Free Radical Substitution IB Chemistry: 6-Mark Mechanism

1. Br₂ → 2Br•, with UV light above the arrow [1].
2. Br• + CH₄ → •CH₃ + HBr [1]; •CH₃ + Br₂ → CH₃Br + Br• [1].
3. Br• + Br• → Br₂ [1]; •CH₃ + Br• → CH₃Br [1]; •CH₃ + •CH₃ → C₂H₆ [1].

Try them at the end of the video first. Questions are original, in the style of the exam; marks are estimates.

A chlorine radical hits methane. Hydrogen radical, or methyl radical?

One of them never forms. In seven minutes you can answer all six marks.

Mix methane and chlorine, and shine UV light on them. One hydrogen atom of methane is replaced by a chlorine atom. That is a substitution. The products are chloromethane and hydrogen chloride.

But this does not happen in one step. Step one is initiation. UV light breaks a chlorine molecule into two chlorine radicals. Count the radicals: zero before, two after.

Careful here. The second product is hydrogen chloride, not hydrogen gas. The hydrogen atom that leaves methane joins a chlorine atom.

Small task. Ethane reacts with bromine in UV light. One product is bromoethane. What is the other product?

Hydrogen bromide.

Step two is propagation. It has two parts. First, a chlorine radical hits a methane molecule and takes one hydrogen atom. This makes hydrogen chloride and a methyl radical.

Second, the methyl radical hits a chlorine molecule and takes one chlorine atom. This makes chloromethane and a new chlorine radical. That chlorine radical does the first part again. So the two parts repeat, again and again. This is a chain reaction.

Count the radicals in each part: one before, one after.

Careful here. A hydrogen radical never forms in this reaction. The chlorine radical takes the hydrogen atom away. It does not take the place of the hydrogen atom.

Small task. Which radical is made again at the end of propagation?

The chlorine radical. That is why the chain continues.

Step three is termination. Two radicals meet. Their two unpaired electrons make one new covalent bond. Count the radicals: two before, zero after. So the chain stops.

There are two kinds of radical here, so there are three ways to do it. Two chlorine radicals make a chlorine molecule. A methyl radical and a chlorine radical make chloromethane. Two methyl radicals make ethane.

Careful here. Ethane has two carbon atoms, but we started with methane. So a little ethane in the products is evidence for methyl radicals.

Small task. A step has two radicals on the left and no radicals on the right. Which step is it?

Termination.

Now the same steps with ethane and bromine. Three times.

The two propagation steps. First, a bromine radical takes a hydrogen atom from ethane. This makes hydrogen bromide and an ethyl radical. Second, the ethyl radical takes a bromine atom from a bromine molecule. This makes bromoethane and a new bromine radical.

A termination step. I write an ethyl radical plus a bromine radical. You finish. What is the product?

Bromoethane. The two unpaired electrons make the bond between carbon and bromine.

Now alone. Write the termination step with two ethyl radicals, and name the product. Pause the video.

Two ethyl radicals join. The product has four carbon atoms and ten hydrogen atoms. It is butane.

Now you are the examiner. A two mark question and three answers. Decide the marks.

Student A made a hydrogen radical. That never happens here. Zero. Student B wrote a correct first step. One mark. But the second line joins two radicals, so it is termination, not propagation. Student C has one radical on each side of both equations. Two marks.

So check every propagation step: one radical in, one radical out.

Back to the question from the start. Pause the video and try every part.

Part a. Methane plus chlorine gives chloromethane plus hydrogen chloride. One mark.

Part b. A chlorine radical plus methane gives a methyl radical plus hydrogen chloride. One mark. A methyl radical plus a chlorine molecule gives chloromethane plus a chlorine radical. One mark.

Part c. Two methyl radicals give ethane. One mark.

Part d. The second propagation step makes a new chlorine radical. One mark. This radical reacts with another methane molecule, so the two steps repeat many times. One mark.

Six out of six.

One screen. Initiation: UV light, and zero radicals become two. Propagation: two steps, one radical in and one radical out, and the chlorine radical comes back. Termination: two radicals join, and the chain stops. Overall: an alkane and a halogen give a halogenoalkane and a hydrogen halide.

Three questions are on screen. Try them today, and again in three days. The answers are in the description.

Next Wednesday, part three: why alkanes are so unreactive, and why this reaction gives a mixture of products. See you there!