Topic 10 Organic Chemistry Answers
10.1 Exercises Part 1
1. Define the following terms:
a) organic compound
A member of any family of compounds that contain carbon as their basic building block.
b) inorganic compound
All compounds other than organic compounds. Includes carbon containing compounds that do not contain a
hydrocarbon unit, e.g. carbonates.
c) homologous series
A series of organic compounds, with a general formula and with the same chemical properties due to the
presence of the same functional group. They show a gradual change in physical properties due to an increase
in molecular weight. In a homologous series, each subsequent molecule differs in formula by a fixed group of
atoms, usually CH2.
d) homologue
A member of a homologous series of a family of compounds.
e) general formula
A formula that can be used for a group of compounds. The alkanes, alkenes and alkynes have predictable
numbers of hydrogen atoms depending on the number of carbon atoms in the compound. These can be defined
by the general formulas for these homologous series. The general formulas are:
Alkane: CnH2n+2
Alkene: CnH2n
Alkyne: CnH2n-2
f)
hydrocarbon
An organic compound that contains only hydrogen and carbon.
g) aliphatic
Aliphatic compounds are a group of hydrocarbons that contain carbon atoms joined together in straight chains,
branched chains or in simple rings. They are not aromatic.
There are two major classes of organic compounds: the aliphatic and the aromatic.
These names come from the early organic chemists. Aliphatic means fat-like, and fats
and vegetable oils were useful sources of organic compounds; aromatic meaning has
an aroma, with benzene being an example. These terms have expanded to cover a
wider range compounds than the original meaning gave. Aromatic today includes
compounds based on benzene, many have no aroma.
h) functional group
A group of atoms that is common to all members of a homologous series. A functional group is the reactive site
of the molecule.
i)
empirical formula
A way of representing the atoms making up of a molecule by giving only the types and numbers of atoms in
their lowest ratio.
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j)
molecular formula
A way of representing the atoms in a molecule by giving the types and actual numbers of atoms present in one
molecule of the compound.
k) structural formula
Represents unambiguously how the atoms are arranged by showing all atoms and all bonds in the molecule.
They can be written in a number ways, e.g. hexanoic acid CH3CH2CH2CH2CH2COOH or CH3(CH2)4COOH
l)
structural isomer
Different compounds with the same molecular formula, but having different structural formulas, are called
structural isomers.
m) carbon backbone
The longest carbon chain in an organic molecule.
2. Describe the features of a homologous series.
The chemical compounds in a homologous series have the same functional groups but differ in formula by a
fixed group of atoms, normally a CH2 group. The members of a homologous series have similar chemical
properties, but their physical properties differ.
3. What is the formula difference between two members of the same homologous series?
A single CH2 group.
4. What is meant by the term “substituted alkane”? What is a substituent?
A substituted alkane is a molecule where one or more hydrogen atoms are replaced by other elements or
groups. A substituent is an atom or group that replaces hydrogen in an organic compound.
5. What is the difference between a straight chain, branched, cyclic and acyclic alkane? Illustrate your
answer using derivatives of hexane as an example.
The connectivity of the carbon atoms in the backbone differs between the different types of alkanes. Hexane is
an example of a straight chain form of an acyclic (i.e. not a cycle/ring) alkane. 2-methylbutane is also an acyclic
alkane, but it has a branched methyl group. Cyclohexane is the cyclic (i.e. ring) form of hexane. All of these
compounds contain the same number of carbon atoms.
H2
C
H2C
CH2
CH3
H2C
CH3
CH2
CH2
CH2
CH2
CH3
CH3
CH2
Hexane
CH2
CH
2-methylbutane
CH3
CH2
C
H2
cyclohexane
6. What do we mean when we say the longest “straight chain”. Why are straight chain alkanes not
“straight” realistically? Hint: Think about the structure, using the VSEPR theory.
When we say the longest “straight chain” we mean the longest chain of successive carbon atoms in the
molecule.
The electron density about a carbon atom is tetrahedral and as a result, the “straight chain” of carbons is
actually a zig-zag.
2
7. Explain with examples, the difference between saturated and unsaturated hydrocarbons.
Saturated hydrocarbons, for example the alkanes, contain the maximum number of hydrogen atoms possible
and no more can be added to the molecule. They only contain C–C single bonds and C–H bonds. Unsaturated
hydrocarbons, for example the alkenes and alkynes, contain at least one carbon to carbon double or triple bond
and do not contain the maximum number of hydrogen atoms possible around the carbon atoms.
8. Explain in terms of intermolecular forces the trend in boiling points of members of a homologous
series.
In the homologous series, the number of carbon atoms increases, so, there will be greater van der Waals’
forces between the molecules. This means that that they are more tightly bound together and that more energy
will be required to break the molecules apart and boil the sample.
9. Structural formulas show which atoms are joined to which; they should indicate clearly the bonding
between atoms. Which of the following are not structural formulas for hexane? For those that are
not structural formulas indicate what type of formula is represented.
a) C6H14
molecular formula
b) CH3(CH2)4CH3
structural formula
c) CH3CH2CH2CH2CH2CH3
structural formula
d) C3H7
empirical formula
10. Arrange the following hydrocarbons in order of decreasing boiling point
CH3
H3C
CH3(CH2)8CH3 > CH3CH2CH2CH2CH3 >
CH
CH2
CH3 > C H
3 8
If the carbon chain of the molecule is longer, there will be greater van der Waals’
interactions between molecules leading to a higher boiling point. Branching of the
molecule makes them more spherical. This will decrease the boiling point due to a
decreased surface area available for van der Waals’ interactions.
11. When does isomerism begin in the alkanes? Draw structural formulas for the isomers of this alkane
and the isomers of the next alkane in the homologous series, and using IUPAC nomenclature, name
these compounds
Isomerism begins in the alkanes at the four carbon structure, butane. The structural formulas for this alkane and
its isomers are:
CH3
CH3
CH2
CH2
CH3
CH3
CH
CH3
2-methylpropane
butane
3
The next alkane in the homologous series is pentane. The structure of pentane and its isomers are:
CH3
CH3
CH3
CH3
CH2
CH2
CH2
CH3
CH3
CH2
CH
C
CH3
CH3
CH3
2-methylbutane
pentane
2,2- dimethypropane
12. Which of the following are structural isomers of butane?
H
A
H
H
H
C
C
C
H
CH3 H
CH3
H
H
B
C
CH3
CH3
HO
C
C
CH3
CH3
CH3
H
D
H
H
H
C
C
C
H
CH3 H
CH3
A and B are structural isomers of butane as they have the same molecular formula as butane. C and D do not
have the same molecular formula as butane.
13. Using IUPAC rules provide names for the compounds in question 12.
A 2-methylpropane (1-methylpropane is actually butane, so the connectivity of the atoms is implied – i.e. the
methyl group MUST be in the number 2 position)
B Methylpropane (the same compound as A, but drawn differently)
C methylpropan-2-ol
D 2-methylbutane
2-methylpropane can also correctly be written as methylpropane. 1-methylpropane is
actually butane, so the connectivity of the name methylpropane is implied, it can only
have the methyl group in the number 2 position. The numbered name, 2methylpropane is the full systematic name and recommended for IB Chemistry.
