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C
C
C
C
C
C
C
C
H
H
H
HH
HH
HH
HH
HH
HH
H
H
H
C
C
C
C
C
C
C
C
H
H
H
HH
HH
HH
CH
HH
CH
H
H
H
H H H HH C H
HH
C C
C
CC
C C
H
H
C
H
H
H
H
H HH
H
H
H
H
H
H
H
octane - unbranched (straight-chain)
4-ethyl-2-methyloctane - branched
ethylcyclohexane - cyclic
Figure 1 - Unbranched, branched, and
cyclic hydrocarbons.
Experiment #3 - Hydrocarbons
Introduction
Organic chemistry is the chemistry of the
compounds of carbon. Currently over twenty
million compounds have been reported in the
chemical literature; about 90% of them are
organic, ie they contain carbon. The remaining
compounds are called inorganic and are formed
from the other elements, of which there are
about 100. That carbon so dominates
compound formation is a result of the fact that it
is almost unique in its ability to form long
chains with other carbon atoms. [Carbon’s
neighbor in the periodic table, silicon, can do
this but rarely does.] These chains with one
carbon joined to a second and the second
joined to a third, etc., can be branched, ie,
chains of carbon atoms can be attached to
carbons in the original chain. It is also possible
for one carbon in a chain to become bonded to
another carbon in that chain, resulting in a
closed ring of atoms. We call these compounds
cyclic.
Since there are so many organic
compounds it is fortunate that we can organize
them into various groups that have some
similarity to each other. For example, one large
group of organic compounds is known as the
hydrocarbons because members of this group
contain only carbon and hydrogen and no other
elements. Figure 1 shows examples of branched, unbranched
and cyclic hydrocarbons.
It is possible to subdivide the hydrocarbon group of compounds
based on the
bonding between the carbons. If all the carbon-carbon bonds
are single, the compound
is an alkane. If at least one of the carbon-carbon bonds in the
compound is a double
bond, and the remaining carbon-carbon bonds are single, the
compound is an alkene. If
at least one of the carbon-carbon bonds in the compound is a
triple bond, and the
remaining carbon-carbon bonds are single, the compound is an
alkyne. If the compound
contains a six carbon ring that has alternating double and single
bonds around the ring
(three double and three single), we say that ring, and the
compound, is aromatic. An
aromatic ring looks like an alkene with three double bonds
because that’s the way we
draw it using Lewis structures. However, the actual bonding in
such a ring is
considerably different from that in alkenes and, consequently,
many of the chemical
properties are different also. Therefore, we place these
compounds in a separate family.
By the way, the term aromatic as used here has nothing to do
with fragrance.
Experiment #3 Hydrocarbons Page 2
Hydrocarbons may be saturated or unsaturated. A saturated
hydrocarbon is one
that is maxed out in terms of the number of hydrogens that can
be present given the
number of carbons in the compound; it is impossible to add
more hydrogen atoms to the
compound so it is saturated with hydrogen. Acyclic (no rings)
alkanes are saturated;
there is no way additional hydrogens can be added while
keeping the same number of
carbons and maintaining normal bonding between the atoms.
Alkenes, alkynes, aromatic
compounds, and cyclic alkanes are unsaturated because
hydrogen can be added to them,
in theory and usually in practice, making them into acyclic
alkanes. Some examples
follow.
Physical Properties
Some molecules carry an electrical charge because there is a
difference between
the number of electrons (each with a -1 charge) and the number
of protons (each with a
+1 charge) in the molecule. We call molecules of this type
ions. The ammonium ion,
NH4
+
, has a +1 charge because it has 11 protons (7 from nitrogen and
4 from the
hydrogens) and 10 electrons (2 in nitrogen’s first shell, and 8 in
its second shell [which
are used to bond the hydrogens to the nitrogen]). So, with 11
plus charges and 10 minus
charges, the ammonium ion has a net charge of +1. Most
molecules are not ions (and we
simply call them molecules); in other words they are electrically
neutral – they have a
charge of 0 because they have the same number of protons as
electrons.
H3C CH CH CH3 H2 H3C CH CH CH3
H H
+
catalyst
H3C C C CH3
catalyst
+ H3C C C CH3
H H
H H
H22
C C
C
CC
C H
HH
H
H H
catalystH23
C C
C
CC
C H
HH
H
H H
H H
H
H
H
H
+ H2
in theory,
if not easily
done in
practice
C C
C
CC
C H
HH
H
H H
H H
H
H
H
H
H H
unsaturated
compounds
saturated
compounds
Experiment #3 Hydrocarbons Page 3
φ
φ
φ
φ
v
v v
m
m m
n = ss =
sin
sin
vacuum
materiallight beam
light beam
Figure 2 - Index of Refraction
In some molecules, even though the net charge is 0, the
distribution of positive
charges (protons) and negative charges (electrons) within the
molecule is not the same.
