Download CHEMISTRY 314-01 MIDTERM # 1 – answer key February 10, 2009

CHEMISTRY 314-01
MIDTERM # 1 – answer key
February 10, 2009
Statistics:
• Average: 80 pts (80%);
• Highest: 97 pts (97%); Lowest: 47 pts (47%)
• Number of students performing at or above average: 29 (62%)
• Number of students performing below 55%: 3 (6%)
1.
(8 pts) Mark as true (T) or false (F) the following statements. Do not explain!
• (F) All deactivating substituents are meta-directing;
• (T) The Wolff – Kishner reduction requires basic conditions;
• (T) More shielded signals in an NMR spectrum have lower chemical shift values;
• (T) Coupling constants in NMR are independent of the operating frequency of the instrument;
• (T) The molecular ion peak in mass spectra is the peak with highest m/z ratio;
• (T) All charged species are observed in mass spectrometry;
• (T) Organometallic compounds are Lewis bases;
• (T) Organometallic compounds are Brønsted bases;
2.
Circle ALL that apply:
A.
(3 pts) Carbenes are:
a. Electron-deficient;
b. Charged;
B.
C.
D.
3.
4.
(3 pts) The following reagents cannot be used in Friedel-Crafts acylations:
a. Alkyl halides;
b. Aryl halides;
Electrophilic;
Nucleophilic
c.
d.
Acid chlorides;
Acid anhydrides;
(3 pts) The following are independent of the operating frequency of the NMR:
c.
a. Gyromagnetic ratio;
d.
b. Chemical shift;
Integral intensity;
Accidental equivalence;
(3 pts) The following compounds are converted to alcohols, when reacted with a Grignard reagent, followed by
acidification:
a. Alcohols;
c. Esters;
b. Ketones;
d. Alkyl Halides;
(3 pts) Provide the structure and name for each of the following:
A.
An activating σ-acceptor, π-donor group;
More than one possibility. Example: -NH2, amino group
B.
A σ-acceptor, π-acceptor group;
More than one possibility. Example: -CN, nitrile group
C.
A σ-acceptor group;
More than one possibility. Example: -CF3, trifluoromethyl group
(4 pts) Provide the correct structural formula for each of the following substances:
B.
A.
Phenyllithium;
Li
5.
c.
d.
Isopropylmagnesium bromide;
MgBr
(2 pts) In each of the following pairs of compounds, indicate (circle) the compound, which will react faster with the indicated
reagent. Do not explain!
A.
Toluene or nitrobenzene, with sulfuric acid upon heating;
6.
or
O
B.
, with a nitric/sulfuric acid mixture;
(3 pts) For each of the following structures, indicate (e.g. circle) the ring carbon at which an SEAr reaction is expected to occur
most readily.
NO2
NH2
OH
CH3
CH3
CH3
Br
I
7.
(4 pts) Draw the major product when each of the following compounds is treated with HNO3 and H2SO4.
OCH3
OCH3
CH3
O2N
CH3
NO2
NO2
Cl
H3CO
8.
C
O
H3CO
C
Cl
NO2
O
Write and complete a chemical equation for each of the following reactions:
(3 pts) Naphthalene and sulfuric acid, upon heating;
A.
SO3H
H2SO4
Δ
B.
(3 pts) Propylmagnesium bromide with benzaldehyde, followed by acidification;
OH
O
MgBr
1)
H
2) H3O+
C.
9.
(3 pts) Cyclobutene with diiodomethane, in the presence of Zn(Cu) alloy;
CH2I2
Zn(Cu),
ether
(12 pts) Indicate the principal organic product of each of the following reactions. If o-,p-mixture is expected, write both
products.
SO3H
O
Cl
Cl2
O
O
+
Fe
H2SO4
heat
Cl
SO3H
O
O2N
NO2
HNO3
O2N
NO2
H2SO4
heat
O
O
O
O
O
O
AlCl3
NO2
+
O
O
CH3
O
CH3
Cl
CH3CH2OH
N
H
H3PO4
CH3
CH3
N
H
SnCl4
O
O
O
H2NNH2
1) HC CNa , liq. NH3
KOH , heat
2) H3O+
CHCl3
(CH3CH2)2CuLi
I
HO
Cl
KOC(CH3)3
Cl
t-butanol
10. (4 pts) Suggest a suitable combination of organometallic compound and a carbonyl compound or ester that could be used to
generate the following alcohols:
ether
OH
O
OH
+
MgBr
+
CH3Li
O
11. (4 pts) Identify the reagents in the following scheme:
Cl
Cl
AlCl3 O
AlCl3
1) CH3MgBr,
ether
2) H3O+
O
OH
12. (4 pts) Suggest a detailed mechanism for the following reaction.
