
FeBr3
D
E
Remember that you CANNOT carry out a Freidel-Crafts Alkylation at a primary
position because the primary cation is too unstable. It rearranges via a 1,2-shift
mechanism to a more stable secondary or tertiary ion, depending on the exact
structure you are working with. In this case, the secondary ion is formed, and
reacts to produce isopropylbenzene.
2. Which set of reagents will perform the following transformation?
O
O
?
H
H
A. Zn(Hg)/HCl (aq.)
B. Na (s), NH3 (l), EtOH
C. H2 (g), Pd (cat.), Quinoline
D. 1) Hg(OAc)2, H2O; 2) NaBH4
E. 1) B2H6, THF; 2) H2O2, NaOH (aq.)
Quinoline is a “catalyst poison” that decreases the ability of the Pd to reduce
alkenes. Thus the alkyne can be hydrogenated to a cis alkene – recall that all
hydrogenations give SYN addition of H2 across the double bond being reduced.
2B
ANSWER KEY
3. Which set of compounds could be used to prepare the following molecule using
the conditions specified?
CN
?
heat
+
A
+
B
CN
CN
NC
NC
etc.
CN
etc.
CN
+
C
+
D
CN
CN
CN
CN
E
+
CH2
CH2
NO REACTION
This is a Diels-Alder cycloaddition. Recall that this type of reaction always
occurs between a diene and a dienophile (alkene), and that the dienophile should
be substituted with an electron-withdrawing group. Also, recall that the product
will be a cyclohexene ring, in which the double bond is located between what
were the C2 and C3 positions of the diene.
Options A and B both will produce mixtures of several Diels-Alder adducts, but
not the desired structure. Option C gives only one adduct, but not the one we
want. Option E leads to no reaction, because there is no diene!
2B
ANSWER KEY
4. What product is obtained from the following 2-step sequence of reactions?
O
N
HNO3, H2SO4
?
Cl
AlCl3
N
N
N
N
N
NO2
NO2
O O
O
O
B
NO2
NO2
NO2
A
O
C
E
D
The N,N-dimethylamino group is a strongly activating ortho/para director. Thus,
nitration occurs at the para position. The second reaction is controlled by both
the nitro and amino groups, and both favor reaction at the same site, ortho to the
amine and meta to the nitro group.
5. Which set of reaction conditions would perform the following transformation?
?
O
A. 1) Hg(OAc)2, H2O; 2) NaBH4
B. H2SO4, H2O
C. OsO4
D. 1) O3, CH2Cl2; 2) Zn, AcOH
E. 1) B2H6, THF; 2) H2O2, NaOH (aq.)
This reaction requires the cleavage of the alkene in order to form the ketone
product. The only set of reagents from the list that can do that is Option D,
ozonolysis followed by reductive workup. The other reagent sets all lead to
various addition reactions forming alcohol products.
2C
ANSWER KEY
2.222 In-class Quiz #2 Friday February 1, 2002.
CIRCLE the correct response for each of the following questions.
1. What is the Major Product of the following reaction?
Br
+
?
FeBr3
Br
OH
A
B
C
D
E
This is a straightforward Freidel-Crafts Alkylation. Note that the electrophile
that is formed when 2-bromobutane is treated with FeBr3 is a stable secondary
cation, and there is no rearrangement pathway to improve upon it.
2. Which set of reagents will perform the following transformation?
O
?
A. Zn(Hg)/HCl (aq.)
B. Na (s), NH3 (l), EtOH
C. H2 (g), Pd (cat.), Quinoline
D. 1) Hg(OAc)2, H2O; 2) NaBH4
E. 1) B2H6, THF; 2) H2O2, NaOH (aq.)
This is a Clemmensen Reduction of the aryl ketone. Option B describes a
dissolving metal reduction, which in this case would actually reduce the benzene
ring, although we did not discuss this aspect of this reaction – as far as you have
been told it would not reduce the alkene or the ketone. Option C is a
hydrogenation using a poisoned catalyst – which reduces alkynes to cis alkenes,
but won’t touch ketones. Options D and E are addition reactions leading to
alcohols.
2C
ANSWER KEY
3. Which set of compounds could be used to prepare the following molecule using
the conditions specified?
H H
?
