Topic 5

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Everything from this topic in one place, for the last read before an exam. If anything here is unfamiliar, go back to the full page.

Definitions to learn word for word

Alkene: a hydrocarbon with a C=C double bond. General formula CₙH₂ₙ for one C=C. Unsaturated (Alkenes).
σ bond: overlap of orbitals along the line between the nuclei. π bond: sideways overlap of p orbitals, above and below the plane of the atoms.
Geometric isomers: compounds with the same structural formula but a different arrangement of groups in space, because rotation around the C=C is restricted (Restricted rotation).
Electrophile: an electron-pair acceptor. Carbocation: an ion with a positively charged carbon atom (Electrophiles).
Addition polymerisation: many monomers, each with a C=C, join to make one long saturated polymer and no other product. Repeat unit: the smallest part of the chain that repeats (Addition polymerisation).
Biodegradable: broken down by microorganisms (Polymer disposal).

The two conditions for geometric isomers

A C=C (it cannot rotate).
Two different groups on each carbon of the C=C.

Name them by priority: higher atomic number of the atom attached to the C=C wins. Z = higher priorities on the same side; E = on opposite sides (E–Z naming). Alkene counts: C₄H₈ has 4 alkene isomers, C₅H₁₀ has 6 (Geometric isomers).

Reactions of alkenes

Reagent	Conditions	Product from ethene	Page
H₂	nickel catalyst, about 150 °C	ethane	5
Br₂ (or Cl₂)	room temperature	1,2-dibromoethane	5
HBr (or HCl)	room temperature	bromoethane	6
steam, H₂O(g)	phosphoric acid catalyst, about 300 °C, 60 atm, ⇌	ethanol	6
acidified KMnO₄	room temperature	ethane-1,2-diol	7

Tests for C=C: bromine water orange to colourless; acidified manganate(VII) purple to colourless. Alkanes give no change (page 7).

Addition has 100% atom economy: one product, all atoms used.

The mechanism: electrophilic addition

For ethene and Br₂ (page 9):

The C=C π electrons induce a dipole in Br–Br. Arrow 1: from the C=C to the Br^δ+. Arrow 2: from the Br–Br bond to the Br^δ−. Product: a carbocation (C⁺ with CH₂Br) and Br⁻.
Arrow 3: from a lone pair on Br⁻ to the C⁺. Product: 1,2-dibromoethane.

For ethene and HBr (page 10): H–Br is already polar (H^δ+), so no induced dipole is needed. Same three arrows, giving CH₃CH₂⁺, then bromoethane.

Every arrow starts at a bond or a lone pair and ends at an atom or between two atoms. Draw dipoles, charges and lone pairs on Br⁻.

Unsymmetrical alkenes: the major product

Primary: C⁺ with one alkyl group. Secondary: two. Tertiary: three. Stability: tertiary > secondary > primary, because alkyl groups release electrons and spread the charge (page 11).
The more stable carbocation forms more easily, so it gives the major product.
Propene + HBr: 2-bromopropane (major, secondary carbocation); 1-bromopropane (minor, primary carbocation). Propene + steam: propan-2-ol (major) (page 12).
2-methylpropene + HBr: tertiary carbocation, 2-bromo-2-methylpropane.

Polymers

Monomer	Polymer	Repeat unit
ethene	poly(ethene)	–CH₂–CH₂–
propene	poly(propene)	–CH₂–CH(CH₃)–
chloroethene	poly(chloroethene), PVC	–CH₂–CHCl–
tetrafluoroethene	poly(tetrafluoroethene)	–CF₂–CF₂–

C=C becomes C–C. Brackets with extension bonds crossing them, n outside. Repeat unit of a monomer with one C=C: two main-chain carbons, side groups attached. Backwards: find two neighbouring carbons, take away the extension bonds, put the C=C back (page 14). Chain length: $M_\text{r}$ (polymer) ÷ $M_\text{r}$ (repeat unit).

Disposal

Polymers are saturated, with strong, almost non-polar bonds, so are not biodegradable.
Biodegradable polymers (for example from starch) are broken down by microorganisms.
Incineration reduces landfill and gives heat, but PVC gives HCl and incomplete combustion gives CO. A scrubber with a base removes the acid gas: CaO + 2HCl → CaCl₂ + H₂O; CaCO₃ + 2HCl → CaCl₂ + H₂O + CO₂.

