Why does carbon make so many things?

Here are the numbers, and the two colours are not one chemistry. The brown comes from a polymer called melanoidin. A polymer is a long molecule built from repeating units. Melanoidin is the polymer made when sugars and proteins react together in heat. Bao and colleagues, 2025, made it at 125 °C. That is bread-crust temperature. By mass it was 38 to 46 per cent carbon and 46 to 57 per cent oxygen. Those ranges come from several samples, so their top ends do not add up to one sample. That is an oxygen-rich solid, and it is not char. Char made by hard heating is 65 to 84 per cent carbon, measured by Chatterjee and colleagues, 2020. Those char figures come from four crop materials heated to 500 to 800 °C, not from bread. Take the middle of each range, and melanoidin sits at 42.0 per cent carbon against char's 74.5. Divide 74.5 by 42.0, and the char holds 1.77 times the carbon share, by mass. Where bread itself crosses from brown to black has not been published, and this page does not guess it. The crossover sits somewhere above the browning range, and that is all the sources allow.

So why is ordinary toast not char? Bread is mostly starch and protein. Starch is about 91 per cent volatile matter. Volatile matter is the part that leaves as gas on heating. That figure is from a 2020 study of six biomass components. The study is in Energy Conversion and Management, and only its abstract could be read. So starch leaves little solid behind. Wood is different. Wood has lignin, the tough polymer that makes wood char so rich in carbon. Bread has no lignin at all. So bread is a poor char-former. That fits the brown being melanoidin, not carbon. When plant material is heated hard, what remains is carbon-rich. That was measured on crop material, not on toast. Toast is bad at charring, which is exactly why its brown is not carbon.

What is the black, then, when it does form? It is carbon bonded to carbon, in a tangle of rings and chains. Pure carbon has tidier shapes of its own. Diamond is every carbon bonded to four others, in one rigid net. Graphite is flat sheets, each carbon bonded to three others. Same element, two arrangements, and you can hold either in your hand. Benzene, a flat ring of six, is the next module's question.

Here is a way to put a number on the trade. Every element on this page lives in a world full of oxygen. So an element's bond to itself has a rival. That rival is its bond to oxygen. Compare the two. Take the energy to break one X–X bond. Divide it by the energy to break one X–O bond. Call that the ratio. A ratio near 1 means the self-bond is worth almost as much as the bond to oxygen. A small ratio means the self-bond is worth much less, so oxygen wins easily.

Every figure in the table is a mean bond enthalpy. That is the average energy to break that bond, over many compounds. It is not the energy of one bond in one molecule. Every number comes from one table, OpenStax Chemistry 2e, Table 7.2. A ratio built from two different tables would mean nothing. Catenation is an element bonding to itself in chains. The table below is a catenation test.

Read the table's own arithmetic first, because the ratio is the whole argument. Carbon's C–C bond is 345 kJ/mol and its C–O bond is 350 kJ/mol. Divide 345 by 350, and carbon's ratio is 0.986. Silicon's Si–Si bond is 230 kJ/mol and its Si–O bond is 370 kJ/mol. Divide 230 by 370, and silicon's ratio is 0.622. Nitrogen's N–N is 160 kJ/mol against N–O at 200 kJ/mol, so 160 divided by 200 is 0.800. Phosphorus's P–P is 215 kJ/mol against P–O at 350 kJ/mol, so 215 divided by 350 is 0.614.

One figure will be challenged by a reader with a different textbook, and it should be. Brown and LeMay's Chemistry: The Central Science gives Si–O as 452 kJ/mol, not 370. This module uses 370 because every other bond on the page comes from the same OpenStax table. But run the sum on the other book's figure. Divide 230 by 452, and silicon's ratio becomes 0.509. That is further from carbon's 0.986 than 0.622 was, not closer. So either textbook makes the case against silicon. This module chose the number that makes it more weakly. The argument is stronger on the figure it did not use.

So the answer to the gate is no, and the table says why. Carbon's self-bond is worth almost as much as its bond to oxygen. Silicon's self-bond is worth only about two thirds of its bond to oxygen. A silicon chain that meets oxygen turns into silicate. Silicate is silicon bonded to oxygen, the stuff of sand and rock. Valency alone said the two elements should match. The ratio says they cannot.

