Chem 21
Homework set 7
End-of-chapter problems from Hornback: Ch 7: 16a-f, 17-19, 20a-g, 21, 22, 26-31. Two graded
sets are attached — the "normal" 2-pager, and an extra optional problem due a week later.
1. Decide whether each of the following compounds is chiral or achiral. If achiral, indicate the
position of the plane of symmetry or inversion center.
Cl
Cl
Cl
F
Cl
Cl
Cl
Br
F
Cl
Br
F
For the following set, don't forget about the possibility of rotation about CC σ-bonds.
H
Br
CH3 Br
OH
Cl
Br
HO
Cl
O
H
Br
Br
O
For the cations below, keep in mind that the structures shown may not tell the whole story.
OCH3
Cl
Cl
CH3O
Remember, some of these critters are really flat, and others aren’t. Maybe it matters, maybe not.
F
Br
Br
Br
Cl
tBu
2. Draw these. No alkyl abbreviations, and no hideously distorted structures, please.
(a) (R)-1-bromo-3-methylhexane
(b) The compound in part a is not cyclic. If it had been cyclic, i.e., same substits attached at the
1 and 3 Cs of a cyclohexane ring, how many stereocenters would there be? And how many R/S
labels?
(c) (1S,2R)-1-tert-butyl-2-(trifluoromethyl)cyclopentane
(d) If one were to accidentally botch the one in part c by making it acyclic, how many
stereocenters and how many labels would there be?
(e) cis-1-((R)-sec-butyl)-3-((S)-1-chloropropyl)cyclobutane
(f) (1R,2S)-1-iodo-2-((R)-3-chlorobutyl)cyclohexane
3. Here's a useful exercise in learning to draw enantiomers. There are several ways to do it that
will all produce exactly the same structure, since an object can only have one mirror image.
(a) Pick any chiral molecule that has at lease two stereocenters and with stereochemistry shown
explicitly at each (i.e. with wedges/dashes).
(b) (i) draw its reflection across a vertical mirror plane; (ii) reflect it across a horizontal mirror
plane; (iii) change the configuration at each stereocenter by swapping wedge/dash substits;
equivalent to a reflection through a plane parallel to the page); ((iv) just for fun, turn it inside-out
by passing it through a point.)
(c) Verify that all three (or four) drawings are indeed the same molecule — the enantiomer of the
original structure.
(d) Now repeat the same exercise starting with an achiral molecule. Pick one that has at least
two stereocenters. (e) Verify that all your reflections (and the “inside-out” structure) are
identical to each other and to the original.
4. Let's revisit some of the isomer drawing exercises from early in the semester. Then, we were
only interested in drawing constitutional isomers — molecules having different connectivities —
now let's think about which of these also have different stereoisomeric forms.
Draw all the isomers of the following molecules, paying attention to all types of stereoisomerism
that we've encountered so far — cis/trans (aka E/Z) on a double bond, cis/trans on a ring, R/S at a
stereocenter. (In parentheses are the numbers I found, which doesn't make it correct!)
(a) C5H10 (10 constitutional isomers; 13 when stereochemistry is included)
(b) C4H7Cl (12 constitutional isomers; 19 when stereochemistry is included)
5. How many stereoisomers of each compound below can exist? How many are achiral?
Cl
Cl
CH
CH2CHBrCH3
OH
OH
Chem 21
Fall 2009
HW set 7
30 points; due Wed, Oct 28
Name _________________________________________
1. Draw these in skeletal notation. Show stereochemistry clearly.
(a) (S)-3,6-dimethyl-1-heptyne
(b) (1R,2S)-1-iodo-2-((Z)-3-pentenyl)cyclobutane
(c) meso-2,7-dibromo-(E)-4-octene
(d) Check that you didn't mix up cyclic and acyclic structures above!
2. Decide whether each of the following compounds is chiral or achiral. If achiral, indicate the
position of the plane of symmetry or inversion center.
Cl
CH3
O
OCH3
1,3-dichlorobutane
O
Br
CO2H
HN
Br
O
O
3. (a) Label each stereocenter of the compound in the box below as R or S, and label the double
bond E or Z, as appropriate
enantiomer:
CH3O
Cl
(b) Draw the enantiomer of the compound in the space provided above.
(c) Look carefully at the compounds below, then label each one as: the same as, the enantiomer
of, a diastereomer of, a resonance structure of, or a constitutional isomer of the compound in
the box.
Cl
CH3O
CH3O
Cl
Cl
OCH3
(d) Draw a constitutional isomer of the compound (C8H13ClO) that is achiral.
(There are many possible answers; if you're working with other students, no two people should
end up with structures that are even remotely similar!)
4. Draw all the stereoisomers of 1,2,3-trimethylcyclopentane. Be sure you don't draw the same
structure twice! Indicate which are enantiomers and which are diastereomers.
Chem 21
Fall 2009
Extra HW problem
10 points; due Wed, Nov 4
Name _________________________________________
Here's an extra HW problem just for fun. You may find this very interesting if you're willing to
put in the time and effort to think carefully about it and do it correctly. If not, it's just a waste of
your time and ours, so please skip it. No partial credit.
Before the early 1900s, the geometry of tetracoordinate carbon was not known. Three postulated
geometries are shown below — square planar, square pyramidal, and tetrahedral. Let's explore
the stereochemical consequences of each of these forms and see how the clever experiment
described in part b establishes the tetrahedral geometry that we all know and love today.
(a) Draw the structures of all the possible stereoisomers of the compound below (the structure
given at left is meant to show connectivity only), assuming the central carbon has (1) square
planar, (2) square pyramidal, and (3) tetrahedral, geometries. For each geometry, indicate
whether the stereoisomers you have drawn are enantiomers or diastereomers. (Use the back of
the page or a separate sheet.) Hint: After you figure out the set of square planar structures, think
about how these might be "perturbed" to make the square pyramidal ones.
CH2CH3
CH3
C CONH2
CO2H
planar
square
pyramidal
tetrahedral
(b) In 1914, Emil Fischer converted the optically active compound below into two optically
inactive ones by two short series of reactions, as shown below. (Again, the drawings are intended
to show connectivities only.) The reactions he used did not break any bonds to the central C, and
thus could not have changed its configuration; only the attached groups were altered.
CH2CH3
CH3
C CO2H
CO2H
optically inactive
CH2CH3
CH3
C CONH2
CO2H
optically active
CH2CH3
CH3
C CONH2
CH3
optically inactive
Show how these results are inconsistent with the planar or square pyramidal geometries, but
consistent with the tetrahedral geometry for C. Keep in mind that Fischer did not know precisely
which of the optically active structures he might have been starting with, so you'll have to work
through the implications of these results for each possibility.