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CHEMISTRY 313-03
MIDTERM # 1 – answer key
October 05, 2010
Statistics:
• Average: 73 pts (73%);
• Highest: 99 pts (99%); Lowest: 31 pts (31%)
• Number of students performing at or above average: 61 (52%)
• Number of students performing below 55%: 8 (7%)
• Number of students performing at or above 90%: 17 (15%)
1.
(12 pts) Mark as true (T) or false (F) the following statements. Do not explain!
• (T) Single bonds are always σ-bonds;
• (F) Resonance structures are always in a state of rapid equilibrium;
• (F) The gauche conformation of butane is a transition state;
• (T) The chair conformation of cyclohexane is a global minimum;
• (F) Cyclopropane does not have any torsional strain;
• (F) Molecules with polar bonds always polar;
• (T) All Arrhenius acids are also Brønsted acids;
• (F) Lewis acids are proton donors;
• (T) Increasing oxidation number indicates an oxidation process;
• (T) Intermediates are not formed in concerted reactions;
• (T) The Hammond postulate relates the energies and structures of two adjacent species on the potential energy profile;
• (T) Carbocations are electron-deficient species;
2.
(5 pts) Provide the structural formula for each of the following molecules.
3.
(2 pts) Provide a structure for each of the following compounds:
a. Bicyclo[1,1,0]butane.
b.
Spiro[2,2]pentane.
4.
(2 pts) Draw curved arrows to rationalize the following conversion.
5.
(2 pts) Draw the structure of the carbocation produced from 2-methyl-2-butanol.
6.
(6 pts) For each of the following molecules, complete the Lewis structure and provide two additional valid resonance forms.
Use the curved arrow formalism to show the flow of electrons. Rank the resultant resonance structures.
a.
Acetonitrile oxide;
b.
1
2
O
C
O
C
3
Methyl acetate;
H3C
O
CH3
H3C
1
7.
H3C C N O
H3C C N O
H3C C N O
O
CH3
H3C
3
O
C
O
CH3
2
(4 pts) Each of the following species has a resonance structure that is higher ranking than the one shown. Show that structure
and very briefly explain why it is higher ranking.
N
CH3
N
CH3
more bonds
O
O
more bonds,
less charge separation
8.
(6 pts) For some of the substances listed below the acid-base equilibrium for the reaction with KOH is shifted to the left,
while for others it is shifted to the right. Using Table 1.8, decide the equilibrium shift for each structure.
9.
(4 pts) Predict the shift of equilibrium (to the left or right) for the following acid – base reactions.
HC CH
+
NH2
+
HCN
HC C
+
+
NH3
to the right
CN
to the left
N
H
N
10. (6 pts) Label the reactants in the following acid – base reactions as Lewis acids or Lewis bases.
Cl
Lewis base
+
AlCl3
Lewis acid
Cl
AlCl3
H
H
H
+
O
O
Lewis acid
Lewis base
CH3NH2
+
N
Cl
H
H
+
Cl
Lewis acid
Lewis base
11. (6 pts) For each of the following reactions, determine if it is a reduction, oxidation or not redox with respect to the organic
compound.
H3C
+
O
hν
Cl2
H3C
oxidation
O
CH3OH
H3O+
OH
O
C
+ HCl
Cl
MCPBA
CH3
H3C
O
C
OCH3
not redox
CH3
oxidation
O
12. (4 pts) Draw an example of each of the following classes of compounds:
a.
An amine with 4 C-atoms;
b.
An acid chloride with a 5-membered ring;
c.
A carboxylic acid with 5 C-atoms;
13. (4 pts) For each of the following compounds, draw an isomer that has the same functional group.
More than one possibility exists in each case!