The ratio is not the whole test, and nitrogen shows why. Nitrogen scores 0.800 on it, better than silicon's 0.622. Yet chains of nitrogen atoms are famously unstable. The second test is the self-bond's own size. Carbon's C–C is 345 kJ/mol and nitrogen's N–N is 160 kJ/mol. Divide 345 by 160, and carbon's self-bond is 2.16 times nitrogen's. A chain needs both tests. Its self-bond must be strong on its own, and it must not lose badly to oxygen. Nitrogen fails the first. Silicon fails the second. Nitrogen also fails an earlier hurdle, because nitrogen makes three bonds, not four. Four bonds was never enough on its own. Two more things have to hold as well.

One warning before you leave this table. It compares carbon's self-bond against a single bond to oxygen. Oxygen can also make a double bond to carbon. What that changes is §6's business.

Bench 1 counts isomers for you to check against. Pick a formula from the list. The bench shows how many arrangements that formula has. An alkane is carbon and hydrogen with single bonds only. Seven formulas are offered. Three more are held back on purpose. They are C₆H₁₄, C₇H₁₆ and C₃H₆. You predict each of those counts first, in the boxes below the chart. Each guess is checked against a published sequence. The alkane counts are checked against OEIS A000602. OEIS is the online catalogue of integer sequences. A stereoisomer has the same bonds but a different arrangement in space. Tick the stereoisomer box, and the check is against A000628 instead. C₃H₆ is not an alkane. It is checked against its own listed structures.

Why hold three back? The bench never tells you how to count. It only checks. So a count you predict is your own, not a rule played back to you. Here is exactly what is hidden. The three held-back counts are hidden. So is the curve between five and fifteen carbons, until both alkane guesses are checked. The 355 and the 4,347 you read above are not among the held-back counts. What does this bench prove? Only the counts it builds and checks against OEIS. A held-back count appears only after you have predicted it.

Why does the count explode? A straight chain has one shape. Add a branch, and there is a choice of where. Add a second branch, and the choices multiply. Longer chains offer more places to branch. So the count does not climb steadily. It runs away. Once both alkane guesses are checked, the curve extends to fifteen carbons. It is drawn on a log scale. A log scale gives each tenfold step the same height. Each extra carbon multiplies the count, it does not add to it.

Oxygen joins the game too. C₄H₁₀O has four alcohols and three ethers. An alcohol carries an –OH group. An ether has an oxygen between two carbons. So 4 + 3 = 7 isomers. One published table says eight. It counted one alcohol twice. Which one, and why, is the next section's business. Tick the stereoisomer box and watch the curve. The two published sequences agree up to six carbons. They first part at seven. That is what the two sequences show, and you can check it. Two of C₇H₁₆'s isomers each have a second form the plain count misses. Where that second form comes from is the next section's question.

There is a second question people ask of a formula, and it is not a count. The degree of unsaturation is how many rings or double bonds a formula must hide. C₅H₁₂ hides none, and neither does C₄H₁₀O. C₃H₆ must hide one, and so must C₄H₈. It is a useful question. It just answers something different from the bench.

Bench 2 draws the carbon and its mirror image side by side. Drag either one to turn it. The bench also searches for you. It tries every rotation and every matching of the groups. It never bends a bond. It never uses a reflection, only turns. A reflection would be cheating. A reflection is what made the mirror image in the first place. The search stops only when its best fit has settled. If the fit has not settled, the bench says so. It gives no verdict then. An unsettled search is its own answer, not a yes and not a no. Switch to the molecule with two identical groups and run it again. What does this bench prove? Only the verdict the settled search gives, for the two molecules it holds. Superposable means one can be laid exactly on the other. The bench reports superposable or not, and nothing in between.

