IB Chemistry HLTopic 6 — Electron Pair SharingPaper 1 & 2Organic~10 min read
Addition to Unsymmetrical Alkenes
With ethene it does not matter which carbon the hydrogen lands on — both ends are identical. Make the two ends different and suddenly there are two possible products, and one of them wins.
📚 What you need to know
An unsymmetrical alkene has different groups on the two carbons of the C=C, such as propene.
The electrophile can add to either carbon, giving two possible carbocations.
Carbocation stability: tertiary > secondary > primary, because of the positive inductive effect.
The route through the more stable carbocation gives the major product.
Markovnikov’s rule: with HX, the halogen ends up on the more substituted carbon.
Put another way, the hydrogen joins the carbon that already has more hydrogens.
With two different halogens (such as ICl), the more electronegative halogen goes to the more substituted carbon.
Why there are suddenly two answers
Propene is CH3CH=CH2. One end of the double bond carries a methyl group and a hydrogen; the other end carries two hydrogens. They are not the same, so the electrophile has a choice.
Add H+ to the middle carbon and you get a primary carbocation. Add it to the end carbon and you get a secondary carbocation. Both are possible — but they are not equally likely.
Carbocation stability decides it
Alkyl groups are electron-donating. They push electron density towards whatever they are attached to, which is the inductive effect. Pushing electron density towards a positively charged carbon spreads the charge out and lowers its charge density.
More alkyl groups means more pushing, means a more stable ion, means a lower activation energy for the slow step. So the reaction mostly goes down that road.
Nothing forces the reaction down one path. The secondary carbocation simply forms more easily, so far more molecules take that route.
Markovnikov’s rule
Rather than reason through carbocation stability every time, you can use a shortcut. It comes to the same answer.
Markovnikov’s rule
In HX addition, the halogen bonds to the more substituted carbon
The more substituted carbon is the one with fewer hydrogens and more carbon groups attached. So the flip side of the rule is the version most people remember: the hydrogen goes to the carbon that already has more hydrogens.
The CH3 group on the left is not part of the double bond, so it takes no part in the decision. Only the two C=C carbons matter.
If you can only remember one version, remember the carbocation one. It always works, including in cases where the shortcut phrasing gets confusing — and it earns the explanation marks that the shortcut does not.
Two different halogens
Something like iodine monochloride, ICl, is polar because chlorine is more electronegative than iodine. That makes iodine the δ+ end and therefore the electrophile.
Iodine adds first, to whichever carbon gives the more stable carbocation. The chloride ion then attacks that positive carbon, which is the more substituted one.
Rule for mixed halogens
the more electronegative halogen ends up on the more substituted carbon
🧩 Working out the major product every time
Find the two carbons of the C=C. Ignore everything further along the chain.
Work out which part of the reagent is the electrophile (the δ+ end or the positive ion).
Try adding it to each carbon in turn and draw both possible carbocations.
Classify each one: primary, secondary or tertiary.
The more substituted carbocation wins. That route gives the major product.
Finish the mechanism by attacking that carbocation with the negative ion, then name the product.
Worked examples
WORKED EXAMPLE
Give the major product when but-1-ene reacts with HBr, and explain your reasoning in terms of the intermediate.
Write the alkene out
But-1-ene is CH3CH2CH=CH2. The C=C carbons are carbon 1 and carbon 2.
Try H+ on carbon 1
That leaves the + on carbon 2, which has two carbon groups on it: a secondary carbocation.
Try H+ on carbon 2
That leaves the + on carbon 1, with only one carbon group: a primary carbocation.
Pick the winner
The secondary ion has two alkyl groups pushing electron density in, so it is more stable and forms faster.
Br– then attacks carbon 2Major product: 2-bromobutane1-bromobutane is still made, just far less of it
WORKED EXAMPLE
2-methylpropene, (CH3)2C=CH2, reacts with HCl. Predict the major product and say why the preference is even stronger here than for propene.
Identify the two ends
One C=C carbon carries two methyl groups; the other carries two hydrogens.
Add H+ to the CH2 end
The positive charge lands on the carbon with two methyls, giving a tertiary carbocation.
The alternative
Adding H+ the other way gives a primary carbocation — barely stable enough to exist.
Why the gap is bigger
Tertiary versus primary is a wider stability difference than secondary versus primary, so the split is more one-sided.
Major product: 2-chloro-2-methylpropanethree alkyl groups pushing in beats one, comfortably
💡 Exam tip
If a question says explain, do not just quote Markovnikov. Talk about the carbocation stability.
Use the phrase positive inductive effect when describing what the alkyl groups do.
Draw both carbocations if the marks allow it. Comparing them is usually where the marks are.
Name the products properly with locants: 2-bromopropane, not “bromopropane”.
Say major and minor rather than “the product” — both are formed.
Check whether the alkene is actually unsymmetrical first. With ethene or but-2-ene there is only one answer.
⚠️ Common mix-up
Putting the halogen on the carbon with more hydrogens. That is backwards. The hydrogen goes there.
Counting hydrogens on the whole molecule. Only the two C=C carbons count.
Saying the minor product does not form. It does, just in smaller amounts.
Confusing “more substituted” with “bigger”. It means more carbon groups attached, fewer hydrogens.
Applying Markovnikov to a symmetrical alkene. There is nothing to decide.
Forgetting which end of ICl is δ+. Chlorine is more electronegative, so iodine is the electrophile.
Up next: Electrophilic Substitution in Benzene — the last page of this topic, and the one place where an electrophile attacks but nothing is added.
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