14. The first three members of the alkene series are ethene, propene and butene
a) Draw structural formulas for ethene, propene and but-2-ene
H2C
CH2
CH2
ethene
CH
CH3
propene
H3C
CH
CH
CH3
but-2-ene
b) Draw a position isomer for but-2-ene, and provide its name
H2C
CH
CH2
CH3
but-1-ene
15. Draw structural formulas for these unsaturated hydrocarbons, and deduce their names
a) C2H4
H2C
CH2
ethene
b) C3H6
CH2 CH
CH3
propene
4
c) C2H2
HC
CH
ethyne
d) C5H10 (all possible isomers)
CH3
CH2
CH2
CH
CH2
CH3
CH3
pent-1-ene
CH
CH
CH2
CH2
CH3
pent-2-ene
CH
C
CH
CH3
3-methylbut-1-ene
CH3
CH3
CH
CH2CH3
CH3
2-methylbut-2-ene
CH3
CH2
C
2-methylbut-1-ene
16. Deduce names for the compounds with the following structural formulas
H3C
C
CH2
H3C
CH3
CH3
a)
2,2-dimethylbutane
CH3
H3C
b)
CH
CH2
CH3
2-methylbutane
CH3
H3C
CH
CH3
c)
2-methylpropane
CH3
H2C
CH
H2C
d) H3C
3-methylpentane
CH3
17. Using IUPAC rules deduce structural formulas for the following
a)
2-methylpropane
CH3
CH3
b)
CH3
CH
CH2
hex-3-ene
CH2
CH
CH
CH2
CH3
5
c) 2-ethylbutane
CH2CH 3
CH3
d)
CH
CH2
CH3
2, 2-dimethylbutane
CH3
CH3
CH2
C
CH3
CH3
e) but-1,3-diene
H2C
f)
CH
CH
CH2
2-methylpent-2,3-diene
CH3
H3C
C
C
CH
CH3
g) 3,5-dimethylhex-2-ene
CH3
CH3
CH
CH3
C
CH2
CH
CH3
18. Draw and name three structural isomers of the alkene pentene, C5H10.
CH3
CH2
CH
CH2
CH2
CH3
CH3
pent-1-ene
CH
CH
CH2
CH3
CH2
pent-2-ene
CH
C
CH
CH2CH3
CH3
2-methylbut-2-ene
CH3
C
CH2
2-methylbut-1-ene
19. Deduce names for the compounds with the following structural formulas
CH
H3C
a)
pent-2-ene
CH2
CH
CH3
3-methylbut-1-ene
CH3
CH3
CH
CH3
b) CH3CH=CH(CH2)2CH3
hex-2-ene
6
CH3
CH3
H3C
C
CH2
CH
CH2
c)
3-methylpent-2-ene
H3C
CH3
C
CH
CH
d) H3C
3,4-dimethylpent-2-ene
CH3
e) CH3CH=CHCH(CH3)2
4-methylpent-2-ene
f)
CH2=CHCH3
propene
g) CH2CH2
ethene
h) CH3CH=CHCH3
but-2-ene
i)
CH3CH=CHCH=CHCH3
hexa-2,4-diene
10.1 Exercises Part 2
1. Define the following terms:
a) R group
Normally a hydrocarbon chain. The letter R is usually used in chemical formulas to indicate the rest of the
molecule, i.e. the part not containing the functional group.
b) functional group
The atom or group of atoms in a compound responsible for the characteristic chemical reactions that the
compound undergoes.
c) bond moment
A measure of the extent to which the average electron charge is displaced towards the more electronegative
atom.
d) dipole moment
The vector sum of all the bond moments within a molecule. This dipole moment means that the molecule has
overall polarity (i.e. a plus and a minus end).
e) polar compound
A molecule which contains a dipole moment.
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f)
non-polar compound
A compound which does not have a dipole moment.
Opposing bond moments can cancel each other out. For example, carbon dioxide, a
linear molecule, contains two polar bonds but these cancel each other out, so the
molecule does not have a dipole moment and is non-polar overall.
g) solubility
The extent to which a substance, the solute, dissolves in another substance, the solvent.
h) volatility
Refers to the speed at which a solid or liquid evaporates to form a vapour.
2. Circle any carbonyl and/or hydroxyl groups, and name any functional groups in the following
compounds:
O
C
H
CH3
aldehyde
O
C
OH
carboxylic acid, benzene
O
H3C
OCH2CH3
ester
H
OH H
H
C
C
C
H
CH3 H
H
alcohol (primary)
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3. Deduce names for the following compounds (draw out expanded structural formulas if this helps):
a) CH3CH2CH(CH3)(CH2)2OH
CH3
H3C
CH2
CH
CH2
CH2
OH
3-methylpentan-1-ol
b) CH3Cl
chloromethane
c)
CH3
H
C
H3C
C
H
CHO
CH3
2,3-dimethylbutanal
d)
CH3
H3COC
C
CH3
CH3
3,3-dimethylbutan-2-one
e) CH3CHO
ethanal
f)
CH2BrCH2CHClCH2CH3
1-bromo-3-chloropentane
g) HOCH2CH(CH3)CH3
2-methylpropan-1-ol
h) CH3CH2CO2H
propanoic acid
i)
(CH3CH2)2CO
pentan-3-one
j)
CH3CH(OH)CH2CH3
butan-2-ol
k) CH3COCH3
propanone
4. Circle and name the functional groups in the following compounds:
O
H3C
OCH2CH3
a)
ester
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R Br
b)
halide (bromine)
R CHO
c)
Aldehyde
d)
NH2
COCH3
amine, ketone, benzene
NH2CH2COOH
e)
amine, carboxylic acid
f)
CH3CH2CH2COOCH2CH3
ester
g)
O
H2N
CH C
OH
CH2
(phenylalanine)
amine, carboxylic acid, benzene
5. Identify as many functional groups as you can in the following compounds:
a) vanillin, the compound responsible for vanilla flavour
HO
O
H3CO
C
H
alcohol, benzene, aldehyde
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b) carvone, the compound responsible for spearmint flavour
O
CH3
H3C
CH2
ketone, alkene
c) zingerone, a compound found in ginger which gives the spicy flavour
H3CO
CH2CH2COCH3
HO
alcohol, benzene, ketone
The H3CO– group shown above is sometimes referred to as the "methoxy group"
(methyl oxygen). The methoxy group may be found as part of an ester functional
group, or as part of an ether. An ether is another functional group with is not part of
the standard level course.