O
heat
O
O
O
A HO2C
CO2H +
B
+ O
O
O
CO2H
O
HO2C
O
O
C
+O
O
O
D
+O
O
O
O
O
O
O + Et2Zn + CH2I2
E
NO REACTION
O
This is a Diels-Alder cycloaddition. Recall that this type of reaction always
occurs between a diene and a dienophile (alkene), and that the dienophile should
be substituted with an electron-withdrawing group. Also, recall that the product
will be a cyclohexene ring, in which the double bond is located between what
were the C2 and C3 positions of the diene.
This example comes right out of lecture and the textbook. Notice that Option E
describes conditions appropriate for a Simmons-Smith cyclopropanation, but that
the organic molecule is actually an AROMATIC compound – it would not react!
2C
ANSWER KEY
4. What product is obtained from the following 2-step sequence of reactions?
O
Br
HNO3, H2SO4
Cl
AlCl3
Br
Br
NO2
O
Br
Br
NO2
Br
NO2
O
O
O
NO2
NO2
O
A
?
B
C
E
D
Bromine is a weakly deactivating ortho/para director. Thus, Freidel-Crafts
Acylation will occur at the para position. The second reaction might actually be
a bit difficult with two deactivators on the ring, but it would occur at the position
favored by both of them – ortho to the bromine and meta to the ketone.
5. Which set of reaction conditions would perform the following transformation?
?
H
OH
A. 1) Hg(OAc)2, H2O; 2) NaBH4
B. H2SO4, H2O
C. OsO4
D. 1) O3, CH2Cl2; 2) Zn, AcOH
E. 1) B2H6, THF; 2) H2O2, NaOH (aq.)
The product clearly arises from the SYN addition of H2O across the alkene, in an
anti-Markovnikov sense. This result could only come from a
hydroboration/oxidation sequence. Option A and B would both give Markovnikov
addition to form the tertiary alcohol. Option C forms a vicinal diol, while Option
D leads to oxidative cleavage of the alkene to form a ketone and an aldehyde
group!
3A
ANSWER KEY
2.222 In-class Quiz #3 Monday March 4, 2002.
CIRCLE the correct response for each of the following questions.
1. Which of the following pairs of reagents could react to form an ENAMINE product?
O
O
H
N
A
O
O
B
+
O
NH +
O
CH3
N
C
H
N
D
+
+
O
H
N
E
O
+
O
Enamines are formed when secondary amines react with enolizable ketones or
aldehydes. Note that when primary or secondary amines react with esters or other
carboxylic acid derivatives, the product is an amide. This arises from substitution of the
leaving group, not of the carbonyl oxygen. Tertiary amines cannot form adducts with
carbonyl compounds.
2. Which of the following carbonyl compounds is most acidic at its C-α position?
O
O
H
H
A
O
O
O
H3CO
H
B
C
D
E
As discussed in class and in the textbook, aldehydes are more acidic at the C-α position
than either ketones or esters. Also as discussed in class, we find that acidity parallels the
stability of the conjugate base, and therefore any structural feature that stabilizes the
anion by delocalization will increase the acidity of the carbonyl compound.
3A
ANSWER KEY
3. Which of the following benzene derivatives would be nitrated most rapidly by
HNO3/H2SO4?
CH3
Br
OH
CN
OAc
A
B
C
D
E
Hydroxyl is a strong π-donor and is thus a powerful activating group for aromatic
substitution. The alkyl groups are weak activators, as are ester groups when linked via
their oxygen. Halogens are weak deactivators, and cyano is a moderately strong
deactivator.
4. Which of the following reactions would NOT proceed as written? A mild acidic workup
can be assumed to follow each proposed reaction.
O
2 CH3CH2MgBr
A
H3C
OCH3
O
ether, 25 oC
2 CH3CH2MgBr
B
ether, 25 oC
H3C
OH
O
CH3CH2MgBr
C
ether, 25 oC
H3C
CH3
O
CH3CH2MgBr
D
ether, 25 oC
H3 C
H
O
2 CH3CH2MgBr
E
ether, 25 oC
H3C
SCH3
HO
H3 C
HO
H3C
HO
H3C
HO
H3C
HO
H3 C
CH2CH3
CH2CH3
CH2CH3
CH2CH3
CH3
CH2CH3
H
CH2CH3
CH2CH3
CH2CH3
You cannot use
Grignard
reagents in the
presence of
acidic functional
groups!
