The Basics
Identification of Functional Group, Intermolecular Forces, Shape and Polarity
Question B1
Give examples of molecules that have each of the functional groups discussed in class without using
any examples I used. If you find an example that has more than one such group, then all qualify
towards the completion of this question. Good places to look are: your text, the CRC handbook, the
Merck Index, and the Aldrich Catalog handbook of Fine Chemicals (reference section of the library).
Reference your answers with source and page.
Answer B1
There are of course, many answers to this question. One thing to watch out for: some more complex
functional groups may contain the same atom sequence as another functional group. For example,
acetic anhydride and ethyl acetate both look somewhat similar but they are different functional groups.
O
O
O
C
C
O
H3C
CH3
Acetic Anhydride
C
H3C
O
CH2 CH
3
Ethyl acetate
Ethyl acetate is an ester. Acetic anhydride is classed as an anhydride, another carboxylic acid
derivative. Beware.
Question B2 (8 marks)
Circle and label by name all of the functional groups present in the following molecules:
H
C
C
CHCl
H3CH 2C C C
OH H
O
A sedative.
C NH2
N
An analgesic.
Answer B2
H
Alkyne
H3CH 2C
C
CHCl
C
Aziridine.
O
C
C
H
OCH 2CH3
C
NH2 Amide
Alkyl Halide
OCH 2CH3
Ether
Alkene
H
Alcohol
A sedative.
H
N
OH
Aromatic ring
An analgesic.
Aziridine. An amine.
1
Question One (four marks)
Give the structures of all of the different possible molecules that have the molecular formula C4H8.
You need only show the connectivity, not the 3-dimensional structure.
Answer
H3C
H
H2C
C
C
H
H
C
C
CH2CH3 H
H
H
C
C
CH3
CH3 H3C
CH3
H
H
CH3
C
CH3
Question (four marks)
Draw all of the possible molecules that have the molecular formula C3H8O and identify to which class
of compound they belong (i.e. the functional group).
Answer
H3C
H2
C
Alcohol
C
H2
OH
CH3
H3C
CH
Alcohol
OH
H3C
H2
C
O
CH3
Ether
Question Three (4 marks)
For each of the following pairs of molecules, indicate which will have the higher boiling point. Explain
your reasoning.
1)
O
H3CCH2
C
O
OH or
H3C
C
OCH 3
2)
H3CH 2C
O
or
CH2CH3
H3C
O
CH3
Answer
1) The first molecule is capable of forming hydrogen bonds with other molecules of its own kind.
These interactions are strong and must be overcome for vapourization to occur. Indeed, carboxylic
acids often exist as dimers even in the gas phase.
The second molecule lacks any X-H bond (X = N, O, S) and cannot form H-bonds.
2) Both molecules are ethers and cannot form H-bonds. They will have similar dipoles so their dipolar
interactions will be comparable. The larger ether will have more substantial London dispersive forces
operating, so it will have the higher boiling point.
Question Four (ten marks)
For each of the following species:
a.) Add lone pairs to all atoms such that their octets are full.
b.) Add formal charges to those atoms that have formal charges.
c.) Indicate the hybridization of all C, N and O atoms.
d.) Redraw each of the species showing the proper geometry. Keep as many atoms in the plane of the
page as possible and use dashes/wedges (where appropriate) to show 3-dimensional character. Building
a model may help.
e.) Draw the net molecular dipole for each molecule, if there is one.
2
O
H3C CH
BF3
H2C=CCl2
H3C
CN
Answer
F
B sp3
F
C sp2 F
sp2
O
HH
C
sp3
H
H
H
sp2 C
Cl
C sp2
H
Cl
H
H
sp3 C
H
C N
sp sp
Question B2
Draw three-dimensional structures of the following and show the net molecular dipole.
ClHC
(both isomers)
CHCl
O
H3C S
CH2Cl2
CH3
Answer B2
H
H
H
C
C
Cl
Cl
C
Cl
+
Cl
C
net dipole
H
O
no net dipole
H3C
S
Cl
CH3
Cl
C
H H
Dimethyl sulfoxide is actually a tetrahedral molecule due to the presence of a lone pair. The net
molecular dipole still runs from S to O.
Methylene chloride has a dipole that bisects the Cl-C-Cl bond angle.
Question B3
Intermolecular H-bonds are those formed between different (i.e. separate) molecules. Intramolecular
H-bonds are formed between functional groups attached to the same molecule. Of the following three
molecules:
Which form intermolecular H-bonds only with water?
Which form intramolecular H-bonds?
Which form intermolecular H-bonds with water and with itself?
HO
OH
CH3CH2
N H
CH3CH2CH2SH
CH2CH3
Answer B3
Intermolecular H-bonds are formed between two or more molecules. These molecules may be the same
type of molecule or different. For example, the first molecule shown (1,3-propanediol) can form Hbonds to other molecules of itself (as is the case when it is in its pure state)
3
HO
O
H
H
H O
O
H
H O
OH
OH
or with water (when a mixture):
Note that BOTH of these cases are examples of intermolecular H-bonds.
This molecule has TWO polar functional groups and is also capable of forming an intramolecular Hbond:
O
H
O
H
Diethylamine can form H-bonds with other molecules that have a polar bond to an H atom; water for
example:
and with other molecules of its own kind:
CH3
H
H3CH 2C
N
H
CH2CH3
N H
CH3
H
H3CH 2C
CH3
O
H
N H
CH3
H
N
CH2CH3
O H
The S-H bond of propanethiol is not very polar (the same as a C-H bond, in fact) and for the purposes
of this course, we do not consider it to donate H-bonds. It can accept H-bonds to the S lone pairs. For
somewhat complex reasons it does form H-bonds in the pure state and with water. Evidence for this is
found in the fact that ethanethiol is a liquid while ethane is a gas.
Question
Show all formal charges on the appropriate atoms in the following molecules. All non-bonding
electron pairs (lone pairs) are shown.
c.) CH3
C
H3C
CH3
Cl
Cl
C
b.)
H
d.) O
e.) N(CH3)4
H3C C CH3
:
H
H N
:
a.)
:
H N
-1 on N
Cl
Cl
C
b.)
no formal
charges
:
H
:
a.)
CH3
C+
H3C
CH3
+1 on C
c.)
4
:
Answer
d.) O
H
H3C C CH3
+1 on O
e.) N(CH3)4
+1 on N
Question
Rationalize in terms of intermolecular forces the trends in the boiling points of the following three
isomeric amines:
H3CH 2CH2C
N
H3CH 2C
H
36°
48°C
N
CH3
CH3
H
H
BP:
N
H3C
CH3
3°
Answer
The ability to form hydrogen bonds is a key factor in determining the physical properties of a
compound. Of these three compounds, the latter (trimethylamine) is incapable of forming hydrogen
bonds because there are no N-H bonds in the molecule. It is capable of forming H-bonds with other
compounds that have N-H or O-H bonds, but not with itself. C-H bonds are so non-polar that they do
not participate in H-bonding.
With the other two compounds there are several arguments that can be made to rationalize the
difference in BP. The first compound (propylamine) has 2 N-H bonds per molecule so you might
expect that it will form a greater number of H-bonds than is possible with the middle compound (NMethyl ethylamine). You might also argue that with the longer hydrocarbon chain of propylamine that
there is a greater degree of London intermolecular forces possible. Both of these are quite acceptable
answers.
Question
For each of the following molecules, show the hybridization of all non-hydrogen atoms and give the no.
of p- and s-bonds in the molecule. Add lone pairs where appropriate.
C N HC
C
O
CH2OH
H3C C O CH3
N
H3C
C
CH3
Answer
sp sp
HC
sp3
sp3
C CH2OH
7 s, 2 p
sp2
sp
sp
sp2
C N
sp2
6 s, 3 p
All 6 C atoms are sp2.
N
H3C
sp
3
C
sp2
CH3
O
sp2
H3C C O CH3
10 s, 1 p
2
3
sp
sp
3
sp
sp3
21 s, 4 p
sp3
5
Question
For the following molecules, do the following:
a.) add lone pairs to all atoms such that their octets are full
b.) determine the formal charge on the atoms in bold type.
O
H3C S
O
H3C N
CH3
C O
O
Answer
O
O
C O
H3C N
H3C S
CH3
O
1) This molecule is known as dimethyl sulfoxide. It is very readily absorbed through the skin and has
been touted as a cure for arthritis. It has also been explored as an alternative to needle injection method
for administering drugs. The formal charges are shown. Note that the S atom is tetrahedral in
geometry.
2) This molecule is nitromethane. We did this in class so it should not be a mystery to anyone. The N
atom is sp2 hybridized and carries a +1 formal charge. The bottom O atom is sp3 hybridized and
carries a -1 formal charge.
3) Carbon monoxide is a fascinating molecule. You will note that the formal charges are -1 on carbon
and +1 on oxygen. CO has an unusual affinity for bonding to metals. When it does bond, it bonds
through the lone pair on carbon, not on oxygen.
Question
Identify the bond dipoles and the net molecular dipoles of the following molecules:
N
H
H3C NO2
H
H
C C
C
C
BF3
C C
C
N
C
N
H
N
Answer
1/2
O
H3C N
O
O
H3C
H3C N
O
O
N
O
1/2
Nitromethane (again) has two resonance structures as shown. These are degenerate (equal in energy
and therefore equal in importance) and tell us that both oxygens have identical character relative to the
center nitrogen. Therefore, there will be significant and identical bond dipoles pointing from nitrogen
to oxygen along both N-O bonds. The vector sum of these two bond dipoles is shown. It is colinear
with the C-N bond and bisects the O-N-O bond angle. (You should recognize that the four atoms: C, N
and the 2 Os are all coplanar.
6
H
N
H
H
C C
C
F
C C
C
N
C
C
N
H
B
F
N
F
The polar bond in this molecule is the C-N triple bond. Bond dipoles are shown and their vector sum
is similar in nature to the previous case. The arrangement of the two CN groups (on the same side of
the double bond) is known as cis. This molecule, you should know, is flat as a pancake. All atoms lie
in the same plane.
