Alkyl, Alkenyl and Alkynyl Substituent Groups
The naming of prefix for alkyl, alkenyl and alkynyl groups are pretty simple. It follows a simple
formula.
We have already explored the appropriate prefix based on the number of carbons in the table above
labelled “Naming unbranched Alkanes”.
 E.g. Hydrocarbons (Alkanes or Alkenes or Alkynes) with one carbon has prefix “Meth”
 E.g. Hydrocarbons with two carbons has prefix “Eth”, three carbons has prefix “Prop”, etc.
Prefix of alkyl substituent groups = Appropriate prefix based on number of carbon atoms + “yl”.
Prefix of alkenyl substituent groups = Appropriate prefix based on number of carbon atoms +
“enyl”.
 Yes, methylene is an exception as it doesn’t follow the same format of adding ‘enyl‘ after meth.
Prefix of alkynyl substituent groups = Appropriate prefix based on number of carbon atoms +
“ynyl”.
NOTE: We can express the organic molecules’ molecular formula using stick structures as depicted
using pink lines in diagram above.
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We deliberately drew yellow dots on the line structure to depict how, at every yellow dot, there
is a carbon atom. You should NOT draw the yellow dots in your line structure as we only did it
here to get the idea through to you.
In line structures, we DO NOT show hydrogen atoms that are attached to the carbon atoms
(yellow dots in diagram) as that’s the purpose of line structures to make it faster and easier to
draw, read and communicate the chemical structure of organic molecules.
Learning Objective #1 - Investigate the IUPAC nomenclature for alkenes
Now let’s move onto naming alkenes! These are organic molecules with a carbon double bond,
(C = C).
Basically, to name alkenes, we use the same procedure that we have outlined above for alkanes.
However, there are few exceptions or things that you need take into consideration when naming
alkene molecules.
These important exceptions are for alkenes are:
The parent carbon chain must contain the double-bond(s) (C=C) functional group.
Like all alkenes molecules, there is an ‘ene‘ after the root-name of the alkene molecule.
The C=C functional group has highest priority in alkene molecules. Therefore you should assign
“1” to the the first carbon you encounter that makes up the C=C bond from one end of the
parent carbon chain.
Example: But-2-ene, where the “2” indicates that the carbon double bond (C=C) is found on the
second carbon atom in butene molecule. Notice that you include the position (“2”) of the carbon
double bond (C=C) before you write the suffix for alkene “ene” and it separates using dash symbols
“-” as per IUPAC conventions.
NOTE: Keep in mind that you must assign the least carbon number as possible to all of your
substituent groups. So, check your work once you are finished!
There are special cases when you are doing with cyclic compounds but these
exceptions NOT necessary HSC Chemistry. You will learn in Higher Chemistry 1A if you decide to
take Chemistry university.
Learning Objective #1 - Investigate the IUPAC nomenclature for alkynes
Moving on to the next station, alkynes! These are organic molecules with a triple carbon
bond, (C ≡ C).
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To name alkynes, we use the same method that we used to name alkanes. However, again, there are
some exceptions that you will need to evaluate.
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The parent carbon chain must contain the triple bond(s) (C ≡ C) functional group.
Like all alkynes molecules, there is an ‘yne‘ after the root-name of the alkyne molecule.
The C≡C functional group has the highest priority in alkyne molecules. Therefore, you should
assign “1” to the first carbon you encounter that makes up the C ≡ C bond from one end of the
parent carbon chain.
 In an organic molecule that contains both double and triple carbon bonds, they have equal
priority. This means whichever of the two functional group appears on first from one side of the
parent carbon chain, it will have the lower assigned carbon number.
 As mentioned in the alkane section, if the double and triple carbon bonds are in at equal carbon
atom distance from each side of the parent carbon chain, we use assign the lowest carbon
number according to alphabetical order, the carbon attached to the double bond (-ene-) will be
assigned a lower carbon number than the carbon atom attached to the triple bond (-yne-)
as e comes before y in the alphabet.
 NOTE: Keep in mind that you must assign the least carbon number as possible to all your
substituent groups. So, check your work once you are finished!
Moving into alcohols, aldehydes, ketones, carboxylic acids, amine, amides & halogenated
compounds
For alkanes, alkenes, alkynes and halogenated compounds, the substituent groups, i.e. alkyl, alkenyl,
alkynyl and halogen groups, have equal priority and thus are named according to the alphabetical
order of their substituent group’s prefix.
However, for the following compounds, the highest priority substituent (or functional group to be
more specific) are based on the following functional group priority naming order:
Highest priority Side
Carboxylic acid group
Amide group
Aldehyde group
Ketone group
Alcohol group
Amine group
Lowest priority Side
Alkene (C=C) and alkynes (C≡C) have equal priority as we have mentioned before and they both
have lower priority than amine groups.
This is, of course, followed by the following substituent groups have equal priority and thus are
named in the order in an alphabetical order which we have mentioned:
Alkyl, Alkenyl, Alkynyl groups and halogen groups
NOTE: Halogens are not treated as functional group in organic chemistry nomenclature. However,
when the question is not concerning nomenclature, halogen are treated as functional group as they
give chemical properties to the organic molecule.
NOTE: There are many more other functional groups that fits between these groups but they
are not relevant for HSC Chemistry purposes as they’re outside the scope of what the syllabus
required us to know.
Learning Objective #1 - Investigate the IUPAC nomenclature for alcohols
A parent carbon chain with hydroxyl group as the highest priority substituent group is known as
an alcohol.
To name an alcohol, we can use the following the steps:
Step 1: Identify the parent carbon chain, i.e. the longest, continuous carbon chain, that comprises of
the highest priority functional group. In this case, the hydroxyl group (- OH) is the highest priority
group.
Step 2: Assign carbon numbers to every atom in the parent carbon chain with the carbon atom that
is bonded to the highest priority group where the highest priority functional group have the the
lowest carbon number.
Step 3: Name any double and triple bonds in a similar fashion as we have discussed previously.
Step 4: Since the parent carbon chain has hydroxyl group as the highest priority group, it is an
alcohol. The organic molecule that is an alcohol has the suffix ‘ol’. Therefore, we need to replace
the ‘e‘ at the end of the root-name of the organic molecule with the relevant suffix. In this case of
alcohols, the suffix after the root-name would be ‘ol’. Again, like other cases, there’s no space
between root-name and suffix. Include the carbon number too before the suffix using dash symbols
“-” to signal where the main functional group is located.
 Finally, use the relevant prefix based on carbon number. E.g. Two carbons = ‘eth’. Three carbons
= ‘Prop’.
Step 5: Name any other other substituent groups (disregarding priority order now) that are present
in the organic molecule in alphabetical order, according to their prefix names, and
assigning lowest carbon position number to every substituent group as possible.
