Hydrocarbons: Question 9

Worked solution

Step 1: How bromine becomes electrophilic

Although $\text{Br}_2$ is a non-polar molecule, as it approaches the electron-rich $\pi$ bond of propene, the nearby electron density repels the electrons within the $\text{Br}-\text{Br}$ bond, inducing a temporary dipole ($\text{Br}^{\delta+}-\text{Br}^{\delta-}$). The bromine atom nearer the double bond becomes electrophilic even though $\text{Br}_2$ has no permanent dipole.

Step 2: The addition mechanism

The $\pi$ electrons of propene attack this electrophilic bromine atom, forming a new C-Br bond and breaking the $\text{Br}-\text{Br}$ bond heterolytically, releasing a bromide ion, $\text{Br}^-$, and leaving a carbocation on the other original double-bond carbon:

$\text{CH}_3\text{CH}=\text{CH}_2 + \text{Br}_2 \rightarrow \text{CH}_3\overset{+}{\text{C}}\text{HCH}_2\text{Br} + \text{Br}^-$

The bromide ion released then bonds rapidly to the carbocation carbon, completing the addition.

Step 3: Why only one product is possible

Unlike the addition of $\text{HBr}$ or steam, where a hydrogen atom and a chemically different group (Br or OH) compete for each end of the double bond, here the same atom (Br) ends up bonded to both of the original alkene carbons regardless of which one first forms the C-Br bond and which becomes the carbocation. So whichever pathway is followed, the final product is always 1,2-dibromopropane, $\text{CH}_3\text{CHBrCH}_2\text{Br}$. There is no equivalent of Markovnikov’s “major/minor product” distinction for the addition of a single, non-polar diatomic halogen like $\text{Br}_2$.

Why the other options are wrong

• A confuses this electrophilic addition, which occurs readily at room temperature in the dark, with the free-radical substitution of alkanes, which does require UV light.
• C proposes a chemically impossible product (1,3-dibromopropane would require bonding to non-adjacent carbons) and wrongly imports Markovnikov-style major/minor reasoning where it does not apply.
• D is incorrect because the induced-dipole mechanism means $\text{Br}_2$ can act as an electrophile even without a permanent dipole; the bromine water is in fact rapidly decolourised.

Final answer

B. Bromine water is decolourised via electrophilic addition with an induced dipole in $\text{Br}_2$; only one product, 1,2-dibromopropane, is possible.