Such a molecule has a lopsided charge distribution – one side of
the molecule is electron
rich (has a partial negative charge) and the other side is proton
rich (has a partial positive
charge). We say that such a molecule is polar or that it has a
dipole moment. The more
lopsided the charge distribution, the larger the dipole moment
and the more polar the
molecule. In some molecules the distribution of positive and
negative charges is the
same; these molecules have no dipole moment and are nonpolar.
When organic
molecules are polar it is usually because there are one or more
highly electronegative
atoms on one side of the molecule. Electronegative atoms are
those that attract electrons
toward themselves; common examples are nitrogen, oxygen, and
the halogens, especially
fluorine and chlorine.
Polar molecules attract each other because the negative side of
one attracts the
postive side of another (opposite charges attract). The more
polar the molecules the
more they attract each other. It is also true that nonpolar
molecules do not attract each
other as strongly as polar molecules do and polar molecules are
not attracted to nonpolar
molecules as much as polar molecules are attracted to each
other.
Hydrocarbons are molecules that have little or no polarity
because they do not
contain electronegative atoms. They are soluble in solvents of
low polarity. They are not
soluble in water, which is very polar, because the water
molecules attract each other
strongly (and are not nearly as interested in attracting nonpolar
molecules).
Density is the mass of a material divided by its volume; it is
often expressed in
terms of grams of mass per cubic centimeter of volume. The
density of water, for
example, is 1.00 gram per cubic centimeter. For the most part,
hydrocarbons are less
dense than water, so, given their insolubility in water, they float
on it. Crude oil and its
derivatives gasoline, kerosine, and fuel oil are mainly
hydrocarbons; none of these is
soluble in water and they float on its surface since they are less
dense than water.
The index of refraction, n, of a compound is
the speed of light in a vacuum, sv divided by the
speed of light in that substance, sm. Since light
travels faster in a vacuum than anywhere else, the
index of refraction is greater than 1.000 for any
substance. It turns out that the index of refraction
can be measured by noting how much a beam of
light bends when it travels from a vacuum (or air,
since the speed of light in air is almost the same as
that in a vacuum) into some substance. See Figure 1.
The index of refraction of liquids is usually measured
using an Abbé refractometer where the liquid is
placed on a glass prism of known index of refraction.
[Glass is easier to work with than a vacuum.] Because
of the large number of electrons located around the
ring in aromatic compounds these compounds usually have
larger indexes of refraction
than other hydrocarbons. Typically, aromatic compounds will
have indexes of refraction
greater than 1.45, while other hydrocarbons will have smaller
values.
Experiment #3 Hydrocarbons Page 4
C 3H 8 O 2 C O 2
H 2O5 3 4+ +
H3C CH CH CH3 H3C CH CH CH3
Br Br
+
H3C C C CH3 + H3C C C CH3
Br Br
Br Br
2
Br2
Br2
Natural gas is composed of alkanes; it is about 90% methane
with small amounts
of ethane and propane. Methane, ethane, propane, and butane
are gasses at room
temperature. Pentane is barely a liquid at room temperature and
the higher molecular
weight alkanes are liquids, the straight-chain versions becoming
solids at about 16 carbon
atoms. Butane boils at about 0
o
C, which is why butane lighters do not function well
below that temperature: the liquid changes to gas too slowly
below its boiling point to
maintain a flame.
Chemical Reactions of Hydrocarbons
1. Combustion.
All of the hydrocarbons undergo combustion – they burn in the
presence of
oxygen. If there is enough oxygen the combustion will be
complete, ie, the products of
the combustion will not burn. In fact, if combustion is complete
the products will be
carbon dioxide and water, and, of course, heat is given off – the
reaction is highly
exothermic – which is ordinarily the purpose of this reaction.
The equation for
combustion of propane (bottle or LP gas, used for cooking and
heating, is mainly
propane) is shown below.
2. Reaction with Bromine.
Hydrocarbons with multiple bonds (unsaturated hydrocarbons
except most
cycloalkanes) react with bromine. Tetrachloromethane (carbon
tetrachloride) or
cyclohexane are usually used as solvents because they are
unreactive toward both
bromine and hydrocarbons that have multiple bonds.
Alkenes and alkynes undergo an addition reaction with bromine.
The double
bond of an alkene becomes a single bond and one bromine atom
becomes bonded to
each of the carbons that had shared the double bond. No other
product is formed; the
alkene and bromine simply add together, which is why it’s
called an addition reaction.
The triple bond of an alkyne also undergoes an addition
becoming a single bond, but in
this case each of the carbons that had been joined by the triple
bond will now hold two
bromine atoms. Examples follow.
This reaction usually occurs quickly at room temperature
without a catalyst.
