THE (SATURATED) ALKANES

IQ2. How can hydrocarbons be classified based on their structure and reactivity?

IQ3. What are the products of reactions of hydrocarbons, and how do they react?

SYLLABUS REARRANGED

A. ALKANES - THE SATURATED HYDROCARBONS:

2.1 construct models, identify the functional group, and write structural and molecular formulae for homologous series of organic chemical compounds, up to C8
2.3 analyse the shape of molecules formed between carbon atoms when a single, double or triple bond is formed between them - the tetrahedral alkanes
2.2 conduct an investigation to compare the properties of organic chemical compounds within an homologous series, and explain these differences in terms of bonding
2.4 explain the properties within and between the homologous series of alkanes with reference to the intermolecular and intramolecular bonding present
3.2 investigate, write equations and construct models to represent the reactions of saturated hydrocarbons when substituted with halogens

2.1 construct models, identify the functional group, and write structural and molecular formulae for homologous series of organic chemical compounds, up to C8

– alkanes

2.3 analyse the shape of molecules formed between carbon atoms when a single, double or triple bond is formed between them

Homologous series are groups of compounds which are characterised by:

a common general formula
a common functional group
similar structures and chemical properties
gradations in their physical properties (as MM increases)

The shape of the distribution of atoms around any carbon atom usually this will fit one of three types:

We often draw methane as four bonds at right angles, but the molecular shape is actually tetrahedral. Another way of drawing this is:

Task 2.1.1

Q1. Use the interactive to

a) model the alkane homologous series https://chemagic.org/molecules/amini.html

b) identify the shape of the molecule wrt the tetrahedral bond angle

c) write names and draw formulas

Q2. Write the molecular formula and draw the structural formula for the following alkanes:

a) methylbutane

b) 2,3-dimethylhexane

c) cyclopentane

Build methane, ethene and ethyne to identify the shapes.

View video:

OC#10 Molecular Shapes https://www.youtube.com/watch?v=6PdKzBrheDk [10.14 mins]

2.2 conduct an investigation to compare the properties of organic chemical compounds within a homologous series and explain these differences in terms of bonding.

2.4 explain the properties within and between the homologous series of alkanes with reference to the intermolecular and intramolecular bonding present

When studying the properties of the hydrocarbons, there are two groups of properties that are important: physical properties and chemical properties.

The physical properties of organic molecules are determined by two factors: the functional group (largest impact if this involves polar bonds) and the chain (length and arrangement).

We will generally be considering two types of physical properties: phase change (melting point, boiling point - state at room temperature) and solubility in aqueous solution.

In very broad terms, the chemical properties of a compound or a series of compounds are associated with the intramolecular bonding (bonding between atoms in the molecule)

1. THE ALKANES (CnH2n+2): PHYSICAL AND CHEMICAL PROPERTIES

For the homologous series of the alkanes, the functional group for each member of the series is the same, so just the chain length and arrangement have significance.

Physical properties

The physical properties of the alkanes are related to the intermolecular forces between molecules. As a result the physical properties are not just based on the non-polarity of the covalent bonds within the molecules, but also on their size and shape. The dominant intermolecular forces are dispersion forces.

The presence of non-polar bonds and non-polar molecules affects the physical properties of the compound. The hydrocarbons are separate (=discrete) molecules with strong covalent bonds in the molecule but only weak dispersion forces between molecules. As the chain length increases, the number of dispersion forces between molecules increases, and hence larger molecules have higher melting and boiling points: small alkanes exist as a gas at SLC (Standard Laboratory Conditions = 25C and 100kPa), butane, octane and several others are liquids, while the longer chains, such as the paraffin waxes (20+ carbon atoms) are solids.

The non-polar nature of the hydrocarbon molecules also makes them insoluble in a polar substance such as water. However as our rule is like dissolves like, they will dissolve in other non-polar solvents, eg CCl4.

(The lack of free-moving electrons also means the hydrocarbons are poor electrical conductors.)

MolyMod Practical

This exercise is mainly an investigation with a practice writing exercise.

A. Make 2 model molecules of ethane

Identify the types of intramolecular forces between atoms in the ethane molecule

1.Ethane is formed from non-polar covalent C-C and C-H bonds.

