Chemistry 217, Problem Set 5
Recommended Problems from the text: 4.2—4.4, 4.8, 4.11-4.17, 4.20-4.21, 4.23-4.24,
4.26-4.32, 4.37-4.38, 4.40-4.44, 4.47-4.49, 4.53-4.60, 4.70, 5.2-5.11
(1st ed.: 4.2-4.5, 4.10-4.16, 4.19-4.20, 4.22-4.24, 4.26-4.31, 4.36-4.37, 4.39-4.42, 4.454.47, 4.51-4.56, 4.57-4.58, 4.68, 5.2-5.11)
Supplemental Problems: Nomenclature: Klein, Ch. 5; Newman projections: Klein, 6.16.2 Cyclohexanes: Klein, Ch. 6.3-6.7; Stereochemistry: Klein, Ch. 7
1.
2.
Draw structures of the following molecules.
(a)
1-tert-butyl-2-methylcyclobutane
(b)
(c)
2-cyclohexyl-4,4-dimethylpentane
(d)
1-isopropyl-3-methylcyclopentane
1,1,2,2,3,3,-hexaethylcyclohexane
3-ethyl-2,4,6-trimethylheptane
(e)
Give the IUPAC names for the following compounds:
a)
1-ethyl-1-methylcyclohexane
c)
1-sec-butyl-2-ethylcyclopentane
e)
3-isopropyl-2,5-dimethylhexane
(b)
(d)
(f)
isopropylcyclodecane
1-cyclobutyl-3-methylpentane
cis-1-isopropyl-3-pentylcyclohexane
3-tert-butyl-1,5-diethylcycloheptane
g)
(h) 3-ethyl-2,4,4,5-tetramethylheptane
3.
Draw an energy diagram for all conformations of the following molecules when
they are rotated about the bond indicated with an arrow.
You may have drawn your Newman diagrams in different way, but the relative energies
should be the same and the number or ecliped/gauche interactions should be the same.
(a)
1 Me-Me eclipsed
1 Me-H eclipsed
1 H-H eclipsed
1 Me-Me eclipsed
1 Me-H eclipsed
1 H-H eclipsed
H3C CH3
H3C CH3
H
H3C
H
H
H
H
CH3
H
3 Me-H eclipsed
H CH3
2 Me-Me gauche
2 Me-H gauche
2 H-H gauche
H3C
H
H 3C
CH3
CH3
H
H
H
1 Me-Me gauche
4 Me-H gauche
1 H-H gauche
H3C
H
CH3
H
H
CH3
1 Me-Me gauche
4 Me-H gauche
1 H-H gauche
H
H
CH3
CH3
H
CH3
CH3
H
(b)
1 Et-Me eclipsed
1 Me-H eclipsed
1 H-H eclipsed
H3CH2C CH3
1 Et-H eclipsed
1 Me-Me eclipsed
1 H-H eclipsed
H
H
CH3
H
H3C CH3
H
H3CH2C
1 Et-H eclipsed
2 Me-H eclipsed
H
H
1 Et-Me gauche
1 Me-Me gauche
1 Et-H gauche
1 Me-H gauche
2 H-H gauche
H3CH2C
CH3
CH3
H
H
H
2 Et-H gauche
1 Me-Me gauche
2 Me-H gauche
1 H-H gauche
H3C
H
CH3
H
H
CH2CH3
H CH3
H
H3C
1 Et-Me gauche
1 Et-H gauche
3 Me-H gauche
1 H-H gauche
H
H
CH3
CH2CH3
H
CH3
CH2CH3
H
(c)
1 iPr-Me eclipsed
1 Me-H eclipsed
1 H-H eclipsed
1 iPr-Me eclipsed
1 Me-H eclipsed
1 H-H eclipsed
H3C
H
H
H3C
H
1 iPr-H eclipsed
2 Me-H eclipsed
H
CHH3
H
H3C
1 Me-iPr gauche
1 iPr-H gauche
3 Me-H gauche
1 H-H gauche
H
H
H3C
H
H
H
CH3
4.
2 iPr-Me gauche
2 Me-H gauche
2 H-H gauche
H
CH3
Draw the flat structure of the following compounds.
H
H
H
(a)
H3C
CH3
CH3
H
(c)
H
H
H
CH3
CH3
CH3
H
H
H
H
H
(b)
CH2CH2CH3
H
(d)
H
H
CH3
1 Me-iPr gauche
1 iPr-H gauche
3 Me-H gauche
1 H-H gauche
H3C
H
CHH3
CH3
H3C
H
H
H
H
H3C
H
CH3
CH3
H
H
H3C
(e)
(f)
5. Name each of the compounds in question 5 (when a benzene ring is a substituent as it
is in part b, it is called a phenyl).