6. Deduce structural formulas for the following compounds
a) ethanol
CH 3
CH 2
OH
b) ethanoic acid (common name acetic acid)
O
CH 3
C
OH
c) ethanal (common name acetaldehyde)
O
CH 3
C
H
d) propanone
O
H 3C
C
CH 3
e) chloroethane
Cl
H 2C
f)
CH 3
pentane-2,4-diol
OH
CH 3
OH
CH
CH 2
CH
CH 3
g) 3-methylbutan-2-one
CH 3
O
CH 3
C
CH
CH 3
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7. Deduce structural formulas for the following compounds:
a) 2,3-dibromopentane
Br
CH 3
C
CH 2
CH 2
CH 3
Br
b) 2,2,4,4-tetramethylhexan-1-ol
OH
CH 3
CH 2
C
CH 3
CH 2
CH 3
C
CH 2
CH 3
CH 3
c) 2-iodo-3,3-dimethylbutane
CH3
I
CH3
CH2
C
CH3
CH3
d) hexan-3-one
O
CH 3
CH 2
C
CH 2
CH 2
CH 3
e) butanoic acid
O
CH 3
f)
CH 2
CH 2
C
OH
methanoic acid (common name formic acid)
O
H
C
OH
g) 3-methylbutanal
O
H
C
CH2
CH
CH3
CH3
h) trichloroacetic acid
Cl
Cl
O
C
C
OH
Cl
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8. Draw the expanded structural formula and label the following alcohols as primary, secondary or
tertiary:
a) CH3CH(OH)CH2CH3
OH
CH3
CH
CH2
secondary
b)
(CH3)3COH
CH3
OH
H3C
C
CH3
CH3
tertiary
c) CH3CH(CH3)CH2OH
H3C
CH3
OH
C
H
CH2
primary
d) CH3OH
OH
H
C
H
H
primary
e) CH3(CH2)2COH(CH3)2
OH
H3C
H2
C
H2
C
C
CH3
CH3
tertiary
9. Label the appropriate carbons as primary, secondary or tertiary carbons in the following
halogenoalkanes:
a) CH3BrCHCH3
secondary
b) ICH2CH2CH3
primary
c) ClCH2C(CH2CH3)3
primary
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10. Consider propane (bp -42 OC), ethanal (bp 20 OC).
a) deduce their structural formulas
Propane
CH 3
CH 2
CH 3
Ethanal
O
H 3C
C
H
b) calculate the relative molecular mass for each
Propane
Molecular formula: C3H8
Relative molecular mass = 3 x 12.01 + 8 x 1.01 = 44.038 g mol-1
Ethanal
Molecular formula: C2H4O
Relative molecular mass = 2 x 12.01 + 4 x 1.01 + 16.00 = 44.06 g mol-1
c) propose a reason for the large difference in boiling points
Ethanal has a much higher boiling point than propane due to the presence of the aldehyde functional group.
This group gives the molecule a dipole moment, therefore ethanal is polar. The secondary forces between polar
ethanal molecules are dipole-dipole interactions and van der Waals’ interactions. Propane is non polar, and only
van der Waals’ interactions are possible. Dipole-dipole interactions are a stronger than van der Waals’ forces
and so more energy is required to break these secondary interactions in ethanal in order to boil the sample.
11. With reference to their structures, (see IB Chemistry Data Booklet, section 21) compare the relative
water solubilities and volatilities of Vitamins C and D.
Vitamin C has 4 polar hydroxyl groups capable of hydrogen bonding with water and a polar carbonyl group that
may take part in dipole-dipole interactions with water. Vitamin C has a relatively small non-polar hydrocarbon
component and due to the many strong interactions it is able to form with water it is very soluble. In contrast,
there is only one hydroxyl group in Vitamin D, which is a considerably larger hydrocarbon, the molecule would
be non polar and therefore not soluble in water. The intermolecular interactions of Vitamin C are strong
hydrogen bonds. These bonds require a lot of energy to overcome and Vitamin C is not volatile. Vitamin D has
only weak van der Waals' interactions
12. Compare the relative volatilities of 2-methylbutane, 2-chlorobutane and 2,2-dimethylbutane.
CH3
CH3
CH3
CH
Cl
CH2
CH3
2-methylbutane
CH3
CH
CH3
CH2
2-chlorobutane
CH3
CH
CH2
CH3
CH3
2,2-dimethylbutane
On inspection of the structures it may be estimated that the difference in volatilities of 2-methylbutane and 2,2dimethylbutane would probably be quite close. 2,2-dimethylbutane has an additional side chain attached. This
14
additional side chain makes the molecule more spherical, which decreases its surface area. The decreased
surface area of the 2,2-dimethylbutane molecule compared to 2-methylbutane means the van der Waals’
interactions between adjacent molecules are decreased. However, the extra side chain also increases the
molecular weight of 2,2-dimethylbutane. Increased molecular weight increases van der Waals’ interactions due
to the increased electron cloud. In terms of the effect on volatility, the larger molecular weight of 2,2dimethylbutane is more significant than its more spherical nature and the molecule with the smaller molecular
mass, 2-methylbutane, is therefore more volatile than 2,2-dimethylbutane.
2-methylbutane and 2,2-dimethylbutane are non-polar molecules with intermolecular interactions consisting of
weak van der Waals' forces only. 2-chlorobutane contains an electronegative chlorine atom and the highest Mr.
The Cl-C bond has a bond moment, and the molecule will have a dipole moment overall. This means that there
are dipole-dipole interactions between 2-chlorobutane molecules, in addition to van der Waals’ interactions.
These dipole-dipole interactions are much stronger than van der Waals' interactions so the intermolecular forces
between molecules of 2-chlorobutane are stronger. This means that this compound is less volatile than 2methylbutane and 2,2-dimethylbutane (has the highest boiling point).
(For your information: literature values from Aldrich fine chemicals catalogue- 2,2-dimethylbutane: bp 50°C, Mr =
86.18, 2-methylbutane: bp 30°C, Mr = 72.15 2-chlorobutane bp 68-70°C, Mr = 92.57)
Dispersion forces become stronger as the atom (or molecule) becomes larger. This
trend is illustrated well by the halogens (from smallest to largest: F2, Cl2, Br2, I2).
Fluorine and chlorine are gases at room temperature, bromine is a liquid, and iodine is
a solid.
13. Chloromethane and ethanol have similar molecular masses, however ethanol is miscible with water
and chloromethane is not. Account for this fact. (Hint: consider the different types of intermolecular
forces present between molecules)
Ethanol has a hydroxyl functional group that is able to hydrogen bond with water molecules in solution. Although
Chloromethane has a dipole moment, it is not capable of hydrogen bonding and there can only be weaker
dipole-dipole interactions between it and water. These dipole-dipole interactions are weaker than hydrogen
bonds so water is not able to easily solvate (dissolve) the chloromethane. The hydrogen bonds between ethanol
and water are strong interactions, so the water molecules can more easily solvate ethanol molecules.
Chloromethane, CH3Cl, was once used as a refrigerant (trade name Freon-40) but its
use was discontinued due to its toxicity.
14. Arrange the following lipids in order of increasing water solubility and decreasing volatility: stearic
acid, palmitic acid, lauric acid (structures may be found in the IB Chemistry Data Booklet, section
22) and give reasons.
These lipids all contain a polar head group and a long non-polar hydrocarbon chain. The length of this
hydrocarbon chain will determine the water solubility of the lipid. The shorter this chain is, the more water
soluble the molecule will be. As the chain gets longer, more van der Waals’ interactions are possible as such
the volatility is decreased. As a result, the order of these in increasing water solubility and decreasing volatility
is: stearic acid, palmitic acid, lauric acid
15
Lipids are fat like organic compounds that are soluble in organic solvents, e.g. ethanol,
but have limited to zero solubility in water.