3A
ANSWER KEY
5. Which of the compounds listed below would be a major product of the following twostep reaction sequence?
OCH3
1) CH3COCl,
AlCl3, CH2Cl2
2) NaOH (aq.), 3 I2
OCH3
OCH3
OCH3
OCH3
OCH3
COOH
CH2COOH
CH2COOH
A
B
COCH3
COOH
C
D
E
Friedel-Crafts acylation will occur either ortho or para to the methoxy group, forming a
methyl ketone. Methyl ketones are converted to carboxylic acids by the iodoform
reaction.
3B
ANSWER KEY
2.222 In-class Quiz #3 Monday March 4, 2002.
CIRCLE the correct response for each of the following questions.
1. Which of the following pairs of reagents could react to form an ENAMINE product?
O
+
A
H3C
O
H
N
B
H3C
OH
CH3
O
H3C
O
O
CH3
H3C
+
H3C
CH3 H3C
CH3
O
CH3
D
CH3
H
N
+
H3C
H3C
CH3
O
C
H
N
+
E
N
+
H3C
CH3
O
H3C
CH3
NH2
Enamines are formed when secondary amines react with enolizable aldehydes or
ketones. Carboxylic acids simply protonate amines and are not nucleophilically attacked
by them. Secondary amines will react with anhydrides to form amides, while the final
two options both lead to no reaction.
2. Which of the following carbonyl compounds is most acidic at its C-α position?
H
A
O
O
O
Br
H3CO
H2N
B
O
O
C
D
H
E
Aldehydes typically have pKas for removal of the C-α hydrogens on the order of 16-17 as
we mentioned in class. Ketones are less acidic, and carboxylic acid derivatives even less.
Amide and ester pKa values at C-α are about 25. The addition of an electron
withdrawing group such as a halogen will certainly increase the acidity of an ester but
not by 8 orders of magnitude.
3B
ANSWER KEY
3. Which of the following benzene derivatives would be nitrated most rapidly by
HNO3/H2SO4?
CH3
Br
H3C
OAc
H3C
H3C
A
B
CN
OCH3
H3C
H3C
E
D
C
All these rings have a mildly activating methyl group, plus one other group which is
either activating or deactivating. The methoxy group is the strongest activator, and thus
this compound will be the one nitrated fastest. The sequence of letters attached to the
answers was incorrect as you can see – don’t worry about that.
4. Which of the following reactions would NOT proceed as written? A mild acidic workup
can be assumed to follow each proposed reaction.
O
2 CH3MgBr
A
H3C
OCH3
O
ether, 25 oC
2 CH3MgBr
B
ether, 25 oC
H3CH2C
OCH3
O
2 CH3MgBr
C
ether, 25 oC
HO
CH2CH3
O
2 CH3MgBr
D
ether, 25 oC
H3 C
H
O
2 CH3MgBr
E
ether, 25 oC
H3C
SCH3
HO
H3 C
HO
H3C
HO
H3C
HO
H3C
HO
H3 C
CH3
CH3
CH2CH3
CH3
CH3
CH2CH3
H
CH3
CH3
CH3
You cannot use
Grignard reagents
in the presence of
acidic functional
groups!
3B
ANSWER KEY
5. Which of the compounds listed below would be a major product of the following twostep reaction sequence?
1) CH3COCl,
AlCl3, CH2Cl2
2) NaOH (aq.), 3 I2
N
COCH3
COOH
CH2COOH
HOOCH2C
N
A
COOH
N
N
N
N
B
C
D
E
Friedel-Crafts acylation will occur ortho or para to the strongly activating
dimethylamino group. The resulting methyl ketone is cleaved to the carboxylic acid by
the iodoform reaction.
3C
ANSWER KEY
2.222 In-class Quiz #3 Monday March 4, 2002.
CIRCLE the correct response for each of the following questions.
