Catenation and the diversity of organic molecules: Exercises

Before we begin, three numbers you will reuse — memorise them:

Level 1 — Recognition

Goal: just name and spot the idea. No calculation.

Q1.1 Define catenation in one sentence, and give one example each of the three shapes it produces.

Q1.2 Which of these elements catenates best: carbon, silicon, or germanium? State the single number that decides it.

Q1.3 True or false: "Catenation means only long straight chains of carbon."

Recall Solution — L1

Q1.1 Catenation is the ability of an element to bond to more atoms of the same element, building chains, branches, and rings.

Straight chain: $C - C - C - C$ (n-butane)
Branched: a carbon hanging off the main chain (isobutane)
Ring: benzene / cyclohexane

Q1.2 Carbon catenates best. The deciding number is the $C - C$ bond energy $\approx 348$ kJ/mol — the largest on the ladder, so its links are hardest to break and its chains can grow longest.

Q1.3 False. Catenation covers any self-bonding pattern: straight and branched and ring. Look at the figure below — all three are catenation.

Figure — Catenation and the diversity of organic molecules

Level 2 — Application

Goal: apply one stated rule to a fresh case.

Q2.1 Using only the bond-energy ladder, predict which will survive as a long chain at room temperature: a 20-carbon alkane $C_{20} H_{42}$ or a 20-silicon silane $S i_{20} H_{42}$ .

Q2.2 Carbon has 4 valence electrons and makes 4 bonds. In a middle carbon of a chain ( $- C H_{2} -$ ), how many bonds go to neighbouring carbons, and how many are "spare" for H or other groups?

Q2.3 The molecular formula of an alkane is $C_{n} H_{2 n + 2}$ . Compute the number of hydrogens for $n = 4, 5, 6$ .

Recall Solution — L2

Q2.1 The alkane. Each $S i - S i$ link (222) is far weaker than $C - C$ (348). A 20-atom chain has 19 links; if each is weak, the whole chain snaps easily. So $C_{20} H_{42}$ is a stable everyday chemical (eicosane) while $S i_{20} H_{42}$ does not survive.

Q2.2 A middle carbon uses 2 bonds to its two chain neighbours, leaving 2 spare bonds — here filled by 2 hydrogens ( $C H_{2}$ ). Those spares are why chains can branch or carry functional groups: carbon is never "used up."

Q2.3 Plug into $2 n + 2$ :

$n = 4 : 2 (4) + 2 = 10 \Rightarrow C_{4} H_{10}$
$n = 5 : 2 (5) + 2 = 12 \Rightarrow C_{5} H_{12}$
$n = 6 : 2 (6) + 2 = 14 \Rightarrow C_{6} H_{14}$

Level 3 — Analysis

Goal: break a problem into cases and count carefully.

Q3.1 Draw and count all structural isomers of $C_{5} H_{12}$ .

Q3.2 Draw and count all structural isomers of $C_{4} H_{10}$ .

Q3.3 From the table below, by roughly what factor does the isomer count jump when $n$ goes from 10 to 20?

$n$ isomers

4 2

5 3

6 5

10 75

20 366319

$n$	isomers
4	2
5	3
6	5
10	75
20	366319

Recall Solution — L3

Q3.1 — $C_{5} H_{12}$ has $3$ isomers. Build them by shrinking the main chain and moving the freed carbon (see figure):

Straight chain (n-pentane): all 5 carbons in a row.
One branch (isopentane / 2-methylbutane): a 4-carbon backbone with 1 carbon attached to C-2.
Most branched (neopentane / 2,2-dimethylpropane): a central carbon carrying 4 methyls.

There is no fourth: a branch on C-3 of a 4-carbon chain is just isopentane relabelled, and you cannot branch on an end carbon (that only lengthens the chain).

Q3.2 — $C_{4} H_{10}$ has $2$ isomers: n-butane (straight) and isobutane (a central carbon with 3 methyls + 1 H, i.e. 2-methylpropane).

Q3.3 $\frac{366319}{75} \approx 4884$ . The count multiplies by roughly five thousand when $n$ merely doubles from 10 to 20 — this super-fast combinatorial blow-up is the mathematical face of "organic diversity."

Level 4 — Synthesis

Goal: combine two or more separate reasons into one argument.

Q4.1 Give the complete reason (all three factors) why carbon forms stable long chains but silicon does not. State each factor and why it matters.