15. Consider the structure of aspirin (see IB Chemistry Data Booklet, section 20) and explain how this
molecule can easily be dissolved in water but can also cross the blood-brain barrier for which it
requires a degree of lipid (i.e. fat) solubility.
Aspirin contains a polar ester functional group and carboxylic acid functional group. The carboxylic acid
functional group can take part in hydrogen bonding with water molecules, allowing the water to solvate the
aspirin, thus allowing it to dissolve. For a molecule to dissolve in a lipid, it requires a non-polar component.
Aspirin has a non polar benzene ring which confers lipid solubility on the molecule, and allows it to cross the
blood brain barrier.
16. Consider the series of alcohols, aldehydes, ketones, carboxylic acids and halides. Why is it that the
first three members of each homologous series are very soluble in water, but as the series is
ascended, solubility decreases rapidly?
Alcohols, aldehydes and ketones are all able to form hydrogen bonds with water molecules in solution making
them soluble. Halides are able to form dipole-dipole interactions with water molecules. As the homologous
series is ascended, there is a much larger non-polar hydrocarbon component in the molecule. This reduces the
effect of the polar groups in allowing the water molecules to solvate the solute.
17. Ethanal has a bp of around 21°°C, while ethanol has a bp of 78°°C. Account for this large difference in
boiling points.
Ethanal contains the aldehyde functional group. The only secondary forces that are able to occur between
ethanal molecules are dipole-dipole forces and van der Waals’ interactions. Between ethanol molecules, in
addition to these two secondary forces, hydrogen bonding can occur between the oxygen of a hydroxyl group
on one molecule and the hydrogen of the hydroxyl group on an adjacent molecule. Hydrogen bonding a very
strong type of secondary interaction, thus more energy is required to break these bonds. This leads to a higher
boiling point.
18. Why does oil float on water?
"Oil" in general, is comprised of long hydrocarbon chains that are non-polar whereas water is polar. As a result,
the two liquids are not miscible. Oil is less dense than water, so when the two liquids are added together, the oil
layer will always be on top of the water layer.
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10.2 Exercises
1. Outline the main chemical properties of the alkanes.
Alkanes are relatively unreactive and require much energy to react. Their main reaction is with oxygen, which is
strongly exothermic. Combustion provides the activation energy required for reaction. They are widely used as
fuels.
2. Explain the chemical inertness of the alkanes in terms of bond enthalpies and bond polarity.
Alkanes are hydrocarbons that contain strong carbon to carbon single bonds and carbon to hydrogen single
bonds only. The bond enthalpy of the C–H bond is 412 kJ mol-1 and for C–C is 348 kJ mol-1 These are high
values and mean that the C–H and C–C bonds are very strong and not easily broken.
Alkanes are non-polar molecules due to the bonds within the molecule having almost no dipole moment.
Carbon and hydrogen have similar electronegativities (2.5 and 2.1 respectively; see IB Data Booklet Section 7).
These similar electronegativities mean that the carbon to hydrogen bond is non polar, as is the carbon to carbon
bond. Non polar bonds are relatively unreactive, as the electron density within the bond is not polarised towards
one atom. This means that neither atom participating in the bond is susceptible to nucleophilic attack or
electrophilic attack and they are quite unreactive. (For an explanation of electrophilic and nucleophilic attack,
see section 10.5)
3. The possibility of silicon-based life has been proposed given that silicon is the second most
abundant element on Earth after oxygen, and silicon is in the same group as carbon and has similar
characteristics, notably its valency. Comment on the validity of this statement with reference to the
average bond enthalpies of the C–C, Si–Si, C–H and Si–H bonds (see IB Chemistry Data Booklet).
What other two main differences between silicon and carbon would affect the stability of
compounds of life derived from these elements?
The complexity of life demands a complex, but strong, molecular structure that only carbon can give. Because
the carbon atom is smaller than the silicon atom C–C bonds and C–H bonds are far stronger than Si–Si and Si–
H bonds. Structures based on silicon would be much weaker than those based on silicon. The chemical
evidence is against a silicon based life.
Bond enthalpies given in the data booklet are: C–C, 348; C–H, 412; Si–Si, 226 and
Si–H, 318 kJ mol-1
Remember that the inverse square law applies to the distance between positive and
negative charges; double the distance quarter the force. The distance between the
carbon nuclei and its shared electrons is shorter than that between silicon nuclei and
its shared electrons. Hence the C bonds are stronger.
Bond enthalpies reflect this. More about bond enthalpies in Topic 5 Energetics.
Despite having the same valency as carbon (it can form the same number of bonds), silicon is hindered by its
size. Carbon atoms are small enough to be able to form strong carbon to carbon double bonds, whereas silicon
can not. Carbon compounds are discrete (meaning individually distinct, separate, discontinuous) molecular
structures. The product from respiration is molecular CO2(g). Silicon compounds have a giant lattice structure.
What would silicon respiration give? Solid SiO2?
17
Consider these two group 4 elements, C and Si. Living things are built from C–C
chains in molecules, but much of the Earth’s crust (non-living) is built from Si–O
chains in giant lattices.
4. Explain, with examples, the difference between straight chain and branched alkanes.
CH3
CH3
CH3
CH2
CH2
CH2
CH2
CH3
hexane
CH2
CH
CH2
CH3
CH3
CH2
C
CH3
CH3
CH3
3-methylpentane
2,2-dimethylbutane
There is both a physical (boiling point) and chemical (structure and slight reactivity) difference. By looking at the
above structures it can be seen that branching makes the molecule more spherical. Spherical molecules have
less surface area than long unbranched molecules, so that there is decreased van der Waals’ forces between
more spherical molecules. This means that in the series of molecules above, volatility increases, i.e. boiling
point decreases. In combustion the surrounding methyl groups protect the central C from exposure to oxygen,
slowing slightly its reaction rate.
In petrol branched chained hydrocarbons are preferred. They, unlike straight chained
hydrocarbons, do not explode in the cylinder. Exploding is called knocking and
reduces the efficiency of a fuel. Branched chained hydrocarbons burn more slowly and
uniformly so maintaining a uniform pressure during the power stroke.
5. 2,2 dimethylbutane (bp 50ºC) and hexane (bp 69ºC) have the same molecular weight. Account for the
difference in boiling points by drawing structural formulas of the compounds.
CH3
CH3
C
CH2
CH3
CH3
CH2
CH2
CH2
CH2
CH3
CH3
2,2-dimethylbutane
hexane
Despite both of these alkanes having the same molecular weight, 2-2-dimethylbutane is a branched alkane
whereas hexane is linear. The branched 2,2-dimethylbutane is more spherical in shape so that van der Waals’
interactions between adjacent molecules are decreased compared to the linear hexane. This means that 2-2dimethylbutane has a lower boiling point.
6. Arrange the following alkanes in order of decreasing boiling points, using their names.
hexane > 2-methylpentane > 3,3-dimethylpentane > 2,2-dimethylbutane
7. Name the products of the complete combustion of an alkane.
Carbon dioxide and water.