Learning Objective #2 - Explore and distinguish the difference types of structural isomers,
including saturated and unsaturated hydrocarbons including: Chain Isomers
Alright, let’s explore the first type of structural isomers – chain isomers!
We can say that two or more molecules are chain isomers if they satisfy the following conditions:
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They must have the same molecular formula.
They must differ in their connectivity of carbon atoms to the parent carbon chain.*
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*This means that these isomers have different parent carbon chain lengths, hence they have
the name being chain isomers.
We are going to the next stop to explore another type of structural isomers, position isomers
In order for two or more molecules to be considered position isomers, they need to meet the
following criteria:
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Like all structural isomers, position isomers must have the same molecular formula.
Unlike chain isomers, position isomers have the same carbon structure in terms of parent
carbon chain length.
One or more substituent and/or functional groups must differ in their connectivity to the carbon
atoms along the parent carbon chain.*
*This effectively means that these isomers needs to have one or more of their substituent
and/or functional groups being in a different position along the carbon structure. Hence, they
have the name being position isomers.
Recall that from the table that we examined in Learning Objective #4, where we showed how
double and triple bonds between carbon atoms are functional groups.
So, isomers that differ in their position of double and triple carbon bonds in their carbon
structure can be considered position isomers (provided that the molecules satisfy all of other
criteria for position isomers outlined above).
Functional isomers:
Again, there are a set of conditions that two or more molecules must satisfy in order for them to be
considered functional group isomers. These are:
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Like all structural isomers, they must have the same molecular formula.
They may or may not have the same parent carbon chain length, depending on arrangement of
atoms.
Due to the isomers differing in their arrangement of atoms, (thus, NOT limited to only
substituent and functional group like for positional isomers), this results in the isomers having
different functional groups.*
*Hence their name, functional group isomers.
Variation in physical and structural properties between different types of structural isomers
Chemical properties
Structural isomers would have the same chemical property if they have the same functional
group(s) in their structural formula.*
Physical properties
The following guidelines will help you compare the melting and boiling points between different
isomer molecules.
As you will explore with polymers which is part of the last Inquiry Question in Module 7, you learn
that molecules (e.g. isomers) that exhibit a less branched (or linear) structural formula have
a higher melting and boiling point.
Therefore, you would expect the more chain branching that a molecule exhibits, the lower its
melting and boiling point will be.
This is because chain branching prevents the isomers from packing their carbon chains closely
together and, thus, the intermolecular forces (e.g. dispersion forces) between them would be
weaker. As we have explored in Preliminary HSC Chemistry, intermolecular forces play an important
role in governing the melting and boiling point of molecules.
As for functional group isomers, the isomers have different functional groups allowed differing
extent of chain branching (thus affecting dispersion forces) and maybe different degree of dipoledipole and hydrogen bonding. Therefore, their melting and boiling point (physical property) may
vary.
Example: Alcohols (- OH) and carboxylic acids (- COOH) have different melting and boiling points due
to different functional groups.
Learning Objective #2 - Conduct an investigation to compare the properties of organic chemical
compounds within a homologous series, and explain these differences in terms of bonding.
General formula for: Alcohol Homologous Series is: Cn(2n+1)OH
General formula for Aldehyde Homologous Series is: CnH(2n+1)CHO
General formula for Ketone Homologous Series is: CnH(2n+2)CO
General formula for Carboxylic acids Homologous Series is: CnH(2n+1)COOH
General formula for organic amide homologous series: R-CO-NH2
General formula for organic amine homologous series: R-NH2
Homologues in the Alkane Homologous Series (without substituent groups)
As we have learnt in the HSC Preliminary Chemistry Course, the type (and thus strength) of
intermolecular forces plays a major role in determining the melting and boiling point of molecules.
This is not limited to just alkanes but for all the compounds in each class of homologous series which
we will explore soon.
 Here, we will talk about alkanes without substituent groups.
We have already explored the structure of unbranched alkanes without substituent groups in last
week’s notes as well as in this week’s learning objective #1. They only have of C-C and C-H bonds
which are non-polar bonds due to the electronegativity difference between C-C is basically zero and
the electronegativity difference between carbon and hydrogen in the C-H bond not being high
enough to be considered a polar covalent bond.
 Recall that we have already learnt the electronegativity difference ranges that allows us to
classify non-polar, polar and ionic bonds in Preliminary HSC Chemistry.
For such reason, as both C-C and C-H bonds are non-polar in addition to the fact that unbranched
alkanes are linear, it allows uniform distribute of any dipole moments, it is justifies why unbranched
alkane molecules without substituent groups are non-polar.
Since they are non-polar, it is only dispersion forces being the intermolecular force that is
determining the melting and boiling points of unbranched alkanes without substituents.
NOTE: We have already talked about dispersion forces in Preliminary HSC Chemistry, so we expect
you to know how they are created. But, just for a quick recap on what they are, dispersion forces are
intermolecular forces created due to the uneven distribution of electrons, thus even electron
density, around the nucleus of an atom. This therefore results in a small dipole that is formed where
one side of the atom is slightly positive and the other side being slightly negative. This dipole induces
a similar dipole in a neighbour atom, resulting in a dispersion force being created between the two
atoms.
This means alkanes with larger molecular mass will have more atoms (carbon and hydrogen) and
hence will have a greater number of electrons. Thus, the degree (and thus strength) of dispersion
force formed between larger alkane molecules would therefore be greater than between smaller
alkane molecules.
So, a higher amount of energy is required to break the intermolecular forces (dispersion forces only
between alkane molecules) in alkanes with higher molecular mass.
Therefore, as the number of carbon and hydrogen atoms increase (i.e. alkane molecule increases in
molecular mass – such as going from methane to octane), the melting and boiling point will
increase.
As you are aware from Preliminary HSC Chemistry, hydrocarbons from C1 to (and including) C4 are
gases at room temperature. Hydrocarbons between C4 – C8 are liquids at room temperature.
Branched vs Unbranched Alkanes
For two alkanes, one being branched and the other unbranched, both with the same molecular
formula and thus molecular mass. The unbranched alkane will exhibit higher melting & boiling point.
The reason for this is because unbranched alkanes are more linear whereas branched alkenes are
less linear (or more spherical or more bulky) in shape.
For such reasons, the carbon chains between unbranched alkane molecules are able to pack more
closely together than the carbon chains between branched alkanes. Therefore, a higher surface
contact area means that the extent of dispersion force is greater between unbranched alkanes as
more atoms’ electrons are able to interact with each other to induce dipoles. This results in
unbranched alkanes having a higher melting & boiling point.
Another differing physical property other than melting & boiling point between alkanes of different
molecular weight is their density. Their density increase as the molecular weight increases, however,
it doesn’t exceeds the density of water. Obviously, another density trend is that, since branched
alkanes have their carbon chains packed more closely with each other, they have higher density than
unbranched alkanes.
With that, we conclude the physical property differences between homologues in the alkane
homologous series!