H3O+
H3O+
+ H3O+
H2O
H
13. Suggest a detailed synthetic sequence for the preparation of each of the following molecules, starting from benzene, and using
any other necessary reagents (For convenience, assume that you can separate o- and p-isomers):
NO2
COOH
A.
(4 pts)
Br
1) KMnO4
AlCl3
O
O
Cl
2) HCl
OH
Br2
NO2 O
OH
Fe
HNO3
OH
H2SO4
Br
Br
SO3H
B.
Cl
(4 pts)
Cl
O
O
O
Cl2
Zn/Hg
H2SO4
AlCl3
Fe
HCl
heat
HO3S
Cl
Cl
Cl
14. (5 pts) Compound A (molecular formula C10H12O) was treated with Zn/Hg in HCl, to yield compound B (molecular formula
C10H14). Based on the 1H NMR spectra of A and B, shown below, propose the structures for A and B.
1
H NMR spectrum of A
3.56 3.54 3.52 3.50 3.48 3.46 3.44 3.42
1H
8.5
8.0
7.5
7.0
6.5
6.0
5.5
5.0
4.5
4.0
3.5
3.0
2.5
O
Zn/Hg
HCl
Δ
Compound A
Compound B
2.0
1.5
1.0
1
H NMR spectrum of B
2.65
7.5
7.0
2.60
2.55
6.5
2.50
6.0
2.45
2.40
5.5
2.35
5.0
1H
4.5
4.0
3.5
3.0
2.5
2.0
1.5
1.0
15. (4 pts) Circle the compounds that would exhibit one singlet in their 1H NMR spectra. Do not explain!
O
O
H3CCH3
O
O
H3C
CH3
Br
Br
Br
Br
Br
Br
Br
16. (4 pts) In each of the cases below, propose a structure matching the molecular formula that would show only one peak in its 1H
NMR spectrum:
A.
C5H12;
CH3
H3C C CH3
CH3
B.
C5H10;
17. (3 pts) Briefly explain how you could use 13C NMR to distinguish among the following isomeric structures with formula C4H8.
all C equivalent
one signal in 13C NMR
Four different C
4 signals in 13C NMR
Two different C
2 signals in 13C NMR
Three different C
3 signals in 13C NMR
18. (3 pts) Bromine is present in nature in the form of two equally abundant isotopes (79Br and 81Br). In class, we found out that this
gives rise, in the case of monobrominated compounds, to two signals for the molecular ion: M+ and (M+2)+, with equal relative
intensities. What do you expect for dibrominated compounds? How many peaks for the molecular ion and what will be their
relative intensity?
Answer: The allowed combinations are:
79
Br and 79Br, giving rise to M+
Br and 81Br OR 81Br and 79Br, giving rise to (M+2)+
81
Br and 81Br, giving rise to (M+4)+
79
One expects three peaks for the molecular ion. Since 79Br and 81Br are equally abundant, the intensity of the M+ and the (M+4)+
should be the same. However, the (M+2)+ signal should have double intensity, because the combination 79Br and 81Br is twice as
likely statistically. So the ratio of M+ : (M+2)+ : (M+4)+ is expected to be 1 : 2 : 1.
19. (4 pts) Propylbenzene and isopropylbenzene are constitutional isomers. One of these compounds has a mass spectrum with
prominent peaks at m/z 120 and 105 (Spectrum A). The other has a mass spectrum with prominent peaks at m/z 120 and 91
(Spectrum B). Which compound has which spectrum? Briefly explain!
Solution: In both cases the most favored mode of fragmentation will be to produce a benzyllic type of carbocation.
+
m/z 120
Spectrum B
CH3CH2
m/z 91
m/z 120
Spectrum A
CH3
+
m/z 105
20. (3 pts) BONUS PROBLEM (In order to receive credit for this problem, it has to be solved entirely!!). When phenylboronic
acid is nitrated, the major product is m-nitrophenylboronic acid. Provide a structural explanation (e.g. resonance analysis) for the
meta-directing effect of the boronic acid functional group (– B(OH)2).
HO
B
OH
HO
B
OH
HNO3
H2SO4
NO2
Solution: Boron is inherently electron-deficient (only 3 valence electrons, hence only three bonds). Thus, the boronic acid
functional group manifests itself as a strong p-acceptor, which can be demonstrated by the following resonance structures:
HO
B
OH
HO
B
OH
HO
B
OH
HO
B
OH
A positive charge appears on several of the ring carbons, indicating that the benzene ring is deactivated. In particular, those
charges show explicitly at the o- and p-positions, from which we should conclude that they would be most disfavored as a
site for attachment of the electrophile. Thus, overall, the boronic acid group is deactivating and m-directing.