18
8. Give the symbol and/or formula of the products for the incomplete combustion of an alkane.
C, CO, CO2, H2O
9. Write balanced equations for the complete combustion of:
a) propane
C3H8 (l) + 5O2 (g) 3CO2 (g) + 4H2O(l)_
b) butane
C4H10 (l) + 6½O2 (g) 4CO2 (g) + 5H2O (l)
c) pentane
C5H12 (l) + 8O2 (g) 5CO2 (g) + 6H2O (l)
d) hexane
C6H14 (l) + 9½O2 (g) 6CO2 (g) + 7H2O(l)
10. Write two equations (for each) to show possible products for the incomplete combustion of:
a) pentane
C5H12 (l) + 5½O2 (g) 2C(s) + CO(g) + 2CO2(g) + 6H2O(l)
C5H12 (l) + 5O2 (g) 2C(s) + 2CO(g) + CO2(g) + 6H2O(l)
b) hexane
C6H14 (l) + 5O2 (g) 4C(s) + CO(g) + CO2(g) + 7H2O(l)
C6H14 (l) + 5½O2 (g) 3C(s) + 2CO(g) + CO2(g) + 7H2O(l)
11. Define the following terms:
a) Halogenation
A reaction of a molecule with a molecular halogen (i.e. Cl2, Br2) to incorporate that halogen into the molecule.
Halogenations are a type of substitution reaction.
b) Chlorination
The reaction of a molecule with molecular chlorine (Cl2) to form a chlorine containing compound.
c) Photolysis
A chemical reaction initiated with the use of light energy.
d) Substitution
Where a functional group within a molecule is replaced with another.
e) Haloalkane
An alkane with one or more halogen atoms linked to it.
f)
Mechanism
A description of the way a particular chemical reaction occurs.
g) Free radical
An atom or group of atoms that has an unpaired valence electron.
h) UV (written above a reaction arrow)
Written above a reaction arrow, UV is an indicator that a photolysis reaction, initiated by UV radiation, is
occurring.
12. A single chlorine radical is represented thus: Cl• . What does the small dot represent?
The small dot represents the unpaired valence electron in the free radical.
19
13. What is the difference between homolytic and heterolytic fission? Use equations for the bond
cleavage reactions of Cl2 and HCl as examples to support your answer.
In homolytic fission, both of the atoms involved in the bond end up with one electron from the bond that is
broken. This is the case in the fission of Cl2:
Cl2 2Cl•
The fission of HCl however, is heterolytic fission. This is where only one atom involved in the covalent bond
receives both of the electrons of the broken bond:
HCl H+ + Cl
-
In heterolytic fission, no free radicals are formed.
In homolytic fission, free radicals are formed.
14. Write equations for the halogenation of methane and ethane with bromine, Br2.
CH4 + Br2 CH3Br + HBr
C2H6 + Br2 C2H5Br + HBr
15. For the reaction of ethane with bromine, explain the reaction in terms of a free-radical reaction by
indicating the homolytic fission of Br2, as well as showing equations for initiation, propagation and
termination steps.
The following mechanism is for a free radical substitution, with the first step being the homolytic fission of Br2.
Br
Br 2Br •
C2H6•
C2H5•
•
+ Br C2H5•
initiation
+ HBr
propagation
•
+ Br2 C2H5Br + Br
propagation
Br• + Br• Br2
termination
16. What may be used as an initiator in free-radical halogenation reactions?
An initiator would be the molecular form of the halogen that undergoes homolytic fission upon the absorption of
ultraviolet light to form free radicals.
17. Label the following equations as initiation, propagation and termination:
•
a) CH2CH + Cl2 CH2CHCl + Cl
•
propagation
b) X• + Y• XY
termination
c) CH3• + H• CH4
termination
UV
d)
Cl2
2Cl
initiation
•
e) X + RY RX + Y
propagation
20
18. CH3Cl in the troposphere is far more stable than CH3I. CH3Cl will last for about a year in the
troposphere and this gives it sufficient time to enter the stratosphere, where it can destroy ozone.
However, CH3I has about a week in the troposphere before being destroyed by light.
a) What is the name given to a light initiated reaction?
A photochemical reaction.
b) Explain why CH3Cl is more stable than CH3I.
The bond enthalpy of the C–Cl bond is 338 kJ mol-1 and the bond enthalpy of the C–I bond is 238 kJ mol-1. This
means, that more energy is required to break the C–Cl bond than the C–I bond. The reason for this is that C–Cl
covalent bond is much shorter and hence stronger than the C–I covalent bond. Thus, the compound CH3Cl is
more stable.
c) Write an equation to show the decomposition of CH3I by light energy.
UV
CH3I
CH3+ + I -
10.3 Exercises
1. Define the following terms
a) addition reaction
The addition of a molecule across the double or triple bond in an unsaturated compound.
b) substitution reaction
Where a functional group or a H atom in a molecule is replaced with another functional group or atom.
c) hydrogenation
The addition of hydrogen across the double bond of an alkene or a triple bond in an alkyne.
d) hydrohalogenation
The reaction of alkenes with hydrogen halides to form an addition product where hydrogen and the halide have
added across the double bond. Also applicable to the alkynes.
e) hydration
The addition of water to form another compound. The addition of water across a double bond to give an alcohol
is an example.
f)
addition polymerisation
The reaction of alkenes with themselves to form a polymer.
g) polymer
A molecule comprised of repeating units, monomers, connected by covalent bonds.
h) monomer
A small molecule that may bond chemically with other monomers to form a polymer.
i)
repeating unit
The shortest sequence that can be found repeatedly in the polymer.
21
2. Why do alkanes typically undergo substitution reactions while alkenes typically undergo addition
reactions?
Alkanes are saturated, alkenes are not. Alkenes typically undergo addition reactions due to the presence of the
double bond, which the reactant may add across. Alkanes do not have this double bond for the reactant to add
across and thus undergo substitution reactions.
3. Hydrogenation is a type of addition reaction.
a) What is the catalyst used in hydrogenation?
A metal catalyst is used, often Ni, Pt or Pd
b) Write a balanced equation to show the reaction for the addition of hydrogen to ethene.
C2H4 + H2 C2H6
c) The reaction in (b) is exothermic. Suggest four ways of shifting the equilibrium to the right.
•
Remove the hydrogenated product as it is produced
•
Cool down the system
•
Add excess hydrogen
•
Add excess ethene
d) How could you check if the reaction in b) was complete?
Add bromine water to the product of the reaction. If the orange colour remains, the reaction has gone to
completion. If the colour disappears, the bromine is adding across a double bond, so starting material must still
be present.
4. The product of the following reaction: CH3CH2CHCHCH2CH3 + HCl ? is,
Answer: B This is addition across the double bond.
5. Using structural formula, write the equation for the reaction between 2-butene and
a) H2
CH 3
CH
CH
CH 3
+
H2
CH 3
H
H
CH
CH
Br
Br
CH
CH
H
Br
CH
CH
H
OH
CH
CH
CH 3
b) Br2
CH 3
CH
CH
CH 3
+
Br 2
CH 3
CH 3
c) HBr
CH 3
CH
CH
CH 3
+
HBr
CH 3
CH 3
d) H2O (with H2SO4 catalyst)
H 2SO 4
CH 3
CH
CH
CH 3
+
H 2O
CH 3
22
CH 3
6. When a trace of a strong acid, eg sulfuric acid, is added to a mixture of an alkene and water an
addition reaction occurs.
a) Give an equation for the addition reaction
C2H4 + H2O C2H5OH
b) What is the role of the sulfuric acid?