In this second molecule, the CN groups are on opposite sides of the double bond. This
arrangement is known as trans. Because rotation around a C=C bond is a very high energy process (it
requires the breaking of the p-bond) cis and trans compounds are not interconvertible at room
temperature. The bond dipoles are the same for this compound but this time cancel out when added
together. This molecules has no net dipole. An important difference with its very similar isomer
above.
Boron trifluoride is an unsual compound because it has an electron deficient atom, B, at its
center. The B atom has only six electrons (three from B and one from each F atom). Boron is sp2
hybridized (as predicted by VSEPR) and is a trigonal planar molecule. There are bond dipoles point
from Boron to Fluorine and since they radiate out in one plane in the manner shown, these cancel such
that BF3 has no net dipole.
Question Four
Give the hybridization of the atoms in each of the following and determine their shape (linear, trigonal
planar, tetrahedral).
CH3
OH
H
C
O
H
1
C
H
H
H
H3C NO2
H2C C O
H3C N CH3
CH3
CH3
N
H3C
CH3
H3C
2
O
H
C C O
H
3
H3C N
4
O
1) Both Oxygen and Carbon atoms are sp2 hybridized and they are trigonal planar. This cationic
species is flat.
2) All of the carbon atoms and the nitrogen atom are sp3 hybridized. This molecule is analogous to the
ammonium cation but with CH3 groups instead of protons.
3) This molecule (known as ketene) is somewhat analogous to carbon dioxide. The C=C=O atoms are
linear. The carbon bearing the two hydrogens is sp2 hybridized, as is the oxygen atom. The central
carbon has no lone pairs attached to it and is attached to two atoms. This carbon must be sp hybridized.
VSEPR predicts that this be linear, and it is indeed a linear molecule (except for the protons).
4) As described in this resonance structure, the top oxygen is sp2 hybridized as is the N atom. The
bottom oxygen atom is sp3 hybridized. If you try to take both resonance structures into account here
you will have trouble attaching a hybridization label to the O atoms. Just goes to show you what an
artificial thing hybridization really is.
Question
Circle and label all of the functional groups on this molecule and classify them as 1°, 2°, 3° or allylic
where appropriate. (Remember that more than one classification may apply, e.g. 2° and benzylic.)
7
O
CH3
H3C
HO
N
CH3
HO
CH3
CH3CH2
O
CH3
CH3
H3C
O
O
O
Picromycin was the first macrolide
(big ring) antibiotic ever isolated.
CH3
b.) Epoxides are cyclic versions of what functional group?
c.)
H
N
O This molecule, caprolactam, is the precursor
C
to Nylon 6. Cyclic versions of this functional
group are called lactams. What is the
functional group?
Answer
O
Alkene
Ketone
CH3
3° Allylic alcohol
2°Alcohol
H3C
HO
CH3
CH3
N
CH3
HO
CH3CH2
O
CH3
H3C
O
Ether
ester (lactone)
3° Amine
O
O
CH3
b.) Epoxides are ethers.
c.) Lactams are cyclic amides.
8
Ketone
Question
Identify and label the functional groups present in these molecules.
Parthenin (a natural
Phenampromide
product that causes
(an analgesic)
contact dermatitus)
CH3
HO
H
CH2
O H3C
O
N
N
CH3
C
C
CH2CH3
O
O
Epichlorohydrin
(used in the manufacture
of adhesives)
O
Cl
Answer
3¡ Allylic alcohol
HO
CH3
Aromatic ring
Alkene
Alkyl halid
Ketone
H
CH2
O H3C
O
C
O
3¡ Amide
3¡ Amine
N
Alkene
N
CH3
Ester (lactone)
9
C
O
CH2CH3
O
Ether
Cl
Question One (5 marks)
Circle and label the functional groups on the following two molecules.
CH3CH3
O
C N
HO
N
O
HO
Fentanyl, aka. Sublimaze,
a relaxant.
C
O
OH
HO
CF3
Fluprostenol, a prostaglandin used
in treating infertility in mares.
Question One (5 marks)
Circle and label the functional groups on the following two molecules.
3° Amide
O
CH3CH2 C N
3° Amine Aromatic ring 2° Alcohol
HO
Alkene
N
HO AlkeneHO
2° Alcohol
2° Allylic alcohol
Carboxylic Acid
OH
C
O
O Ether
Aromatic ring
CF3
Alkyl halide
Question (4 marks)
H2C O C
HC O
H2C
O
C
O
O C
O
The structure above is a triglyceride or lipid. This lipid is an oil. Fats are also lipids but they have
fewer carbon-carbon double bonds and tend to be solids.
a.) What is it about the structure of fats that makes them solids while oils tend to be liquids? Explain
by making reference to the intermolecular forces (i.e. between molecules) involved. You may wish to
add to the drawing above to help make your point.
b.) How are oils transformed into fats and why is this an important industrial process? Especially in
Southern Alberta?
Answer
a.) The intermolecular forces that keep non-polar molecules such as lipids together are London forces the instantaneous dipole-dipole interactions that occur between chains. The strength of these forces
depends on the ability of the chains to lie closely to other chains. The occurence of a double bond
introduces a "kink" in the chain which makes close approach impossible and thereby weakens the
magnitude of the London force and lowering the boiling point. Unsaturated lipids (those with many
kinks) tend to be liquids as a result.
10
b.) Oils are transformed into fats by catalytic hydrogenation. This is how margerine is made. Canola
margerine is an important product in S. Alberta.
Question Two (4 marks)
Of the following two molecules:
a.) which will be more soluble in water? Explain.
b.) which will have the higher boiling point? Explain.
O
O
C
H3C
OH
2 Ethanoic Acid .
(Acetic Acid)
C
CH3CH2CH2CH2
OH
1 Pentanoic Acid
(Valeric Acid)
Answer
a.) Acetic acid, which all of you know is very soluble in water, is the more soluble of the two. Valeric
acid has the same polar functional group (COOH) which can interact with water, but also has a large
non-polar group (CH2CH2CH2CH3) which will be hydrophobic, and render this molecule much less
water soluble. We went throught these properties in class using butane, butanol, and ethanol for
comparisons.
b.) The general trend is that larger, higher molecular weight compounds tend to have higher boiling
points. You must realize that both of these molecules have the same polar functional group. Any
difference in their boiling point must be due to the alkyl part of the molecule. Van der WŠals forces in
the alkyl chain of valeric acid serve to increase its boiling point. Such dipole-dipole interactions are not
possible in acetic acid which has no alkyl chain.
Question Three (3 marks)
a.) Draw three dimensional structures of the following two molecules. Include the lone pairs, which do
not appear on these diagrams.
O
C N
H3C
Acetonitrile
H
C CH CH2
Acrolein
b.) Label all non-hydrogen atoms with their hybridization (do not draw the hybrid orbitals).
H
3
180°
sp
sp
C C N
sp
H
H Acetonitrile
O
C
H
Acrolein
H
All atoms are sp2.
C
All are trigonal planar with
C
H
120° bond angles around them.
H
A few comments based on students answers. I asked for three dimensional drawings of these
molecules, so you need to show the linearity of the CCN bonding in acetonitrile, despite the way it was
shown in the question. You also need to show the tetrahedral nature of the CH 3 group. Acrolein is a
two-D molecule, so the easiest way to draw it is in the plane of the page! All you needed to do then is
get the 120° angles correct and you're all set. When discussing the bonding in ethylene in class, I often
show this molecule (which is also a flat 2D molecule) in a perpendicular orientation to the page. But I
do this only so that the orbitals making up the p-bond would be in the plane of the page.
11
Question Six (3 marks)
H
H
H C
C
H
C
C
H3C
CH3
H
H
a.) Rank these three molecules in order of decreasing C-H bond length. Explain.
b.) Rank these molecules by their relative acidity and explain your reasoning.
Answer
a.) Rank these three molecules in order of decreasing C-H bond length. Explain.
This was discussed in class. Recall that s orbitals are closer to the nucleus and do not have a node at
the nucleus, as do p orbitals. Since the C-H bond in acetylene is made from an sp hybrid orbital (50% s
character), these bonding electrons are held more tightly by the carbon nucleus. This tends to make the
C-H bond a little stronger and shorter than the corresponding bond in alkenes and alkanes.
b.) Rank these molecules by their relative acidity and explain your reasoning.
The same explanation applies to acidity. When acetylene loses a proton, the carbon bears a negative
charge which is localized in an sp orbital. Carbon doesn't like to bear a negative charge but it is easiest
to do so in an orbital with more s character because the charge can interact more strongly with the
positively charged nucleus. This stabilization of the conjugate base of acetylene means that acetylene is
relatively acidic, as hydrocarbons go. In fact, the difference in acidity between acetylene and ethane
ismore than 20 orders of magnitude. That's a lot.
This was a difficult question, but several students managed to get full marks.
Question Two (3 marks)
Match the following isomeric compounds with their correct boiling points and melting points. Explain
your reasoning.
Compounds: 2,2,3,3-tetramethylbutane and octane
Boiling point: 106.5°, Melting point: -118.3°.
Boiling point: 125.7°, Melting point: -100.7°.
H3C CH3
H3C
C
C
CH3
H3C
H3C CH3
2,2,4,4-Tetramethylbutane
H2
C
C
H2
H2
C
C
H2
H2
C
C
H2
CH3
Octane
Answer
The only intermolecular forces that apply to these molecules are London forces - the instantaneous
dipole-dipole attractions. These forces operate best through a small separation between the C-C bonds,
aligned parallel. The branching in the first molecule prevents close approach of the C-C bonds and
diminishes the London Forces as a result. The second set of physical data correspond to octane.
Question One (8 marks)
Circle and label the functional groups present in these molecules and classify them as 1°, 2°or 3° as is
appropriate.