Experiment #3 Hydrocarbons Page 5
C C
C
CC
C H
HH
H
H H
H H
H
H
H
H
+
C C
C
CC
C Br
HH
H
H H
H H
H
H
H
H
Br2+ HBr
heat or
ultraviolet
light
aromatic
+ alkene
aromatic
+ alkane
aromatic
only
(FeBr3)
H BrBr2 HBr+ +
catalyst
CH2 CH3 + Br2
CH CH2 Br2+ CH CH2
Br Br
addition reaction to
alkene double bond,
bromine color
dissipates quickly
reaction quite slow in the
absence of catalyst,
bromine color fades
quite slowly
reaction slow in the
absence of catalyst,
bromine color fades slowly
CH2 CH3
Br
CH2 CH3Br
+
catalyst
(FeBr3)
Bromine is a reddish-brown color. All of the other substances
in these reactions are
colorless. So, when bromine is added to an alkene or alkyne the
red-brown color
dissipates quickly, often almost instantly.
Bromine can react with an alkane, but this reaction requires heat
or ultraviolet
light to be successful, and the reaction is a substitution, not an
addition: a hydrogen is
replaced by a bromine and hydrogen bromide is a byproduct.
An example follows.
Since this reaction does not take place in the absence of
ultraviolet light or heat, if
bromine is added to an alkane under these conditions (room
temperature and no sunlight
or other source of uv) the reddish-brown color of bromine will
persist.
Aromatic rings react with bromine in a substitution reaction;
this reaction is slower
than the addition of bromine to an alkene or alkyne and may
require a catalyst. Iron (III)
bromide is a good catalyst for this reaction. If iron filings are
added to a mixture of the
aromatic compound and bromine, iron (III) bromide will form.
Note that hydrocarbon
groups that may be attached to the aromatic ring will react as
described above.
Experiment #3 Hydrocarbons Page 6
+
H3C CH CH CH3
H O
S OO
O
H
H3C CH CH CH3 OH S
O
O
O H
Nonpolar and not
capable of hydrogen
bonding; does not
dissolve in H2SO4.
Polar and capable
of hydrogen bonding.
Polar
and
capable
of hydrogen bonding;
soluble in H2SO4.
3. Reaction with Concentrated Sulfuric Acid.
Alkenes react with concentrated sulfuric acid at room
temperature in an addition
reaction that produces an alkyl sulfonic acid. The alkyl sulfonic
acid is highly polar and
is soluble in the polar sulfuric acid with sufficient stirring. The
non-polar alkene is not
soluble in the polar concentrated sulfuric acid, but dissolves as
it reacts and the mixture is
stirred. [A second, related issue here is that the alkene is not
capable of forming hydrogen
bonds, while sulfuric acid and the alkyl sulfonic acid can.
Hydrogen bonding is an
especially strong type of dipole-dipole attraction. In order for
H-bonding to occur at least
one of the molecules must have a hydrogen (directly) bonded to
a fluorine, oxygen or
nitrogen. This hydrogen will be strongly attracted to a fluorine,
oxygen or nitrogen in
another molecule. Hydrogen bonding is a very important
phenomenon, especially in
biological systems.] An example follows.
Alkynes react slowly or not at all with concentrated sulfuric
acid unless a catalyst
is present (HgSO4). They may darken but likely will not
dissolve in the sulfuric acid.
Alkanes do not react. Aromatic hydrocarbon rings react quite
slowly at room temperature
(undergoing a substitution reaction in which a hydrogen bound
to a carbon in the ring is
replaced by an -SO3H group).
4. Reaction with Potassium Permanganate.
Dilute aqueous potassium permanganate oxidizes alkenes to
geminal diols. [Diol
means two -OH groups. Geminal means on adjacent carbons].
Alkynes are oxidized to
geminal diketones. In these processes the purple potassium
permanganate is reduced to
a brown precipitate of manganese dioxide. Since the potassium
permanganate is soluble
in water, but neither the water nor the potassium permanganate
are soluble in the
hydrocarbon, reaction takes place at the water-hydrocarbon
interface, and is somewhat
slow. Consequently, it may take several minutes for a brown
precipitate to form.
Examples follow.
Experiment #3 Hydrocarbons Page 7
a geminal diol
brown
purple
++
++
223
423
KOHMnO2
H2OKMnO4
CHH3C CH CH3
OH OH
CHH3C CH CH3
a geminal diketone
+ 4KOH4+ MnO2
brown
2+ H2O4+ KMnO4
purple
3
3
CH3C C CH3
O O
CH3C C CH3
Caution!
The organic liquids are highly flammable. Keep them away
from flames.
Concentrated sulfuric acid and bromine are both very corrosive.
Wear goggles and
avoid getting them on yourself. If you do get any on you, wash
the affected area with
lots of water. Potassium permanganate will stain your skin and
clothing.
Under more vigorous conditions the geminal diol (from an
alkene) or geminal
diketone (from an alkyne) that is initially produced will be
further oxidized by the
permanganate, producing more manganese dioxide.
Alkanes and aromatic rings are unreactive toward dilute
aqueous potassium
permanganate.
Objectives of the Experiment –
1. To investigate three physical properties of some
hydrocarbons: density, solubility and
refractive index.
2. To compare the chemical reactivities of an alkane, alkene,
alkyne, and an aromatic
compound.
