Major and minor products

When two products are possible

• An asymmetric alkene is one where the two carbons of the C=C are not the same.
  • Propene, $CH_3CH{=}CH_2$ — one C=C carbon carries two H atoms, the other carries one H and a $CH_3$. Asymmetric.
  • Ethene, $CH_2{=}CH_2$ — both carbons carry two H atoms. Symmetric.
• When you add an unsymmetrical reagent ($HX$, or $H_2O/H^+$) to an asymmetric alkene, the reagent can attach two ways round, so two different products form.
• One forms in the larger amount — the major product. The other forms in the smaller amount — the minor product.
• Both are made. A question asking about an asymmetric alkene expects both to be named.

Markovnikov's rule

When $HX$ or $H_2O$ adds to an asymmetric alkene, the hydrogen adds to the double-bond carbon that already carries more hydrogen atoms. The X or the –OH goes to the other carbon.

• That route gives the major product.
• The other way round gives the minor product.
• A quick way to remember it: "the rich get richer" — the carbon with more hydrogens gains another one.

The method

• Check the alkene is asymmetric. If both C=C carbons are identical, there is only one product and none of this applies.
• Check the reagent is unsymmetrical. $HBr$ and $H_2O$ are; $H_2$ and $Br_2$ are not — $Br_2$ gives one product either way round.
• Count the hydrogens already on each of the two double-bond carbons.
• Put the new H on the carbon with more hydrogens.
• Put the X or –OH on the other carbon → this is the major product.
• Swap them over → this is the minor product.
• Name both and label which is which.

Worked through: propene + HBr

• The C=C is between carbon 1 ($CH_2$, 2 H) and carbon 2 ($CH$, 1 H).
• The new H goes to carbon 1 (more hydrogens), so the Br goes to carbon 2.
• Major: $CH_3CHBrCH_3$ = 2-bromopropane.
• Minor: $CH_3CH_2CH_2Br$ = 1-bromopropane.

The same rule for adding water

• Hydration of an asymmetric alkene works identically — the H goes to the more-hydrogenated carbon and the –OH to the other.
• Propene + $H_2O/H^+$:
  • Major: propan-2-ol ($CH_3CH(OH)CH_3$).
  • Minor: propan-1-ol ($CH_3CH_2CH_2OH$).

When the rule does not apply

• Symmetric alkene (ethene, but-2-ene) — both carbons are the same, so there is only one product.
• Symmetric reagent ($H_2$, $Cl_2$, $Br_2$) — it does not matter which way round it adds, so again one product.
• Only when both the alkene and the reagent are unsymmetrical do you get a major and a minor.