2. Identify the types of intermolecular forces between atoms in the ethane molecule

Ethane is non-polar, so it only experiences weak London dispersion forces between molecules.

Background explanation: The London dispersion force is the weakest intermolecular force. It is a temporary attractive force that results when the electrons in two adjacent atoms occupy positions that make the atoms form temporary dipoles. Because of the constant motion of electrons, an atom or molecule can develop a temporary (instantaneous) dipole when its electrons are distributed non-symmetrically about the nucleus. A second atom or molecule, in turn, can be distorted by the appearance of the dipole in the first atom or molecule (because electrons repel one another) which leads to an electrostatic attraction between the two atoms or molecules. Dispersion forces are present between any two molecules (even polar molecules) when they are almost touching.

3a. Describe the strength of these forces (how polarised/electron cloud size)

Ethane has a relatively small electron cloud, so it is only moderately polarisable, and the temporary dipoles formed are weak.

Background explanation: : Electron cloud = total electrons + how dispersed they are.

Ethane has a relatively small electron cloud because it contains a small number of electrons (18) across only two C atoms, so there is little electron density. Those electrons are held close to the nuclei because it is a small molecule with short sigma bonds. This makes the molecule less polarisable, so only very weak temporary dipoles form, resulting in weak London dispersion forces.

London dispersion forces depend on how easily the electron cloud can distort (be polarised)

A small, tightly held electron cloud is harder to distort

→ produces weaker instantaneous dipoles

→ results in weaker intermolecular attractions

3b. Describe the strength of these forces (surface area)

Ethane's limited surface area further reduces the extent of intermolecular contact between molecules. As a result, the dispersion forces between ethane molecules are weak.

Background explanation: Electron cloud = total electrons + how dispersed they are. Dispersion forces are present between any two molecules (even polar molecules) when they are almost touching. A small molecule will have lower surface area and so fewer/lower dispersion forces.

4. Explain the reason that ethane is a gas at room temperature (using 1-3)

Ethane is a non-polar molecule and therefore its intermolecular forces are only weak London dispersion forces. These arise from temporary fluctuations in electron density that create instantaneous dipoles. However, ethane's relatively small electron cloud means that it is only moderately polarisable, and so the temporary dipoles formed are weak. In addition, its limited surface area reduces intermolecular contact between molecules. As a result, only a small amount of energy is required to overcome these weak dispersion forces, resulting in a low melting and boiling points.

B. Now make two molecules of hexane.

Compare the melting points of ethane and hexane (electron cloud size and surface area)

Hexane has a larger electron cloud, and therefore stronger temporary dipoles. This leads to stronger London dispersion forces. In addition, longer molecules have larger surface area, allowing more intermolecular contact. As a result, more energy is required to overcome these forces, so melting and boiling points are higher for hexane than for ethane.

2. Describe the trend for increasing C-chain length and melting point.

As the number of carbon atoms in the molecule increases, the size of the electron cloud increases. This leads to stronger London dispersion forces because larger electron clouds form stronger temporary dipoles. In addition, longer molecules have greater surface area, allowing more intermolecular contact. As a result, more energy is required to overcome these forces, so melting and boiling points increase with increasing carbon chain length.

C. Now make two molecules of dimethyl propane.

Compare the melting points of hexane and 2,2-dimethylbutane (electron cloud size and surface area).

Both hexane and 2,2-dimethlybutane have the same molecular formula (C6H14), so they have the same number of electrons and therefore the same electron cloud size.

Hexane has a larger surface area due to its straight-chain structure. This allows for greater intermolecular contact and stronger London dispersion forces. The compact, branched structure of 2,2-dimethylbutane reduces surface area and intermolecular contact, resulting in weaker dispersion forces. Therefore, hexane has a higher melting point than 2,2-dimethylbutane.

2. Explain the trend in melting point for straight-chain and branched isomers of alkanes.

Although straight-chain and branched isomers of alkanes have the same molecular formula and therefore the same electron cloud size, the straight-chain molecule has a larger surface area. This allows for greater intermolecular contact and stronger London dispersion forces. The more compact structure of the branched isomer reduces surface area and intermolecular contact, resulting in weaker dispersion forces. Therefore, the straight-chain alkane isomer has a higher melting point than the branched form.