3. To use the above properties to determine whether an
unknown is an alkane, alkene,
alkyne, or aromatic.
Procedure –
Experiment #3 Hydrocarbons Page 8
1
0
x
7
5
m
m
t
e
st
t
u
b
e
,
ac
tu
al
s
iz
e
General
1. The hydrocarbons hexane (alkane), cyclohexene (alkene), 1-
hexyne (alkyne), and
toluene (aromatic) are available in dropper bottles.
2. The test reagents 1% bromine in cyclohexane, 1% aqueous
potassium permanganate,
and concentrated sulfuric acid are available in dropper bottles,
also.
3. Unknowns are in dropper bottles labeled A, B, C, and D.
One of these is an alkane,
another an alkene, another an alkyne, and another an aromatic
compound.
4. Record observations in the appropriate places on the report
sheet. If you
don’t record it, it’s like you never did it.
Physical Properties
Step #1. Label four 10x75mm test tubes, labeling each with the
name of a
different hydrocarbon as listed above.
Step #2. Place 10 drops of the appropriate hydrocarbon in each
of these test
tubes.
Step #3. Test for water solubility and density relative to water.
Add 10 drops of water to each test tube. Holding the test tube
by the neck with
one hand, strike the bottom of the test tube horizontally with a
finger of the other
hand several times to mix the two liquids. Allow the test tube
to sit undisturbed
for a minute. Are the compounds miscible (soluble in each
other) or immiscible
(insoluble)? If they are immiscible, which compound is on the
bottom? Can you
determine which compound is more dense? If so, which one is?
Rinse each test tube with acetone and draw air through it, using
an aspirator, to
dry it.
Step #4. Repeat step #2.
Step #5. Test for solubility in hexane and density relative to
hexane.
Add 10 drops of hexane to each test tube. Holding the test tube
by the neck with one
hand, strike the bottom of the test tube horizontally with a
finger of the other hand
several times to mix the two liquids. Allow the test tube to sit
undisturbed for a minute.
Are the compounds miscible (soluble in each other) or
immiscible (insoluble)? If they are
immiscible, which compound is on the bottom? Can you
determine which compound is
more dense? If so, which one is?
Will any of the above tests be useful in distinguishing one type
of hydrocarbon from
another? Is there any point in running these tests on unknowns
A-D?
Experiment #3 Hydrocarbons Page 9
Step #6. Your instructor will have shown you how to use the
Abbé refractometer.
Measure the index of refraction of each of the known
hydrocarbons and of each of the
unknowns, A-D. If you trust the person next to you, you can
work as partners. Each of
you should measure the refractive index of two of the known
hydrocarbons and two of
the unknowns. Be sure to wipe the prisms dry with a Kimwipe
tissue between
measurements.
Chemical Properties
1. Reaction with Bromine –
Wear gloves and, as usual, goggles!
Step #1. Label four 10x75mm test tubes, labeling each with the
name of a different
hydrocarbon as listed above.
Step #2. Place 10 drops of the appropriate hydrocarbon in each
of these test tubes.
Step #3. Add 15 drops of 1% bromine in cyclohexane. Shake
the test tube as above,
briefly, after the addition. Record how long it takes, in
seconds, for the bromine color to
dissipate; in other words, how long it takes for the solution to
go from brown-orange to
colorless or almost colorless. If this happens immediately, say
that. If it takes more than
120 seconds, say that.
Step #4. Label four test tubes, labeling each with the name of a
different unknown, A-D.
Step #5. Repeat Step #2, using the unknowns A-D.
Step #6. Repeat Step #3, using the unknowns A-D.
2. Reaction with Potassium Permanganate –
Step #1. Label four test tubes, labeling each with the name of a
different hydrocarbon as
listed above.
Step #2. Place 10 drops of the appropriate hydrocarbon in each
of these test tubes.
Step #3. Add 10 drops of 1% aqueous potassium permanganate
to each tube. Shake the
test tube as above several times after the addition. The purple
aqueous permanganate
solution will not mix to any obvious extent with the colorless
hydrocarbons because of
the difference in polarity. However, there is a small solubility
between the two liquids
and, if reaction is possible, the permanganate and hydrocarbon
can react at the interface
between the liquids. After 5 minutes of intermittent agitation
examine the tubes. If no
reaction has taken place, the aqueous solution will still be
purple. If reaction has taken
place there will be a brown precipitate of manganese dioxide in
the tube.
Experiment #3 Hydrocarbons Page 10
Step #4. Label four test tubes, labeling each with the letter of a
different unknown, A-D.
Step #5. Repeat Step #2, using the unknowns A-D.
Step #6. Repeat Step #3, using the unknowns A-D.
3. Reaction with Concentrated Sulfuric Acid –
Wear gloves and, as usual, goggles!
Step #1. Label four test tubes, labeling each with the name of a
different hydrocarbon as
listed above.
Step #2. Place 10 drops of the appropriate hydrocarbon in each
of these test tubes and
place the test tubes in an ice bath to cool them. Do not allow
ice or water to get into the
tubes.