The sulphuric acid acts as a catalyst.
7. Which of the following compounds will discolour bromine water?
Answer: D - I and III These are the only unsaturated compounds listed.
8. Give a structural equation and conditions for ethene reacting with:
a) water with H2SO4 catalyst
CH2
CH2
+
H2O
H
OH
CH2
CH2
H
H
CH 2
CH 2
H2SO4
b) hydrogen
CH 2
CH 2
+
H2
c) bromine
CH 2
CH 2
+
Br 2
Br
Br
CH 2
CH 2
H
Br
CH 2
CH 2
d) hydrogen bromide
CH 2
CH 2
+
HBr
e) with itself in a polymerization reaction
CH 2
CH 2
+ CH 2 CH 2
CH 3
CH 2
CH 2
9. Polythene and polyvinylchloride (PVC) are two common plastics.
a) give an equation to show the formation of polythene.
n CH 2
CH 2
CH 2
CH 2
n
b) give an equation to show the formation of polyvinylchloride.
H
Cl
H
Cl
n C
C
C
C
H
H
H
n
H
23
CH 3
c) draw the repeating unit in PVC.
H
Cl
C
C
H
H
10. Propene is the monomer used to make polypropene, which may be used in carpets and clothes.
Using structural formulas, give the equation for this polymerization.
H
n C
CH3
H
CH3
C
C
C
n
H
H
H
H
11. Complete this equation:
a) CH2CH=CHCH2CH3 + Br2(aq) CH2CH=CHCH2CH3 + Br2(aq) CH2CH2CH2CH2CH3
b) Describe the colour change seen during the reaction
The orange solution will change to a colourless one.
12. Write an equation to show how you could prepare 1,2-dichlorobutane from an alkene.
CH2
CH
CH2
CH3
+
Cl2
Cl
Cl
CH2
CH
CH2
CH3
13. What is the main difference between fats and oils?
Fats have higher melting points whereas oils have lower melting points. This is due to the presence of double
bonds which change the shape of the molecule, resulting in “kinks” in the hydrocarbon chains of fats. These
“kinks” make the molecules less able to pack together so the van der Waals’ forces between the molecules are
decreased. This means that these fats have higher melting points.
14. The polyunsaturated oil, linoleic acid and the saturated fat, stearic acid have the same number of
carbon atoms and yet have very different melting points. Explain, with reference to their structures.
CH3
H2C
CH2
O
H2C
H2C
CH2
H2C
HC
HC
H2C
CH2
H2C
HC
CH
CH2
C
OH
CH2
CH2
linoleic acid, C18H32O2 (mp -5ºC)
24
O
CH2
H3C
CH2
CH2
CH2
CH2
CH2
CH2
CH2
CH2
CH2
CH2
CH2
CH2
CH2
CH2
C
CH2
OH
stearic acid, C18H36O2 (mp 69ºC)
Linoleic acid contains alkene groups in the hydrocarbon chain, whereas stearic acid does not. These carbon to
carbon double bonds cause “kinks” in the chain. These "kinks" do not allow the linoleic acid molecules to align
close together. This is unlike stearic acid, where regular “straight” chain of carbon atoms allow the molecules to
pack closely together resulting in increased van der Waals, forces. The increased van der Waals’ forces
between stearic acid molecules are responsible for its higher boiling point over linoleic acid.
15. Indicate which of the following statements are TRUE for the industrial scale production of ethanol:
a) Ethene is used as the starting material
true
b) The product would discolour bromine water
false
c) An acid catalyst is used
true
d) Water is added across a double bond
true
16. The reactions of alkenes have economic importance. For the three key manufacturing processes
that you have learnt about in this topic:
a) Give the name of each reaction
Hydrogenation, hydration and addition polymerisation
b) Illustrate the chemistry behind each reaction by providing a reaction equation for each process
using ethene as the starting material
Hydrogenation: C2H4 + H2 C2H6
Hydration: C2H4 + H2O C2H5O
Addition polymerisation:
H
H
H
H
n C
C
C
C
H
H
H
n
H
c) Give the desired product of each manufacturing process
Hydrogenation: to manufacture margarine
Hydration: to produce ethanol from ethene
Addition polymerisation: to manufacture plastics
25
d) Explain the need or importance of the processes by indicating uses of the products.
The hydrogenation of liquid oils to make solid margarine provides a more appropriate melting point substance
for use in the food industry, to manufacture other food products or in the home as a polyunsaturated substitute
for butter.
Ethanol is widely used as a solvent and a precursor in the manufacturing of so many other chemicals.
Polymers for use as plastics are vital in our everyday lives. Polymers may be used as synthetic fibres in
clothing, packaging and technology. Plastics are highly versatile!
10.4 Exercises
1. Write equations for the complete combustion of propanol and butanol.
C3H7OH(l) + 5O2 3CO2(g) + 4H2O(l)
C4H9OH(l) + 6½O2 4CO2(g) + 5H2O(l)
2. Using your IB Chemistry data booklet, compare the standard enthalpies of combustion of ethanol
and octane (major constituent of petrol).
a) Comment on the relative energy densities of ethanol and octane, and propose a reason for the
difference.
The standard enthalpy of combustion of octane is ∆HoC = -5512 kJ mol-1.
The standard enthalpy of combustion of ethanol is ∆HoC = -1371 kJ mol-1.
So, upon combustion, octane releases much more energy than ethanol per mole. The reason being that ethanol
is already partially oxidised (it contains O) and octane is a larger molecule (4 x more C).
Energy density- energy per gram of fuel:
Octane C8H16 Mr = 114.26
Number of moles in 1 g =
n = 1g/114.26 g mol-1 = 8.75 x 10-3 moles
Energy content per gram =
8.75 x 10-3 moles x –5512 kJmol-1 = -48.24 kJ
Ethanol C2H5OH Mr = 46.08
Number of moles in 1g
n = 1g/46 g mol-1 = 0.0217 moles
Energy content per gram =
0.0217 moles x –1371 kJmol-1 = -29.75 kJ
Octane has the higher energy density.
In deciding what fuel is best to use the chemist considers, among other factors, the
energy density of the fuel. The energy density is the heat energy given out per g or per
kg of fuel. Fuels are purchased by the L or kg, not by the mol. Hence energy density
will give a better comparison of the heat energy available from fuels.
26
b) Considering your answer to (a) why is ethanol being used as a fuel?
As well as the energy content per gram of fuel the other factors that should be taken into account when sourcing
fuel include how much energy is needed to extract or mine the substance, how much energy is required to
transport the fuel, what are the by-products of combustion and how efficient is the fuel (i.e. a fuel may have a
high energy content but may be very inefficient, very little of the fuel may be converted to useful energy). The
main advantage of ethanol is that it is renewable (extracted from corn or sugar cane) while octane is sourced
from oil which is non-renewable and carbon intensive.
c) What are the likely impacts of production of biofuels such as ethanol from biomass?