(Extension note: Molecules that are more symmetrical, linear, or compact pack more tightly in the solid state, leading to higher melting points. Even-numbered alkanes pack better in the solid state than odd-numbered chains, leading to a higher melting point trend. Branching generally reduces the surface area, decreasing the melting point. However, highly branched, symmetrical molecules (e.g., neopentane) can have anomalously high melting points.)

Past Qs

Q1. Within a homologous series such as the alkanes, boiling points:

A. are all approximately the same.

B. increase as the molar masses increase.

C. are independent of intermolecular forces.

D. decrease as the molecules become larger.

ANS: B. Within a homologous series, boiling points increase as molecules get larger due to an increase in dispersion forces.

2022 HSC Q24

Q2. The following graph shows the boiling points of some 1-chloroalkanes. (treat as alkanes)
Explain the trend shown in the graph. (3 marks)

Basic answer below: Rewrite to our standard

As the molar masses increase, the boiling point of alkanes increases. This is due to molecules with a larger molar mass containing more electrons. This allows for a higher likelihood of unequal electron distribution, thus, causing stronger dispersion forces. Since more energy is required to break stronger dispersion forces, alkanes with higher molar masses have higher boiling points.

Extra for experts: Intramolecular bonding of alkanes

hybridorbitals_01.pdf

Example 1: methane

If one of the electrons in the 2s orbital jumps to a 2p orbital, the excited carbon atom can hybridise this state to form 4 identical hybrid orbitals which are equivalent in energy, size and shape.

The formation of these hybrid orbitals slightly lowers the energy, giving some added stability. The orbitals are arranged in an overlapping organisation which accounts for the tetrahedral arrangement of hydrogen atoms around a central carbon atom.

The single bonds representing the four sp3 hybrid orbitals are called sigma (σ) bonds. The sigma bond forms when the heads of two atomic or molecular orbitals overlap. As a result “you can have a sigma bond between an sp3 orbital on a carbon atom and a 1s orbital on a neighbouring hydrogen atom as well as from the overlap of two sp3 orbitals between neighbouring C atoms”. (Walker, 2019)

“A sigma bond is a covalent bond formed by overlap of atomic orbitals and/or hybrid orbitals along the bond axis (i.e., along a line connecting the two bonded atoms)”. (Walker, 2019)

Example 2: ethane

3.2 write equations and construct models to represent the reactions of saturated hydrocarbons when substituted with halogens.

Chemical properties

Alkanes are chemically unreactive for a number of reasons:

The electronegativity values for carbon and hydrogen are very similar, and so the covalent bonds (and therefore the molecules) are non-polar.
The non-polar bonds have no region of high electron density or polarity, so they don’t attract polar or charged species (like ions or full dipoles) that drive many reactions.
The single C–C and C–H (sigma) bonds are strong and require a lot of energy to break, so reactions don’t occur easily under normal conditions.
The paired electrons are in a tetrahedral arrangement which gives stability to the compounds

Alkanes: Substitution Reactions

I think his spelling error means he needs to be there himself!

The low reactivity of alkanes means they do not readily react with halogens unless in the presence of ultraviolet light. Even so, the reaction is not an addition reaction, it is a substitution reaction. One of the halogen atoms substitutes for (or replaces) a hydrogen atom and a haloalkane is formed as a result. A second product will also form from the combination of the substituted hydrogen with the remaining halogen atom.

Ethane + chlorine → chloroethane

CH3—CH3 + Cl2 → CH3—CH2Cl + HCl

The chlorination of methane does not necessarily stop after one substitution (chlorination). It may actually be very hard to get a mono-substituted chloromethane. Instead di-, tri- and even tetra-chloromethanes are formed.