Step #3. Add 5 drops of concentrated sulfuric acid to each test
tube. Carefully stir the
contents of the test tubes with a thin glass rod. Is heat evolved?
Is there a color change?
Did the mixture become homogeneous or are there two layers?
Step #4. Label four test tubes, labeling each with the letter of a
different unknown, A-D.
Step #5. Repeat Step #2, using the unknowns A-D.
Step #6. Repeat Step #3, using the unknowns A-D.
Based on the information you have recorded on your report
sheet, you should be able to
figure out the class of each of the unknown hydrocarbons A-D.
BK 7/03, Rev. 9/03

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  • 3. H H H H H HH C H HH C C C CC C C H H C H H H H H HH H H H
  • 4. H H H H octane - unbranched (straight-chain) 4-ethyl-2-methyloctane - branched ethylcyclohexane - cyclic Figure 1 - Unbranched, branched, and cyclic hydrocarbons. Experiment #3 - Hydrocarbons Introduction Organic chemistry is the chemistry of the compounds of carbon. Currently over twenty million compounds have been reported in the chemical literature; about 90% of them are organic, ie they contain carbon. The remaining compounds are called inorganic and are formed from the other elements, of which there are about 100. That carbon so dominates compound formation is a result of the fact that it is almost unique in its ability to form long chains with other carbon atoms. [Carbon’s neighbor in the periodic table, silicon, can do this but rarely does.] These chains with one carbon joined to a second and the second joined to a third, etc., can be branched, ie, chains of carbon atoms can be attached to carbons in the original chain. It is also possible
  • 5. for one carbon in a chain to become bonded to another carbon in that chain, resulting in a closed ring of atoms. We call these compounds cyclic. Since there are so many organic compounds it is fortunate that we can organize them into various groups that have some similarity to each other. For example, one large group of organic compounds is known as the hydrocarbons because members of this group contain only carbon and hydrogen and no other elements. Figure 1 shows examples of branched, unbranched and cyclic hydrocarbons. It is possible to subdivide the hydrocarbon group of compounds based on the bonding between the carbons. If all the carbon-carbon bonds are single, the compound is an alkane. If at least one of the carbon-carbon bonds in the compound is a double bond, and the remaining carbon-carbon bonds are single, the compound is an alkene. If at least one of the carbon-carbon bonds in the compound is a triple bond, and the remaining carbon-carbon bonds are single, the compound is an alkyne. If the compound contains a six carbon ring that has alternating double and single bonds around the ring (three double and three single), we say that ring, and the compound, is aromatic. An aromatic ring looks like an alkene with three double bonds because that’s the way we draw it using Lewis structures. However, the actual bonding in such a ring is considerably different from that in alkenes and, consequently,
  • 6. many of the chemical properties are different also. Therefore, we place these compounds in a separate family. By the way, the term aromatic as used here has nothing to do with fragrance. Experiment #3 Hydrocarbons Page 2 Hydrocarbons may be saturated or unsaturated. A saturated hydrocarbon is one that is maxed out in terms of the number of hydrogens that can be present given the number of carbons in the compound; it is impossible to add more hydrogen atoms to the compound so it is saturated with hydrogen. Acyclic (no rings) alkanes are saturated; there is no way additional hydrogens can be added while keeping the same number of carbons and maintaining normal bonding between the atoms. Alkenes, alkynes, aromatic compounds, and cyclic alkanes are unsaturated because hydrogen can be added to them, in theory and usually in practice, making them into acyclic alkanes. Some examples follow. Physical Properties Some molecules carry an electrical charge because there is a difference between the number of electrons (each with a -1 charge) and the number of protons (each with a +1 charge) in the molecule. We call molecules of this type ions. The ammonium ion,
  • 7. NH4 + , has a +1 charge because it has 11 protons (7 from nitrogen and 4 from the hydrogens) and 10 electrons (2 in nitrogen’s first shell, and 8 in its second shell [which are used to bond the hydrogens to the nitrogen]). So, with 11 plus charges and 10 minus charges, the ammonium ion has a net charge of +1. Most molecules are not ions (and we simply call them molecules); in other words they are electrically neutral – they have a charge of 0 because they have the same number of protons as electrons. H3C CH CH CH3 H2 H3C CH CH CH3 H H + catalyst H3C C C CH3 catalyst + H3C C C CH3 H H H H H22 C C
  • 8. C CC C H HH H H H catalystH23 C C C CC C H HH H H H H H H H H
  • 9. H + H2 in theory, if not easily done in practice C C C CC C H HH H H H H H H H H H H H unsaturated
  • 10. compounds saturated compounds Experiment #3 Hydrocarbons Page 3 φ φ φ φ v v v m m m n = ss = sin sin vacuum materiallight beam light beam Figure 2 - Index of Refraction
  • 11. In some molecules, even though the net charge is 0, the distribution of positive charges (protons) and negative charges (electrons) within the molecule is not the same. Such a molecule has a lopsided charge distribution – one side of the molecule is electron rich (has a partial negative charge) and the other side is proton rich (has a partial positive charge). We say that such a molecule is polar or that it has a dipole moment. The more lopsided the charge distribution, the larger the dipole moment and the more polar the molecule. In some molecules the distribution of positive and negative charges is the same; these molecules have no dipole moment and are nonpolar. When organic molecules are polar it is usually because there are one or more highly electronegative atoms on one side of the molecule. Electronegative atoms are those that attract electrons toward themselves; common examples are nitrogen, oxygen, and the halogens, especially fluorine and chlorine. Polar molecules attract each other because the negative side of one attracts the postive side of another (opposite charges attract). The more polar the molecules the more they attract each other. It is also true that nonpolar molecules do not attract each other as strongly as polar molecules do and polar molecules are not attracted to nonpolar molecules as much as polar molecules are attracted to each other.