1. Which of the following pairs of reagents could react to form an IMINE product?
O
O
H
N
A
H3C
CH3 + H3C
O
CH3
H3C
+
H2N
CH3
Cl
O
CH3
C
H3C
+
B
D
N
CH3 H3C
+
H3C
CH3
O
H3C
CH3
NH2
O
E
+
H3C
H2 N
CH3
CH3
Imines are formed when primary amines react with ketones or aldehydes. Secondary
amines will form enamines if the carbonyl compound is enolizable. Tertiary amines do
not form adducts with carbonyl compounds, while the reaction of a primary or secondary
amine with an acid chloride will lead to an amide.
2. Which of the following carbonyl compounds is least acidic at its C-α position?
O
O
H
A
B
C
Br
H
H3CO
H2N
O
O
O
D
E
The acidity at C-α in carbonyl compounds was discussed in class and also in the
textbook. We noted that carboxylic acid derivatives were all less acidic than ketones or
aldehydes, and that this paralleled their decreased positive charge character at the
carbonyl carbon. Amides are much less positively charged than are esters because the
nitrogen is a better π donor than is oxygen.
3C
ANSWER KEY
3. Which of the following benzene derivatives would undergo nitration by HNO3/H2SO4 the
slowest?
CH3
Br
H3C
OAc
H3C
H3C
A
B
CN
H3C
OCH3
H3C
E
D
C
All these compounds have at least one weakly activating methyl group, so their relative
reactivities will be controlled by whether the other group is an activator or a deactivator.
The cyano group is the strongest deactivator in this list of compounds. NB: the letters
attached to each compound are out of sequence – don’t worry about that.
4. Which of the following reactions would proceed as written? A mild acidic workup can
be assumed to follow each proposed reaction.
O
2 CH3MgBr
A
H3C
ether, 25 oC
OH
O
2 CH3MgBr
B
HO
CH3
H3C
CH3
HO
CH2CH3
o
ether, 25 C
H3CH2C
OCH3
O
CH3MgBr
C
OH
H3C
CH3
HO
CH3
o
ether, 25 C
H3C
OH
H3C
CH3
CH3
O
D
ether, 25 oC
H3C
OCH3
O
H3C
ether, 25 oC
SCH3
CH3
O
CH3MgBr
E
H3C
O
CH3MgBr
H3C
CH3
In A and C, the
carbonyl compound
contains an acidic
functional group.
Grignard reagents
are incompatible
with acidic groups.
In D and E, the
problem is that you
cannot get a single
addition to an ester
or a thiolester – the
ketone that is
formed is more
reactive and will
add a second
Grignard.
3C
ANSWER KEY
5. Which of the compounds listed below would be a major product of the following twostep reaction sequence?
O
OCH3
1) CH3COCl,
AlCl3, CH2Cl2
2) NaOH (aq.), excess I2
O
O
O
OCH3
OCH3
OCH3 HO2C
HO2C
OCH3
OCH3
CO2H
HO2C
HO2C
O
A
B
C
O
D
E
Friedel-Crafts acylation will occur either ortho or para to the methoxy group, and meta
to the ketone group. Any of these structures could be derived from the initial acylated
product. However, the second step will convert both methyl ketones to carboxylic acids,
and so only response C is correct.
4A
NAME:
STUDENT NO:
2.222 In-class Quiz #4 Friday March 15, 2002.
CIRCLE the correct response for each of the following questions.
1. Which of the following proposed reactions is CORRECT?
O
O
H 3C
(CH3)2CHMgBr
CuI
Ether
H3 C
A
(workup with NH4Cl aq.)
CH3
CH3
O
O
LDA
o
THF, -78 C
H3 C
B
H3CH2C
then CH 3CH2I
O
CH 3
C
OCH3
H 3C
O
NaOCH3
CH 3OH
O
CH3
H3CO
OCH 3
O
O
O
O
D
+
NaOH
H2O, heat
H
O
O
E
OH
O
Mg, ether
Br
OCH3
(workup with NH4Cl aq.)
OCH3
The first reaction is correct – conjugate addition of a Grignard in the presence of copper(I)
iodide. The second reaction should generate the kinetic enolate, but the product shown would
arise from alkylation of the thermodynamic enolate. In C, the product shown lacks the acidic
hydrogen between the two carbonyls – recall that enolization is what provides the driving force
for this reaction to go to completion. Reaction D is a crossed aldol, but you would not get
enolization of the ketone in the presence of the aldehyde. In reaction E, the conditions would
suggest formation of a Grignard reagent, but you can’t form a Grignard from a molecule that
contains a C=O group.
