Hydrocarbons: Question 3

Worked solution

Part (a): Naming the mechanism

Hydrogen bromide adding across the carbon-carbon double bond of but-1-ene is an example of electrophilic addition.

Part (b): The mechanism and Markovnikov’s rule

The $\text{H}-\text{Br}$ bond is polarised ($\text{H}^{\delta+}-\text{Br}^{\delta-}$), so the hydrogen end behaves as an electrophile. The electron-rich $\pi$ bond of but-1-ene attacks this electrophilic hydrogen atom, and the $\text{H}-\text{Br}$ bond breaks heterolytically, releasing a bromide ion, $\text{Br}^-$, and leaving a positively charged carbon (a carbocation) at whichever carbon did not bond to the new hydrogen atom.

Two outcomes are possible, depending on which end of the double bond the incoming hydrogen atom bonds to:

• If hydrogen bonds to C1 (which already carries two hydrogens), the positive charge is left on C2, giving the secondary carbocation $\text{CH}_3\text{CH}_2\text{CH}^+\text{CH}_3$.
• If hydrogen bonds to C2 (which already carries only one hydrogen), the positive charge is left on C1, giving the primary carbocation $\text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_2^+$.

The secondary carbocation is more stable than the primary one, because the two alkyl groups attached to the positively charged carbon donate electron density towards it (an inductive, electron-releasing effect), spreading out and reducing the positive charge; the primary carbocation has only one such alkyl group nearby. Since the secondary carbocation forms more readily and is lower in energy, it is produced in much greater amount, and the bromide ion released in the first step then bonds to this carbocation to complete the addition:

$\text{CH}_3\text{CH}_2\text{CH}^+\text{CH}_3 + \text{Br}^- \rightarrow \text{CH}_3\text{CH}_2\text{CHBrCH}_3$

So the major product is 2-bromobutane, with only a small amount of the minor product, 1-bromobutane, formed via the less stable primary carbocation. This outcome (the hydrogen atom adding to the carbon that already bears more hydrogen atoms, giving the more stable carbocation and hence the major product) is exactly what Markovnikov’s rule predicts.

Part (c): The bromine water test for unsaturation

Shaking but-1-ene with bromine water causes the orange/red-brown colour of the bromine water to fade to colourless almost immediately, as bromine molecules add across the carbon-carbon double bond by the same type of electrophilic addition mechanism, forming 1,2-dibromobutane:

$\text{CH}_3\text{CH}_2\text{CH}=\text{CH}_2 + \text{Br}_2 \rightarrow \text{CH}_3\text{CH}_2\text{CHBrCH}_2\text{Br}$

A saturated compound (containing only C-C and C-H single bonds) has no reactive $\pi$ electrons available to attack bromine in this way, and so does not rapidly decolourise bromine water at room temperature in the absence of light. The rapid loss of colour therefore confirms that but-1-ene contains a reactive carbon-carbon double bond, i.e. that it is unsaturated.

Final answers

• (a) Electrophilic addition.
• (b) Major product is 2-bromobutane, via the more stable secondary carbocation $\text{CH}_3\text{CH}_2\text{CH}^+\text{CH}_3$ (Markovnikov’s rule); minor product is 1-bromobutane, via the less stable primary carbocation.
• (c) Bromine water is decolourised (orange/red-brown to colourless) as $\text{Br}_2$ adds across the C=C bond. Confirms but-1-ene is unsaturated.