  • 12. Hydrocarbons are molecules that have little or no polarity because they do not contain electronegative atoms. They are soluble in solvents of low polarity. They are not soluble in water, which is very polar, because the water molecules attract each other strongly (and are not nearly as interested in attracting nonpolar molecules). Density is the mass of a material divided by its volume; it is often expressed in terms of grams of mass per cubic centimeter of volume. The density of water, for example, is 1.00 gram per cubic centimeter. For the most part, hydrocarbons are less dense than water, so, given their insolubility in water, they float on it. Crude oil and its derivatives gasoline, kerosine, and fuel oil are mainly hydrocarbons; none of these is soluble in water and they float on its surface since they are less dense than water. The index of refraction, n, of a compound is the speed of light in a vacuum, sv divided by the speed of light in that substance, sm. Since light travels faster in a vacuum than anywhere else, the index of refraction is greater than 1.000 for any substance. It turns out that the index of refraction can be measured by noting how much a beam of light bends when it travels from a vacuum (or air, since the speed of light in air is almost the same as that in a vacuum) into some substance. See Figure 1. The index of refraction of liquids is usually measured using an Abbé refractometer where the liquid is placed on a glass prism of known index of refraction. [Glass is easier to work with than a vacuum.] Because
  • 13. of the large number of electrons located around the ring in aromatic compounds these compounds usually have larger indexes of refraction than other hydrocarbons. Typically, aromatic compounds will have indexes of refraction greater than 1.45, while other hydrocarbons will have smaller values. Experiment #3 Hydrocarbons Page 4 C 3H 8 O 2 C O 2 H 2O5 3 4+ + H3C CH CH CH3 H3C CH CH CH3 Br Br + H3C C C CH3 + H3C C C CH3 Br Br Br Br 2 Br2 Br2 Natural gas is composed of alkanes; it is about 90% methane with small amounts of ethane and propane. Methane, ethane, propane, and butane
  • 14. are gasses at room temperature. Pentane is barely a liquid at room temperature and the higher molecular weight alkanes are liquids, the straight-chain versions becoming solids at about 16 carbon atoms. Butane boils at about 0 o C, which is why butane lighters do not function well below that temperature: the liquid changes to gas too slowly below its boiling point to maintain a flame. Chemical Reactions of Hydrocarbons 1. Combustion. All of the hydrocarbons undergo combustion – they burn in the presence of oxygen. If there is enough oxygen the combustion will be complete, ie, the products of the combustion will not burn. In fact, if combustion is complete the products will be carbon dioxide and water, and, of course, heat is given off – the reaction is highly exothermic – which is ordinarily the purpose of this reaction. The equation for combustion of propane (bottle or LP gas, used for cooking and heating, is mainly propane) is shown below. 2. Reaction with Bromine. Hydrocarbons with multiple bonds (unsaturated hydrocarbons except most
  • 15. cycloalkanes) react with bromine. Tetrachloromethane (carbon tetrachloride) or cyclohexane are usually used as solvents because they are unreactive toward both bromine and hydrocarbons that have multiple bonds. Alkenes and alkynes undergo an addition reaction with bromine. The double bond of an alkene becomes a single bond and one bromine atom becomes bonded to each of the carbons that had shared the double bond. No other product is formed; the alkene and bromine simply add together, which is why it’s called an addition reaction. The triple bond of an alkyne also undergoes an addition becoming a single bond, but in this case each of the carbons that had been joined by the triple bond will now hold two bromine atoms. Examples follow. This reaction usually occurs quickly at room temperature without a catalyst. Experiment #3 Hydrocarbons Page 5 C C C CC C H HH
  • 16. H H H H H H H H H + C C C CC C Br HH H H H H H H H
  • 17. H H Br2+ HBr heat or ultraviolet light aromatic + alkene aromatic + alkane aromatic only (FeBr3) H BrBr2 HBr+ + catalyst CH2 CH3 + Br2 CH CH2 Br2+ CH CH2 Br Br addition reaction to alkene double bond, bromine color dissipates quickly
  • 18. reaction quite slow in the absence of catalyst, bromine color fades quite slowly reaction slow in the absence of catalyst, bromine color fades slowly CH2 CH3 Br CH2 CH3Br + catalyst (FeBr3) Bromine is a reddish-brown color. All of the other substances in these reactions are colorless. So, when bromine is added to an alkene or alkyne the red-brown color dissipates quickly, often almost instantly. Bromine can react with an alkane, but this reaction requires heat or ultraviolet light to be successful, and the reaction is a substitution, not an addition: a hydrogen is replaced by a bromine and hydrogen bromide is a byproduct. An example follows. Since this reaction does not take place in the absence of ultraviolet light or heat, if bromine is added to an alkane under these conditions (room