One way to avoid this problem is to use a much higher concentration of methane in comparison to chloride. This reduces the chance of a chlorine radical running into a chloromethane and starting the mechanism over again to form a dichloromethane. Through this method of controlling product ratios one is able to have a relative amount of control over the product.

dichloromethane CH3—CH2Cl + Cl2 → CH2Cl—CH2Cl + HCl

View video:

OC#15 https://www.youtube.com/watch?v=09mXAqtEohQ&list=UUkdi7YOXBGAapx_dR40UoFQ&index=19 [5.46 mins]
Reactions of Alkanes https://www.youtube.com/watch?v=T0H1yNN5uOM [13.08 mins]

3.2 Extra for experts

Halogenation mechanism. In the methane molecule, the carbon‐hydrogen bonds are essentially non-polar covalent bonds. The halogen molecule has a totally non-polar covalent bond. UV light contains sufficient energy to break the weaker non-polar chlorine‐chlorine bond (∼58 kcal/mole), but it has insufficient energy to break the stronger carbon‐hydrogen bond (104 kcal/mole).

The breaking of the chlorine molecule leads to the formation of two highly reactive chlorine free radicals (chlorine atoms). A free radical is an atom or group that has a single unshared (unpaired) electron. It is represented by the element symbol with a dot to the centre right (the unpaired electron).

Cl·

In the bond that is ruptured, each of the originally bonded atoms receives one electron. The chlorine free radicals that form are in a high‐energy state and react quickly to complete their octets and liberate energy. (Once the high‐energy chlorine free radicals are formed, the energy source (UV light or heat) can be removed. The energy liberated in the reaction of the free radicals with other atoms is sufficient to keep the reaction running.)

When a chlorine free radical approaches a methane molecule, a fission of a carbon‐hydrogen bond occurs. The chlorine free radical combines with the liberated hydrogen free radical to form hydrogen chloride and a methyl free radical. The methyl free radical then combines with another chlorine free radical and forms chloromethane.

TASK 3.2.1

Q. Alkanes are relatively unreactive. What is the best explanation for this?

A) They contain only single carbon–carbon bonds
B) They lack reactive functional groups
C) They have low melting and boiling points
D) Their intramolecular bonds have tetrahedral angles

Ans B.

Q2. Excess fluorine is added to methane in the presence of UV light. What would be the expected products? Justify your response. (3 marks)

Q3. Intense conditions are required to undergo halogenation of an alkane. Which of the following do not contribute to these conditions?

A) High temperatures (400oC)

B) UV light

C) High halogen concentration

D) Concentrated acid catalyst

Q4. Explain why the chlorination of methane, if not carefully controlled, can produce tetrachloromethane. In your answer, include relevant reactions. (3 marks)

ANS

dichloromethane CH3—CH2Cl + Cl2 → CH2Cl—CH2Cl + HCl

Investigation 2.2.1

On one graph, plot the information from Table 10.1 for melting points for alkanes, alkenes, alkynes

2. On one graph, plot the information from Table 10.1for boiling points for alkanes, alkenes, alkynes

3. Explain the graphs in terms of number of carbons in the series for state at room temperature

4. From the data in the second table below

a) explain the change in enthalpy of combustion for methane to hexane

b) predict whether the complementary alkene/alkyne would have a lower or higher enthalpy. Justify your answer.

Investigation 2.2.2 Experiment: Properties of Hydrocarbons

Introduction

Hydrocarbons are compounds which contain only carbon and hydrogen. They can be classified into several types, depending on their structure.

Chain hydrocarbons are divided into three classes:

Alkanes which have only single bonds and are said to be saturated.
Alkenes which have at least one carbon-carbon double bond and alkynes which have at least one carbon-carbon triple bond are said to be unsaturated.

In this experiment you will perform experiments to illustrate some of the properties of saturated and unsaturated hydrocarbons.

Method

Solubility and Density of Hydrocarbons

1. Collect a 5mL sample of each of the following hydrocarbons; hexane (H-), hex-1-ene (H-e), cyclohexane (CH), cyclohexene (CH-e).

2. Collect four test tubes and place them in a test tube rack. Label the test tubes H-, H-e, CH, CH-e.

3. Add 5 mL of water to each test tube.

4. Test the solubility of the four hydrocarbons in water by adding 1mL of one hydrocarbon with a dropper to the 5mL of water in the test tube marked with its identifier (ie, add 1mL of hexane to the test tube of water marked H-).

