More Practice: Drawing Isomers
1. Draw and name all compounds with molecular formula C5H10.
1-pentene
(E)-2-pentene
(Z)-2-pentene
2-methyl-1-butene
3-methyl-1-butene
2-methyl-2-butene
cyclopentane
ethylcyclopropane
methylcyclobutane
1,1-dimethylcyclopropane
1
1,2-dimethylcyclopropane
Chiral or Achiral?
1. Determine whether each of the following molecules is chiral or achiral, and further specify if any are
meso.
F
Br
(a)
(b)
F
Br
achiral (meso)
chiral
Cl
Cl
Cl
(c)
(d)
Cl
achiral (meso)
chiral
O
(e)
Et
(f)
H
C
Me
H
achiral
chiral
F
(g)
F
O
OH
H
(h)
O
O
HO
H
achiral (meso)
chiral
(i)
Cl
(j)
Cl
Br
Br
achiral (meso)
chiral
(k)
OH
HO
O
OH
OH
OH
chiral
2
Enantiomers and Diastereomers
1. For each of the following pairs of stereoisomers, indicate if they are enantiomers, diastereomers, or
identical compounds.
diastereomers
identical
enantiomers
identical
CH3
CH3
Br
Br
CH3
H3C
Br
Br
H3C
diastereomers
CH3
3
Identifying Isomers
1. For each of the following pairs of isomers, indicate whether they are Identical, Enantiomers,
Diastereomers, or None of the above by circling the appropriate term for each.
HN
Me
HN
H
H
HN
Identical
Enantiomers
Diastereomers
None
Identical
Enantiomers
Diastereomers
None
Identical
Enantiomers
Diastereomers
None
Identical
Enantiomers
Diastereomers
None
Identical
Enantiomers
Diastereomers
None
Identical
Enantiomers
Diastereomers
None
Me
HN
H
H
Me
Me
H
H
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Stereochemical Concepts
1. Which of the following are true? Give counterexamples for those that are false.
(a) In some cases, constitutional isomers are chiral.
True
(b) In every case, a pair of enantiomers have a mirror-image relationship.
True
(c) Mirror-image molecules are in all cases enantiomers. False
are identical
(d) If a compound has an enantiomer, it must be chiral. True
H
F
(e) Every chiral compound has a diastereomer. False H C
3
has no diastereomer
Cl
are diastereomers
and
(f) If a compound has a diastereomer it must be chiral. False
Br
(g) Any molecule containing an asymmetric carbon must be chiral. False
is meso and thus is
achiral
Br
(h) Any molecule with a stereocenter must have a stereoisomer. True
(i) Some diastereomers have a mirror image relationship. False
(j) Some chiral compounds are optically inactive. False
By def inition, NO diasteromers
have that relationship!
ALL pure chiral compounds are optically
active.
(k) All chiral molecules have no plane of symmetry. True
(l) If a compound has a stereocenter, it has an enantiomer. False
See answer to part (g) above.
(m) A meso compound will necessarily have at least two diastereomers.
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True
Working with Stereoisomers
1. Consider the species 2,3-butanediol, whose structural formula is:
HO
OH
2,3-butanediol
H3C
CH3
a) There are three distinct, pure, configurational stereoisomers of 2,3-butanediol, all of which have the same
structural formula above. Using any unambiguous notation, draw the three different stereoisomers in the
boxes below. (The boxes have been arbitrarily labeled A, B, and C.)
A
B
C
HO
OH
HO
OH
HO
OH
H3C
CH3
H3C
CH3
H3C
CH3
b) By circling the appropriate terms below, indicate whether each of the three species you drew above is
chiral or achiral.
A:
chiral
achiral
B:
chiral
achiral
C:
chiral
achiral
c) Two of the above species have the same melting point (19°C). The other species has a different melting
point (34°C). Circle the letter of the species which has a melting point of 34°C:
A
B
C
What characteristic of the other two stereoisomers causes them to share the same melting point (19°C)?
B and C are enantiomers, which always have identical physical properties.
6
Cyclic Compounds and Reactions
1. You will learn next semester that ketones can react with water in the presence of catalytic acid to form
hydrates. The equilibrium constant for ketone hydration is usually quite small. Explain, then, why the equilibrium
constant for the hydration of cyclobutanone is so much greater than 1. In other words, why does the equilbrium
in the reaction shown below lie so far to the right? (Hint: Focus on the hybridization of the atoms in the
reactants and the products. You do not need to know anything about how the reaction proceeds
mechanistically, although working through it would make an excellent FMO exercise!)
O
HO
H3O
OH
+
Keq >> 1
cyclobutanone
hydrate
cyclobutanone
T he carbonyl carbon in cyclobutanone is sp2 hybridized, and sp2 hybridized atoms have ideal
bond angles of about 120o. However, the carbons in a f our-membered ring are f orced, by the
geometry, to adopt bond angles near 90o. T hus, the carbonyl carbon is f orced to adopt a bond
angle that is much less than ideal, creating a large amount of strain.
In contrast, the carbon in the hydrated product is sp3 hybridized, and thus its ideal bond angle is
only 109o. T his angle is much closer to 90o, and so there is f ar less ring strain in the hydrate
product than in the starting material. T his is why the equilbrium constant lies so f ar to the right.
In case you tried to f igure it out f or yourself as an exercise, here is the correct mechanism. T his
is not required knowledge, but it is analogous to acid-catalyzed alkene hydration, and would
make a nice HOMO/ LUMO question!