  • 19. temperature and no sunlight or other source of uv) the reddish-brown color of bromine will persist. Aromatic rings react with bromine in a substitution reaction; this reaction is slower than the addition of bromine to an alkene or alkyne and may require a catalyst. Iron (III) bromide is a good catalyst for this reaction. If iron filings are added to a mixture of the aromatic compound and bromine, iron (III) bromide will form. Note that hydrocarbon groups that may be attached to the aromatic ring will react as described above. Experiment #3 Hydrocarbons Page 6 + H3C CH CH CH3 H O S OO O H H3C CH CH CH3 OH S O O
  • 20. O H Nonpolar and not capable of hydrogen bonding; does not dissolve in H2SO4. Polar and capable of hydrogen bonding. Polar and capable of hydrogen bonding; soluble in H2SO4. 3. Reaction with Concentrated Sulfuric Acid. Alkenes react with concentrated sulfuric acid at room temperature in an addition reaction that produces an alkyl sulfonic acid. The alkyl sulfonic acid is highly polar and is soluble in the polar sulfuric acid with sufficient stirring. The non-polar alkene is not soluble in the polar concentrated sulfuric acid, but dissolves as it reacts and the mixture is stirred. [A second, related issue here is that the alkene is not capable of forming hydrogen bonds, while sulfuric acid and the alkyl sulfonic acid can. Hydrogen bonding is an especially strong type of dipole-dipole attraction. In order for H-bonding to occur at least one of the molecules must have a hydrogen (directly) bonded to a fluorine, oxygen or nitrogen. This hydrogen will be strongly attracted to a fluorine,
  • 21. oxygen or nitrogen in another molecule. Hydrogen bonding is a very important phenomenon, especially in biological systems.] An example follows. Alkynes react slowly or not at all with concentrated sulfuric acid unless a catalyst is present (HgSO4). They may darken but likely will not dissolve in the sulfuric acid. Alkanes do not react. Aromatic hydrocarbon rings react quite slowly at room temperature (undergoing a substitution reaction in which a hydrogen bound to a carbon in the ring is replaced by an -SO3H group). 4. Reaction with Potassium Permanganate. Dilute aqueous potassium permanganate oxidizes alkenes to geminal diols. [Diol means two -OH groups. Geminal means on adjacent carbons]. Alkynes are oxidized to geminal diketones. In these processes the purple potassium permanganate is reduced to a brown precipitate of manganese dioxide. Since the potassium permanganate is soluble in water, but neither the water nor the potassium permanganate are soluble in the hydrocarbon, reaction takes place at the water-hydrocarbon interface, and is somewhat slow. Consequently, it may take several minutes for a brown precipitate to form. Examples follow. Experiment #3 Hydrocarbons Page 7
  • 22. a geminal diol brown purple ++ ++ 223 423 KOHMnO2 H2OKMnO4 CHH3C CH CH3 OH OH CHH3C CH CH3 a geminal diketone + 4KOH4+ MnO2 brown 2+ H2O4+ KMnO4 purple 3
  • 23. 3 CH3C C CH3 O O CH3C C CH3 Caution! The organic liquids are highly flammable. Keep them away from flames. Concentrated sulfuric acid and bromine are both very corrosive. Wear goggles and avoid getting them on yourself. If you do get any on you, wash the affected area with lots of water. Potassium permanganate will stain your skin and clothing. Under more vigorous conditions the geminal diol (from an alkene) or geminal diketone (from an alkyne) that is initially produced will be further oxidized by the permanganate, producing more manganese dioxide. Alkanes and aromatic rings are unreactive toward dilute aqueous potassium permanganate. Objectives of the Experiment – 1. To investigate three physical properties of some hydrocarbons: density, solubility and refractive index. 2. To compare the chemical reactivities of an alkane, alkene, alkyne, and an aromatic
  • 24. compound. 3. To use the above properties to determine whether an unknown is an alkane, alkene, alkyne, or aromatic. Procedure – Experiment #3 Hydrocarbons Page 8 1 0 x 7 5 m m t e st t u b e , ac tu al s
  • 25. iz e General 1. The hydrocarbons hexane (alkane), cyclohexene (alkene), 1- hexyne (alkyne), and toluene (aromatic) are available in dropper bottles. 2. The test reagents 1% bromine in cyclohexane, 1% aqueous potassium permanganate, and concentrated sulfuric acid are available in dropper bottles, also. 3. Unknowns are in dropper bottles labeled A, B, C, and D. One of these is an alkane, another an alkene, another an alkyne, and another an aromatic compound. 4. Record observations in the appropriate places on the report sheet. If you don’t record it, it’s like you never did it. Physical Properties Step #1. Label four 10x75mm test tubes, labeling each with the name of a different hydrocarbon as listed above. Step #2. Place 10 drops of the appropriate hydrocarbon in each of these test tubes. Step #3. Test for water solubility and density relative to water. Add 10 drops of water to each test tube. Holding the test tube by the neck with
