Investigate the structural formulae, properties and reactions of alkanes, alkenes and alkynes, including combustion and addition reactions of alkenes

What this dot point is asking

NESA wants you to distinguish alkanes, alkenes and alkynes by structure, predict and write equations for their characteristic reactions (combustion, addition, substitution), and explain why a $C=C$ bond makes alkenes much more reactive than alkanes. The dot point also covers the trends in physical properties down each series and the test that distinguishes saturated from unsaturated hydrocarbons.

The answer

Structural comparison

A double bond is one $\sigma$ plus one $\pi$ bond; a triple bond is one $\sigma$ plus two $\pi$ bonds. The $\pi$ electrons are loosely held and are the site of attack in addition reactions.

Physical properties

Across all three series, increasing chain length raises both melting point and boiling point because the dispersion forces between molecules grow with molecular size. C1 to C4 hydrocarbons are gases at room temperature, C5 to about C16 are liquids, and longer chains are waxy solids. All hydrocarbons are non-polar, immiscible with water, and float on water because their density is below 1 g/mL.

Comparing series at the same carbon number: an alkane and the corresponding alkene or alkyne have very similar boiling points, because the molecules differ only in two or four hydrogens. The double bond does not introduce significant polarity.

Combustion (all hydrocarbons)

Complete combustion (excess $O_2$) gives carbon dioxide and water. The general equation for any $C_xH_y$:

Combustion is highly exothermic and is the basis for using hydrocarbons as fuels. For methane:

Incomplete combustion (limited $O_2$) gives carbon monoxide $CO$ or soot $C$ plus water. Alkenes and alkynes burn with a sootier flame than alkanes because they have a higher carbon to hydrogen ratio, so less oxygen reaches the inner part of the flame.

Substitution reactions of alkanes

Alkanes are unreactive towards most reagents at room temperature. With halogens (chlorine or bromine) in UV light, they undergo free radical substitution:

The mechanism has three steps: initiation ($Cl_2 \rightarrow 2Cl \cdot$ under UV), propagation ($Cl \cdot + CH_4 \rightarrow HCl + CH_3 \cdot$, then $CH_3 \cdot + Cl_2 \rightarrow CH_3Cl + Cl \cdot$), and termination (radicals combine).

Addition reactions of alkenes

Addition reactions break the weaker $\pi$ bond and add two new groups across the former double bond, leaving a saturated product.

1. Hydrogenation (H₂, Ni catalyst, heat). Alkene plus hydrogen gives the corresponding alkane:

2. Halogenation ($Br_2$ or $Cl_2$, room temperature, no catalyst). Alkene decolourises bromine water (orange-brown to clear) instantly. This is the standard test for unsaturation:

3. Hydrohalogenation (HX, e.g. HCl or HBr). Alkene plus a hydrogen halide gives a haloalkane. Asymmetric alkenes follow Markovnikov's rule: H adds to the carbon already carrying more hydrogens, X adds to the more substituted carbon.

4. Hydration ($H_2O$, dilute $H_2SO_4$ catalyst, heat). Alkene plus water gives an alcohol. Markovnikov also applies.

This is industrially how ethanol is made from ethene.

Reactions of alkynes

Alkynes undergo combustion and addition like alkenes, but each $\pi$ bond can be added across in turn. So ethyne plus excess bromine gives 1,1,2,2-tetrabromoethane:

Ethyne $C_2H_2$ burns at very high temperatures with oxygen, which is why oxyacetylene torches are used for welding and metal cutting.

The bromine water test

To distinguish a saturated hydrocarbon (alkane) from an unsaturated one (alkene or alkyne), add a few drops of bromine water and shake.

• Alkene or alkyne: orange-brown colour rapidly disappears (clear/colourless) due to addition.
• Alkane: colour persists, unless exposed to UV light in which case it fades slowly with HBr fumes (substitution).

A second confirmatory test is acidified $KMnO_4$: alkenes and alkynes decolourise purple permanganate at room temperature; alkanes do not react.

Examples in context

Example 1. Cracking at the Kurnell refinery (legacy operations). Until its 2014 conversion to an import terminal, Caltex's Kurnell refinery cracked long-chain alkanes from Bass Strait crude into shorter alkenes and alkanes for the petrol pool. Steam cracking ethane gave ethene plus hydrogen: $C_2H_6 \rightarrow C_2H_4 + H_2$. The ethene was the feedstock for the adjacent ethanol and polyethene units. Combustion of the residual fuel gas in the cracker furnace followed the standard alkane equation $C_xH_y + (x + y/4)O_2 \rightarrow xCO_2 + (y/2)H_2O$ in excess oxygen. The HSC framework for combustion and addition reactions is the same chemistry the plant ran at industrial scale.

Example 2. Bromine water test in NSW HSC depth study. A school lab issued unlabelled samples of hexane and hex-1-ene asks students to identify them. Adding bromine water (orange-brown) to hex-1-ene gives instant decolourisation as the bromine adds across the double bond: $CH_2=CH(CH_2)_3CH_3 + Br_2 \rightarrow CH_2BrCHBr(CH_2)_3CH_3$. Hexane does not decolourise bromine water in the absence of UV. The contrast is the canonical HSC test for unsaturation. The same bromine number method is used at Hunter Valley wineries to measure unsaturated lipids in grape-seed oil and at Manildra to check ethanol purity for fuel-grade specifications.

Try this

Q1. State the general molecular formula for an alkane, an alkene and an alkyne, and give one observation that distinguishes an alkene from an alkane chemically. [3 marks]

• Cue. Alkane $C_nH_{2n+2}$, alkene $C_nH_{2n}$, alkyne $C_nH_{2n-2}$; alkene decolourises bromine water instantly without UV.

Q2. Write a balanced equation for the complete combustion of octane ($C_8H_{18}$) and calculate the volume of $CO_2$ at 25 degrees C and 100 kPa produced from 5.00 g of octane (molar volume 24.79 L mol$^{-1}$). [3 marks]

• Cue. $C_8H_{18} + 12.5 O_2 \rightarrow 8 CO_2 + 9 H_2O$; $n(\text{octane}) = 5.00 / 114.23 = 0.0438$; $n(CO_2) = 0.350$; $V = 0.350 \times 24.79 = 8.69$ L.

Q3. But-1-ene reacts with HBr to give a major and minor product. (a) State Markovnikov's rule. (b) Identify the major product. (c) Explain why no second product forms with the alkane equivalent unless UV is supplied. [1+2+2 marks]

• Cue. (a) H adds to the carbon with more H atoms. (b) 2-bromobutane (major). (c) Alkane requires free-radical substitution initiated by UV homolysis of $Br_2$.