4A
2. Which of the following anions is the strongest base?
O
CH3O
(CH3)2N
A
B
O
O
HO
C
D
E
The dialkyl amide ion shown in B is the most basic. Recall that we use the diisopropylamide ion
in LDA to quantitatively deprotonate carbonyl compounds – this indicates that such ions are
significantly more basic than enolates (option E). The carboxylic acid in C is the conjugate base
of a moderately strong acid, and is thus a weak base. Hydroxide (D) and methoxide (A) have
similar basicities, and are the conjugate bases of water and methanol respectively. Alcohols are
quite a bit more acidic than amines, which we can understand in terms of oxygen’s greater
electronegativity over nitrogen.
In terms of pKa, the following table may be helpful. Remember that the strongest acid
corresponds to the weakest conjugate base and vice versa:
Compound Type
Carboxylic Acids
Alcohols
Water
Carbonyl α-positions
Amines
pKa
4 to 5
17 to 20
14
17 to 25
ca. 42
3. What set of reagents would be required to complete the following transformation?
H3CO
OCH3
HO
H
O
?
HO
H
A. LiAlH4, THF then H2O.
B. KMnO4, H2O/NaOH.
C. SOCl2, heat.
D. H2SO4, H2O.
E. NaOCH3, CH3OH.
This reaction is taken from the textbook, see page 613. Acetals are hydrolysed to aldehydes by
aqueous acid.
4A
4. Which of the following reaction processes involves cationic intermediates?
A. Epoxidation of alkenes by Peroxycarboxylic acids.
B. Formation of cyclopropyl rings using metal carbenoids.
C. Hydrogenation of alkenes in the presence of Pt catalysts.
D. Nitration of benzene with nitric acid and sulfuric acid.
E. Reduction of ketones to secondary alcohols with NaBH4.
Cationic intermediates are usually associated with multistep reactions occurring under acidic
conditions. Epoxidation by peroxyacids is a concerted reaction which has no intermediates.
Cyclopropanation by carbenoids (Simmons-Smith reaction) is also concerted. Hydrogenation on
a catalyst is basically a type of free radical reaction, while reduction by borohydride involves
basic, anionic intermediates. Electrophilic aromatic substitution (D) proceeds via the NO2+ ion
and the arenium ion intermediates.
5. Which of the following organic compounds could be used in making a Wittig Reagent (a
phosphonium ylide)?
O
Br
P
P
O
A
B
C
D
E
Wittig reagents are prepared from an alkyl triphenylphosphonium ion, which is made by SN2
displacement of a primary or secondary alkyl halide by triphenylphosphine (B). The tertiary
halide (A) cannot undergo this reaction. Acetone (C) may be attacked by a Wittig reagent, but it
is not involved in making the actual reagent. Triphenylphosphine oxide (D) is a product from
Wittig reagents, as is the alkene (E).
4B
NAME:
STUDENT NO:
2.222 In-class Quiz #4 Friday March 15, 2002.
CIRCLE the correct response for each of the following questions.
1. Which of the following proposed reactions is INCORRECT?
O
O
H 3C
H3 C
(CH3CH2 )2CuLi
A
Ether
(workup with NH4Cl aq.)
CH3
O
B
O
LDA
THF, -78 oC
H 3C
H 3C
CH2CH 3
then CH3CH2I
O
C
OCH3
OH
NaOCH3
CH 3OH
O
H3CO
OCH3
O
O
O
O
NaOH
H2O, heat
+ H
D
O
O
E
OH
O
Zn, ether
Br
OCH3
(workup with NH4Cl aq.)
OCH 3
Option C is incorrect – the product would not arise from Dieckmann cyclization of a diester but
instead would have to come from a reaction involving an aldehyde as electrophile. All of the
other reactions are straightforward examples of processes covered in class and/or in the text.
4B
2. Which of the following compounds is the weakest Bronsted acid?
O
CH3OH
A
O
OH
(CH3)2NH
B
H2O
C
D
E
The weakest acid corresponds to the strongest conjugate base. See the answer to question 2 on
Quiz 4A for details.
3. What set of reagents would be required to complete the following transformation?
O
O
O
?