  • 26. one hand, strike the bottom of the test tube horizontally with a finger of the other hand several times to mix the two liquids. Allow the test tube to sit undisturbed for a minute. Are the compounds miscible (soluble in each other) or immiscible (insoluble)? If they are immiscible, which compound is on the bottom? Can you determine which compound is more dense? If so, which one is? Rinse each test tube with acetone and draw air through it, using an aspirator, to dry it. Step #4. Repeat step #2. Step #5. Test for solubility in hexane and density relative to hexane. Add 10 drops of hexane to each test tube. Holding the test tube by the neck with one hand, strike the bottom of the test tube horizontally with a finger of the other hand several times to mix the two liquids. Allow the test tube to sit undisturbed for a minute. Are the compounds miscible (soluble in each other) or immiscible (insoluble)? If they are immiscible, which compound is on the bottom? Can you determine which compound is more dense? If so, which one is? Will any of the above tests be useful in distinguishing one type of hydrocarbon from another? Is there any point in running these tests on unknowns A-D?
  • 27. Experiment #3 Hydrocarbons Page 9 Step #6. Your instructor will have shown you how to use the Abbé refractometer. Measure the index of refraction of each of the known hydrocarbons and of each of the unknowns, A-D. If you trust the person next to you, you can work as partners. Each of you should measure the refractive index of two of the known hydrocarbons and two of the unknowns. Be sure to wipe the prisms dry with a Kimwipe tissue between measurements. Chemical Properties 1. Reaction with Bromine – Wear gloves and, as usual, goggles! Step #1. Label four 10x75mm test tubes, labeling each with the name of a different hydrocarbon as listed above. Step #2. Place 10 drops of the appropriate hydrocarbon in each of these test tubes. Step #3. Add 15 drops of 1% bromine in cyclohexane. Shake the test tube as above, briefly, after the addition. Record how long it takes, in seconds, for the bromine color to dissipate; in other words, how long it takes for the solution to go from brown-orange to colorless or almost colorless. If this happens immediately, say that. If it takes more than
  • 28. 120 seconds, say that. Step #4. Label four test tubes, labeling each with the name of a different unknown, A-D. Step #5. Repeat Step #2, using the unknowns A-D. Step #6. Repeat Step #3, using the unknowns A-D. 2. Reaction with Potassium Permanganate – Step #1. Label four test tubes, labeling each with the name of a different hydrocarbon as listed above. Step #2. Place 10 drops of the appropriate hydrocarbon in each of these test tubes. Step #3. Add 10 drops of 1% aqueous potassium permanganate to each tube. Shake the test tube as above several times after the addition. The purple aqueous permanganate solution will not mix to any obvious extent with the colorless hydrocarbons because of the difference in polarity. However, there is a small solubility between the two liquids and, if reaction is possible, the permanganate and hydrocarbon can react at the interface between the liquids. After 5 minutes of intermittent agitation examine the tubes. If no reaction has taken place, the aqueous solution will still be purple. If reaction has taken place there will be a brown precipitate of manganese dioxide in the tube.
  • 29. Experiment #3 Hydrocarbons Page 10 Step #4. Label four test tubes, labeling each with the letter of a different unknown, A-D. Step #5. Repeat Step #2, using the unknowns A-D. Step #6. Repeat Step #3, using the unknowns A-D. 3. Reaction with Concentrated Sulfuric Acid – Wear gloves and, as usual, goggles! Step #1. Label four test tubes, labeling each with the name of a different hydrocarbon as listed above. Step #2. Place 10 drops of the appropriate hydrocarbon in each of these test tubes and place the test tubes in an ice bath to cool them. Do not allow ice or water to get into the tubes. Step #3. Add 5 drops of concentrated sulfuric acid to each test tube. Carefully stir the contents of the test tubes with a thin glass rod. Is heat evolved? Is there a color change? Did the mixture become homogeneous or are there two layers? Step #4. Label four test tubes, labeling each with the letter of a different unknown, A-D. Step #5. Repeat Step #2, using the unknowns A-D. Step #6. Repeat Step #3, using the unknowns A-D.
  • 30. Based on the information you have recorded on your report sheet, you should be able to figure out the class of each of the unknown hydrocarbons A-D. BK 7/03, Rev. 9/03