4.1.3 Alkenes

Alkenes are unsaturated hydrocarbons containing a C=C double bond. That double bond consists of one σ bond and one π bond.

The π bond comes from sideways overlap above and below the line between the carbon nuclei. If you tried to rotate one carbon around the C=C, you would destroy that overlap, so rotation is restricted. That is why geometric stereoisomerism can exist around a double bond.

When you look at either carbon in a C=C, there are three regions of electron density around it. Those regions arrange themselves trigonal planar with bond angles of about 120°, so the atoms directly attached to the two alkene carbons lie in the same plane.

Compare that with an alkane carbon: four bonding regions give tetrahedral 109.5°. I find that contrast a useful way to stop the two shapes getting mixed up.

Stereoisomers have the same structural formula but a different arrangement of atoms in space. For an alkene, restricted rotation around C=C can lock two different spatial arrangements in place if each carbon of the double bond has two different groups attached.

E/Z isomerism is the general naming system. Cis–trans is a special case you can use when there is a suitable identical group on each alkene carbon: same side is cis, opposite sides is trans.

For E/Z naming, rank the two groups on each alkene carbon using the Cahn–Ingold–Prelog rules. Start with the atom directly attached to C=C: the higher atomic number gets higher priority.

If those atoms are the same, move outward until you reach the first point of difference. Once you have the higher-priority group on each carbon, compare their positions: Z means the two higher-priority groups are on the same side; E means they are on opposite sides.

When you are deciding whether E/Z isomerism is possible, inspect each carbon of the C=C separately. If either carbon has two identical groups attached, there is only one arrangement and no E/Z pair.

If both carbons have two different groups, draw the two locked arrangements and then use CIP priorities to name them E or Z. Use cis/trans only when the structure really allows that simpler description.

The C=C contains a strong σ bond and a weaker π bond. The π electrons sit above and below the bond axis, where they are relatively exposed, so the π part is the one that reacts in addition reactions.

You should not say the whole double bond is simply “weak”. The σ bond remains between the carbons while the π bond breaks and its electrons are used to make new σ bonds.

When you do an alkene addition reaction, the C=C becomes C–C and each alkene carbon gains a new bond. Work across the double bond and attach the atoms supplied by the reagent.

For OCR Year 12, know the familiar routes: H₂/Ni gives an alkane; Br₂ at room temperature gives a dibromoalkane; HX gives a haloalkane; steam with H₃PO₄ catalyst gives an alcohol. With an unsymmetrical alkene, HX or steam can give more than one structural product.

• H₂, Ni → alkane
• Br₂, room temperature → dibromoalkane
• HX → haloalkane
• steam, H₃PO₄ catalyst → alcohol

An electrophile is an electron-pair acceptor. In an alkene reaction, the π bond is electron rich, so it can donate an electron pair to an electrophile.

When you look for the electrophilic atom, look for positive charge or δ+ character. The curly arrow then starts from the electron-rich C=C bond and points towards that electrophilic centre.

Start by identifying the electron-rich π bond and the electrophilic end of the reagent. Your first curly arrow must start at the C=C π bond because that is where the moving electron pair is. If a reagent bond breaks heterolytically, draw a second arrow from that bond to the atom taking the pair.

For HBr, the first step gives a carbocation and Br⁻; Br⁻ then donates a lone pair to the positively charged carbon. For Br₂, show the induced polarity in Br₂ rather than inventing a permanent dipole on the alkene.

If you add an unsymmetrical reagent such as HBr to an unsymmetrical alkene, two carbocation intermediates may be possible. The route through the more stable carbocation is favoured, so it gives the major product.

I would draw the two possible carbocations if the answer is not obvious. Compare the intermediate carbocations, not the final products: tertiary carbocations are generally more stable than secondary, which are more stable than primary.

When you draw an addition-polymer repeat unit, open the C=C bond of the monomer and use those two carbon atoms in the polymer backbone. Put the repeating section in brackets and continue the backbone bonds through the brackets.

To work backwards from a polymer, take a two-carbon backbone repeat and restore a C=C between those carbons. Keep every substituent attached to the same carbon it occupied in the repeat unit.

There is no single ideal route for polymer waste. Combustion can recover energy but produces CO₂ and, depending on the polymer and conditions, may produce harmful gases that need removing. Feedstock recycling breaks polymers into smaller useful molecules that can become chemical raw materials again.

In an evaluation, you can compare energy/resource recovery with emissions, separation requirements and the chemistry of the particular polymer. Avoid claiming that all plastics behave identically.

A biodegradable polymer can be broken down by biological action, while a photodegradable polymer is designed to break down when exposed to light. Both can reduce the persistence of plastic waste under the conditions where degradation actually happens.

Be careful with the environmental claim: breaking down more readily does not automatically remove every impact. You still need to think about where the material ends up, what products form and whether the conditions for degradation are present.