In the case of ethanol, which is often extracted from cornfields, we must also consider the cost of farming the
land for fuel, as in this case of many countries in South America; the growing of crops for ethanol may come at
the great price of deforestation. Forests are "carbon sinks", they absorb CO2 from the atmosphere by
photosynthesis. Deforestation may therefore increase CO2 levels in the atmosphere, contributing to global
warming. The carbon offset obtained from using biofuels such as ethanol may not be significant!
3. Oxidation of primary alcohols can involve two steps.
a) Write equations to illustrate the two step oxidation process of propan-1-ol.
OH
O
H+/Cr2O72-
CH2
CH2
[O]
CH3
O
H+/Cr2O72-
CH
CH2
[O]
CH3
HO
C
CH2
CH3
b) How could the product of the first oxidation step be isolated in good yield from the reaction?
Refer to intramolecular forces in your answer.
Add the oxidant mixture drop wise into the propanol at the same rate as the propanal (bp 48 OC) distils over.
The boiling points of propan-1-ol, propanal and propanoic acid are different. Propan-1-ol and propanoic acid are
both capable of hydrogen bonding whilst propanal is not. Only weaker dipole-dipole forces are able to operate
between propanal molecules. As a result, propanol and propanoic acid have much higher boiling points than
propanal. Watch the temperature as you distil. If it exceeds 48 OC then stop. Collect only the lower bp fraction.
Restricting the amount of oxidant also prevents further oxidation.
c) Outline the reaction conditions required in order to produce a good yield of propanoic acid.
To ensure that all the propan-1-ol is oxidised to propanoic acid, the reaction should be carried out by heating
under reflux with excess oxidant. This will not allow the propanal to escape, as it will condense back in to the
reaction mixture where it can be further oxidised to propanoic acid.
4. Write an equation to show the conversion of butanol to the corresponding aldehyde.
OH
CH2
H+/Cr2O72CH2
CH3
CH3
[O]
O
CH
27
CH2
CH3
CH3
5. Write half equations to show the oxidation of ethanol to ethanal by potassium permanganate in acid
solution. (Hint: you may need to refer back to Topic 9 if you have forgotten how to write half
equations!)
8H + 5e- + MnO4- Mn2+ + 4H2O
+
C2H5OH C2H4O + 2H+ + 2e6. Determine the products, if any of the following reactions:
a)
H3C
H
CH3
C
C
H+/Cr2O72-
OH
CH3
H
[O]
No reaction – the reactant is a tertiary alcohol
b)
CH3
H3C
CH
CH2
CH3
H+/Cr2O72H3C
[O]
OH
C
CH2
O
but-2-one
c)
CH3(CH2)3OH
H+/Cr2O72-
CH3(CH2)2COOH
[O]
heated at reflux
2 hrs
butanoic acid
d)
OH
CH3
CH2
CH
O
H+/Cr2O72CH2
CH3
[O]
CH3
CH2
C
CH2
pent-3-one
e)
O
CH2
C
OH
H
H+/Cr2O72[O]
Distill lower boiling
product from mixture
aldehyde (benzaldehyde)
28
CH3
7. Suggest what alcohol and reaction conditions you might use to obtain the following products after
oxidation:
a) CH3CH2CHO
Reactant: propan-1-ol
Reaction conditions: Distillation of low boiling point material from reaction mixture. Limit addition of oxidant.
b)
O
Reactant: cyclohexanol
OH
Reaction conditions: Heat under reflux with oxidant. This ketone can not be further oxidised sos there is no need
for a distillation.
c)
OH
C
H3C CH2
O
Reactant: propan-1-ol
Reaction conditions: Heat under reflux. Excess oxidant.
d)
H3C
O
CH2
CH
H3C
C
CH2
OH
Reactant: 2-ethylbutan-1-ol
H3 C
OH
CH2
H 3C
CH CH 2
CH 2
Reaction conditions: Heat under reflux. Excess oxidant.
e)
H3C
OH
CH
H3C
C
O
Precursor: 2-methylpropan-1-ol. Reaction conditions: Heat under reflux. Excess oxidant.
29
f)
CH3(CH2)3CHO
Reactant: pentan-1-ol
Reaction conditions: Distillation of low boiling point material from reaction mixture. Limit oxidant.
10.5 Exercises
1. Define, with an example, the following terms:
a) Electrophile
An ion or molecule that is electron deficient and can accept electrons. They may do this by reacting with (or
"attacking") parts of molecules that are rich in electrons in order to gain electrons. An example is a carbon atom
attached to an electronegative group, i.e. the Cδ+ end of a polar covalent bond. H+ is another.
b) Nucleophile
An ion or molecule that can donate electrons. They are either negative ions (eg OH-) or molecules that have
electron pairs (lone pairs or double bonds) eg NH3. They are electron rich and in reactions tend to attack
positively charged parts of a molecule.
Electrophiles are electron ‘loving’; they seek out electrons.
Nucleophiles seek out the positive; they are nucleus ‘loving’.
2. What do curly arrows indicate when used in a reaction mechanism?
Curly arrows are used to show the movement of pairs of electrons in the reaction mechanism. The arrow begins
on the electron pair and the head of the arrow shows where the electrons have moved to.
3. What do δ+ and δ- indicate?
Partial positive charge and partial negative charge. These symbols indicate the polarity of a bond moment or
dipole moment. The δ- end of the polar bond is the atom which has the greatest share of the electron pair in the
bond (the more electronegative atom) whilst the δ+ end is the atom which does not have the greatest share of
the electron pair in the bond (less electronegative).
4. Consider the following halogenoalkanes: C2H5I, C2H5Br, C2H5Cl and C2H5F. Which is the least
reactive? Which is the most reactive? Give reasons for your answer.
C2H5F is the least reactive; C2H5Br is the most reactive. The C–F covalent bond is the strongest because the
small F atom is closest to the C atom. Because the two atoms are closer together, increased orbit overlap
results in a stronger bond. In this case the distance factor overrules the electronegativity factor (ie
electronegativity decreases down the group, and lower electronegativity differences would normally result in a
stronger bond). As a result of the increasing bond strength, the order of these compounds in decreasing
reactivity is: C2H5I, C2H5Br, C2H5Cl and C2H5F.
30
5. Explain the difference between homolytic and heterolytic bond cleavage. What is the difference
between fission and fusion?
Homolytic bond cleavage is the breaking of a chemical bond into two free radicals, with each radical taking one
electron from the broken chemical bond. Heterolytic cleavage is the breaking of a compound where the two
fragments are oppositely charged ions, with one fragment taking both electrons from the bond.
Fission means splitting, e.g. in nuclear fission the atom is split.
Fusion means joining, e.g. in nuclear fusion H atoms join to make He atoms.
6. Formal charge
a) How many valence electrons does the carbon atom of a carbocation have?
The carbon of a carbocation has three valence electrons. A C atom has 4 outer shell electrons: to become a C+
it must lose one electron, leaving three valence electrons.
Carbocations have a positive C atom, C+, they are intermediates of some organic
reactions.
b) Draw the Lewis structure for the hydroxide ion.