SH
O
O
SH
O
O
14. (7 pts) Circle and name all functional groups in the following structures.
amine amide
sulfide
S
sulfide
NH2 H
N
benzene
ring
carboxyl
O
C
OH
NH2
S
O
alkene
N
O
amide
amine
methionine, an amino acid
O
C
OH
carboxyl
cephalexin, an antibiotic
15. (6 pts) Give the relationship between the following pairs of structures. There are four (4) possible relationships: same
compound, constitutional isomers, cis-trans isomers, not isomers (i.e. different molecular formula).
and
and
a.
Br
and
b.
not isomers
d.
same compound
Br
Br
Br
same compound
e.
same compound
and
O
O
and
HO
and
c.
HO
constitutional isomers
f.
cis-trans isomers
16. (4 pts) Draw the structure of the principle organic product of each of the following reactions.
OH
PBr3
SOCl2
Br
OH
heat
Cl
17. (6 pts) Using Newman projections, draw the conformations arising upon a full 360o turn (in 60o steps) around the C1 – C2
bond in isobutyl bromide. Represent the energy changes on a qualitative potential energy/dihedral angle diagram. Assign the
proper term (i.e. minimum, transition state, etc.) to each conformation.
See next page!!
Br
θ = 0o
CH3
Br
H
H
CH3
H
eclipsed,
one Br --- CH3
eclipsing
interaction
E
Br
H
CH3
H
staggered,
two Br --- CH3
gauche
interactions
H
H
CH3
Br
eclipsed,
one Br -- CH3
eclipsing
interaction
transition
state
H
H
CH3
Br
staggered,
one Br -- CH3
gauche
interaction
BrH
60o
CH3
H
H
CH3
H
staggered,
one Br --- CH3
gauche
interaction
transition
state
180o
H
H
CH3
H
eclipsed,
one Br --- CH3
eclipsing
interaction
transition
state
global
minimum
120o
CH3
Br
H
Br
eclipsed,
zero Br -- CH3
eclipsing
interactions
transition
state
θ = 360o = 0o
CH3
CH3
H
H
local
minimum
0o
θ = 300o
θ = 240o
CH3
CH3
H
CH3
H
θ = 180o
θ = 120o
θ = 60o
global
minimum
240o
300o
360o
θ
18. Consider menthol:
CH3
HO
a.
(3 pts) Draw the most stable chair conformation of menthol.
CH3
OH
H3C
a,a,a
OH
e,e,e
most stable conformation
b.
(4 pts) How many stereoisomers of menthol are possible? Draw their structures using a bold-and-dashed wedge
representation.
19. Consider 1-methylcyclohexanol and cyclohexylmethanol.
a. (2 pts) Does the energy below better describe the reaction of 1-methylcyclohexanol or cyclohexylmethanol with
HBr? Briefly explain!
The reaction with HBr would result in substitution and formation of the corresponding alkyl bromide. The process could
occur following either an SN1 or an SN2 mechanism. Based on the structure of each of the alcohols, one can make a
prediction on the mechanism of its reaction with HBr, as shown above to the right. The profile, on the other hand, definitely
reflects a stepwise reaction, i.e. one that follows an SN1 mechanism. The conclusion is that the profile must describe the
reaction of 1-methylcyclohexanol with HBr.
b.
(5 pts) Assign the correct chemical structure(s), corresponding to each minimum in the diagram.
A=
+
H Br
+
Br
E=
+ H2O
+
Br
OH
C=
+ H2O
G=
Br
OH2
20. (2 pts) BONUS PROBLEM (In order to receive credit for this problem, it has to be solved entirely!!). Ethylene glycol
(see below!) is unusual in that the gauche conformation is more stable than the anti conformation. Offer an explanation.
H
H H
HO C C OH
H H
ethylene glycol
intramolecular hydrogen bond
O
OH
H
H
H
OH
H
H
H
H
OH
anti conformation
H
gauche conformation
The gauche conformation of ethylene glycol is strongly stabilized by hydrogen bonding. Ethylene glycol is a good example
of a species whose conformational preference is affected by intramolecular hydrogen bonding.