General physical properties of all alkanes share are:
 Alkanes are not soluble in polar substances but soluble in non-polar substances (the reason for
this is already covered in HSC Preliminary Course, so we won’t re-explain here).
 Alkanes are colourless and odourless.
NOTE: As we have mentioned previously, the chemical properties shared between homologues in a
homologous series such as alkane are the same since there is no different atoms or functional
groups.
Homologues in the Alkene Homologous Series (without substituent groups)
The differences between alkene molecules of increasing molecular mass is similar to the case of
alkanes that we described above already. So, we will only do a very brief summary.
Very brief summary (as the case of alkenes is similar to alkanes which we explained in detailed
above): As the molecular mass of alkenes increases, the strength of dispersion forces increases and
thus results in a trend of increasing melting & boiling point with increased molecular mass.
It is important to note that alkenes have a lower melting and boiling point than their corresponding
alkanewith the same number of carbon atoms. This is because alkenes have a lower molecular mass,
lowering the strength of dispersion force.
For the same reason, alkenes have a lower density than corresponding alkanes with the same
number of carbon atoms as alkenes have a lower molecular mass.
However, both alkanes and alkenes have less density than water as well as being insoluble.
Similar to alkanes, the alkenes from C1 up to and including C4 exist in the gas state at room
temperature whereas alkenes that is C5 – C8 are liquids.
In general, apart from melting & boiling point, alkenes share similar physical properties to
alkanes are they are comprised of the same types of atoms, carbon and hydrogen only.
The chemical properties shared between alkene homologues are same as there is no difference in
functional groups.
Homologues in the Alkynes Homologous Series (without substituent groups)
The physical properties are the same as alkenes and alkanes. However, alkynes have lowest melting
& boiling point out of the three due to lower molecular mass than corresponding alkane and alkene
hydrocarbons with the same amount of carbons.
Alkyne homologues share similar with each other chemical properties as they all have a triple bond
between carbon atoms.
Difference in chemical properties between Alkanes, Alkenes & Alkynes withoutsubstituent groups
As alkenes and alkynes have double and triple bonds between carbon atoms, they are more reactive
than alkanes. This is because there is higher electron density in a double and triple bond region
compared to single bonds meaning that breaking a C-C bond from double bond and triple bond
between carbon atoms require less energy to break compared to breaking the one and only single
carbon-carbon bond between carbon atoms.
This therefore allows alkenes and alkynes to undergo addition reaction whereby a carbon-carbon
bond is broken so that a substituent group or atom can be added to each carbon.
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Alkanes cannot undergo addition reactions but instead undergoes substitution reactions as they
only have a single bonds, i.e. C-C and C-H.
A common test for alkanes and alkenes is bromination (a type of halogenation reaction whereby Br2
is added to the organic compound).
What happens is that alkanes will not react with the bromine molecule unless under the presence of
high temperature or U.V light as discussed last week under substitution, halogenation reaction
conditions.
Comparatively, addition reaction occurring between alkenes and alkynes with bromine does not
need high energy due to lower bond energy of breaking of one of the several C-C in a double bond or
triple bond between carbons atoms. Therefore, the bromine molecule will be incorporated into the
alkenes and alkyne molecules via addition reaction at room temperature.
Since bromine has a red-brown colour and the hydrocarbons (e.g. alkane, alkene, alkynes) are
colourless, the effect is that when bromine undergoes addition reaction or substitution reaction,
there will no longer be bromine molecules in solution and thus the solution will become colourless
after reaction.
For alkynes, it basically is the same as alkenes in terms of chemical properties but they are more
reactive than alkenes (and alkanes) as two C-C bonds can be broken with lower energy in its triple
carbon bonds between carbon atoms. Therefore, alkynes can undergo addition reaction twice to be
compared into an alkene which can only undergo only one addition reaction.
In general, this means that alkynes are less stable than alkenes which are less stable than alkanes.
In order of acidity, alkynes are the most acidic: i.e. alkynes > alkenes > alkanes.
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That is alkynes are the strongest acid as they are readily donate their hydrogen atom in the form
of H+.
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The reason towards this trend of acidity is because the resulting lone electron pair on the carbon
atom, after a hydrogen atom is donated is situated in the sp orbital compared to the sp2 in
alkenes and sp3 in alkanes.
Recall from Preliminary HSC Chemsitry:
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Sp orbital have 50% s character and 50% p character.
Sp2 has 1/3 s character and 2/3 p character.
Finally Sp3 orbitals have 25% s character and 75% p character.
As electrons that are closer to the nucleus (more s character) are more stable than electrons found
further away from the nucleus (less s character) due to higher effective nuclear charge, the reaction
of alkynes acting as a Bronsted-Lowry Acid by donating a H atom occurs more readily compared to
alkenes and alkanes as the conjugate base formed is more stable. This is because the conjugate
base’s lone pair of electron have greater s character in alkynes than alkenes and alkanes which
experiences a higher effective nuclear charge and thus is more stable as the electrons are closer to
the nucleus which is positively charged.
Finally, alkanes have higher chance of undergoing complete combustion than alkenes which have
higher chance than alkynes. This is because the ratio between carbon to hydrogen is higher in
alkynes than alkenes which is higher than alkanes so more oxygen is required to fully oxidise the
carbon atoms.
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Most of the oxygen are used to form water and any remaining is used to form either carbon
dioxide (two oxygen per carbon) or carbon monoxide (one oxygen per carbon), where carbon
monoxide is favoured if there is limited availability of oxygen.
Homologues in the Alkyl halides (or Haloalkane) Series
The electronegativity difference between the halogen and the carbon atoms makes the C-X bond
polar, where X represents a halogen atom. Due to this, there are dipole-dipole forces between
haloalkane molecules as well as dispersion forces compared to only dispersion forces present
between molecules in the three hydrocarbon homologous series (alkane, alkenes and alkynes).
For such reasons, haloalkanes have stronger intermolecular forces than corresponding
alkanes/alkenes or alkynes with the same number of carbon atoms. Again, similar to other
homologous series, as haloalkanes increases in alkyl group chain length, the molecular mass will
increase. This results in higher extent of dispersion forces due to greater number of electrons,
resulting in stronger intermolecular forces.
Another similar trend is that, as the length of the non-polar alkyl group increases, the solubility of
haloalkanes in polar substances decreases due to increasing non-polar nature of the overall alkyl
halide.
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Do note alkyl halides cannot perform hydrogen bonding with water molecules so they are not
readily soluble in water. So, alkyl halides are less soluble than corresponding ketones, aldehydes,
alcohols, amides and amines with the same number of carbon atoms (similar molecular mass)
that can perform hydrogen bonding with water molecules.
A thing to note about alkyl halides is that as its molecular mass increases (i.e. moving from C1 to C8),
the density of the homologues in the series decreases (measured at a fixed temperature) as the ratio
between mass to volume for density decreases.