H3CO
H
H3CO
H3CO
OCH3
H
A. NaBH4, H2O.
B. CH3OH, H2SO4 (cat.).
C. NaOCH3, CH3OH.
D. H2SO4, H2O.
E. SOCl2, heat.
The product is an acetal, the result of nucleophilic attack on an aldehyde by two molecules of
alcohol. This reaction is only favourable under acidic conditions. See the textbook pages 609613.
4. Which of the following reaction processes does not involve cationic intermediates?
A. Addition of Br2 to alkenes in CH2Cl2 solvent.
B. Nitration of benzene with nitric acid and sulfuric acid.
C. Conversion of terminal alkynes to methyl ketones using Hg(OAc)2/H2SO4.
D. Epoxidation of alkenes by Peroxycarboxylic acids.
E. Diazotization of arylamines with nitrous acid.
Epoxidation by peracids is a concerted cycloaddition process that has no intermediates at all.
All other reactions are multi-step processes. Addition of Br2 involves the cyclic bromonium ion,
a 3-centre, 2-electron cation. Nitration of benzene is an electrophilic aromatic substitution that
proceeds via an arenium ion. Hydrolysis of alkynes to ketones in the presence of Hg(II) involves
several cationic species, including a cyclic mercurinium ion. Diazotization with nitrous acid
involves several cationic species as well.
4B
5. Which of the following organic compounds could be the product of a Wittig Reaction?
O
P
Br
A
Br
B
C
D
Wittig reactions form alkenes by the reaction of a phosphonium ylide with a carbonyl.
E
4C
NAME:
STUDENT NO:
2.222 In-class Quiz #4 Friday March 15, 2002.
CIRCLE the correct response for each of the following questions.
1. Which of the following proposed reactions is INCORRECT?
O
O
H3CH2C
H3 C
H 3C
NaOH/H2O
A
CH3 CH2I
HO
O
H 3C
(CH3CH 2)2CuLi
B
H3 C
Ether
(workup with NH 4Cl aq.)
O
O
C
+
O
NaOH
H2 O, heat
H
O
O
Br2, PBr3
D
Br
OH
(workup with H2O)
OH
O
O
NaOCH3
CH3OH
E
O
Reaction B is incorrect – lithium dialkyl cuprates are soft nucleophiles that add 1,4 to α,βunsaturated ketones.
2. Which of the following compounds is the weakest base?
O
CH3O
A
O
O
B
Cl
(CH3)2NH
C
D
E
Chloride ion is the conjugate base of hydrochloric acid, an extremely strong acid. Thus, chloride
is the weakest base of this set. See the answer to question 2 on Quiz 4A for details.
4C
3. What set of reagents would be required to complete the following transformation?
O
O
O
?
H3CO
H
H3CO
OH
A. NaBH4, H2O.
B. CH3OH, H2SO4 (cat.).
C. H2NNH2, KOH, heat.
D. Zn, HCl.
E. H2 (g), Pt.
The reaction shown is a reduction, formally involving the addition of a hydride nucleophile to
the aldehyde. Sodium borohydride will reduce aldehydes or ketones, but will have difficulty
reducing esters. Option B is not a reduction process. Options C and D are specific reactions for
the complete removal of a C=O group, leading to a CH2 group. Option E will not reduce
saturated aliphatic aldehydes, although it could reduce a benzylic carbonyl group.
4. Which of the following reaction processes involves bridged cationic intermediates?
A. Addition of Br2 to alkenes in CH2Cl2 solvent.
B. Friedel-Crafts acylation of aromatic rings.
C. Conversion of terminal alkynes to geminal dibromides with HBr.
D. Conversion of alkenes to vicinal cis diols using OsO4.
E. Diazotization of arylamines with nitrous acid.
Addition of Br2 to alkenes proceeds through a cyclic bromonium ion, in which a bromine atom
bridges the two alkene carbons in an electron-deficient 3-centre, 2-electron structure. The other
reactions do not involve bridged structures – in fact reaction D does not involve cationic
intermediates at all.
5. Which of the following compounds could react with a Wittig Reagent?
O
O
O
P
NH2
A
OCH3
B
C
Wittig reagents react with aldehydes or ketones to form alkenes.
D
E
5A
ANSWER KEY
2.222 In-class Quiz #5 Monday April 1, 2002.
CIRCLE the correct response for each of the following questions.