O
H3C O C
O
CH2CH3
N C
N
Lofentanyl CH3
12
O
H
H
C
C
H
CH2OH
Glycidol
Answer
Ester
O
H3C O C
ether (or epoxide)
3° Amide
O
CH2CH3
C
N
O
H
H
N
Aromatic
Ring
3° Amine
CH3
C
C
H
CH2OH
1° alcohol
Question Three (5 marks)
Examine the following two compounds.
H3C
H2
C
1
H3C
N H
H
2
C
H2
H2
C
C
H2
H2
C
C
H2
H2
C
N H
H
i.) Which of the two would you expect to be more soluble in water? Explain your reasoning.
ii.) Which of the two would you expect to have a higher boiling point? Explain your reasoning.
iii.) These molecules are examples of what functional group?
Answer
i.) Both molecules are amines and can form strong H-bonds to water. Water will solubilize the smaller
molecule 1 more effectively, however, because it has a smaller hydrophobic chain. The 7-C chain of 2
will have a greater affinity for other molecules of 2 than it will for water.
ii.) Both amine functional groups will have similar intermolecular attractive forces. The larger 2,
however, will have more London forces operating due to its longer chain therefore 2 will have the
higher boiling point.
iii.) They are amines.
Question Five (12 marks)
For each of the following molecules:
i.) draw the molecule in three dimensions (using dashed and wedged bonds). Note: no geometry is
implied by the representations given.
ii.) if the molecule has a substantial net dipole, indicate this by drawing a dipole vector. If it does not,
state: no net dipole.
iii.) indicate the hybridization of each non-hydrogen atom.
iv.) name the functional group.
O
1) H C
3
2)
C H
H3C OH
3) H3C
C
C H
13
Answer
sp3
O
3
H sp C
C sp 2 H
H
H
3
H sp
C
HH
Aldehyde
H
sp3 C
O
H
H
sp sp
C
H
non-polar
H
Alcohol
C
Alkyne
v.) Which of the molecules in the Question Five will form hydrogen bonds with other molecules of its
own kind?
Only the alcohol.
vi.) Which will form hydrogen bonds with H2O?
The alcohol and the aldehyde.
Question Five (12 marks)
For each of the following molecules:
i.) draw the molecule in three dimensions (using dashed and wedged bonds). Note: no geometry is
implied by the representations given.
ii.) if the molecule has a substantial net dipole, indicate this by drawing a dipole vector. If it does not,
state: no net dipole.
iii.) indicate the hybridization of each non-hydrogen atom.
iv.) name the functional group.
O
1.) H3C C O H
2.) H3C C H
3.) H3C C N
H C H
v.) Which of the molecules in the above question will form hydrogen bonds with other molecules of its
own kind?
vi.) Which will form hydrogen bonds with H2O?
Answer
sp3
H
H C sp 2 O
H
H sp 3 C
O
sp2
Carboxylic Acid
2
H sp H
C
3
H sp C
C sp 2 H
H
H
v.) Only the carboxylic acid.
vi.) The carboxylic acid and the nitrile.
non-polar
Alkene
14
H
sp3 C
sp sp
C
H
H
Nitrile
N
Bonding and Moleculal Orbital Theory
Question
i) Construct an MO diagram for the Li2 molecule using the valence electrons only and show the atomic and
molecular orbitals involved.
ii) What is the bond order for the Li2 molecule?
ii) Compare your result with the MO diagram of H 2. Comment on similarities and differences.
Answer
i)
Li
Li
s*2s
Li2
2s
2s
s2s
ii) 1
iii) It looks exactly the same and the orbitals are the same shape. They will be different in energy, though, and
since they are made from 2s orbitals, they will be larger than the corresponding orbitals of H2. In principle,
there are also p-orbitals just like all of the other second row diatomics. All such orbitals will be empty, though.
Question
As carbocations go, allyl carbocations are relatively stable while vinyl carbocations are relatively unstable.
i.) For the allyl carbocation, how is the carbocation carbon hybridized?
ii.) How is the carbocation stabilized by the adjacent double bond?
iii.) For the vinyl carbocation, how is the carbocation carbon hybridized?
iv.) How does the carbocation interact with the double bond?
H
H
C C H
H2C
H
Allyl carbocation
C C
H
H
Vinyl carbocation
Answer
i.) The carbocation carbon is sp2 hybridized and has an empty atomic p orbital. This empty orbital can align
itself parallel to the double bond and be stabilized by overlap with it.
H C
H
C
H
C H
H
iii.) In this case the carbocation carbon is sp hybridized, as predicted by VSEPR. The HCC bond is linear. The
carbocation carbon uses its two hybrid orbitals to form the s-bonds to H and C. It has two unhybridized p
orbitals, the 2py and the 2pz WHICH ARE ORTHOGONAL. One of these orbitals makes up part of the p-bond.
1
The other is vacant, and because it is orthogonal to the p-bond IT CANNOT OVERLAP WITH IN ANY
WAY and therefore cannot derive any stabilization.
p-bond in plane of page
H
C
C H
H
C
C H
H
H
p-orbital perpendicular to page
Question (4 marks)
Sketch on the diagram below the atomic orbitals (on the right), and the molecular orbital (on the left) that make
up the C-O p-bond of formaldehyde. Oxygen is more electronegative than carbon and this should be
taken into account in your drawing.
H
H
H
C
O
H
C
O
Answer
H
H
H
C
O
C
H
O
The resulting p -bond is polarized toward the O atom as you'd expect on the basis of O's electronegativity
despite the smaller size of the atomic p-orbital.
Question (3 marks)
The molecule shown below is called allene. It is in a class of molecules known as cumulenes.
i.) What is the hybridization of each of the three carbon atoms?
ii.) Draw a picture of allene (using wedge/dash bonds if necessary) to indicate its structure. You probably
want to build a model of this molecule.
iii.) How are the p-bonds of allene oriented with respect to one another?
H2C C CH2
Answer
i) The central C atom is sp hybridized - the other C atoms are sp2 hybridized.
ii)
H
H
H
C C C H
H
H
C C C H
H
iii) They are at right angles to one another. In the diagram above the p-bond on the right is in the plane of the
page - the one on the right is orthogonal to the page.
2
H
H
H
Csp
H
2
Csp
Csp
2
H
The s-framework.
H
C
C
C
H
H
The atomic orbitals making up
the p-bonds. Note that the two p-orbitals
that result must be orthogonal to each other.
Question (8marks)
The carbon-carbon double bond of ethylene consists of two different types of bond, given the symbols s and p.
H
H
C C
H
H
i.) How are the carbon atoms hybridized?
ii.) Sketch the orbitals (atomic and/or hybrid orbitals) that are used to form the s bond and sketch the bonding
molecular orbital that results from their overlap.
iii.) Sketch the orbitals (atomic and/or hybrid orbitals) that are used to form the p bond and sketch the bonding
molecular orbital that results from their overlap.
iv.) Which of the two types of bonds (s or p) is stronger and why?
v.) Rotation of the carbon-carbon bond in ethylene is not possible. Explain making reference to the bonding.
Answer
i) They are sp2 hybridized.
ii)
H
H
C
C
H
H
1s atomic orbital
H
C
sp2 hybrid
C
H
H
H
iii)
p-bond
Unhybridized atomic p orbitals
H
H
C
C
H
H
H
H
C
C
H
H
iv) Because s-bonds result in enhanced electron density in the region of space directly between the two positive
nuclei, they will be stronger. The enhancement of electron density in the p-bonds is above and below the plane
- less direct, therefore a p-bond is not quite as strong. Do not confuse the strength of a double vs. a single bond
with this question. Double bonds are stronger than single bonds because they are composed of a s- and a pbond.
3
v) Rotation of one end of ethylene destroys the overlap of the atomic p-orbitals necessary for the formation of a
p-bond. Effectively, the bond must be broken for rotation to occur.
Question (8marks)
The carbon-carbon triple bond of acetylene consists of two different types of bond, given the symbols s and p.
H C C H
i.) What is the hybridization of the carbon atoms?
ii.) Sketch and label the atomic orbitals of carbon that are used to form the two hybrid orbitals and sketch the
hybrid orbital(s) that result from them.
iii.) Sketch the C-C s bonding molecular orbital that results from the overlap of the two hybrid orbitals.
iv.) Sketch the atomic orbitals that are used to form one of the p bonds and sketch the bonding molecular
orbital that results from their overlap.
Answer
i) They are sp hybridized.
ii)
Usually omitted
for clarity.
2s
C
C
2p
C
2 sp hybrids
C
2p
2s
iii)
C
C
C
C
s-bond
2 sp hybrids
iv)
p-bond
Unhybridized atomic p orbitals
H
C
C
H
H
C
C
4
H
Question
Shown below is the MO diagram for the p-bond in cis-2-butene.
i.) On the left diagram, sketch the two molecular p-orbitals and the atomic orbitals from which they are
derived.
ii.) Add the electrons to the diagram.
iii.) If cis-2-butene is irradiated with UV light, then a facile bond rotation is possible:
H
H
hn
H
C C
H3C
CH3
C C
CH3
H3C
H
The light converts cis-2-butene into its first excited state (an electron excited state). In the right diagram, show
what has happened to the electron distribution in the p-MOs upon excitation.
v.) Why is bond rotation facilitated by irradiation?
Answer
ii and iii)
C
C
p*
p*
p
C
p
Ground state:
p bond order = 1
C
Excited state, p bond
order = 0
v) This excitation does not effect the s-bond. But the p-bond order becomes zero since there are now equal
numbers of bonding and antibonding electrons. The C-C bond effectively becomes single in character.
5
Question
Sketch the bonding molecular orbital (MO) formed from the overlap of a N sp3 hybrid and hydrogen 1s atomic
orbital.
N
H
Answer
N
H
Question (four marks)
Metal carbonyls such as Fe(CO)5 are compounds made from transition metals and carbon monoxide. It is
believed that part of the metal-carbon bond results from the overlap of a filled (with electrons) dxy Fe orbital
with the p* orbital of carbon monoxide. These orbitals are shown below.
