Chemistry • Year 12 • Module 7 • Lesson 4
Hydrocarbon Structure & Bonding
Apply bond geometry, hybridisation, and bond-data trends to real scenarios, a data table, and a graph-based stimulus.
1. Interpret bond-length and bond-energy data
The table below shows average bond-length and bond-energy data for carbon–carbon bonds of different orders. Use it to answer the questions that follow. 7 marks
| Bond type | Bond order | Average bond length (pm) | Average bond energy (kJ mol−1) | Hybridisation of each C |
|---|---|---|---|---|
| C–C (single) | 1 | 154 | 347 | sp³ |
| C=C (double) | 2 | 134 | 614 | sp² |
| C≡C (triple) | 3 | 120 | 839 | sp |
Data adapted from: Atkins & Jones, Chemical Principles, 6th edn (2012), Table 7B.1.
1.1 Describe the relationship between bond order and bond length using data from the table. 2 marks
1.2 Using the data, explain why a C=C double bond is not simply twice as strong as a C–C single bond. What does this tell us about the relative strengths of sigma and pi bonds? 3 marks
1.3 Predict the hybridisation state of the carbon atoms in propyne (CH3–C≡C–H) at the triple bond, and state the expected bond angle at those carbons. 2 marks
2. Graph interpretation — bond length vs bond order
The graph below plots average C–C bond length against bond order for the three bond types in the table above. 6 marks
Figure 2.1. Average C–C bond length (pm) as a function of bond order. Values from Atkins & Jones (2012).
2.1 Describe the trend shown in the graph. 1 mark
2.2 Explain, at the level of electron orbital overlap, why increasing bond order decreases bond length. 3 marks
2.3 Using the graph, estimate the expected bond length (in pm) for a C–C bond in a benzene ring, where the bond order is approximately 1.5. Justify your estimate. 2 marks
3. Cause-and-effect chain — why cis-trans isomers exist in alkenes but not alkanes
Complete the cause-and-effect chain by filling in the missing effect boxes. 5 marks
| CAUSE: An alkene contains a C=C double bond, made of one σ and one π bond. | → | EFFECT 1: _______________________________________________ |
| CAUSE: The pi (π) bond requires parallel p-orbital alignment across both carbons. | → | EFFECT 2: _______________________________________________ |
| CAUSE: Rotation around C=C is restricted. | → | EFFECT 3: _______________________________________________ |
| CAUSE: For cis-trans isomerism to exist, each double-bond carbon must bear two different substituents. | → | EFFECT 4: _______________________________________________ |
| Overall outcome (so…): | ||
4. Apply to a new scenario — margarine production
During the industrial production of margarine, unsaturated vegetable oils (containing many C=C double bonds) are partially hydrogenated. The process adds H2 across C=C double bonds, converting some double bonds to single bonds. A by-product of incomplete hydrogenation is the formation of trans fatty acids — fats in which naturally occurring cis C=C bonds are converted to the trans configuration. 6 marks
4.1 Using your knowledge of hybridisation, explain how the hydrogenation of a C=C double bond changes the hybridisation state of the carbon atoms involved. 2 marks
4.2 A cis C=C bond has substituents on the same side of the double bond; a trans bond has them on opposite sides. Using lesson content, explain why the two forms do not interconvert at room temperature, meaning the trans fatty acid product is a permanent structural change. 3 marks
4.3 Predict-and-justify: if the trans fatty acid produced during hydrogenation were fully hydrogenated (all C=C bonds converted to C–C), would cis-trans isomerism still be possible in the product? Justify your answer in one sentence. 1 mark
Q1.1 — Bond order vs bond length trend
As bond order increases from 1 (154 pm) to 2 (134 pm) to 3 (120 pm), average bond length decreases. The C≡C triple bond (120 pm) is 34 pm shorter than the C–C single bond (154 pm). [1 for correctly describing the inverse relationship; 1 for quoting at least one data point to support it.]
Q1.2 — C=C strength vs twice C–C
A C=C double bond (614 kJ mol−1) is less than twice as strong as a C–C single bond (2 × 347 = 694 kJ mol−1) [1]. This indicates that the pi (π) bond component of the double bond (614 − 347 = 267 kJ mol−1) is significantly weaker than the sigma (σ) bond component [1]. Side-on overlap of p orbitals (pi bond) is less efficient than head-on overlap (sigma bond), resulting in a weaker pi bond [1].
Q1.3 — Propyne hybridisation
The two carbons in the C≡C triple bond of propyne are sp hybridised [1]. The expected bond angle at these sp carbons is 180° (linear geometry) [1].
Q2.1 — Graph trend
As bond order increases, C–C bond length decreases in a non-linear fashion (the decrease is greater from order 1 to 2 than from 2 to 3).
Q2.2 — Why higher bond order → shorter bond
Each additional bond between two carbon nuclei introduces more electron density into the internuclear region [1]. Greater electron density between the nuclei pulls both nuclei closer together, reducing the internuclear distance (bond length) [1]. A triple bond has one sigma + two pi bonds, placing more shared electron density between the atoms than a double bond (1 sigma + 1 pi) or a single bond (1 sigma only), hence each successive increase in bond order shortens the bond [1].
Q2.3 — Benzene bond length estimate
For bond order 1.5 (midway between 1 and 2), interpolating linearly: 154 − (1.5 − 1)/(2 − 1) × (154 − 134) = 154 − 0.5 × 20 = 144 pm (accept 140–148 pm) [1]. This is justified because the bond order of 1.5 sits midway between single and double, so the bond length should be approximately midway between 154 pm and 134 pm [1]. (The experimental benzene C–C bond length is 140 pm.)
Q3 — Cause-and-effect chain
Effect 1: The double bond contains a pi bond formed by side-on overlap of unhybridised p orbitals on adjacent carbons. Effect 2: Rotation about the C=C bond would break the pi bond by destroying this parallel p-orbital alignment, which requires significant energy input. Effect 3: The groups attached to the double-bond carbons are locked in fixed spatial arrangements relative to each other — they cannot swap sides. Effect 4: If both conditions are met (each C has two different groups, and rotation is restricted), two distinct stable compounds (cis and trans) can exist. Overall outcome: Geometric (cis-trans) isomers exist as separate, isolable compounds in alkenes but are impossible in alkanes, which have only sigma bonds and freely rotating C–C bonds.
Q4.1 — Hybridisation change on hydrogenation
Each carbon at the C=C double bond is initially sp2 hybridised. When H2 adds across the double bond, the pi bond breaks and two new C–H sigma bonds form. Each carbon is now bonded to four atoms by single bonds only, so it becomes sp3 hybridised with tetrahedral geometry [1 for sp2 → sp3; 1 for linking to bond change].
Q4.2 — Why cis and trans do not interconvert
Converting a cis C=C to a trans C=C would require rotating about the double bond [1]. Rotation about C=C is prevented by the pi bond, which is formed by side-on overlap of p orbitals; rotation would break this overlap and destroy the pi bond, requiring substantial energy input [1]. At room temperature, thermal energy is insufficient to break the pi bond (∼267 kJ mol−1), so once the trans configuration is formed during hydrogenation it is permanent until a chemical reaction (e.g. further hydrogenation or reaction with heat/catalyst) occurs [1].
Q4.3 — Predict: no isomers after full hydrogenation
No, cis-trans isomerism would not be possible, because full hydrogenation converts all C=C double bonds to C–C single bonds, removing all pi bonds and all restricted rotation — without a double bond there is no structural basis for geometric isomerism. [1]