5. Repeat step 4 for each hydrocarbon.

6. Look for the formation of any separate layers and determine which is the water layer in each case.

7. Collect another four test tubes and place them in a separate test tube rack. Label the test tubes as before.

8. Add 5 mL of dichloromethane to each test tube.

9. Test the solubility of the four hydrocarbons in dichloromethane by adding 1mL of one hydrocarbon with a dropper to the 5mL of water in the test tube marked with its identifier (ie, add 1mL of hexane to the test tube of water marked H-).

10. Repeat step 8 for each hydrocarbon.

11. Test the solubility of the four hydrocarbons in dichloromethane. If separate layersare formed, determine which is dichloromethane in each case.

12. Record your results.

Pentane, methyl butane and dimethyl propane all have the same molecular formula and hence the same molar mass (72.146 g/mol) however, the boiling points are as follows:

Pentane 36oC

Methyl butane (isopentane) 27.8oC

Dimethyl propane (neopentane) 10oC

As the compound becomes more branched, the available surface area is less, the dispersion forces holding the molecules together become slightly weaker, and so the highest melting point is for the most linear compound (pentane).

View Videos:

OC#9 https://www.youtube.com/watch?v=vyBn-MOzkVU [11.23]
OC#11 https://www.youtube.com/watch?v=OSHWS9-vLwk [14.09]

ALKENES, ALKYNES

TASK 2.2.1 Research to understand the answers to the following questions.

1. In general, the alkene homologous series with increasing number of carbons show little variation in what? (Shenfield & Silove, 2018, p. 72)

A) Freezing point

B) Strength of inter-molecular forces

C) Density at room temperature

D) Flammability

TASK 2.2.1 Research to understand the answers to the following questions.

1. Alkanes are generally unreactive species. This can be explained by which of the following? (Shenfield & Silove, 2018, p. 72)

A) Consist only of highly stable carbon-carbon bonds

B) Lack of functional groups

C) Highly non-polar nature

D) Relatively low boiling points

2. Dipole-Dipole Forces

These forces occur in molecules that are permanently polar versus momentarily polar. In this type of inter-molecular interaction, a polar molecule such as water or H2O attracts the positive end of another polar molecule with its negative end of its dipole. The attraction between these two molecules is the stronger dipole-dipole force.

3. Hydrogen "bonding"

A hydrogen bond is a special type of dipole-dipole attraction which occurs when a hydrogen atom bonded to a strongly electronegative atom is close to another electronegative atom with a non-bonding pair of electrons.

These bonds are generally stronger than ordinary dipole-dipole and dispersion forces, but weaker than true covalent andionic bonds.

Introduction

For a hydrogen bond to occur there must be a hydrogen atom covalently bonded covalently bonded with a strongly electronegative atom such as N, O, or F. The electronegative ion or molecule must posses a non-bonding electron pair (O has two in water, shown as twin sets of dots in the diagram below left) in order to form a hydrogen bond.

Since the N, O or F atom is strongly electronegative, it pulls the covalently bonded electron pair closer to its nucleus, and away from the hydrogen atom. The hydrogen atom is then left with a partial positive charge, creating a strong dipole-dipole attraction (shown as dotted lines in the diagrams below) between the hydrogen atom bonded to the donor, and the non-bonding electron pair on the electronegative atom.

Note: Chlorine has a relatively low ability to form hydrogen bonds because it is a large atom. When the radii of two atoms differ greatly in size or are both large, their nuclei cannot get close enough when they interact, so they have only a weak interaction.

SUMMARY

Start at the green box on the left halfway down.

Note: We just call "London dispersion forces" dispersion forces

No need to learn detail. This is to give you an idea of the energy required to overcome the attraction and move it to a freer state (ie solid to liquid or liquid to gas).

Important for us are:

Dispersion
Dipole-dipole
H "bonds" (alkanols, alkanoic acids etc)

Solubility is another physical property affected by intermolecular forces.

Consider the table below of solubility of alcohols in water (polar) or hexane (non-polar) solvents

The term miscibility refers to the ability of a liquid solute to dissolve in a liquid solvent.

Solubility is more often used to mean the ability of a solid solute to dissolve in a liquid solvent. Miscibility is used when talking about the solubility of -- specifically -- liquid solutes.

Miscible liquids are also defined as liquids that can mix to form a homogeneous solution. Miscible liquids generally mix without limit, meaning they are soluble at all amounts.

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