SL/HL IB Chemistry - Reaction of Alkanes

Using integers, what is the smallest sum of all the coefficients when the equation for the complete combustion of octane, C₈H₁₈ in oxygen is correctly balanced?

• 59

• 31

• 30

• 61

Answer explanation

2 C₈H₁₈(l) +   25 O₂(g) → 16 CO₂(g) + 18 H₂O(l)

Smallest sum of all coefficients = 61

A possible equation for the incomplete combustion of one mole of butane in air is:

C₄H₁₀(g)  + 5O₂(g)  →  xCO₂(g)  + yCO(g) + 5H₂O(l)

Which are the correct values for x and y?

• x = 2 and y = 2

• x = 2 and y = 3

• x = 1 and y = 3

• x = 3 and y = 1

Answer explanation

x + y must equal 4 as a molecule of butane contains 4 carbon atoms. 2x + y must equal 10 − 5. So, y = 5 − 2x. Substituting for y: x + (5 − 2x) = 4 so x = 1 and y = 3.

Which factors contribute to the explanation as to why alkanes are not very reactive?

I. Unlike the other elements in group 14, carbon cannot expand its octet.

II. C−H and C−C bonds are relatively strong compared to other covalent single bonds.

III. Alkanes either are non-polar or have very low polarity.

• I and II only

• I, II and III

• II and III only

• I and III only

Answer explanation

All three are important. Because alkanes are non-polar, electrophiles and nucleophiles are not attracted to them. It requires considerable energy to overcome their bond strengths and molecules such as water cannot use their lone pair to form coordinate bonds with carbon atoms in alkanes as they have no empty available orbitals. This last point explains why CCl₄ is immiscible and does not react with water whereas SiCl₄ reacts violently with water to form fumes of HCl.

Methane reacts with chlorine in the presence of ultraviolet light to form chloromethane and hydrogen chloride. What is the function of the ultraviolet light?

• It breaks the Cl−Cl bond in chlorine heterolytically.

• It breaks the C−H bond in methane homolytically.

• It breaks the C−H bond in methane heterolytically.

• It breaks the Cl−Cl bond in chlorine homolytically.

Answer explanation

The Cl−Cl bond has a bond enthalpy of 242 kJ mol⁻¹ which, unlike the C−H bond (414 kJ mol⁻¹), can be dissociated homolytically by ultraviolet light to form free radicals. 

Which is a propagation step in the substitution reaction between ethane and bromine in the presence of ultraviolet light?

• H₃C−CH₂· + H₃C−CH₂· → H₃C−CH₂−CH₂−CH₃ 

• H₃C−CH₂· + Br· → H₃C−CH₂Br

• H₃C−CH₃ + Br· → H₃C−CH₂· + HBr

• H₃C−CH₃ + Br₂ → H₃C−CH₂Br + HBr

Answer explanation

A propagation step must produce more radicals.

What will be the product of the reaction between pent-2-ene and bromine in the dark?

• 1,2-dibromopentane

• 2-bromopentane

• 2,3-dibromopentane

• 2-bromopent-2-ene

Answer explanation

Bromine adds across the double bond in the 2- and 3- positions.

The first step in the reaction between methane and chlorine in the presence of ultraviolet light is the homolytic fission of the chlorine-to-chlorine single bond, Cl-Cl(g), in gaseous chlorine.

i) Explain the meaning of the term homolytic fission.

Answer explanation

During homolytic fission the bond breaks symmetrically so that one electron forming the bond between two atoms becomes attached to one of the atoms and the other electron becomes attached to the other atom resulting in the formation of two radicals.

The first step in the reaction between methane and chlorine in the presence of ultraviolet light is the homolytic fission of the chlorine-to-chlorine single bond, Cl-Cl(g), in gaseous chlorine.

ii) Explain why homolytic fission occurs with the Cl-Cl bond in chlorine and not the C-H bond in methane.

Answer explanation

The ultraviolet light provides the energy to break the Cl-Cl bond homolytically. This energy, 242 kJ mol-1 , is much less than the 414 kJ mol-1 of energy required to break a C-H bond.

The first step in the reaction between methane and chlorine in the presence of ultraviolet light is the homolytic fission of the chlorine-to-chlorine single bond, Cl-Cl(g), in gaseous chlorine.

iii) Explain why only a few homolytic fission reactions involving chlorine need to be successful in order to bring about the complete reaction between chlorine and methane to form chloromethane and hydrogen chloride.

Answer explanation

Once a chlorine free radical is formed it reacts with a methane molecule to produce hydrogen chloride and a methyl radical. This methyl radical can react with another chlorine molecule to form chloromethane and generate a new chlorine radical which can then repeat the process. This is a propagation step. The formation of new radicals will only stop when a termination reaction occurs. (A termination step may be between two radicals or between a radical and an impurity or the walls of the reaction vessel).