CO
OC
Fe
CO
CO
Fe
C
O
Fe
C
O
CO
Fe(CO)5
a.)
b.)
c.)
d.)
Sketch the bonding molecular orbital that results from the overlap of these two orbitals.
Does this MO resemble more closely a p-type or a s-type orbital? Why?
Draw lines to show the position of all nodes.
Which is longer: the C-O bond length in free carbon monoxide or the C-O bond length in Fe(CO)5 ?
Explain your answer.
Answer
a)
Fe
C
O
Fe
C
O
Nodes
b) It is not radially symmetrical and has a node in the plane of the Fe-C bond. It is a p-type of orbital.
c) Shown above.
d) The orbital drawn above is bonding with respect to the Fe and C atoms, but anti-bonding with respect to the
C and O atoms. If this orbital is populated with electrons, then you'd expect to see a weakening and lengthening
of the CO bond.
6
Resonance Structures
Arrow Pushing Practice
The following is a collection of ions and neutral molecules for which several resonance structures can be drawn.
For the ions, the charges can be delocalized implying greater stability; for the neutral molecules, electrons can
be shuffled around to show areas of high or low electron density. Use electron arrows to show how one
resonance structure can be transformed into another.
In this example, a lone pair of electrons on an O atom is delocalized onto another O atom through the psystem as shown. Note that the arrows start at a pair of electrons. These resonance structures are used to
explain the acidity of the carboxylic acid functional group.
O
H 3C
For example:
O
C
H 3C
O
C
O
The resonance structures below show how a charge-separated resonance structure can be drawn for a
neutral molecule. This additional resonance structure suggests that the electron density is high on the carbon
that bears a negative charge. Indeed, there is spectrascopic evidence that confirms this and this carbon acts as a
nucleophile in many reactions. Note that the electron pushing is identical to that in the example above.
H 3C
O
H 3C
C
O
CH2
C
CH2
H
H
The following resonance structures depict the allyl cation. Allyl cations are relatively stable (as cations
go) due to delocalization of the positive charge. In this case, the p-electrons move towards the electron
deficient carbon atom. Again, the arrow starts at an electron pair - the p-electrons in this case.
H 2C
CH2
CH2
H 2C
Look for the similarities in these electron movement processes. There are only three basic kinds of electron
movements in all of these examples: movement of a lone pair of electrons towards a p-bond, movement of a pbond towards a carbocation or movement of a lone pair towards a carbocation.
The next example shows how an electron-deficient carbocation is stabilized by an adjacent atom that
bears a lone pair. Again, note that the arrow starts at an electron pair. The resulting resonance structure is a
very important one since, unlike the original (left) structure, all atoms have their octets.
OH
H
OH
C
H
H
C
H
In the vinyl cation, the positive charge cannot be delocalized by the p-electrons because the empty porbital on C is orthogonal to the electrons in the p-bond.
H
C C H
H
1
Armed with these examples, you should be able to discover all of the resonance structures of the following
molecules. Enjoy.
The allyl anion:
The conj. base of nitromethane:
The pentadienyl cation:
H
O
H
H 2C
C
H 2C
N
H 2C
O
CH2
The phenoxide anion Aniline:
H
C
C
C
H
The benzyl cation:
Conj. base of
acetaldehyde (ethanal)
O
CH2
NH2
(Conj. base of phenol)
CH2
O
H 2C
The propargyl cation: Protonated acetone:
C
C
CH2 H 3C
CH3
O
O
C
H
H
N
H
O
O
C
H 3C
C
C
OH
C
H 3C
C
H
OCH3 H C
3
H
H
N
H 3C
O
C
H
O
CH3
H
Conj. base of propionitrile:
OH
H
C
C
N
H
2
C
NH2
Question
In each of the following, judge which of the resonance structures are more or less important contributors than
the others. Explain your reasoning.
:
+
H2C CH CH2
_
O
:
:
O
H3C C +O CH3
:
:
CH3
H3C C
+ O
:
:
:
: :
H3C C O CH3
_
: :
O
c.)
CH2
:
+
b.) H2C CH CH2
+
H3C O
:
: :
+
CH2
a.) H3C O
Answer
a) The left-hand structure has an electron deficient carbon where the right-hand structure has complete octets
for all atoms. This makes the right-hand structure more favourable, even though it puts a positive charge on
oxygen.
b) These structures are degenerate (equal). Please note that these are DIFFERENT RESONANCE
STRUCTURES NONETHELESS. These structures imply that the charge is delocalized over 2 carbon atoms,
and indeed, this kind of carbocation (called an allylic carbocation) is resonance stabilized.
c) The first structure is most favoured. It has no electron-deficient atoms, nor does it have any charge
separations (as the third structure has). The third structure is favoured over the second (middle) structure for the
reasons outlined in (a) above.
Question Six
For the following sets of resonance structures, check to see that all structures are valid according to the rules set
out in your text. If they are not valid, indicate why not.
H
i.)
H
O
C
O
C
H
H
C
H
H
O
ii.)
H
C
O
H
iv.)
H 3C N
(Formal charges
O
are omitted.)
H 3C N
O
O
C
H
H
H
iii.)
C
O
O
H
C O
C
C
.
H
H
O
C O
O C
O
Answer
i.) These are not resonance structures because a proton is shifting its position. This violates Rule 2. The
process shown does take place but it represents a chemical equilibrium between two distinct chemical species.
This is known as a tautomerism; in this case an enol-keto tautomerism.
ii.) This looks a lot like (i.) but without the proton shift. These are valid resonance structures for a species
known as the enolate anion, very important in carbonyl chemistry.
iii.) If you count the electrons around the O atom with the negative formal charge you will note that it comes to
10, 2 more than an octet. This is not valid for 2nd row elements (see Rule 4).
iv.) The structure on the left has two more electrons that the one on the right. This violates Rule 3. If you
worked the formal charges out you probably noticed that these two species have different overall charges.
That's not allowed either.
3
H3C
O
H3C
CH2
O
CH2
:
a.)
: :
Question Four (four marks) Explain your reasoning.
For the two species shown, evaluate the resonance structures and determine:
i.) which atom bears most of the charge in the species?
ii.) does the oxygen-CH2 bond length correspond more closely to a single or a double bond?
Answer
The right hand resonance structure contributes more to the character of this species since all atoms have their
octets. As a result, the O atom clearly bears most of the charge and the O-CH2 bond length will be closer to
double bond length.
Question Five (six marks)
The following are all valid resonance structures for methyl isopropenyl ether. All non-bonding electrons are
shown.
H3C
O
H3C
C
H3C
O
CH2
C
H3C
1
H3C
O
CH2
C
H3C
2
CH2
3
a) Rank them in decreasing order of importance, i.e. which structure contributes most to the character of this
molecule, etc. Explain your reasoning.
b) What do these structures imply about the C=C bond length?
Answer
Structure 1 is best since it is uncharged and all atoms have their octets. Structure 3 is next best because all
atoms have octets. Structure 2 is the least important since carbon is electron deficient.
These structures imply that there is a small amount of double bond character in the carbon-carbon bond making
it a little shorter.
:
:
Question One
Give all reasonable resonance structures of the following.
a) [H3COCHOCH3]+
b) [CH2NO2]c)
d)
O
O
C
H
N
H3C
S
O
:
H3C
H
O
4
H3C
C
CH3
O
H3C
C
H
b)
H2C
H
N
:
:
O
C
H3C
H3C
H
O
S
H
N
H
d)
CH3
H
O
H3C
:
C
C
O
:
:
H3C
H3C
N
O
O
CH3
O
O
N
c)
O
H
O
H2C
:
:
O
:
:
O
C
N
H
This structure
not always shown.
O
H3C
O
O
S
H
:
O
:
:
:
:
a)
:
:
Answer
O
H3C
O
O
S
O
O
Question
The following pairs of resonance structures describe the molecule acetone (I) and protonated acetone (II).
O
I
H3C
C
O
II
H3C
C
O
CH3
H3C
C
H
O
CH3
H3C
C
CH3
H
CH3
Examine the two pairs of resonance structures. Which the these two species (I or II) has the most electrophilic
character? (In both cases, the right-hand structure has a carbocationic carbon that is capable of acting as an
electrophile. You must keep in mind the rules for determining the "favourability" of certain resonance
structures over others and also that molecules are hybrids (sort of weighted averages) of their resonance
structures.)
5
Answer
This is an important point but a tricky thing to perceive. We are comparing two species (I and II) each
of which is represented by two resonance structures. In each pair of resonance structures, the left-hand structure
is the one that contributes the most character (because all atoms have full octets). Comparing the right-hand
structures, we note that for I we have a destabilizing charge separation and an electron deficient carbon while
for II we have no charge separation, only the electron-deficient carbon. As a result, the right-hand structure of
II contributes more to the character of II than the right-hand structure of I contributes to the character of I.
Since the right-hand structure in each case is the one which has the electrophilic carbocation character, this
means that II will be more electrophilic than I.
Question (5 marks)
a.) Rank the following resonance structures for this molecule
in order of importance
_
_ and state your reasoning.
O
O
H
H C N+
2
H
H
H C N
1
O
H
C N H
H +
3
.
H
b.) The molecule above can act as a Bršnsted base. Will it accept a proton on the O atom or the N atom?
Explain your reasoning.
Question Seven (5 marks)
a.) Rank the following resonance structures for this molecule
in order of importance
_
_ and state your reasoning.
O
O
H
H C N+
2
H
H
H C N
1
O
H
C N H
H +
3
.
H
b.) The molecule above can act as a Bršnsted base. Will it accept a proton on the O atom or the N atom?
Explain your reasoning.
Answer
a) The structure 1 is the most important contributor to the character of formamide. All atoms have octets and
(unlike the latter two structures) there are no charge separations (see Rule 5).
The next most important contributor is the middle structure. While it has a charge separation (making less
important than the first structure) all atoms have complete octets (unlike the third structure). Another way of
noting this is that it has one more bond than the last structure (three lines as opposed to only two). See Rule 4.
