Chapter 09: Alkanes
Chapter 09: Alkanes
Chapter 9
Alkanes
This is probably one of the simplest chapters in the whole of organic chemistry. However, there are certain concepts that students tend to fumble and overlook, hence the following information will help you to keep in mind the important concepts to note for this chapter.
Alkanes contain C-C and C-H bounds. Due to the difference between electronegativities of C and H atoms being negligible, the C-H bond is non-polar, hence alkanes are non-polar in nature.
Here are some important properties of alkanes that students should take note of:
| Physical properties |
|---|
|
Boiling point
(i) The boiling point of alkanes increase with increasing number of carbon atoms
(ii) Boiling points of isomeric alkanes decrease with increasing degree of branching
|
|
Solubility
(i) Insoluble in water
(ii) Soluble in organic solvent
|
| Chemical properties |
|---|
|
Alkanes are relatively UNREACTIVE due to
(i) relatively strong C-C and C-H bonds
(ii) its non-polar nature
|
| Physical properties | Chemical properties |
|---|---|
|
Boiling point
(i) The boiling point of alkanes increase with increasing number of carbon atoms
(ii) Boiling points of isomeric alkanes decrease with increasing degree of branching
|
Alkanes are relatively UNREACTIVE due to
(i) relatively strong C-C and C-H bonds
(ii) its non-polar nature
|
|
Solubility
(i) Insoluble in water
(ii) Soluble in organic solvent
|
Alkanes are generally unreactive but undergo 2 major reactions with non-polar reagents.
1. Combustion
It is important to note that alkanes are good sources of fuel that burn readily (hence reaction is very exothermic). Students are expected to know and apply the combustion equation in novel context
The equation is as such:
Note: In excess oxygen, alkanes burn with a non-sooty flame.
(In comparison to saturated hydrocarbons that burn with a sooty flame)
2. Free radical substitution (FRS)
Alkanes can react with halogens such as Br₂ and Cl₂, in the presence of UV light or heat, to form halogenoalkanes or alkyl halides. Students are expected to memorise both the conditions for reaction as well as the reaction mechanism. This should be the common expectation for students regarding the subsequent reactions in the rest of the chapters of organic chemistry.
- REAGENTS/CONDITIONS : Limited Cl₂(g)/Br₂(l) in UV light (for mono-substituted products)
- OBSERVATIONS : Either decolourisation of greenish-yellow Cl₂ or decolourisation of reddish brown Br₂
The free radical substitution mechanism is as such:
Checklist when drawing FRS Mechanism:
❏ Name the mechanism
❏ List the steps in its sequential order
❏ Insert the necessary half arrows
❏ Labelling of radicals on the right Carbon atom
Further Substitution
If chlorine is present in excess, further FRS can take place. This is also known as ‘continuous propagation’.
IMPORTANT questions to note:
1) Why is the synthesis of halogenoalkane from alkane not ideal using Free Radical Substitution?
Reasons:
- Difficult to control the extent of substitution (with presence of excess chlorine)
- Formation of other by-products
- Difficult to control the position of substitution of halogen
Production of isomeric products
Isomeric products can be obtained as where the halogen will be substituted will vary.
The different products formed depend on the different chemical environments of the hydrogen atoms. Identifying the different chemical environments in the compound will allow us to calculate the ratio of isomeric products formed.
For example, looking at 2-methylpropane below, it has two groups of hydrogen atoms with different chemical environments, Hᵃ and Hᵇ. Substituting Hᵃ and Hᵇ gives two isomers, which is 1-chloro-2-methylpropane and 2-chloro-2-methylpropane respectively.
This concept is commonly tested in paper one as a multiple-choice-question (MCQ). Hence let’s go through a worked example to better understand this concept.
Worked Example 1:
A mixture of 2-methylbutane and bromine is allowed to react in the presence of sunlight. How many products are formed and what is the ratio in which they are formed?
a) 2 products; 1:3
b) 3 products; 3:6:2
c) 4 products; 6:1:2:3
d) 5 products; 4:3:1:2:2
Answer: c) 4 products; 6:1:2:3
More questions to take note:
1) Why do alkanes not react with I₂?
Reason:
- Weak C-I and H-I bonds formed causes the 1st propagation step to be endothermic, preventing the formation of iodoalkanes.
2) Why do alkanes not react with aqueous Cl₂?
Reason:
- The presence of polar water molecules in aqueous Cl₂ will polarise Cl₂ to break.
heterolytically, causing a disproportionation reaction where HOCl and HCl are formed instead of the desired chloroalkane.
Stability of Radicals
More alkyl groups attached to the carbon radical will result in an increase in stability of the carbon radical:
Alkyl groups are generally electron-donating, thus more alkyl groups attached to the carbon radical causes it to be less electron deficient and the radical is more stable.
Continue Reading on Chapter 10: Alkenes