Alkenes, geometric isomers and addition

What you’ll learn

• Explain the double bond and E/Z isomerism.
• Describe electrophilic addition using electron-pair movement.
• Relate alkene monomers to addition-polymer repeat units.

Understand the idea

A carbon–carbon double bond contains a sigma bond and a pi bond. The pi bond restricts rotation. E/Z isomerism is possible only if each double-bond carbon has two different substituents. Rank the two substituents on each carbon by the priority rules; higher-priority groups together give Z and on opposite sides give E. For the first directly attached atoms, greater atomic number means higher priority.

An alkene's electron-rich pi bond attracts an electrophile, an electron-pair acceptor. In addition of hydrogen bromide, draw a curly arrow from the pi bond to the hydrogen and another from the H–Br bond to bromine. The resulting bromide donates a lone pair to the carbocation. For an unsymmetrical alkene, the pathway through the more stable carbocation generally gives the major product; tertiary carbocations are more stabilised by alkyl groups than secondary, then primary ones.

Bromine addition removes the alkene double bond and decolourises bromine solution. The pi electrons polarise the approaching bromine molecule: draw an arrow from the pi bond to the partially positive bromine and from the Br–Br bond to the other bromine. Bromide then donates a lone pair to an electron-deficient carbon in the intermediate, completing addition. Other additions include hydrogen to form an alkane and steam to form an alcohol; controlled mild oxidation can give a diol. Harsher oxidation can give different products.

Addition polymerisation opens the pi bond and joins many monomers without eliminating a small molecule. Show two backbone carbons in the repeat unit, retain their substituents and draw bonds through the brackets. Polymer durability helps products last but complicates disposal; reuse, recycling, degradable designs and controlled treatment of combustion gases address different parts of that problem.

Key formulae

$$\text{acyclic monoalkene: }\mathrm{C_nH_{2n}}$$

$$\mathrm{CH_2{=}CH_2+HBr\rightarrow CH_3CH_2Br}$$

$$\mathrm{CH_2{=}CH_2+Br_2\rightarrow CH_2BrCH_2Br}$$

$$n\mathrm{CH_2{=}CH_2}\rightarrow\left[\mathrm{-CH_2-CH_2-}\right]_n$$

Worked example

Predict the major constitutional product when propene undergoes electrophilic addition of hydrogen bromide, and explain the preference.

1. The pi bond donates an electron pair to hydrogen; the H–Br bond breaks heterolytically.
2. Addition of hydrogen to the terminal carbon leaves a secondary carbocation on the middle carbon, more stable than the alternative primary carbocation.
3. Bromide donates a lone pair to the positively charged middle carbon.

Answer: The major product is 2-bromopropane.