The least important contributor is the last structure. It has a charge separation and also has an electron deficient
carbon atom (having only six electrons; the carbon bearing the positive charge).
b) To figure this out, you might have looked at the resonance structures of the conjugate acid of this compound
that result from protonation on the O atom vs. protonation on the N atom. For O atom protonation there are
three possible structures:
+ OH
OH
H
H C N
1
H
H C N+
2
H
H
For N atom protonation there are only two:
O
OH
C N H
H +
3
.
H
_
O H
C N H
H +
+
3
.
H
H
H
H C N+
1
H
6
As a result, it is clear that the conjugate acid which results from protonation on the O atom is more resonance
stabilized than that resulting from N protonation. This will make the O atom effectively more basic. Note, that
this is opposite to the case described in question 4, where resonance factors are not involved.
Question One (6 marks)
i.) What functional group does this molecule contain?
ii.) Rank the following resonance structures in order of importance by saying which one contributes most to the
character of this molecule, and which contributes least. Explain your reasoning.
H3C
O
O
C
C
1
O CH3
H3C
O
O
CH3
2
H3C
C
O
CH3
3
Answer
i.) It is an ester.
ii.) Structure 1 is best since there are no charges and no electron deficient atoms. Structure 2 is the smallest
contributor since it has an electron deficient C atom and a charge separation. Structure3 is intermediate between
1 and 2 because it has a charge separation but more bonds (no electron deficient atoms) as does 2.
Question (6 marks)
The following resonance structures are used to describe the molecule diazomethane. It is a bright yellow,
volatile, toxic, explosive and useful material.
H2C N N
H2C N N
i.) Indicate the presence of any formal charges on these resonance structures by placing + and/or - signs
above the appropriate atoms. Show all such charges in both structures.
ii.) What is the hybridization of the central N atom?
iii.) Which resonance structure contributes more to the character of diazaomethane: the one on the left or
the one on the right?
Explain your reasoning.
iv.) Indicate in which direction the net molecular dipole of diazomethane points by labeling one of the
structures with a dipole vector.
v.) Where would a proton, H+, attack this molecule: on the C atom, the central N atom or on the terminal N
atom?
Answer
i.)
H2C N N
H2C N N
ii) It is sp hybridized.
iii) Both structures are the same in terms of charge and the number of bonds. The left hand structure is best, in
that the negative charge is located on the more electronegative atom.
iv) Based on which structure dominates the character of this molecule, one would draw a dipole pointing from
carbon to nitrogen.
v) Dizaomethane protonates on carbon since C- is more basic than N- (C less electronegative than N).
7
Question (6 marks)
i.) Rank the following resonance structures in order of importance by saying which one contributes most to the
character of this molecule, and which contributes least. Identify any degenerate (equivalent) structures.
Explain your reasoning.
O
H3C O
C
O CH3
1
H3C O
O
C
O CH3
H3C O
2
O
C
O
C
O CH3
H3C O
O CH3
4
3
Answer
Structure 2 is the least important contributor (has fewer bonds - electron deficient carbon and bears charges).
Structures 3 and 4 are degenerate (contribute equally). They have the same number of bonds as structure 1 but
have charges present.
Structure 1 is the most important contributor. No charges and no electron deficient atoms.
Question (6 marks)
The following resonance structures are used to describe the molecule methyl isocyanate. It is the toxic gas that
was at the root of the Bhopal disaster.
H3C N
H3C
C O
N C
H3C N
O
C O
i.) Indicate the presence of any formal charges on these resonance structures by placing + and/or - signs
above the appropriate atoms. Show all such charges in all structures.
ii.) What is the hybridization of the central C atom?
iii.) Of the two, which resonance structure contributes more to the character of methyl isocyanate: the one
on the left or the one on the right?
Explain your reasoning.
iv.) Indicate in which direction the net molecular dipole of methyl isocyanate points by labeling one of the
structures with a dipole vector.
Answer
I)
H3C N
H3C
C O
N C
H3C N
O
C O
ii) It is sp hybridized.
iii) Both structures have the same number of bonds and same charges. The best structure will have the negative
charge on the more electronegative atom. That would the the left hand structure.
iv) The dipole should point towards the O atom.
Question
N2O is a linear molecule in which the atoms are ordered NNO. Draw the resonance structures for this molecule
and indicate which contribute more to the character of this molecule. On the basis of this, what net dipole
would you expect for this molecule?
:N
_
+
N O:
: :
: :
+
N N O
:
_
:
Answer
The two resonance structures are:
In both cases all atoms have their octets. In both cases their are two charges, one of them being a positive
charge on the central N atom. The difference, then is the placement of the negative charge. This charge will
prefer to sit on the more electronegative oxygen atom, therefore the right-hand structure will be major resonance
structure and the net molecular dipole will point toward the O atom. Other valid resonance structures can be
written but they involve large charge separations or incomplete octet atoms.
8
Alkanes/Nomenclature
Question
Draw the following molecules and, if necessary, give correct names to them.
a.) 3-ethyl-4-methylpentane
b.) 5-neopentyldecane
c.) 2-ethyl-2,4,6-trimethylheptane
d.) 3,4-dimethyl-1-isopropylcyclohexane (ignore cis/trans isomers)
e.) 3-butyl-2,2-dimethylhexane
(f.) cis-3-methyl-3-pentene
Answer
a)
b)
CH2
H3C
C
c)
3,3,5,7-tetramethyloctane
CH3
3-Ethyl-2-methylpentane
CH3
CH3
e)
d)
f)
4-t-butyloctane
cis-3-methyl-2-pentene
Question (7 marks)
a.) Give systematic names for the following compounds. Do not include any stereochemical descriptors (i.e.
configurational symbols R or S, E or Z).
(a.)
H2
C
H3C
CH
CH3 (b.)
F
(c.)
H3C
CH3
CH3
CH3
Answer
a) sec-butylbenzene (or s-butylbenzene)
b) 2-fluoro-3,5-dimethylhexane
c) 2-methylisopropylcyclopentane
1
H3C
Question
Give acceptable IUPAC names for the following eight molecules.
b)
a) H3CH2C CHCH
CH3
CH2CH3
CH3
3
H3CH2CH2 C CH CH2CH3
C C C CHCH2CH3
H2
H2
CH3
CH2CH2CH3
H
f)
CH3CH2HC
d)
BrH2CH2CH2 CHCH3CH2 CH2C
C
C
c)
H
C
C
ClH2 CH2CH2C
e)
H
C
C
H
H
H3CH2CH2 C
H
C
g) H3CH2CH2 C
CH2CH(CH3)2
C
CH2CHBrCH2 CHCH3
CH2CH3
Answer
a) 4-ethyl-4-methylheptane
b) 3-ethyl-5,5,7-trimethylnonane
c) cyclopropylethyne
d) 8-bromo-6-methyl-1-octyne
e) 2-propyl-5-chloro-1-pentene
f) trans-3-isobutyl-6-n-propylcyclohexane
g) cis-7-bromo-9-methyl-4-undecene
Question Two (7 marks)
a.) Give systematic names for the following compounds. Do not include any stereochemical descriptors (i.e.
configurational symbols R or S, E or Z).
(a.)
(b.)
CH2CH3
CH2CH3
H3C
(c.)
CH
CH
CH2 CF2 CH3
CH3
C CH3
CH3
CH2CH3
Answer
a) ethylcyclobutane or cyclobutylethane
b) 4-ethyl-2,2-difluoro-5-methylheptane
c) t-butylbenzene
Question (4 marks)
Name the following compounds.
H Br
CH3
a.)
H3CH 2CH2C C CH CH(CH 3)2
H
CH2CH2CH3 c.)
b.)
2
d.)
H3C
CH2Cl
Answer
a) 3-bromocyclohexene
b) 4-isopropyl-5-methyloctane
c) vinylbenzene or phenylethene or phenylethylene
d) 1-chloro-1-butyne
Question (7 marks)
a.) Give systematic names for the following compounds. Do not include any stereochemical descriptors (i.e.
configurational symbols R or S, E or Z).
(a.) CH2CH(CH3)2
CH3
(b.)
F
(c.) CH3
H3C
CH3
Br
CH3
Answe
r
a) isobutylbenzene
b) 5-fluoro-2,3-dimethylhexane
c) 1-bromo-2-methylcyclobutane
Question (7 marks)
a.) Give systematic names for the following compounds. Do not include any stereochemical descriptors (i.e.
configurational symbols R or S, E or Z).
(a.)
H2
C
H3C
CH3 (b.)
CH
F
H3C
(c.)
H3C
CH3
CH3
CH3
Answer
a) sec-butylbenzene or s-butylbenzene
b) 3,5-dimethyl-2-fluorohexane
c) 2-methylisopropylcyclopentane
Question (3 marks)
Give systematic names for the following compounds:
Br
F
Answer
a) 5-methyl-1-fluoroheptane
b) isopropylcyclopropane
c) 4-ethyl-2,2-dimethylhexane
d) 3-bromocyclohexene
Question (2 marks)
Draw the most stable and least stable conformers of 2,3-dichlorobutane using Newman projections. Methyl
groups are bigger than chlorines, by the way.
3
Answer
CH3
Cl
H
H
H
H
Cl
CH3
CH3
CH3
Cl Cl
Most stable
Least stable
Question
Shown below are two isomeric decalins. Which one is more stable? Explain.
H2
C
H2C
H2C
C
H2
H
H2
C
H2
C
C
CH2
H2C
C
CH2
H2C
H
C
H2
C
H2
H
H2
C
C
CH2
C
CH2
H
C
H2
cis-Decalin
trans-Decalin
Answer
H
H
H
H
H
trans-Decalin
H
cis-Decalin
Note that this C-C bond is axial. This has the same
effect as an axial substituent in that it interacts sterically
with the 1,3-hydrogens. This destabilizes this isomer
relative the the trans-isomer which lacks any 1,3-diaxial
interactions.