(a) 2-methylbutane
(b) 2-phenylbutane
(c) hexane
(d) cyclopentane
(e) 1,1-dimethylcyclopentane
(f) 1,2-dimethylcyclopentane
6. For each compound below, convert the chair structure into a flat drawing.
6. For each compound below, convert the chair structure into a flat drawing.
Br
Br
HO
OH
(a)
Br
(b)
OH
Br
Br
Br
(c)
OH HO
HO
Br
OH
OH
Br
7. Draw both chair conformations of the following molecules, and determine which one
is most stable.
(a)
(b)
(c)
8. For each compound below, draw the opposite chair conformation and determine
which of the two is most stable.
(a)
2 ax Me
1 eq. Et
1 ax Et
2 eq. Me
1 ax t-Bu
1 eq. Me
1 eq. t-Bu
1 ax. Me
(b)
3 eq. Me
1 ax. Me
(c)
3 ax Me
1 eq. Me
9. The structures of five sugars are drawn below. Rank them from least to most stable.
Some of them may have very similar stabilities.
OH
OH
O
OH
HO
OH
O
OH
OH
OH
O
OH
OH
OH
HO
HO
least
stable
galactopyranose
HO
O
OH
1 axials
OH
O
OH
HO
HO
HO OH
no axials
OH
OH
OH
O
OH
OH
HO
gulopyranose
HO
2 axials
OH
OH
mannopyranose
OH
O
OH
HO
OH
O
OH
HO HO
OH
HO
glucopyranose
OH
10.
OH
OH
OH
O
OH
altropyranose
HO
OH
HO
OH
OH
OH
O
2 axials
altropyranose = gulopyranose < mannopyranose = galactopyranose < glucopyranose
1 axials
most stable
Define the following terms, and give an example of each.
a. conformers: orientations of a molecule which result from rotation about single
bonds. Three examples:
CH3
Cl
Cl
H
and
HO
HO
Cl
H
HO
CH3
CH3
CH3
and
H HO
and
H
Cl
(notice that these Newman diagrams are the same
conformations as the compounds shown on the left).
b. enantiomers: two stereoisomers which are mirror images of each other, but are
non-superimposable. Examples:
and
and
d. chiral molecule: a molecule which has an enantiomer. Examples:
e. stereocenter (stereogenic center): a carbon which has four different things
attached to it. Examples (all stereocenters are starred).
Cl
HO
11.
*
*
*
*
*
Locate all stereocenters, if any, in the following molecules:
HO
O
*
*
O
O
O
O
F3 C
*
O
*
*
HN
O
*
*
*
N
H
Prozac
*
*
Zocor
(a)
Ritalin
12.
Determine if the following molecules are chiral or achiral. If it is chiral, draw the
enantiomer of the compound. If it is achiral, locate the mirror plane.
a)
(b)
(b)
The mirror image of this
compound is the same compound.
Therefore, it is achiral. The mirror
plane bisects the molecule, and contains
the methyl group and the hydrogen.
The symmetry of this molecule is now
broken, and the mirror images are
not superimposable. There is no longer
a mirror plane, so the molecule is chiral.
The enantiomer is the mirror image. The
enantiomer can also be drawn like this:
(c)
same
redraw to see mirror plane:
H
carbon has:
1 methyl, 1 hydrogen, 2 ethyl groups
Since it does not have four different
groups, it is achiral. The mirror
image is superimposable. The mirror
plane contains the hydrogen and the
methyl group.
(c)
non-superimposable
The symmetry of this molecule is now
broken, and the mirror images are
not superimposable. There is no longer
a mirror plane, so the molecule is chiral.
The enantiomer is the mirror image. The
enantiomer can also be drawn like this:
not a stereocenter
stereocenter
(d)
Br
Br
Br
Br
Even though this molecule contains
two chiral carbons, the molecule is achiral
because there is a mirror plane (it is meso).
(e)
(f)
There is no longer a mirror plane, so this
molecule is chiral. The enantiomer is:
Br
Br
13.
The 72 students taking organic chemistry right now and their parents represent 16
different countries. The flags of those countries are shown below. Examine each flag for
planes of symmetry. Some of them have no symmetry plane, some have a vertical plane,
some have horizontal plane, and some have both a vertical and horizontal plane
(disregard the symmetry plane in the plane of the page that all flags have).
Canada
(vertical)
China
(no plane)
Germany
(vertical)
India
(vertical)
Iraq
(vertical w/o writing
incl. writing: no
plane)
Mauritius
(vertical)
Nepal
(no plane)
Pakistan
(no plane)
Poland
(vertical)
Singapore
(no plane)
Sweden
(horizontal)
United Arab
Emirates
(no plane)
Vietnam
(vertical)
Zambia
(no plane)
United Kingdom
(no plane)
United States
(no planee)