 However, if the alkyl chain is the same length between two haloalkanes, the one with the heavier
or higher molecular mass halogen will result in the haloalkane being more denser.
Most alkyl halides exists as a liquid at room temperature except for few of those with low molecular
mass such as fluoromethane and chloromethane.
Homologues in the Ketone Homologous Series
Compared to aldehydes, ketone molecules have a pleasant and sweet odour.
Due to the electronegativity difference between the oxygen and carbon atom in the C=O group, the
carbonyl group is polar. This essentially gives ketone molecules their polar properties.
The main intermolecular force that determines the melting & boiling point of ketone molecules are
dipole-dipole forces. Obviously, dispersion forces are present amongst the molecules too.
Ketones have a higher boiling point than aldehydes because ketones have two alkyl (R) groups
attached a central carbon atom where each alkyl group has a greater electron density than a
hydrogen atom. Comparatively, aldehyde only has one alkyl (R) group and one hydrogen atom
attached to a central carbon atom.
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Recall that since the C=O bond is polar due to the oxygen atom being more electronegative than
carbon atom. This means that oxygen is partially positive and the carbon atom is partially
negative.
 This means that two alkyl groups (each has higher electron density than a hydrogen atom) will
attracted and be located closer to the central carbon atom. This effectively makes the central
carbon atom in ketones more negatively charged, resulting ketones having a more polarised C=O
bond compared to that C=O bond in aldehydes.
 A more polarised C=O bond in ketones means that the dipole-dipole forces between ketone
molecules are stronger than those between aldehyde molecules. This means that more energy is
required to break the dipole-dipole intermolecular forces between ketone molecules than
between aldehydes.
 For such reason, ketones have a higher boiling point than aldehydes.
Although the ketones with one to three carbons (C1 to C3) are very soluble in water, as the
molecular mass increases (i.e. moving into C4 and beyond), the length of the non-polar hydrocarbon
chain increases. This results in ketone molecules experiencing a gradual decrease in water solubility
as molecular mass increases, resulting in the non-polar alkyl chain/group increasing in length.
Like all of the previous cases, as the number of carbon and hydrogen atoms increases (i.e. molecular
mass) for homologues in the ketone homologous series, the strength of dispersion force increases.
Thus, the melting & boiling point increases.
Homologues in the Aldehyde Homologous Series
Compared to ketones, aldehydes have an unpleasant and strong/sharp odour.
Aldehydes are polar for the same reason (for the most part) as Ketone as explained in the ketone
section.
The main intermolecular force between aldehyde molecules are dipole-dipole forces, followed by
dispersion forces.
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If you pursue higher chemistry at university, you will learn that aldehydes can in fact hydrogen
bond with themselves at certain configurations but that’s outside the scope of HSC Chemistry. To
limit to the scope of HSC Chemistry scope, aldehydes and ketones cannot form hydrogen
bonds with each other. They can form hydrogen bonds with other molecules such as water
though!
You can try draw out the structures to see how ketones and aldehyde can hydrogen bond with
water as we have learnt about how hydrogen bonding can occur in different situations in
Preliminary HSC Chemistry.
Aldehydes have a lower melting and boiling point than ketones where the reason why has been
explained in the previous section on ketones.
Although the aldehydes with one to three carbons are very soluble in water, as the molecular mass
increases (i.e. moving into C4 and beyond), the length of the non-polar hydrocarbon chain increases.
This results in aldehydes experiencing a gradual decrease in water solubility as molecular mass
increases, resulting in the non-polar alkyl chain/group increasing in length.
Like all of the previous cases, as the number of carbon and hydrogen atoms increases (i.e. molecular
mass) for homologues in the aldehyde homologous series, the strength of dispersion force increases.
Thus, the melting & boiling point increases.
Homologues in the Alcohol Homologous Series
Like homologues in any homologous series, the homologues in the the alcohol homologous series
differ by one carbon and one CH2 unit. This difference in molecular size that gives homologues in the
alcohol homologous series their different physical properties. Since all homologues in the alcohol
homologous series have one OH functional group, their chemical properties are the same.
The main and strongest intermolecular force between homologues in the alcohol homologous series
is hydrogen bonding. Of course, as you have learnt in Preliminary HSC Chemistry, molecules of a
compound that are capable of hydrogen bonding amongst each other means that they will also have
dipole-dipole as well as dispersion forces.
Here, the learning objective is concerned about the properties between homologues in the alcohol
homologous series. We know that the intermolecular forces between molecules of a compound
plays a major role in determining the physical compound’s properties from Preliminary HSC
Chemistry.
Hydrogen bonding is stronger than dispersion forces as the electronegative difference between the
atoms involved in creating the intermolecular force is greater than in generic dipole-dipole forces.
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Quick recap from Preliminary HSC Chemistry on hydrogen bonding: For a hydrogen bond to be
formed, a highly electronegative atom must be added to a hydrogen, creating a slightly negative
charge on the more electronegative atom and slightly positive charge on the hydrogen. The
slightly positive charged hydrogen atom can perform hydrogen bonding with another
electronegative atom (slightly charged) from another molecule as long as the electronegativity
difference is strong enough.
For such reason, alcohols have higher melting & boiling point than the hydrocarbon homologous
series (i.e. alkanes, alkenes and alkynes) that do not have hydrogen bonds and only dispersion
forces.
Another factor that contributes towards alcohol’s higher boiling point is due to the existence of a
hydroxyl group (OH) that replaced a hydrogen in the hydrocarbon homologous series (e.g. in
alkanes).
As you have learnt in Preliminary HSC Chemistry, the strength of dispersion force is proportional to
electron number, an hydroxyl group (OH) has an extra oxygen atom (thus more electrons) compared
to a hydrogen atom. Therefore, the strength of dispersion forces is also greater in alcohols than in
homologues in hydrocarbon homologous series like alkanes.
NOTE: Since homologues in the Aldehyde and Ketone categories only has dipole-dipole forces and
not hydrogen bonding, corresponding alcohol molecules with the same number of carbon atoms (i.e
similar molecular mass) would therefore have higher melting & boiling points.
Although the alcohol molecules with one to three carbons (C1 to C3) are soluble in water, as the
molecular mass increases (i.e. moving into C4 and beyond), the length of the non-polar hydrocarbon
chain increases. This results in alcohols experiencing a gradual decrease in water solubility as
molecular mass increases due to the increasing non-polar alkyl chain/group.
Like all of the previous cases, as the number of carbon and hydrogen atoms increases (i.e. molecular
mass) for homologues in the alcohol homologous series, the strength of dispersion force increases.
Thus, the melting & boiling point increases.
Lastly, as the molecular mass increases (due to increase carbon and hydrogen atoms) between
alcohol homologues, the viscosity of the homologues increases. This is because the the larger
molecular size of the alcohol molecules increases the extent of dispersion forces as well as the large
size provide resistance in hindering alcohol molecules from escaping the liquid surface. This trend of
viscosity can be applied to other applicable organic molecules that are in the liquid state in different
homologous series.