Question Two (nine marks)
a.) Give systematic names for the following compounds. Include any stereochemical descriptors (i.e.
configurational symbols R or S, E or Z) where possible. Part (c) is not a Fischer projection.
(a.)
F
H
(b.)
H2
C
H3C
CH3
CH
C
H2
CH3
CH
CH
CH3
H3C
CH2
Answer
a) trans-1-fluoro-3-methylcyclopentane
b) 6-ethyl-3,5-dimethylnonane
c) sec-butylbenzene or s-butylbenzene
4
H
H3C C CH2CH3
CH3
H2
C
H2C
H
(c.)
Question
a.) Sketch the conformational energy diagram for rotation about the C1-C2 bond of 1,1- dichloropropane.
b.) Label the diagram with Newman projections corresponding to the relative minima and maxima.
c.) Identify the eclipsed and the staggered conformations.
H H
1,1-Dichloropropane
H3C C CHCl2
Answer
Eclipsed
Eclipsed
Cl H
Staggered
Cl
∆E H
H
H
Staggered
Cl
H
H
Eclipsed
Cl H
Cl CH3
Cl
CH3 H
Cl
H CH3
CH3
H
Cl
H
H
Staggered
Cl
H
Cl
H
Cl
Cl
H
H
CH3
CH3
H
H
Question Four (3 marks)
Draw the Newman diagrams (looking down the C2-C3 bond) for:
a.) the lowest energy staggered conformation
b.) the lowest energy eclipsed conformation
of 2,3-dimethylbutane.
CH3
H3C
CH
CH
CH3
2,3-Dimethylbutane
CH3
c.) Draw the most stable conformation of cis-1-t-butyl-4-methylcyclohexane.
t-Bu
Me
Answer
a)
CH3
H3C
H
H
b)
c)
H3C
H
CH3
H
CH3
CH3
H3C CH3
Lowest energy
staggered.
Lowest energy
eclipsed.
CH3
H
C(CH3)3
H
5
The t-butyl group is much
larger and will prefer to
be equatorial. This forces
the CH3 group to be axial.
If both groups to be axial,
one would
need
a
different molecule.
Question (4 marks)
a.) Give a systematic name for the following molecule and, also, in addition, draw it in its most stable
conformation.
CH3
H3C C CH3
Br
b.) Give a systematic name for the following molecule and, also, in addition, draw the Newman projection
(along the C2-C3 bond) of its most stable conformer.
Answer
a)
H
C(CH3)3
Br
H
H
H
b) 2-Methylbutane
CH3
CH3
H
CH3
Question
In a study of the acidities of alcohols, it was desired to compare the acidities of hydroxyl groups in the
equatorial and axial positions. The equatorial case is easily measured since, in cyclohexanol, the hydroxyl
group prefers to adopt an equatorial position. What structurally modified cyclohexanol would be useful in
determining the acidity of an axial hydroxyl group? Explain your reasoning.
Answer
A couple of possibilties, both of which involve placing another substituent on the ring that has a strong
preference for the equatorial position.
OH
OH
H
t-Bu
H
H
a trans-decalin
6
Stereochemistry
Question
Consider cis-1,2-dimethylcyclohexane, shown below in its chair form.
CH3
H cis-1,2-dimethylcyclohexane
CH3
H
i.) Is this conformation chiral?
ii.) Is this molecule chiral? Why or why not?
Answer
Yes, this conformation is chiral because its mirror image is non-superimposable. The molecule is not
considered chiral because these mirror image conformations can be converted into each other simply by ring
inversion. You can also orient this molecule in the following conformation
H3C
CH3
H
H
mirror
plane
.
In this conformation it is obvious that this is a meso compound. A good rule to remember is that if a molecule
has any conformation in which it is achiral, then it is an achiral molecule.
Question
You dissolve 65.3 mg of camphor in 100. mL of methanol and measure (in a 1.00 dm sample tube) the optical
rotation to be -2.88°. What is the specific rotation of camphor? Watch your significant figures, folks.
Answer
[α ]oD =
α
lc
[α ]oD =
−2.88°
= −44.1°
0.0653g / 100mL ·1.00dm
Question
For each of the following pairs of molecules, identify the stereochemical relationship. Are they enantiomers,
diastereomers, constitutional isomers (isomers with differing connectivity), or the same molecule. Building
models of these will be a real help.
ALSO IDENTIFY ANY MESO COMPOUNDS AS SUCH.
a.)
CHO
H3C
OH
H
CH3
H
CHO
OH
As you can tell, these are Fischer projections.
c.)
H
HO
Cl
HO
Cl
H
d.)
Br Br
Br Br
H
b.)
H
1
H
H H
Cl
Cl
Cl Cl
H
e.) H
F
H
F
CH2OH HOCH2
HO
OH
H
g.)
H3C
H3C
Cl H3C
H
H3C
Cl
H
Answer
a.)
CHO
H3C
OH
H
H
F
HO
H
H
H
f.)
h.)
H
CH3
H
CH3
HO
H
F
CH3
H
F
F
CH3
H
CH3
H
CHO
OH
These are enantiomers. You may have noticed
that one diagram can be converted into the other
simply by rotating 90°. You can't do that with
As you can tell, these are Fischer projections. Fischer projections. Remeber what how these
projections are defined.
CHO
H3C
OH
H
CHO
CH3
CHO
Rotate 90°
H
CHO
H3C
OH≠ H3C
OH
OH
H
H
b.)
HO
c.)
These are constitutional isomers. In the first molecule,
the double bond is closer to the HO group; in the
second it is closer to the Cl group. The connectivities
Cl are therefore different.
HO
Cl
H
H
These are enantiomers. Note that the other
chair conformers of these molecules have both
bromines axial, but these conformers are also
going to be enantiomers.
Br Br
Br Br
H
H
d.)
H
H H
Cl Cl
Cl
e.)
H
F
H
Cl One is cis and the other trans-1,3-dichlorocyclobutane.
They are stereoisomers in that the connectivities are
the same. They are obviously not enantiomers, so
H they must be diastereomers.
H I love this question and use it every year. As
F drawn these clearly are mirror images of each
CH2OH HOCH2
OH
H
HO
H
other. But if you look carefully, there is no
stereocenter in this molecule. This is obvious
H if you draw it another way:
H
F
CH2OH
These represent
the same molecule. This
CH2OH
is a wonderful example of how
drawings can mislead you. If you got this wrong,
and most of you will, take comfort: I blew it too.
2
f.)
H
F
HO
H
F
top stereocenter is different, the bottom one
CH3 The
is the same. These are diastereomers.
H
CH3
CH3 HO
H
CH3
g.)
H These are planar molecules and have no
Cl H3C
H3C
H3C
stereocenters. E/Z isomerism is not possible
since the left end of the double bond has two
H3C
Cl identical substituents. If you flip one of these
over, it will be obvious that these are the same
molecule.
These are different drawings of the same molecule.
F
H This molecule lacks stereocenters and has an internal
mirror plane of symmetry that runs through the CF bond
and chops the ring in half.
H
h.)
H
F
Question
Give the configurational symbols (E/Z, R/S) for all of the stereocenters in each of the following molecules:
a.)
CHO
H3C
b.) H C
3
Cl
OH
H
H
CH2CH3
Answer
2
a.)
CHO
H
c.)
Br
Br
d.)
HO
O
H
H
The Parity rule works well for Fischer projections.
Recall that the horozontal bonds are coming towards you. This will
H3C
OH give the sequence 1432 as one possibility and 3421 as the other.
3
This molecule is S.
H4
1
b.)
1
1
H3C
Cl
H
After prioritizing each end of the double bond, we find both
top priority groups on the same side. This molecule is Z.
CH2CH3
2
H R
c.)
The best way here is to build a model and figure this out.
Note that for this molecule, both symbols had to be the same.
If they were opposite, then this would be a meso-compound.
Br
Br
R
H
The problem here is distinguishing between the 2nd and 3rd
priority groups. I guess the easiest way to explain this is to point out
4
that in one case you follow the sequence
H R 2 C-O-C (as you go around the ring)
HO 3
and the other way you go
C-O-H. Since C has priority over H, the former path has higher priority.
d.)
1
O
3
Question
Assign configurational symbols to the following.
CH3
H 3C
N
H
CH
2
H 3C
C
C
H 2C
H
H
O
OH
H
OCH3
C
H 3C
Darvon, a painkiller
OH
H
O
O
H 3C
CH3
H
CH3
Isomenthol: a terpene, whatever
that is.
O
ClCH2CH2
C
C
COOH
H
C
N
H
CH3CH2
H 2N
COOH
H
OH
H
CH2OH
HO
(-)-Serine, one of
H
those nifty amino acids
L-Proline
H Cl
CH2CH3
OH
H Br
H 3C
O
H
H
CH3
H
H
OH
O
O
H
HO
H
C
CH3
H
NHCOCHCl2
F
F
CH2OH
Chloramphenicol, an antibiotic
used in the treatment of typhoid.
4
CH=CH-CH=CH2
Answer
H 3C
N
H
CH3
CH2
H 3C
CR
C
S
H 2C
H
H
R
O
OH
H
Darvon, a painkiller
OCH3
C
H 3C
H
R S
O
O
H 3C
CH3
S
R
CH3
Isomenthol: a terpene, whatever
that is.
H
O
ClCH2CH2
COOH
Z
C
C
S
N
H
H
L-Proline
H Cl
H 3C
Z
H 2N
COOH
H
CH3CH2
O
R
R
H
OH
C
S
OH
H
CH2OH
(-)-Serine, one of
HO
those nifty amino acids
H
OH
R
S
H
Z
H
R
CH2CH3
H Br
R
CH3
H
OH
O
O
H
HO
H
R
C
R
H
R
R
NHCOCHCl2
R
CH2OH
F
F
Chloramphenicol, an antibiotic
used in the treatment of typhoid.