1. What would be the expected product of the following reaction?
O
O
?
KOH (aq.)
(acid workup)
OH
COOH
CHO
O
COOH
A
B
OH
COOH
C
O
D
E
This is a Benzilic acid rearrangement, in which a 1,2-diketone is converted to an α-hydroxyacid
on treatment with hydroxide ion. In fact, this example is identical to the one presented in the
textbook and in class, except that the two phenyl rings are linked by an additional bond.
2. Which of the following compounds could NOT result from a C–C bond forming reaction
of an electrophile with a malonate ester enolate?
O
O
O
O
O
H3CO
H3C
OCH3
CH3
A
H3CO
H
B
OCH3
O
O
H3CO
OCH3
C
O
O
H3CO
H
D
OCH3
O
H3CO
OCH3
E
Options A-C are simply alkylation products – resulting from SN2 attack of an enolate on
appropriate primary or secondary alkyl halides. Option E would arise from the Knoevenagel
condensation, which is just a special case of the Aldol condensation. Option D could not be
formed by enolate chemistry because you CANNOT displace a leaving group from a phenyl ring
in an SN2 process or any other nucleophilic attack involving an enolate.
5A
ANSWER KEY
3. Which of the following compounds could arise from alkylation of the KINETIC enolate
of an unsymmetrical ketone?
O
O
O
H3C
H3C
CH3
OH
O
CH3
CH3
A
B
C
D
E
The kinetic enolate is formed by irreversible deprotonation of a ketone at low temperature by a
very strong base. It will be formed on the less-substituted side of the ketone. Option B could not
arise from enolate chemistry of the type we have been discussing. Options C and D are the
result of thermodynamic enolates, while Option E is simply irrelevant. It would arise from a
Claisen rearrangement and not from a ketone enolate at all.
4. What would be the product of the following rearrangement reaction, an example of the
Hofmann Rearrangement? (HINT: Look closely at the reaction conditions.)
O
Br2
NaOCH3/CH3OH
NH2
?
O
O
C
H
N
NH2
NH2
N
Br
OCH3
O
Br
A
B
C
D
E
In the Hofmann Rearrangement, an amide is transiently converted to an N-bromoamide. This
rearranges to an isocyanate (Option A). If the reaction is conducted in aqueous NaOH, the
isocyanate is attacked by OH- to form an unstable carbamic acid, which loses CO2 to form an
amine (Option C). However, in this example the solvent and base are methanol and sodium
methoxide – there is no water present. Thus, isocyanate A will be attacked by methoxide to form
the carbamate D, which cannot decarboxylate. The hint was directing you to think about the
mechanism and the role of the base in this reaction.
5A
ANSWER KEY
5. Which set of reagents would be the best choice to complete the following reaction?
?
O
OH
H
a. NaOH/H2O (acid workup)
b. (CH2=CH)2CuLi, Ether (acid workup)
c. [(C6H5)3P+–CH3]Br–/n-BuLi/THF
d. CH2=CHMgBr, Ether (acid workup)
e. i) SOCl2; ii) CH2=CH2; iii) LiAlH4/THF (acid workup)
The reaction shown is nucleophilic addition to the carbonyl group, by a 1,2-addition
pathway. This requires a HARD nucleophile such as a Grignard reagent.
5B
ANSWER KEY
2.222 In-class Quiz #5 Monday April 1, 2002.
CIRCLE the correct response for each of the following questions.
1. What would be the expected product of the following reaction?
O
?
CH3CO2H
H2O2/H2SO4
O
O
COOH
OH
COOH
O
COOH
A
B
O
C
D
E
This is a Baeyer-Villiger rearrangement, which converts a ketone into an ester. Option A is a
reduction product, whereas these reaction conditions are clearly oxidizing. Option B and C both
have more carbons than the starting material and are thus impossible under these conditions.
Note that Option D would require epoxidation of an aryl ring, with destruction of the aromatic
system, which will NOT happen with peroxycarboxylic acid reagents.
2. Which of the following compounds could result from an intramolecular condensation
reaction of a diester, under basic conditions?