:OH
2
O
H
H
O
H
H
O
:OH2
HO
O
H
H
Donor: nO
Acceptor: *HO+
(Acid-base rxn)
Donor: nO
Acceptor: *HO+
(Acid-base rxn)
Donor: nO
Acceptor: *CO
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HO
OH
Cyclopropyl Rings
1. Due to the exceedingly high strain between the carbons in a cyclopropyl ring, the bonding orbitals are forced
to pucker outward at an angle of approximately 104 degrees, and the CC bonds are as a consequence "bent"
outward away from the plane of the ring, as shown in the diagram below.
104o
It has been observed that cyclopropylmethyl carbocations are roughly as stable as the corresponding allyl
carbocations (both shown below), despite lacking any lone pairs or pi bonds to participate in resonance
stabilization of the positive charge. Using your knowledge of orbitals and the information provided above,
propose an explanation for this observed stability. Your answer should incorporate clearly drawn "cartoon"
orbitals, and making a model might help. (Hint: consider the nature of the bonding orbitals -- what orbital
character do they seem to have?)
H
H
H
roughly equal
in stability to:
H
Due to the strained bonds between the carbons in cyclopropyl rings, the bonding orbitals have no choice but to
pucker outward away from the ring at an angle of approximately 104 This gives the orbitals more p character
than typical sp3 orbitals, and this allows them to overlap with the vacant p orbital of a carbocation in a manner
similar to that of proper pi bonds or p orbitals!
H
H
Two filled-unfilled orbital interactions help
to stabilize the carbocation like a bond.
8
Cyclohexyl Rings
1. Look at the chair conformation of cyclohexane drawn below. Notice the axial and equatorial substituents.
Pay close attention to the parallel lines!
2. Draw each of the following dimethyl cyclohexanes (on the planar ring), then identify whether the
substituents would be either: "Axial and Equatorial" or "Both Equatorial"
1,2-cis
Axial & Equatorial
1,2-trans
Both Equatorial
1,3-cis
Both Equatorial
1,3-trans
Axial & Equatorial
3. Draw each of the following in the most stable chair conformation:
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1,4-cis
Axial & Equatorial
1,4-trans
Both Equatorial
Chair Conformations
1. Consider the following molecule:
CH3
Cl
H3C
CH3
a) This molecule is expected to have two relatively stable chair conformations. Draw clear representations of
the two different conformations in the boxes below. (The boxes have been arbitrarily labeled A and B.)
A
B
H3C
H
H3C
H
Cl
H
CH3
H
H
Cl
H3C
H3C
H
H
H
b) Which of the two chair conformations is more stable? (circle one of the statements below)
1) A is more stable than B.
2) Both conformations are equally stable.
3) B is more stable than A.
No explanation is necessary.
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CH3
Chair Conformations and "A Values"
1. For the following six-membered ring system, draw each of the two chair conformations in the boxes below
and circle the one that is more stable. In two sentences or less, explain your reasoning.
Cl
Cl
Cl
Cl
Cl
Cl
Both structures have one equatorial and one axial chlorine atom, but the
lef t-hand structure is more stable because the smaller methyl group is
axial whereas in the right-hand structure the larger isopropyl group is
axial.
(b) Consider the following "A values" (energy differences between the axial and equatorial conformations) of
cyclohexanes substituted with a single halogen atom.
Cl
H
Cl
H
G = +0.53 kcal/mol
H
G = +0.48 kcal/mol
H
G = +0.47 kcal/mol
Br
H
Br
I
H
I
It may come as a surprise that the G values are all nearly equal. Identify two competing factors
that, taken together, may account for the similarity among the three G values. Explain briefly.
Factor 1: Size
Iodine is much larger than bromine, and bromine is much larger than chlorine. All else being
equal, then, iodine would be expected to have the greatest pref erence f or the equatorial position
(largest G value) and chlorine would be expected to have the least pref erence f or the equatorial
position (smallest G value).
Factor 2: Bond Lengths
T he C_I bond is much longer than the C_Br bond, and the C_Br bond is much longer than the C_Cl
bond. Longer bond lengths mean the halogen substituents are f urther away f rom the ring, and
this decreases the severity of potential 1,3-diaxial repulsions. Evidently, this bond length f actor
compensates f or the size f actor above.
11
Bicyclic Compounds and Bredt's Rule
1. Draw each of the following bicyclic compounds in a good "perspective" drawing.
CH3
CH3
CH3
CH3
CH3
H
CH3
H3C
CH3
2. The molecule trans-cyclooctene is known to exist. (It is chiral, by the way). Why is the analogous molecule
trans-cyclohexene unstable? A trans alkene would be impossibly strained in a six-membered ring!
(Build a model to see why this is so.)
H
H
H
H
=
H
H
trans-cyclooctene
(planar representation)
trans-cyclooctene
(perspective representation)
trans-cyclohexene????
3. Draw each of the following bicyclic alkenes in a good "perspective" representation. Only one of these
three compounds actually exists. Which one, and why?
Does not exist
Does not exist
The sp2-hybridized carbons cannot be planar in these two
molecules -- they would be much too highly strained to be
stable. Build a model to see this. Also, each has a trans
alkene in a small ring, which we have seen is impossible.
12
Exists
Challenging Problems in Fused-Ring Polycyclic Compounds
1. Draw each of the following compounds in a good "perspective" drawing of its lowest-energy conformation.
Your models may help immensely in visualizing these!
N
N
H
H
H
OH
OH
H
H
H
H
H
H
H
Cl
H
Cl
H
H
H
H
H
Cl
Cl
H
H
H
NOT
H
Cl
H
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