5
CH3
R
CH=CH-CH=CH2
Question
What is the stereochemical relationship between the following pairs of molecules?
i)
ii)
Br
Cl
H
H
CH 3
CH3CH 2
Cl
H
iii)
Br
CH 2CH 3
CH 3
H
iv)
Cl
CH 3
H
Cl
H
COOH
CH 3
COOH
OH
H
HO
HO
H
v)
H
H
OH
vi)
H
H
H
H
H
H 3C
CH3
H 3C
F
F
H
CH 3
H
F
viii)
O
O
CH3
CH3
O
Cl
H
Cl
Cl
H
CH 3
H
H
H
H
O
H
H 3C
Cl
ix)
H
x)
Cl
Cl
Cl
H
Cl
CH3
CH 3
CH 3
H
H
H
H
H 3C
H
H
CH3
Cl
H
H
CH2Cl
xi)
H 3CH 2C
H 3C
CH 3
CH 2CH 3
H 3C
CH 2CH 3
H 3CH 2C
CH 3
Answer
i) identical molecules ii) enantiomers
iii) enantiomers
v) same molecule (a meso compound)
vi) diastereomers
viii) rotational isomers of the same molecule
ix) enantiomers
x) constitutional isomers
xi) identical molecules
6
H
F
H
H
vii)
F
iv) enantiomers
vii) enantiomers
F
Question (7 marks)
Examine the following pairs of molecules.
a.) What is the stereochemical relationship between the following pairs of molecules? Are they constitutional
isomers, the same molecule (note: changing the conformation of a molecule does not make it a different
molecule), enantiomers or diastereomers.
b.) Clearly identify all meso compounds.
a.)
H3CH 2C
H
c.)
CH3
OH
CH3
H
H3CH 2C
CH3
Br
b.)
OH
H
CH3
H3C
H
Br
H
Br
H
Br
Cl
Cl
H
CH3
H3C
H
CH3
d.)
H
Cl
CH3
CH3
Br
e.)
H
Cl
H
H
Br
Answer
a) same molecule
b) enantiomers
c) diastereomers, the left hand molecule is a meso compound
d) same molecule, a meso compound
e) same molecule
Question
The isolation of Cordycepic acid from a fungus was reported in the Journal of the American Pharmaceutical
Association, 46, 114, (1957). It was found to be optically active with [a]26 = +40.3°. It was assigned the
structure shown below in which all four hydroxyl groups are equatorial.
HO
HO
HO
O
OH
OH
Should you believe everything you read? Elaborate making reference to the above example.
7
Answer
This compound has several (two) chiral centers but also has an internal mirror plane of symmetry. It is achiral
and cannot possibly be optically active.
HO
HOCO2H
O
OH
HO
OH
HO
HO
OH
OH
Question
1.) Convert the following structure of Sorbitol, a sugar derivative and artificial sweetener, into its
corresponding Fischer diagram.
CH2OH
OH
H
OH
H
CH2OH
HO
CH2
H
OH
H
OH
CH2OH
Is Sorbitol chiral, yes or no? (three marks)
Answer
CH2OH
OH
H
OH
H
HO
CH2
H
OH
H
H
CH2OH HO
H
OH
H
CH2OH
OH
H
or
HO
H
HO
H
OH
H
OH
HO
CH2OH
OH
H
CH2OH
These are the same: take one and rotate by 180°.
Yes, it is chiral.
8
Question
a) Complete the Fischer projection diagrams for the two compounds shown below.
H
OH
O
HOCH2
H
H
OH
H
Reduction
H
OH
HOCH2
OH
CH2OH
H
D-(+)-Xylose
OH H
"Xylitol"
CHO
OH
CH2OH
CH2OH
CH2OH
b) Reduction of D-(+)-Xylose (a sugar, [a] = -9.3°) gives Xylitol (an artificial sweetener) which has no
measurable optical activity. Why not? The product belongs to a special stereochemical class. What might that
be?
Answer
Xylitol has an internal mirror plane of symmetry that xylose does not. Xylitol is a meso compound and
therefore lacks any optical activity.
24
D-(+)-Xylose
CHO
H
OH
HO
H
H
OH
CH2OH
"Xylitol"
CH2OH
H
HO
OH
H
H
OH
CH2OH
Question Five (3 marks)
a/) When looking at the structure of a molecule, how does one determine the maximum number of possible
stereoisomers?
b.) When is the actual number of stereoisomers less than that predicted by the procedure outlined in (a)?
Answer
a) One counts all of the chiral centers and double bonds that are E or Z. If we let n = the sum of these two
numbers, then the maximum number of stereoisomers is 2n.
b) If some of the stereoisomers are meso compounds, then this will reduce the total number of possible
stereoisomers.
Question Six (4 marks)
a.) Examine the molecule below (drawn in 2-D). Is this molecule really flat? If not, draw it in three
dimensions using dash/wedge bonds to present a more accurate descriptiton. Building a model is highly
recommended.
c.) Is this molecule chiral?
b.) Label (single examples will suffice; you
CH3
H
need not label all examples):
i.) a primary (1°) H atom
ii.) an sp2 hybridized C atom
iii.) a vinyl H atom
H
H3C
iv.) an allylic C atom
9
Answer
vinyl H
1° H atoms
sp2 C
H
CH3
H3C
H
allylic C
c) This molecule is not superimposable on its mirror image, therefore it is chiral (even though it has no
chiral centers - remember chirality is a property of objects).
Question
Assign configurational symbols (R/S or E/Z) to the following molecules but do not name them.
H
OH
H3C
Br
H
H
O
H
H3CH 2C
H
C
C
H
C
CH2CH3
H
H
CH3
Answer
H
1
3
H3C R
H
1
OH
Br
H
4
PR: 3124 = 2 = R
H
H
S
H3CH 2C
2
3
2
H3C
4
O
2
PR: 4123 = 3 = S
1
H
C
H
1C
Z
C
H
2
CH2CH3
H
2 CH3
1
R Br
H4
3
PR 4321 = 6 = R
PR means using the parity rule. You may have gotten a different sequence BUT the first number
should have been the same.
10
Question Three (7 marks)
Examine the following pairs of molecules.
a.) What is the stereochemical relationship between the following pairs of molecules? Are they
constitutional isomers, the same molecule (note: changing the conformation of a molecule does not
make it a different molecule), enantiomers or diastereomers.
b.) Clearly identify all meso compounds.
H
OH
a.)
b.)
H
H
H H
H3C
CH3 H3CH 2C
H3CH 2C
H
c.)
OH
CH3
CH3
CH3
H3C
d.)
HCl
CH3
Br
e.)
Br
Br
ClH
Br
Br
HCl
CH3 H3C
CH3
H Cl
H
H Br
Answer
a.)
H3CH 2C
H
H
OH
b.)
H
H H
H3C
CH3 H3CH 2C
Br
Br
Br
Br
These are mirror images of the same
molecule, a meso compound.
OH
Enantiomers
c.)
H
CH3
CH3
H3C
ClH
Cl
H
CH3
d.)
CH3
HCl
CH3 H3C
H Cl
Diastereomers. The molecule on the
right is a meso compound.
CH3
Enantiomers
e.)
Br
H
H
Br
These are conformational isomers of the same achiral
molecule. This molecule has an internal mirror plane
of symmetry, but that does not make it a meso
compound.
11
Question Six (2 marks)
Optically pure R-Glycidol has a specific rotation, [a] of +12° (neat, i.e. without solvent). What would
be the measured rotation of a sample of R-glycidol that is contaminated by its enantiomer such that
25% of the sample is S-glycidol?
Answer
CH2OH
R-Glycidol
H
O
You could reason this out. When pure the rotation is 12°.
When 75% pure, the rotation would drop to 75% of 12 which
is 9°. But since the impurity is its enantiomer, the impurity
will rotate the plane polarized light 25% of -12° or -3°.
9°-3°= 6°, the correct answer. Another way to reason this
out: 50% of each is racemic (0°). 100% is +12. Half way
between the two must be 6°.
Question Ten (3 marks)
The enzyme aconitase catalyzes the hydration of aconitic acid to two products, citric acid (which is
achiral) and isocitric acid (which is chiral).
O
O
C
C OH
HO
Aconitic acid
O
C
H
OH
The respective products result from the Markovnikov and anti-Markovnikov addition of water to the
carbon-carbon double bond. Identify the structures of citric and isocitric acid.
Answer
O
O
C
C OH
HO
Aconitic acid
O C
OH
H2O
O
O
C
HO HO
O C
OH
O
O
C
C
HO HO
OH
O C
Citric Acid
OH
H
H2O
O
C OH
H
H
O
C
HO
H
O C
*
C OH
*
OH
H
OH
Isocitric Acid
There are two possible regioisomeric structures that result from the addition of water to the double bond
in aconitic acid. They can be distinguished by the fact that the one on the left (citric acid) has no
assymetric atoms and is therefore achiral. The other (isocitric acid) has two assymetric centers and is
chiral. You cannot predict the absolute configuration of the two assymetric centers.
12
Question Three (seven marks)
For each of the following:
i) assign group priorities and configurational symbols (R or S, E or Z) to all stereocenters where
possible and
ii) indicate whether the molecule is chiral or achiral. Part (c) is a Fischer projection of C-13 labeled
glycerol.
a)
Br
Br
H
CH3 c) O
b) Br
d)
OH
CH3
C
H
H3CH 2C
H
T
D
C
T = 3H
H D = 2H
H = 1H
H3C
Answer
a)
Br R
S Br
H
b) Br
Z
CH3
c) O
d)
OH
CH3
C
H
C
H3CH 2C
H
Achiral
Achiral
(meso)
T
D R
Chiral
T = 3H
H D = 2H
H = 1H
H3C
Chiral
Question Five (eight marks)
Examine each of the following pairs of molecules and indicate whether they are constitutional isomers,
enantiomers, diastereomers or identical.
i)
CH2OH
CH2OH
H
OH
HO
H
H
OH
HO
H
CH2OH
Fischer
ProjectionS
iii)
F
F
ii)
HO
CH3
H3C
OH
CH2OH
F
F
iv)
H3C
H
C
Br
F
C
C
Answer
i) enantiomers ii) enantiomers iii) diastereomers iv) constitutional isomers
13
H3C
F
H
C
Br
Question Six (two marks)
Optically pure S-Glycidol has a specific rotation, [a] of -12° (neat, i.e. without solvent). What would
be the measured rotation of a sample of S-glycidol that is contaminated by 25% of R-Pinene ([a] =
+50.7° (neat))? Explain your answer.