Amide Homologous Series
Apart from the amide that has only one carbon (C1) in the homologous series existing as a liquid at
room temperature, i.e. methanamide, all the rest of the other homologues in the amide homologous
series are solids at room temperature.
The main intermolecular force that determines the melting & boiling point of amides is hydrogen
bonding.
Compared to alcohols and carboxylic acids with the same number carbon of atoms, the
corresponding amide has a higher melting & boiling point. This is because amides have the C=O
bond as well as the two N-H polar bonds that can hydrogen bonds.
As you will learn soon, all of the homologous series that we will talk and have talked about in this
learning objective will have a lower density than water except for amides! This isn’t surprising
considering amides are solids at room temperature except the amide with one carbon atom (C1), i.e.
methanamide.
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An identical trend to homologues in other series, the density of amides will increasing molecular
mass.
In terms of solubility, this is similar to other cases we have talked about. As the molecular mass
increases, the non-polar alkyl group will increase resulting in a gradual decline in solubility in polar
substances.
Like all of the previous cases, as the number of carbon and hydrogen atoms increases (i.e. molecular
mass) for homologues in the amide homologous series, the strength of dispersion force increases.
Thus, the melting & boiling point increases.
Amine Homologous Series
Amines in the gas phase have an ammonia odour. Comparatively, liquid amines have a “rotting fish”
odour.
Amines with one and two carbons (C1 and C2) are gas at room temperature – i.e. methanamine and
ethanamine. The other amines from C2 to C8 are liquids at room temperature.
The primary intermolecular force that governs the melting and boiling point of amines is hydrogen
bonding.

This is because the N-H bond is a polar covalent bond as the nitrogen atom is much more
electronegative than the hydrogen atom.
For such reasons, hydrogen bonding exists between amine molecules. If you draw out an amine
molecule, you can see that one amine molecule has the capacity to perform hydrogen bonding with
three water molecules.
In terms of melting and boiling point, alcohols and carboxylic acids have a higher melting & boiling
point than amines. This is because the O-H bond is more polar than the N-H bond as oxygen is more
electronegative than nitrogen.
Similar to alcohols, as the molecular mass of the amine molecules increases, the length of the nonpolar alkyl chain increases. Thus, the solubility of amines molecules in polar substances (or solvents)
such as water decreases with increasing molecular size.
Like all of the previous cases, as the number of carbon and hydrogen atoms increases (i.e. molecular
mass) for homologues in the amine homologous series, the strength of dispersion force increases.
Thus, the melting & boiling point increases.
Carboxylic Acid Homologous Series
The main intermolecular force that determines the melting & boiling point between homologues in
the carboxylic acid homologous series is hydrogen bonding.
As seen from the carboxylic acid functional group (COOH) structure we have explored in last week’s
notes, we can see that there is a carbonyl group (C=O) and a hydroxyl group (O-H). So, each
carboxylic acid group has the capacity to form two hydrogen bonds with another carboxylic acid
molecule.
Comparatively, when we compare carboxylic acids to alcohols, ketone and aldehydes, they can only
form one hydrogen bond. Therefore, carboxylic acids have higher melting & boiling point compared
to their corresponding counterparts in other homologous series with the same number of carbon
atoms except for amide (which has the highest).
Similar to alcohol, although the carboxylic acids with one to three carbons are soluble in water, as
the molecular mass increases (i.e. moving into C4 and beyond), the length of the non-polar
hydrocarbon chain increases. This results in carboxylic acids experiencing a gradual decrease in
water solubility as molecular mass increases due to the increasing non-polar alkyl chain/group.
That being said, when comparing carboxylic acid to their corresponding alcohol with the same
number of carbon, the carboxylic acid is more soluble in water due to the capacity to form two
hydrogen bonds compared to only one for alcohol.
Like all of the previous cases, as the number of carbon and hydrogen atoms increases (i.e. molecular
mass) for homologues in the carboxylic acid homologous series, the strength of dispersion force
increases. Thus, the melting & boiling point increases.
Summary of common trends amongst homologues in various homologous series
We talked about the trend that melting & boiling point increases with the size of homologues (C1 to
C8) within and between homologous series.
General trend of decreasing melting & boiling point:
Amide > Carboxylic Acid > Alcohol > Ketone > Aldehyde > Amine > Haloalkane > Alkyne > Alkane >
Alkene.
We also talked about solubility decreasing with increasing molecular mass.
We also talked about density increasing with increasing molecular mass (except for alkyl halides).
We also differentiated amides against amines in terms of odour. Likewise, aldehydes with ketone
with odour.
We differentiated some homologues in homologous series by their states at room temperature.
Learning Objective #3 - Analyse the shape of molecules formed between carbon atoms when a
single, double or triple bond is formed between them
Single carbon-carbon bond
An example of a molecule that comprises the single carbon-carbon (C-C) bond can be found in
ethane, C2H6.


All alkanes have single carbon-carbon bonds except for methane.
The shape of carbon atoms in ethane have a tetrahedral shape according to VESPR theory since
each carbon in ethane has four chemical bonds. Each carbon makes three C-H bonds and one C-C
each with a 109.5 degree bond angle.
 NOTE: Drawing lewis structures is important to understanding and predicting molecular shapes
which we have touched on when exploring the VESPR theory in the HSC Preliminary Course.
 NOTE: Since we are dealing with molecules with more than one central atom (two or more atoms
bonded to it) such as ethane, sometimes there may not be no single shape that can be used to
describe the shape of the full molecule if the central atoms have differing amount of atoms
attached to them. In our case, we only have two central carbon atoms in ethane and both carbon
atoms have the same amount (4) of atoms attached to it so they have the same shape. So,
ethane can be described using one shape, it has a tetrahedral shape.
 NOTE: If you decide to take a first year chemistry course at university, you will realise that there
is a difference between geometry and shape when describing molecules where geometry takes
into account of lone electron pairs which we won’t and don’t have here.
Double carbon-carbon bond
Ethylene is an example of a molecule that consists of a double carbon-carbon (C=C) bond, C2H4.


All alkenes have double carbon-carbon bonds.
The shape of the central carbon atoms in ethylene have a trigonal planar shape, according
to VESPR theory, since each carbon atom makes two C-H bonds and one C-C, each with are
approximately 120 degree bond angle.

Since both central carbon atoms have the same shape, ethylene has a trigonal planar shape.
Triple carbon-carbon bond
The molecule Ethyne is an example of a molecule that consists of a triple carbon-carbon bond,
C2H2.



All alkynes have triple carbon-carbon bonds.
The shape of the central carbons ethyne is linear according to the VESPR theory since each
carbon atom has one C-H and one C-C bond which each has 180 degrees bond angle.
