Alkanes are the most unreactive organic chemicals. They react mainly by Free Radical Reaction. The CC σ-bond are very stable at SATP or even heated to 100�C. But if heated to about 400�C the bond will break up. This is the chemistry used in the cracking of petroleum.

Consider a molecule CH₃(CH₂)_nCH₃. When heated to 1000�C in the cracking process it breaks up to;

These are known as free radicals; species that have a valence electron eagerly looking for a partner to pair off. (Technospeak: Free radicals are species that have an unpaired valence electron). The radicals can then recombine to give; CH₃CH₃, CH₃CH₂CH₃, etc.

The other possible reaction is on the CH σ−bond. These bonds are much stronger than the C−C σ−bond. The strength of the CC σ−bonds ranges from 380 kJ mol‾� to 350 kJ mol‾�, while that for CH σ−bonds are from 440 kJ mol‾� to 390 kJ mol‾�. But being small and sticking out from the carbon atom, it is exposed and can be approached very closely by reactive free radicals.

Example. At SATP the C−H bonds are easily attacked by free radicals present in the environment (like the oxygen radical, O�, generated by UV light);

By pulling off the proton it leaves behind a carbon free radical.

One interesting discovery is that the reactivity of the C−H bonds are very dependent on the carbon it is attached to. One important commercial process is the reaction of alkanes with halogens. When a mixture of propane and chlorine is irradiated with UV light the products are 57% (by mole) of 2−chloropropane and 43% of 1−chloropropane. The mechanism for the reaction is;

	Cl2	→	2 Cl�	(1)
	Cl� + RH	→	R� + HCl	(2)
	R� + Cl2	→	RCl + Cl�	(3)
	Cl� + RH	→	R� + HCl	(4)
	R� + Cl2	→	RCl + Cl�	(3)
	Cl� + RH	→	R� + HCl	(4)

Tutorial 2.1

Give a name to the reaction mechanism for the chlorination of alkanes, and state the initiation and the propagation steps.    Answer

The chlorine free radical can attack either the proton at C1 or C2 (or both when excess chlorine is used). The telling reactions would be;

To compute the reactivity we focus on the concentrations of H(1) and H(2) protons in the propane;

[H(2)]		k(2)		[2-chloropropane]
	x		=
[H(1)]		k(1)		[1−chloropropane]

Tutorial

When a mixture of 2−methylpropane and chlorine is irradiated with UV light the products are; 64% (by mole) of 1,1−dimethyl−propyl chloride and 36% of 2−methyl−propyl chloride. Compute the relative reactivity of the protons at C(1) and C(2).    (Answers=16)

So the relative reactivity of the protons towards free radical attack is of the order;

Experiment

Place 1 ml of hexane in a test-tube and add 4 ml of bromine water. Stir the mixture, stopper and place it in sunlight. Repeat this with isopropylhexane. Observe the discolouration of the mixture. Which compound reacts faster?
Repeat the experiment (simultaneously) and keep the test-tube in the dark (covering it with a box). Do the colour of the bromine disappear?

REACTIVITY of the PROTONS

Every time when we come across a reactivity issue we would have to look at the theory of bonding. In this case the electrons in the CC σ-bond are equally shared by the two carbon centres, since both have similar electronegativity. This is not so with the CH σ-bond. The hydrogen has a naked proton not shielded by any wall of electrons so the electrons can orbit around it and enjoy the friendship of the positive charge. Just like some youths like spending more time at the home of their pal, rather than your own. So with the CH σ-bond the carbon is not getting its fair share of "valence octet stabilisation" and it would hold on to the hydrogen atom desperately. When a free radical approaches the hydrogen to negotiate for its release;

Carbon atom with more hydrogen atoms attached to it would be more electron deprived and so more positive. Consequently it will be more reluctant to release the proton. (Technospeak: the alkyl group has an electron-donating-inductive effect on the carbon).

This concept that gives rise to the electron-donating-inductive effect can be better illustrated by the difference in C-H bond distance for alkanes, alkenes and alkynes. For alkanes, alkenes and alkynes we have three different hybridisations of the valence atomic orbitals of the carbon. We would expect the hybrid atomic orbital with decreasing p component (or increasing s component) to have a shorter orbital radius, since s-orbitals are nearly to the nucleus. The distance between the carbon and hydrogen atoms for the various hybrid carbons are shown below.

If the carbon has less s-character than the atom it is sharing the electron with, then it has less ownership over the electron in the bond, since the electron will be spending more time in the s-orbital. The C−H distance would be greater.

Similarly a carbon with an alkyl group attached to it should be less deprived of electron then a carbon with a proton attached to it. Conversely we said that the alkyl group has an electron-donating-inductive effect on the carbon.

Tutorial 2.2

State the two factors in a chemical reaction that determine the type of products obtained.

In the bromination of propane, 92% of 1−methyl−ethyl bromide is obtained. What can you say about reactivity and selectivity? (Note that the yield is understood to be in terms of mole percent)      Answer

Tutorial 2.3

Polypropylene has the molecular structure CH₃CH(CH₃) − − {CH₂CH(CH₃)}_n − − CH₂CH(CH₃). If you place a sheet of polyethylene and a sheet of polypropylene in the sun, which one do you expect to degrade faster? Why?    Answer

Because of this difference in chemical reactivity chemist have often communicate in items of primary, secondary and tertiary proton. The proton in methyl (CH₃−R) are known as the primary protons; in R-CH₂−R' as the secondary proton and in R-CH(R')−R" as the tertiary proton.

Also 2-methylpropane is often known as tert-butane (tert = tertiary for short) and 1-methylbutane as sec-pentane (sec = secondary for short). We have also created a term iso− for the unit (CH₃)₂CH− for the same reason. (isos in Greek means equal; like two equal groups attached to the carbon). So sec-pentane can also be known as iso-pentane. Remember one name can only be for one particular molecule, but one particular molecule can have more than one name.

Isopropyl

Isobutyl

sec-Butyl

tert-Butyl

neopentyl

Tutorial 2.4

We have sec-pentane, can we have sec-hexane?    Answer

Tutorial 2.5

If you were a Chemical Process Engineer designing a plant to prepare methyl chloride from the chlorination of methane, what will be the basic concepts to consider? What would you do with the hydrogen chloride produced?    Answer