H
CH3
CH2OH
O
(S)-(- )- Glycidol
(R)-(+)-Pinene
Answer
The net rotation is: (0.75 x -12°) + (0.25x50.7°) = 3.7°
Question Nine (eight marks)
i) Using the following molecules, explain how classical resolution (the separation of enantiomers)
works.
F3C
CH
CH3
OCH 3
C
OH
C
NH2
O
R-Mosher's Acid
(Optically pure)
Racemic
ii) Why is the preparation of optically pure compounds important?
Answer
i) The racemic mixture of the amine consists of molecules that are R or S. When mixed with the
optically pure R-Mosher's acid one obtains a 1:1 mixture of salts:
F3C OCH3
R CH
S CH
3
3
OH
C
C
C
C
H2N H
H2N H
O
Racemic
C
H3N
CH3
R-Mosher's Acid
(Optically pure)
F3C
OCH3
F3C
O
C
C
H
O
CH3
C
H3N
H
OCH3
C
C
O
O
SS salt
RR salt
The RR and SR salts are diastereomeric and can therefore be separated (in principal) by crystallization
etc.
ii) Phthalidomide is an excellent example of the importance of separating enantiomeric compounds.
Often, one enantiomer has a desireable biological activity and the other has an undesireable or unknown
effect.
14
Question Three (four marks)
For reactions that are stereoselective (i.e. they form an asymmetric center but not a racemic mixture),
the degree of selectivity is usually reported in terms of enantiomeric excess.
measured [ a ]o of mixture
enantiomeric excess =
x100%
[a ]oD for pure enantiomer
D
[a ]oD =
For the following reaction:
O
C
H3C
OH
H
CH2CH3
Chiral reducing agent
HO
+
C
H3C
a
l×c
CH2CH3
H3C
H
C
CH2CH3
For the enantiomerically pure R isomer, a °D = -13°
Given that the measured specific rotation of the mixture is -6.5°:
a) What is the enantiomeric excess?
b) What is the % of the mixture that is R and what is the % of the mixture that is S?
c) What would be the measured specific rotation of the reaction mixture if an achiral reducing agent
had been used? Explain.
Answer
a) 50%.
b) 75% is R and 25% is S. Recall that the S isomer rotates plane polarized light in the opposite
direction (+3.25° in this case). The 75% R rotates plane polarized light -9.75°. The sum of the
rotations is -6.5°.
c) An achiral reducing agent would give a racemic mixture: equal proportions of the R and S
products. It would not have any net rotation of plane polarized light.
Question Two (3 marks)
a) Label, where appropriate, all of the stereocenters/double bonds in the following molecules with the
correct configurational symbol. Write down your priority assignments if you want part marks.
b) How many possible stereoisomers are there for this molecule?
H Cl
4H Cl 1
1
1 H C
3
H3C
2
2
H
3
H
2
The stereocenter is R.
The double bond is Z.
To get the number of possible stereoisomers you must find the number of double bonds capable of
being E or Z and the total number of asymmetric atoms capable of being R or S. No. stereoisomers =
2n: four in this case.
15
Question Four (five marks)
Examine the following pairs of molecules.
a.) What is the stereochemical relationship between the following pairs of molecules? Are they
constitutional isomers, the same molecule, enantiomers or diastereomers.
b.) Clearly identify all meso compounds, if any.
Answer
a: In these two molecules, one asymmetric center is the same and one is different. These are
diastereomers.
b: These molecules lack asymmetric carbon atoms, but then, so do golf clubs. These two molecules
are non-superimposable mirror images of one another so they must be enantiomers.
a.) CHO
CHO
OH
H
OH H
OH
HO
H H
CH2OH
CH2OH
(Fischer Projections)
b.)
H
H
H
C C C
C C C Cl
Cl
H
Cl
Cl
Question
a.) Give the structures of all possible stereoisomeric products of the following reaction.
b.) What is the stereochemical relationship between all of the products?
c.) Are the products formed in equal proportions?
H3C
O
O
C
C
1) NaBH4, CH3OH
2) H+, H2O
CH3
Answer
All products will have the same connectivity, but there there are different stereochemical outcomes.
H OH HO H
H3C
HO H
CH3 H3C
H
OH
H OH H
OH
CH3 H3C
CH3
A meso-diastereomer
Enantiomers, formed in
exactly equal amounts.
Three different stereoisomers are formed: a racemic mixture of enantiomeric compounds and a meso
diastereomer. The amounts of the two enantiomers will be equal to each other, but the meso-compound
can, and will likely be formed in some differing amount.
Question
What is/are the product(s) produced upon the catalytic hydrogenation of Z-3,4-dimethyl-3-hexene?
Compare them to the product(s) formed when E-3,4-dimethyl-3-hexene is reduced under the same
conditions.
H3C
H3C
CH3
CH3
16
Answer
Hydrogenation is a syn-addition and H2 is equally likely to add to the top or bottom face of the double
bond. The resulting products are enantiomers of each other and are formed in equal amounts - in other
words the reaction gives a racemic mixture.
H H
H3C
H3C
H
Top face
CH3
H3C
H3C
CH3
CH3 Bottom face
H3C
H3C
H H
H
CH3
CH3
CH3
H3C
H3C
CH3
CH3
H
H
If the E isomer is hydrogenated, a single product is produced - a meso compound.
H3C
H H
H
Top face
CH3
H3C
H3C
CH3
CH3
Bottom face H3C
H3C
CH3
CH3
CH3
H3C
CH3
H3C
H3C
CH3
H
H H
H
H
H
H
H3C
H3C
CH3
CH3
Internal mirror plane of symmetry.
Question
Ethylene and many other alkenes react with bromine in an ionic reaction to form vicinal dibromides
shown in the following mechanism. The central intermediate is called a bromonium ion. Due to
symmetry, attack on either end of this ion is equally likely though only attack on the right end is shown
the example below.
Br
H
H
Br
H
H
Bromonium ion
Br+
H
H
Br
H
H
H
Br
-
H
H
H
Br
This mechanism also applies to substituted olefins. Keeping this in mind, show the importa
stereochemical differences in the nature of the products obtained when one reacts bromine with Zbutene and (separately) with E-2-butene to get 2,3-dibromobutane.
17
Answer
This is quite complicated due to the number of possibilities. Bromine can add to either side of the
molecule to form a bromonium ion. The two possible bromonium ions are enantiomers of each other.
Each bromonium ion can undergo attack by bromide to either carbon (the two possibilities are
represented by solid and dashed arrows) with equal likelihood. So, there are four possible paths in this
case. Interestingly, they all give exactly the same product, a meso-compound.
Br
Bromonium ion
Br
Br
Br+
CH3
H
H3C
H
CH3
H
H3C
CH3
H
H
H3C
Br
H
Br-
H
Br
H3C
Br
H
CH3
H
Br
CH3
H3C
-
Br
CH3
H
H 3C
H
H
Br
H
H 3C
H
CH3
+
Br
Br
H 3C
H
Br
H
CH3
Br
Br
Cis-2-butene gives an identical bromonium ion no matter which side the bromine attacks. Subsequent
attack by bromide can, again, take two possible paths. In this case, however, one obtains a racemic
mixture of dibromides (diastereomeric to the dibromide formed from the trans-2-butene.
Br
Bromonium ion
Br
Br
H
CH3
+
Br
H
H3C
H
CH3
H
H
H3C
H
H3C
R,R
Br
CH3
BrHas an internal plane
of symmetry.
H
H3C
Br
18
Br
S,S
H
CH3
Question
A 1.00 M solution of 2-chloropentane in chloroform in a 10 cm cell gives an observed a of +3.64°.
Calculate [a]D the specific rotation. The molar mass of chloropentane is 106 g/mol.
Answer
[a ]oD =
a
1.0 mol / L x 106 g / mol
c =
= 0.106g / mL
l×c
1000mL / L
The specific rotation is 34.3°.
Question
A mixture of diastereomeric butenes, 60% E and 40% Z, is treated with cold, dilute, basic KMnO 4.
Knowing what you do about the stereochemistry of the formation of 1,2-diols from alkenes,
a.) What are the products produced and in what proportions?
b.) What is the stereoisomeric relationship between all of the products? (6 marks)
H3C
H3C
H
H
CH3
H
CH3
H
Z-2-Butene
E-2-Butene
Answer
This reaction results in the syn-addition of two hydroxyl groups to the double bond of the alkene.
Addition to either face of Z-2-butene gives a meso-diol. Since the meso-diol is derived from the Zisomer, it constitutes 40% of the final mixture.
O
O
Mn
O
H
C
O
Mn
H
H3C
O
O
C
CH3
H
O
O
C
H
C
H3C
CH3
HO
OH
H
H
C C
CH3
H3C
meso-diol
In contrast, addition to the top face of E-2-butene gives one enantiomeric product, while addition to the
bottom face gives the opposite enantiomer. The two enantiomers will be formed in equal amounts and
each will constitute 30% of the entire mixture of three stereoisomeric products.
19
O
O
Mn
O
O
O
H
C
Mn
CH3
C
H3C
H
H3C
H
C
C
H
CH3
O
O
Mn
O
O
O
O
H
C
H3C
H3C
C
CH3
H
H
C
C
H
O
O
Mn
O
O
O
20
CH3
HO
H
C
H3C R
OH
CH3
C
R H
H3C
S
S
C
H
HO
C
H
CH3
OH