The ten traps

Treating but-1-ene as having geometric isomers: it has two H on one carbon of the C=C.
Using cis/trans when the four groups are all different: use E/Z.
Deciding priority from the mass or the size of the group, not the atomic number of the attached atom.
Drawing a curly arrow that starts at an atom instead of a bond or lone pair, or has no tip at the right place.
Forgetting the induced dipole in Br₂, or the partial charges on HBr.
Writing "Markovnikov" without naming the more stable carbocation and giving the reason.
Counting the total carbon atoms instead of the alkyl groups on C⁺.
Writing "water" for steam, or leaving out ⇌ for the hydration of ethene.
Writing "clear" for colourless in the bromine water test.
Keeping the C=C in a repeat unit, or putting a CH₃ in the main chain.

Mixed test

Try these with no notes, then check the answers.

Explain why but-2-ene has geometric isomers but but-1-ene does not.
Name the isomer of CHBr=CHCl in which Br and Cl are on opposite sides of the C=C.
State the reagents and conditions for converting propene into propane, and write the equation.
A student adds bromine water to a colourless liquid and the mixture stays orange. What can she conclude, and what can she not conclude?
Draw the mechanism for the reaction of ethene with hydrogen bromide, showing curly arrows, dipoles, charges and lone pairs.
Explain why 2-methylpropene reacts with HBr to give mainly 2-bromo-2-methylpropane.
4.20 g of propene reacts completely with hydrogen bromide to give 2-bromopropane. Calculate the maximum mass of 2-bromopropane. A student obtains 9.8 g. Calculate the percentage yield. ( $A_\text{r}$ : H 1.0, C 12.0, Br 79.9)
Draw the repeat unit of poly(propene) and give the monomer of a polymer with the repeat unit –CH₂–CCl₂–.
A sample of poly(propene) has an average relative molecular mass of 105 000. Calculate the average number of monomer units in each chain.
Explain why burning poly(chloroethene) is a problem, and how the problem is limited.
Ethene reacts with steam to make ethanol. Give the conditions and explain the sign used in the equation.

Answers cover these until you have tried the questions

Geometric isomers need a C=C and two different groups on each carbon of the C=C. But-2-ene, CH₃CH=CHCH₃, has H and CH₃ on each carbon, so the two CH₃ groups can be on the same side or opposite sides, and rotation is restricted. But-1-ene, CH₂=CHCH₂CH₃, has two H atoms on one carbon, so swapping them gives the same molecule.
On each carbon the higher-priority group is the one whose attached atom has the higher atomic number: Br (35) over H (1), and Cl (17) over H. Br and Cl are on opposite sides, so this is E-1-bromo-2-chloroethene, the E isomer.
Hydrogen with a nickel catalyst at about 150 °C. CH₃CH=CH₂ + H₂ → CH₃CH₂CH₃.
The liquid does not contain a C=C (it could be an alkane). She cannot conclude which compound it is, only that there is no C=C: other compounds also leave bromine water orange.
H^δ+–Br^δ− above CH₂=CH₂. Arrow 1: from the C=C to H^δ+. Arrow 2: from the H–Br bond to Br. Intermediate: CH₃CH₂⁺ and Br⁻ (with a lone pair). Arrow 3: from a lone pair on Br⁻ to the C⁺. Product: CH₃CH₂Br, bromoethane.
H bonds to the CH₂ carbon, leaving C⁺ on the other carbon with three alkyl groups: a tertiary carbocation, (CH₃)₃C⁺. The alternative is a primary carbocation. Alkyl groups release electrons and spread the charge, so the tertiary carbocation is more stable, forms more easily and leads to the major product, 2-bromo-2-methylpropane.
$M_\text{r}$ (propene) = 3 × 12.0 + 6 × 1.0 = 42.0. n = 4.20 ÷ 42.0 = 0.100 mol. CH₃CH=CH₂ + HBr → CH₃CHBrCH₃ (1 : 1). $M_\text{r}$ (C₃H₇Br) = 36.0 + 7.0 + 79.9 = 122.9. Maximum mass = 0.100 × 122.9 = 12.3 g. Percentage yield = 9.8 ÷ 12.29 × 100 = 79.7% (about 80%).
Poly(propene): –CH₂–CH(CH₃)–, in brackets with extension bonds and n outside. For –CH₂–CCl₂–, put the C=C back: CH₂=CCl₂, 1,1-dichloroethene.
$M_\text{r}$ (C₃H₆) = 42.0. n = 105 000 ÷ 42.0 = 2500.
Poly(chloroethene) contains chlorine, so burning it releases toxic hydrogen chloride. The waste gases are passed through a scrubber containing a base, such as calcium oxide, which neutralises the HCl: CaO + 2HCl → CaCl₂ + H₂O.
Steam, phosphoric acid catalyst, about 300 °C and 60 atm. CH₂=CH₂(g) + H₂O(g) ⇌ CH₃CH₂OH(g). The ⇌ sign shows the reaction is reversible: ethanol can break down again into ethene and steam.