Since both central carbon atoms have the shape, ethyne has a linear shape.
Learning Objective #5 - Describe the procedures required to safely handle and dispose of organic
substances.
Below are some common laboratory practices when dealing with organic substances. We will go
through ten laboratory practices which you probably only need to write down three for HSC
purposes.
Handling – i.e. storing, using and treating spills of organic substances
I. Store “like material with like” – so that flammable materials should be stored together and away
(in a different cabinet) from materials that are corrosive, toxic, etc to prevent hazardous reactions.
II. Oxidising agents must be stored separately from organic substances as they can ignite organic
solvents or acids.
III. Organic nitrates and peroxidies are shock-sensitive and should be handled with care as shock can
result them in generating large volume of gases in a short period of time, resulting in a high
increasing pressure. This would result in an explosion.
IV. Some organic acids such as acetic and propanoic acid are corrosive which can cause burns upon
contact with skin where severity depends on concentration of the acid.
 To minimise this risk, a laboratory coat is worn to as an additional protective layer against
corrosive substances for the skin (arms, body & thighs).
 Safety gloves are worn to protect skin from being in physical contact with corrosive substances in
case of spill.
 Safety googles are worn to prevent corrosive organic acids from being in contact with eyes in
case of splashes.
 Enclosed leather shoes are worn to protect feet in case of accidental spills rather than open
shoes.
V. Ensure that a well-ventilated area such as a fume hood is used when handling flammable, volatile
organic substances (e.g. alcohols). This is because the inhalation of alcohol can cause dizziness and
could result in brain damage depending on duration of exposure.
 Ensure that the fume hood is closed when not in used to prevent escape of volatile, flammable
vapour from the organic substance.
 Ensure that there is no open naked flame near the flammable organic substances like alcohols.
VI. When you wish to diluting concentrated acids, you should always slowly add acid to water (of
larger volume) and not the other way around. That is, never attempt to dilute acid by adding water
to the acid. This is because adding acid to water (of larger volume) allows the water molecules to
absorb the heat generated in the dissolution reaction (water also has a higher specific heat capacity
than acids). So, it minimises the concentration of heat in a particular area which would otherwise
result in splashing of water that contains the acids. The acid may end up in your eye, skin, etc which
is undesirable and dangerous.
 If you do it the other way around (adding water to acid), the first water molecule that makes
contact with the acid will react form hydronium ions which generates a big release of heat. This
would cause the water to vapourise or splash which contains the dissolved acid coming at you
which you don’t want. Therefore, never add water to acid when diluting but the other way
around adding acid to water. “A comes before W”
VII. To clean up an organic substance spill (e.g. organic vapour from, a half mask air-purifying
respirator is used. A full-face air-purifying respirator is required if the spilled organic substance is a
known eye irritant.
VIII. Wash your hands before leaving the laboratory, even if gloves are worn as a safety
precaution especiallyafter cleaning up a spill. This is because organic solvents can strip off your
skin’s natural oil layer resulting in skin irritation whereby your skin would be more susceptible to
absorbing toxic chemicals.
Disposing organic substances
IX. Ensure that all flasks containing organic substances are labelled such that they can be disposed
into the right waste container. All waste containers must also be labelled appropriately such that the
correct disposable method can be used.
 Liquid and solid organic waste are required to be segregated into different waste containers.
Also, inorganic and organic liquid wastes are required to be disposed in separate waste
containers.
X. Do not dispose organic substances through the sink as they will contaminate in the waters in
which aquatic organisms reside. Organic substances are toxic when ingested by aquatic organism
and potentially humans or any terrestrial mammals that consume the water.
 Ideally, separate organic and aqueous wastes (mainly water) when disposing organic liquid
wastes to reduce the cost of disposal.
Learning Objective #6 - Examine the environmental, economic and sociocultural implications of
obtaining and using hydrocarbons from the Earth
Source of hydrocarbons
Hydrocarbons, as we have defined already, as compounds that consists of only carbon and hydrogen
atoms.
There are many two classes of hydrocarbons. These are acyclic and cyclic hydrocarbons.
Acyclic hydrocarbons are primarily derived from petroleum and natural gas. These hydrocarbons
have a linear structure. Alkanes, Alkene, Alkynes that we have explored in last week’s and this
week’s notes are examples of acyclic hydrocarbons.
 Comparatively, cyclic hydrocarbons are generally sourced from coal. These hydrocarbon have a
cyclic or circular structure as they have a hexagonal-shaped phenyl ring, C6H5, which has three
alternating C=C bonds.

Petroleum and Natural Gas can be found at different levels deep in the Earth’s crust. Petroleum and
Natural Gas formed as dead aquatic organism are covered in sand, clay and mud at the bottom of
the sea. Due to the environment consisting of the lack of oxygen, high heat and high pressure, the
dead matter from the organisms are converted into petroleum and natural gas.
 As petroleum is lighter than water, it is situated above water. Comparatively, the natural gas is
formed above the petroleum layer but then becomes trapped between the impervious rocks that
is above it. Therefore, natural gas can be sourced from the impervious rocks that are located
deep in the Earth’s crust.
Coal is formed by the decomposition of dead plants and trees on land covered in soil and mud. Over
time, the dead matter is trapped and subjected to an environment that has no oxygen (anaerobic),
high heat and high pressure. Thus, coal is sourced deep within the Earth’s curst.
Now with that introduction out of the way, let’s address the hydrocarbons that are derived from in
petroleum, natural gas and coal!
Petroleum mainly consists of hydrocarbons from C1 to C30 that are used in everyday life or
industrial processes. For example:
 Petrol used in motor vehicles have hydrocarbons mainly consisting of 8 carbons (octane)
 Paraffin wax used in candles & machine lubricants consists of hydrocarbons with 20 carbons or
higher.
Natural gas mainly consists light weight hydrocarbon such as methane (~80%) which has the
molecular formula, CH4. Other light weight hydrocarbons include ethane, propane and butane.

Yes, these unbranched hydrocarbons from C1 to and including C4 (e.g. butane) are all gases at
room temperature as we have pointed out in Preliminary HSC Chemistry.
Coal mainly consists of carbon rich compounds which can be burnt in an anaerobic environment to
produce methane.
Environmental Implications
As we have mentioned previously, petroleum has a wide range of hydrocarbons with differing
molecular weights from C1 to C30 (or even higher). Depending on the hydrocarbon’s physical &
chemical properties, amount of hydrocarbon and time of contact, the hydrocarbon’s effect of the
environment can vary upon contact.

It is important to note that only a small portion of the wide range of hydrocarbons in petroleum
actually do yield harmful effects (e.g. toxic) to the environment.
Let’s first explore how obtaining hydrocarbons can result in environmental implications!