O H
c) What is the formal charge on the carbon atom of a carbocation and the oxygen of the hydroxide
ion?
+1 and -1
7. Number the following carbocations in order of increasing stability.
H3C
C
H 3C
CH 2
H3C
CH 3
CH3
CH3
4
C
1
3
H3C
C
CH3
H
2
The order of stability for carbocations is tertiary > secondary > primary. The more
groups other than hydrogen attached to the C+ the more stabilised the positive
charge.
31
8. Would the following halogenoalkanes undergo SN1 or SN2 reactions, or both?
Compound
A
1°°, 2°° or 3°° halogenalkane
mechanism
primary
SN2
tertiary
SN1
primary
SN2
secondary
SN1 or SN2
H3C-Cl
Br
H3C
C
CH3
B
C
CH3CH2F
D
CH3CH(Cl)CH3
Tertiary halogenoalkanes favour SN1 mechanism.
9. For SN1 and SN2 reactions, which molecular entities do the terms unimolecular and bimolecular refer
to?
The terms unimolecular and bimolecular refer to the number of molecules involved in the rate determining step
of the reaction. For the SN1 reaction, there is only one molecule involved in the rate determining step, whereas
for the SN2 reaction, there are two.
10. 1-bromobutane reacts with sodium hydroxide in a nucleophilic substitution reaction.
a) Write an equation for the reaction.
C4H9Br + NaOH C4H9OH + NaBr
b) Classify the halogenoalkane as primary, secondary or tertiary.
Primary
c) Explain whether the reaction will proceed by a SN1 or SN2, mechanism, give reasons for your
answer.
As a primary alkane is undergoing this substitution reaction, it will proceed via the SN2 mechanism. This is
because, a primary carbocation intermediate would form if this reaction proceeded by the SN1 mechanism. This
is very unstable.
d) Use curly arrows showing electron movement to illustrate the mechanism of the reaction, and
label the electrophile and nucleophile.
Br
OH
H
C
OH
CH2
CH2
CH3
H
H
nucleophile
C
CH2
CH2
CH3
+ Br
H
electrophile
11. 2-chloro, 2-methylpropane reacts with sodium hydroxide in a nucleophilic substitution reaction.
a) Write an equation for the reaction.
CH3(Cl)C(CH3)CH3 + NaOH CH3(OH)C(CH3)CH3 + NaCl
32
b) Classify the halogenoalkane as primary, secondary or tertiary.
tertiary
c) Explain whether the reaction will proceed by a SN1 or SN2, mechanism, give reasons for your
answer.
The reaction will proceed by an SN1 mechanism. The halogenalkane is tertiary, so when the C-Cl bond is
broken, a tertiary carbocation will be formed readily, as the three R groups are able to stabilise the positive
charge (inductive effects). In terms of steric effects: the tertiary carbocation is sterically hindered so the
nucleophile will not be able to attack the electrophilic carbon atom of the halogenoalkane, until the leaving
group, Cl leaves, ie an SN1 mechanism must occur.
d) Use curly arrows showing electron movement to illustrate the mechanism of the reaction, and
label the electrophile and nucleophile
Cl
H 3C
C
H3C
CH3
C
CH3
CH3
CH3
OH
H 3C
OH
C
CH3
H3C
C
CH3
CH3
CH3
electrophile
nucleophile
12. Suggest products for the following reactions and indicate whether the reaction is SN1 or SN2:
a)
CH3 CH3
NaOH
C
CH 3
CH3 CH3
Cl
C
CH2
CH 3
OH
+ NaCl
CH2
SN2 reaction mechanism
(primary halogenoalkane)
b)
NaOH
CH3CH2Br
CH3CH2OH + NaCl
SN2 reaction mechanism
(primary halogenoalkane)
10.6 Exercises
1. Complete the reaction pathways scheme on page 335.
Answers shown at end of this section.
33
2. Suggest starting materials and reaction conditions given the following products:
CH3 CH2
OH
H+/Cr2O72-
CH2
[O]
OH
CH3 CH2
C
O
a)
heat at reflux, excess oxidant
OH
O
H+/Cr2O72[O]
CH3
CH3
b) CH3 CH2 CH
heat at reflux (this material can not be oxidised further)
CH2
CH3
CH3
C
C
Cl + NaOH
H3 C
C
CH3
CH3
C
C
CH3
C
CH2
H3C
+ HCl
H
CH3
d)
gaseous HCl may be used
HC
CH3
CH3
CH2
CH3
c)
dilute NaOH(aq), SN1 as tertiary halogenoalkane
CH3
C
+ Br2
OH
CH3
H
Cl
C
C
H
CH3
Br
Br
CH3
HC
C
C
CH3 H
CH3 H
CH3
e)
add Br2 drop by drop, with shaking
CH3
CH3
CH3
3. Deduce reagents and reaction conditions for the conversions A-G. What are the structures of
products 1 and 2? Provide a mechanism for reactions involving reagents and conditions F and G.
1
2
H
H
A
OH
C
C
Br 2 (aq)
B
C
C
Br
H 2, nickel
catalyst
H
C
H
CH3
H
CH 3
CH3
CH 3
C Br 2
E HBr
D H2O,
conc.
H 2SO 4
F
Br 2
H
C
C
H3C
Br
H
Br
G NaOH, SN1
C
CH3
H3 C OH
34
CH 3
C
Br
CH 3
4. Deduce reaction pathways given the following reactants and products. No more than two steps
should be used. Include equations, reagents and reaction conditions.
a) acetone from propene
OH
CH3
CH
CH2
+ H2O
CH3
CH
H
H+/Cr2O72-
CH2
CH3
[O]
O
H
C
CH2
b) acetic acid from ethanol
O
CH3
CH2
OH
H+/Cr2O72[O]
CH3
C
OH
heat at reflux
c) carbontetrachloride from chloromethane
Cl
Cl
Cl2
H
C
H
Cl
C
H
Cl
Cl
(heat and a catalyst)
d) formaldehyde from chloromethane
Cl
O
OH
+
H
C
H
+ NaOH
H
(warm, dilute NaOH(aq)
H /Cr2O7
H
C
H
2-
C
[O]
H
H
limited oxidant, methanal, bp –21
35
H
O
C, collected by distillation
Draw structural formulas for compounds 1-7 and provide reactants and conditions for steps A-J.
1
2
H
H
H
C
C
H
H
H
alkane
3
H
Cl
Cl
C
C
H
H
A Cl2, UV
H
H
Cl
C
C
H
H
Cl
trihalogenoalkane
(or tetrahalogenalkane)
dihalogenoalkane
2 , Nickel
B Hcatalyst
Cl
Cl2, UV
D Cl2
C Cl2, UV
4
H
36
H
Cl
C
C
H
H
H
H
C
F H 2C
CH 2
C
H
H
alkene
halogenalkane
H
G
H
E HCl
H
H
C
C
H
H
n
poly(alkene)
H 2O, conc.
H 2SO 4
NaOH,
SN2
5
6
H
H
I H +/Cr 2O72-
OH
C
C
H
H
7
H
H
alcohol
H
[O]
C
H
O
H
J H+/Cr2O72H
C
H
aldehyde
[O]
C
H
O
C
OH
carboxylic acid