H3C
O
O
O
O
OCH3
OCH3
O
OCH3
OCH3
O
O
OH
O
A
B
C
D
E
When a diester undergoes an intramolecular condensation reaction (a Dieckmann
condensation), the product is a β-ketoester. The enolate formed from one ester group attacks the
carbonyl group of the other, displacing an alcohol and forming a C-C bond. Moreover, the
process is driven by the highly exothermic deprotonation of the ketoester product and thus
products such as Option A are NOT observed. Options C and E would both have to arise from a
crossed Aldol between an ester and an aldehyde. Option B is not a condensation product at all.
5B
ANSWER KEY
3. Which of the following compounds could arise from alkylation of the
THERMODYNAMIC enolate of an unsymmetrical ketone?
O
O
O
H3C
OH
O
H3C
CH3
CH3
A
B
C
D
E
The thermodynamic enolate is formed by equilibrium (reversible) deprotonation of a ketone by a
relatively weak base. It will be formed on the more-substituted side of the ketone, so that the
enolate C=C bond will be more highly substituted. Note that Option A would arise from the
kinetic enolate (see Quiz 5A), while Option C would not come from an unsymmetrical ketone.
4. What would be the product of the following rearrangement reaction, an example of the
Beckmann Rearrangement?
OH
N
H2SO4, Heat
N
C
O
O
O
NH2
N
H
A
?
B
C
O
N
H
E
D
In the Beckmann rearrangement, the group anti to the OH in the starting oxime will be the one
that migrates, to form an amide product.
5. Which set of reagents would be the best choice to complete the following reaction?
?
O
O
CH3
CH3
a. NaOH/H2O (acid workup)
b. (CH2=CH)2CuLi, Ether (acid workup)
c. [(C6H5)3P+–CH3]Br–/n-BuLi/THF
d. CH2=CHMgBr, Ether (acid workup)
e. i) SOCl2; ii) CH2=CH2; iii) LiAlH4/THF (acid workup)
The product arises from 1,4-addition of a nucleophile to the starting enone. This requires a
SOFT nucleophile such as a lithium dialkyl cuprate.
5C
ANSWER KEY
2.222 In-class Quiz #5 Monday April 1, 2002.
CIRCLE the correct response for each of the following questions.
1. What would be the expected product of the following reaction?
OH
HO
?
H2O/H2SO4
O
CHO
CH2OH
CH2
OH
O
A
B
C
D
E
This is an example of a Pinacol Rearrangement. Note that the reaction leads to contraction of
the relatively strained cycloheptyl ring into the less-strained cyclohexyl system.
2. Which of the following compounds could result from an intramolecular condensation
reaction of a diketone, under basic conditions?
O
O
O
O
OH
O
O
O
A
B
C
D
E
Options A and C would arise from reactions of a ketone enolate with an aldehyde, while option
D is the result of attack of the ketone enolate on an ester. Option B is not the product of a
condensation of a diketone, although it could arise from attack of a ketone enolate on an ester.
5C
ANSWER KEY
3. Which of the following compounds COULD NOT arise from alkylation of a betaketoester, followed by decarboxylation?
O
H3C
H3C
O
O
CH3
O
O
H3C
CH3
A
CH3
B
C
D
E
You cannot obtain tetrasubstituted carbon centres from the acetoacetic ester synthesis. When
decarboxylation occurs, the C-C bond to the carboxyl group is replaced by a C-H bond.
4. What would be the product of the following rearrangement reaction, an example of the
Claisen Rearrangement? (HINT: Look at each possibility carefully!).
heat
O
?
O
OH
O
O
O
A
B
C
E
D
Options A and E are identical to the starting compound, although drawn in different
orientations. In the Claisen rearrangement, an allyl vinyl ether is converted to an unsaturated
carbonyl compound – the driving force is the conversion of a pair of C-O single bonds to the
C=O double bond. Only Option B has the correct functional group pattern.
5. Which set of reagents would be the best choice to complete the following reaction?
?
O
CH2
CH3
a. NaOH/H2O (acid workup)
b. (CH2=CH)2CuLi, Ether (acid workup)
c. [(C6H5)3P+–CH3]Br–/n-BuLi/THF
d. CH2=CHMgBr, Ether (acid workup)
e. i) SOCl2; ii) CH2=CH2; iii) LiAlH4/THF (acid workup)
This is a Wittig reaction – conversion of a ketone into an alkene.
CH3