Obtaining petroleum and natural gas from the Earth results drilling through rocks deep in the Earth’s
crust whereby hydrocarbons from the drill machine’s lubricants can be dispersed into surrounding
water pollutes the surrounding seawater or ocean. These hydrocarbons are toxic to aquatic
organisms that reside in the sea.
These rocks are also returned into the ocean which often contain barium ions from the traces of
lubricant that remain. These barium ions are toxic as they interfere with enzyme activities which can
result in death of living organisms.
Furthermore, there are potassium ions in the machine lubricants in extracting the hydrocarbons (e.g.
petroleum & natural gas) where high levels of potassium can result in uncontrollable algae growth
leading to a state of eutrophication. This essentially allows algae to grow on the water surface which
blocks the sunlight reaching to the plants beneath the water as well as oxygen gas that is dissolved in
the water. This will reduce in the death of plants which decreases the oxygen availability in the
water for aquatic animals like fish as well as bacteria that further uses more oxygen to decomposes
dead matter of demised plants. Overall, it destroys the aquatic ecosystem turning a habitable
environment into a toxic one for the original species.
Other than eutrophication, an increase in ion concentration (e.g. K+ ions) in the surrounding water
can disrupt the osmotic balance in living organisms, resulting in the deformations in aquatic fauna or
flora.
Another environmental implication would involve the noise pollution through the sending of sound
waves to detect potential hydrocarbon deposits for drilling. This would disturb local aquatic
organisms, such as whales, as well as any humans that near the area. The sound waves can
disorientate whales which can result in large scale whale stranding on beaches. This results in their
death through multiple means on is due to dehydration.
To obtain hydrocarbons for use, it must be transported from the sea to land to oil refinery so that
petroleum can be split into their components. The mode of transportation is through the use of
ships. Oil spillages due to various accidents such as collisions with rocks, other ships, etc have
resulted in environmental damages similar to those which we have mentioned already. These
hydrocarbons can be washed ashore which can pollute beaches whereby humans have contact
with.
Hydrocarbons that enter the human body could cause severe respiratory irritation. The long term
effects of hydrocarbon exposure is also currently not fully understood.
Economic Implications
Through potential implications such as leaking toxic hydrocarbons, potassium ions and
eutrophication as we have discussed earlier as an environmental implication, they could all result in
the death and reduction in biodiversity of aquatic organisms. It is important to preserve biodiversity
for many countries as aquatic organisms serves a major source of economic revenue.
For example, aquatic organisms sold from Philippines accounts for over $550 billion USD dollars
annually. The reduction of Philippines’ aquatic organisms would result in a major hit to the country’s
economy and have many social implications such as poverty (which is relevant for the following
section – Sociocultural Implications of obtaining hydrocarbons).

Philippines recognises the importance of preserving biodiversity of its aquatic organisms as it is
first south eastern asian country to regulate the risk of biotechnology as the technology could
reduce the country’s rich aquatic organisms biodiversity.
Sociocultural Implications
Workers involved in the drilling process to obtain hydrocarbons will be exposed to drilled rocks that
are covered in toxic hydrocarbon lubricants, as well as the lubricants in the machinery itself, which
can be accidentally or voluntarily inhaled as oil mists in the air.
We have already mentioned how the leaking of toxic hydrocarbons, potassium ions and
eutrophication during the extraction of hydrocarbons can result in a decline in aquatic organism’s
biodiversity. This has many implications in removing or limiting the diverse range of food in which
humans can enjoy in our everyday lives. Also, as the supply of aquatic organisms decreases due to
hydrocarbon pollution, the price of seafood would increase which means that it would be less
affordable for the global population in general.
Furthermore, the presence of toxic hydrocarbons would result in our everyday potable (drinkable)
water derived from the sea being toxic. The treatment of water would thus be more thorough which
would incur additional cost to consumers.
Learning Objective #1 - Write equations and construct models to represent the reactions
of unsaturated hydrocarbons when added to a range of chemicals
Hydrogenation: Adding hydrogen (H2) to unsaturated hydrocarbons
Hydrogenation is an example of an addition reaction whereby, in our case, hydrogen gas reacts with
unsaturated hydrocarbons such as alkenes and alkynes to form one combined product.
In hydrogenation specifically, the hydrogen reacts with and breaks the C=C bond such that one
hydrogen atom is added (i.e. covalently bonded) to each carbon atom whereby the alkene is
converted into an alkane product.
That is, the carbon that was originally unsaturated became saturated after hydrogenation.
Halogenation: Adding Halogens (X2) to unsaturated hydrocarbons
Another type of addition reaction is halogenation whereby diatomic halogen molecules reacts with
the unsaturated carbon bond, e.g. C=C bond, resulting in one halogen atom being added to each
carbon atom.
The resulting product is a haloalkane (alkane containing halogen atom(s))
This effectively converts the unsaturated carbon in an alkene into an saturated carbons in an alkane.
Hydrohalogenation: Adding Hydrogen Halides (HX) to unsaturated hydrocarbons
Hydrohalogenation is the third type of addition reaction in which a hydrogen halide (HX) molecule
breaks the C=C bond so that a hydrogen is added to one carbon atom and a halogen atom is added
to the other carbon.
The result is a haloalkane being produced.
In HSC Chemistry, we do not need to need to understand which carbon atom the hydrogen is added
to and which carbon atom the halogen is added to. It is because it is outside the HSC Chemistry
Syllabus.
However, we will explore the Markovnikov’s rule in this week’s Youtube Video so that we can
create a more accurate model for haloalkane that is produced via hydrohalogenation reaction.
 Again, you can simply add to any of the two unsaturated carbon as you wish for HSC Chemistry
purpose if you decide to not learn the Markovikov rule as it’s not part of HSC Chemistry syllabus.
This is because Markovnikov’s Rule is outside the scope of HSC Chemistry.
Hydration: Adding Water (H2O) to unsaturated hydrocarbons
The final addition reaction is hydration. In hydration, there is a hydroxyl group (-OH) that is added to
the unsaturated carbon atom (i.e. part of the C=C bond) and a hydrogen atom is added the other
carbon atom which was part of the original C=C bond.
The resulting product is a saturated alcohol.
The second type of unsaturated hydrocarbons is alkyne. So let’s explore how alkynes can undergo
additional reaction.
NOTE: Alkynes can double the number of chemical reaction with halogens compared to alkenes. This
is because alkynes have triple carbon bonds while alkenes have double carbon bonds.
We will only explore halogenation and hydrohalogenation addition reactions for alkynes as the other
reactions we have explored for alkenes will follow the same principle.
That is, alkynes will have double the number of addition chemical reactions that alkenes can
perform.
Halogenation of unsaturated hydrocarbons - Alkynes
Hydrohalogenation of unsaturated hydrocarbons - Alkynes
Learning Objective #1 - Investigate the structural formulae, properties and functional group
including:
Primary Alcohols
Secondary Alcohols
Tertiary Alcohols
We have already talked about the physical shared by all alcohols in general in last week’s notes,
except for the boiling points between primary, secondary and tertiary alcohols which we will explore
in this week’s note.