Q4.2 Silanes ( $S i_{n} H_{2 n + 2}$ ) burst into flame in air and hydrolyse in water; alkanes just sit there. Explain using orbital reasoning — do not use bond strength for this one.

Q4.3 Why is benzene (a 6-carbon ring with alternating double bonds) common and stable, while a comparable single-bonded nitrogen ring is not? Tie your answer to why catenation includes rings.

Recall Solution — L4

Q4.1 Three independent factors, all favouring carbon:

Bond strength — $C - C$ (348) $>$ $S i - S i$ (222). Strong links survive long chains.
Tetravalency — carbon makes 4 bonds; after 2 to chain neighbours, 2 remain to add H, branch, or carry groups. (But note: Si is also tetravalent, so this alone can't explain the gap.)
Kinetic inertness — carbon has no valence $d$ -orbitals and no lone pairs, so there is no low-energy pathway for water or air to attack. Silicon has empty $3 d$ orbitals that let nucleophiles (like $O H^{-}$ ) attack, so its chains hydrolyse. The decisive pair is strength + inertness; valency is necessary but not sufficient.

Q4.2 Silicon's valence shell has empty $3 d$ orbitals. These accept an incoming lone pair from a nucleophile ( $O H^{-}$ , or $O$ from air), opening a low-energy attack route → oxidation and hydrolysis. Carbon's valence shell (2s, 2p only) has no accessible $d$ -orbitals, so no such route exists — alkanes are kinetically inert. This is a reactivity argument, independent of how strong the bond is.

Q4.3 Carbon is small, so it forms strong $σ$ bonds and strong $π$ bonds (good sideways $2 p$ – $2 p$ overlap). A ring of carbons with alternating double bonds (benzene) is therefore highly stable. Catenation is not only chains — it explicitly includes rings, which is why aromatic and cyclic compounds hugely enlarge the structure count.

Level 5 — Mastery

Goal: push the reasoning to limits, degenerate cases, and prediction.

Q5.1 Rank $C$ , $S i$ , $G e$ for expected maximum chain length and justify using both the bond-energy trend and the atomic-size trend as you go down the group.

Q5.2 Degenerate case. What is the "chain" for $n = 1$ ( $C H_{4}$ ) and $n = 2$ ( $C_{2} H_{6}$ )? How many $C - C$ links does each have, and what does this say about the smallest molecule that can display catenation?

Q5.3 Prediction. If a hypothetical element X had a strong X–X bond (like carbon) but empty low-lying $d$ -orbitals (like silicon), would you expect its long chains to be common in nature? Reason through both properties.

Q5.4 Using the CIMB mnemonic, list the four diversity engines and state which single one is responsible for the isomer explosion you computed in L3.

Recall Solution — L5

Q5.1 Chain length: $C > S i > G e$ . Going down group 14, atomic radius increases ( $C \approx 77$ pm, $S i \approx 117$ pm, Ge larger still). Bigger atoms → longer bonds → poorer orbital overlap → weaker links (348 → 222 → 188 kJ/mol). Weaker links snap sooner, so the maximum stable chain shrinks as you descend. Both trends (size up, strength down) point the same way.

Q5.2 For $n = 1$ ( $C H_{4}$ ): 0 $C - C$ links — a lone carbon can't catenate with itself. For $n = 2$ ( $C_{2} H_{6}$ ): 1 $C - C$ link — the first genuine self-bond. So the smallest molecule that actually shows catenation is ethane ( $C_{2} H_{6}$ ); methane is the degenerate "no-chain" case. (Number of links in a straight chain of $n$ carbons $= n - 1$ .)

Q5.3 No, not common. Strong X–X bonds satisfy the thermodynamic requirement (links hard to break), but the empty low-lying $d$ -orbitals give a kinetic attack pathway — nucleophiles and air would react with the chains and destroy them over time. You need both strength and inertness. Carbon is special precisely because it has both; X would fail on inertness.

Q5.4 CIMB = Catenation, Isomerism, Multiple bonding, Bonding to many elements. The isomer explosion of L3 is driven by Isomerism (same formula, different connectivity) working on top of catenation.

Recall One-line self-test before you leave

Which two properties (not one) make carbon the champion of chains? ::: Strong $C - C$ bonds and kinetic inertness (no valence $d$ -orbitals). Smallest alkane that shows a $C - C$ bond? ::: Ethane, $C_{2} H_{6}$ . Isomer count of $C_{5} H_{12}$ ? ::: $3$ .