Practice Exam #2
CHEM 255 – Organic Chemistry I
Prof. Bastin
Summer 2014
Name ___________________________
Provide clear, concise answers using unambiquous, carefully drawn structures (where
appropriate) for all of the questions. Good luck and enjoy!
1) _______/6 pts
2) _______/6 pts
3) _______/12 pts
4) _______/10 pts
5) _______/10 pts
6) _______/10 pts
7) _______/8 pts
8) _______/10 pts
9) _______/10 pts
10) _______/10 pts
11) _______/8 pts
Total: ________/100 pts
Bastin –Exam #2
Chem 255 - Summer 2013
1) (6 pts) Provide IUPAC names for the following compounds.
(a)
H 3C
H
(b)
CH 3
CH 3
2) (6 pts) Provide structures for the following compounds.
a) trans-1,3-dimethylcyclobutane
b) (R)-3-ethyl-4-methylhexane
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Bastin – Exam #2
Chem 255 - Summer 2013
3) (12 pts) Provide the absolute configuration (R/S) of each stereogenic center in each of the
following molecules.
(a)
(b)
(c)
H
Br
CH 3
H 3C
H
CH 3
H 3CH2C
(d)
H 3CH2CH2C
Cl
H
OH
CH2CH 3
(e)
OH
H
I
H
OH
Br
CH 3
(f)
4) (10 pts) Draw all the constitutional isomers for C4H8Cl2.
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Bastin – Exam #2
Chem 255 - Summer 2013
5) (10 pts) Given that a Me–Me eclipsing interaction costs 2.5 kcal/mol, a Me–H eclipsing
interaction costs 1.6 kcal/mol, a H–H eclipsing interaction costs 0.9 kcal/mol, and a Me–Me
gauche interaction costs 0.9 kcal/mol, draw an energy versus dihedral angle plot for the
conformations of 2,2,3-trimethylbutane about the C2–C3 bond.
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Bastin – Exam #2
Chem 255 - Summer 2013
6) (10 pts) (a) Indicate whether each molecule is chiral or achiral. (c) Provide the relationship
between all pairs (enantiomers, diastereomers, identical, configurational isomers).
H 3C
H 3C
H
H
CH 3
H
H
CH 3
A
H
B
CH 3
H
CH 3
CH 3
H
H
C
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CH 3
D
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Bastin – Exam #2
Chem 255 - Summer 2013
7) (8 pts) Which of the following are chiral? Indicate “chiral” by putting an “X” in the space
provided.
(a) A fork
_____
(b) A paper clip
_____
(c) A mirror
_____
(d) trans-1-Isopropyl-4-methylcyclohexane
_____
8) (10 pts) (a) Draw the curved arrow mechanism for the following reaction. (b) Indicate the
Bronsted-Lowry acid, base, conjugate acid, and conjugate base of the reaction. (c) Predict to
which side the equilibrium lies in the reaction and provide the Keq of the reaction. (d) Would
this reaction occur as written if water was the reaction solvent? What about diethyl ether? Be
sure to add any needed lone pairs to the structures. Just add the arrows and lone pairs to the
provided drawing.
a)
H2
C
H 3C
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H
O
+
NH2
H2
C
H 3C
O
+
NH3
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Bastin – Exam #2
Chem 255 - Summer 2013
9) (10 pts) Answer the following questions concerning cis-1,2-dimethylcyclohexane
a) Draw the two chair conformations and label both methyl groups as axial or equatorial.
b) Given the following 1,3-diaxial interaction energies, calculate theΔG of the two chair
conformations. A CH3–H 1,3-diaxial interaction costs 0.9 kcal/mol and a CH3–CH3 1,3diaxial interaction costs 3.6 kcal/mol?
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Bastin – Exam #2
Chem 255 - Summer 2013
10) ???(10 pts) In aqueous solutions, three forms of glucose are in equilibrium (shown below).
Given the following equilibrium ratios, calculate the relative energies of the three isomers at
25°C and sketch a reaction coordinate diagram. Is the transformation of α-D-Glucopyranose
into β-D-Glucopyranose exothermic or endothermic?
CHO
CH2OH
CH2OH
H
O
H
H
OH
OH
H
H
H
H
OH
H
OH
OH
OH
H
O
H
HO
H
OH
H
H
OH
OH
OH
H
OH
CH2OH
α-D-Glucopyranose (36.04%)
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D-Glucose (0.02%)
β-D-Glucopyranose (63.94%)
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Bastin – Exam #2
Chem 255 - Summer 2013
11) ???(8 pts) Which OH group in ascorbic acid (vitamin C, structure below) is the most acidic?
Explain (be sure to include structures of each conjugate base).
HO
O
HO
O
O
H
HO
OH
ascorbic acid
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OH
acetic acid
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Bastin – Exam #2
Chem 255 - Summer 2013
Some useful pKa (water) values
Compound
Alkanes/Alkenes
NH3
Alkynes
O
H3C
pKa
>40
38
25
25
Compound
HOOH
RN+H3
CH3SH
PhOH
pKa
12
~10
10
10
24
24
HCN
9
9
NH2
CH3CN
O
H3 C
O
OCH2CH3
O
R
O
C
H2
R
~19
CH3COOH
4.8
18
HF
3.2
(CH3)3COH and
(CH3)2CHOH
17
ClCH2COOH
2.9
O
17
HNO3
-1.3
16
16
16
RO+H2
H2SO4
HCl
-2
-3
-8
15
HBr
-9
13
HI
-10
[α ] sample
x100
[α ] enant
pure
α =
R
CH3
O
R
NH2
H3C
H
CH3CH2OH
H2 O
O
R
H
CH2
O
RO
O
C
H2
OR
Enantiomeric excess (ee) = %R - %S =
ΔG = −RT lnK
R=1.986 cal/mol•K
€
€
€
DU =# C −
€
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#H #X #N
−
+
+1
2
2
2
K=
α
cl
[ products]
[reac tan ts]
K eq = 10 ΔpKa = 10 (
K = C + 273
pKa( pdt )−pKa(react ))
€
10