So please revisit last week’s notes, if you need to revise on the other physical properties of
alcohols homologues.
That being said, we will explain some differences in the chemical properties between primary,
secondary and tertiary alcohols towards the end of this learning objective as well as in Learning
Objective #4 and #6 which we haven’t explored in last week’s notes.
Anyways, returning to this learning objective. This question we want to address is what are the
differences between primary, secondary and tertiary alcohols?
Well, primary alcohols are alcohols whereby the carbon atom that is bonded to the hydroxyl group
(OH) is only bonded to one R group.
On the other hand, secondary alcohols are organic molecules whereby the two R groups (can same
or different) are attached the carbon atom that is bonded to the hydroxyl group.
Lastly, in the case of tertiary alcohols, they have three R groups (can be same or different) are
bonded to the carbon atom which is also bonded to the hydroxyl group.
NOTE: The ‘R’ groups used here to differentiate primary, secondary and tertiary molecules are
groups that contain at least one carbon atom attached to hydrogen atoms. The R groups here
cannot be hydrogen atoms.
Check out the following diagram that compares the differences between the three categories of
alcohols.
NOTE: The direction in which you draw the R-C bond(s) does not matter. As you can see in the
diagram, the red arrows are drawn to remind and show you that the chemical bonds can be rotated
(clockwise or anti-clockwise)
NOTE: Some sources include methanol as a primary alcohol. Some do not. It is up to you to decide. It
is probably more accurate to exclude it from the primary alcohol class.
So let’s explore in terms of melting & boiling point shared between primary, secondary and tertiary
alcohol with similar molecular masses (i.e. same number of carbon atoms).
So, in the order of increasing melting & boiling points: Tertiary, secondary, primary alcohols and
finally methanol.
An important factor that is contributing towards this trend is that an alkyl group is bigger in size than
a hydrogen atom. But let’s explore why this is important in dictating the trend above.
As tertiary alcohol has more alkyl groups than secondary alcohols which has more alkyl groups than
primary, the hydroxyl group is less exposed due to more alkyl groups. This hinders the extent at
which hydrogen bonding can occur. This means that there would be less hydrogen bonding occurring
in tertiary alcohol compared to secondary alcohols due to the hinderance of alkyl groups.
Therefore, tertiary alcohols have lower melting & boiling point than secondary alcohols and also the
reason for why secondary alcohols have lower M.P and B.P than primary alcohols.
Needlessly, to say by now, methanol does not have any alkyl groups, hence, it has higher melting
and boiling point than corresponding primary, secondary and tertiary alcohols.
As promised at the start of this learning objective, we will discuss about a difference in chemical
property between primary, secondary and tertiary alcohols here.
This difference in chemical property amongst the three different types of alcohol is their acidity. In
order of increasing acidity (from low to high): tertiary alcohol -> secondary alcohol -> primary
alcohol -> methanol
For HSC Chemistry purposes, we can explain this trend by understanding that alkyl groups are better
electron donators than hydrogen atoms.
Therefore, as the number of alkyl groups increases from primary to tertiary alcohol, more electron
density is pushed towards the central carbon atom which passes on some of these electron towards
the oxygen atom in the hydroxyl group. As a result, the electrons that oxygen attracts from hydrogen
atom in the OH group is less extensive. Therefore, the energy required to break the OH bond
increases. This makes it harder to break the hydroxyl bond and release a H+ ion.
Another way to explain this is that as the number of electron donating alkyl groups increases, the
more unstable the conjugate base will be (due to increased number of electrons which the
conjugate base carries) after the alcohol donates its hydrogen atom. Therefore, tertiary alcohols
form the least stable conjugate base, thus, making it the weakest acid of the three.
Lastly, in terms of chemical properties differences, due to their different number of alkyl groups
present in the different types of alcohols, primary alcohol be oxidised to aldehyde and carboxylic
acid but not to ketone.
In the case secondary alcohols, they can be oxidised to ketones but not to aldehyde or carboxylic
acid (unless in extreme oxidative conditions).
Comparatively,, tertiary alcohol does not readily oxidise unless in extreme oxidative conditions
involving the breaking of C-C bonds such as in combustion.
Learning Objective #4 - Write equations, state conditions and predict products to represent the
reaction of alcohol, including but not limited to:
- Combustion
- Dehydration
- Substitution with HX
- Oxidation
Combustion of alcohols
Learning Objective #4 - Write equations, state conditions and predict products to represent the
reaction of alcohol, including but not limited to:
- Combustion
- Dehydration
- Substitution with HX
- Oxidation
Combustion of alcohols
Dehydration of alcohols
Reaction condition: Heating the alcohol in the presence of a concentrated acid catalyst (e.g. H2SO4
or H3PO4) at 170 degrees celsius. The product as a result of the dehydration of alcohol is an alkene
as well as water as a byproduct.
Since alkene is formed as a product, the dehydration of alcohol can only occur if the alcohol has two
or morecarbon atoms.
 NOTE: If the temperature is not suffice then ethers would be produced rather than alkenes. This
is a condensation reaction as water is released as a by-product and the product is larger than the
reactants, i.e. alcohol. Dehydration is a type of condensation reaction.
Substitution reaction involving alcohols with HX (hydrogen halide)
What happens in substitution reaction is that the HX ‘attack’s (interacts) and react with the C – OH
bond to free the hydroxyl group and substitute the original alcohol molecule with the halogen atom
“X”.
The free, negatively charged hydroxide ion can bond with the positively charged hydrogen ion (from
the HX molecule) to form water as a by-product.
If we wish to determine the alcohol is primary, secondary or tertiary, we can use the Lucas
Test which involves the use of a Lucas’ reagent which is essentially a solution comprised of an
anhydrous zinc chloride catalyst dissolved in concentrated hydrochloric acid solution.
The products produced is a chloroalkane (R-Cl) and water. To determine whether the alcohol is
primary, secondary or tertiary, the rate of reaction can be used.
Tertiary alcohol is able to completely react with the hydrochloric acid via substitution reaction
in less than one minute to form the cloudy, white haloalkane precipitate and water as a by-product.
 This means that two layers are formed (originally there is only one layer of alcohol solution at
start of experiment). The bottom layer is the precipitate that is more dense and the top layer is
the water alongside any excess alcohol and the zinc chloride catalyst.
For the case of secondary alcohol, the reaction typically requires 2-5 minutes. Again, the products
are precipitate haloalkane and water.
Lastly, it would take more than 6 minutes for primary alcohol to react with the hydrochloric acid in
the case of primary alcohols. When the reaction is complete, the haloalkane product is formed as a
precipitate as well as water.
 You may need to heat the solution containing primary alcohol and hydrogen halide (HX) to speed
up the rate of reaction.