Here is what the difference does in the world, with matched figures. The two mirror forms of a chiral molecule are called enantiomers. Polarised light is light whose waves all vibrate in one plane. Pass it through a chiral liquid and that plane turns by an angle you can measure. Enantiomers turn it by equal amounts in opposite directions. Carvone is the example, and both of its enantiomers were measured under the same conditions. Those conditions are neat liquid, the sodium D line, and 20 °C, from Sigma-Aldrich product specifications 124931 and 818410. (R)-(−)-carvone turns the light by −61°, and it smells of spearmint. (S)-(+)-carvone turns it by +57.0 to +62.0°, and it smells of caraway. The second figure is a range because that supplier states its specification as a range. The first supplier states a single value. Both are the same measurement under the same conditions. Leitereg and colleagues, 1971, checked the purity of both samples and ruled contamination out.

Next comes the case where counting mirror forms over-reaches, which is tartaric acid. A stereocentre is a carbon whose four groups are all different. Tartaric acid has two stereocentres, so the simple rule predicts 2 to the power 2, which is 4 stereoisomers. It has 3. One of the four is its own mirror image, so that pair collapses to one. That single form is called the meso form, and it is where 2 to the power n over-counts. Two sources give its melting point about twenty degrees apart. Neither could be checked against a primary source. So no melting point is printed here.

Turn to the second molecule on the bench. It has two identical groups. Its mirror image can be turned to fit. Why? The search may match the two identical groups either way round. That freedom is what a full set of different groups takes away. So on a single centre like this one, chirality needs four different groups. Every bond angle in the bench's model stays 109.47° through the whole search. The shape never changed. Only the labelling of the corners did. That is the whole of the mirror problem. And it answers the published table from §3. Its eighth C₄H₁₀O isomer was butan-2-ol's mirror twin. Butan-2-ol has a stereocentre, so its mirror twin is a real second molecule. But it is a stereoisomer, not a new arrangement of bonds. The table counted a §4 answer inside a §3 question.

Bench 3 is worked out from published constants, not measured here. Pick ethane first. Slide the dihedral angle and read the energy. The curve is continuous. The bench names each conformation from the angle itself. At 60°, 180° and 300° the hydrogens sit as far apart as they can. The bench calls that staggered. The energy there is at its minimum, 0 kJ/mol. At 0°, 120°, 240° and 360° they line up behind each other. The bench calls that eclipsed. The energy there is at its maximum, 12.2 kJ/mol. That 12.2 kJ/mol is the barrier. It is from the NIST database, citing Gurvich and colleagues, 1989. Now pick butane or methylcyclohexane. Each shows only two fixed states and a bar for each. No curve is drawn between the states, on purpose. Four secondary sources give four different heights for butane's eclipsed peaks, and none of the primary papers could be reached. So the bench draws nothing it cannot cite. What does this bench prove? Only that the barrier and the populations are read live from the cited constants. The populations are ratios computed from an energy difference, and nothing more.

Here is the arithmetic, and the sentence you keep at the end rests on it. The mean C–C bond enthalpy is 345 kJ/mol, from §2's table. That is the energy to pull the two carbons apart. The barrier to turning that same bond is 12.2 kJ/mol, from the ethane curve above. Divide 345 kJ/mol by 12.2 kJ/mol, and the units cancel, so the bond is 28.28 times stronger than the barrier. So turning costs about one twenty-eighth of breaking, and the two are different events. Now compare the barrier with the thermal energy a mole of molecules has at room temperature. R is the gas constant, 8.314462618 J per mole per kelvin, exact by the 2019 definition of the SI. Multiply R by 298.15 K, and RT is 2478.957 J per mole. That is 2.478957 kJ per mole. Divide the barrier, 12.2 kJ per mole, by 2.478957 kJ per mole, and the barrier is 4.92 times RT. Both are energies per mole, so the units cancel and 4.92 is a pure number. Module 15 gave the Boltzmann tail, the share of molecules with more than a given energy. A barrier under five times RT is crossed constantly by that tail, not rarely. The bond is strong, and it turns all the time. Neither fact touches the other.

The butane bars hide a classic error, so take them slowly. Butane has two low-energy conformations about its middle bond. The anti form puts the two end carbons opposite each other. The gauche form puts them 60° apart. Gauche sits 3.18 kJ/mol above anti, from Chen, Wilhoit and Zwolinski, 1975. In joules that is 3179.84 J/mol, from the paper's 760 cal/mol. That figure is an energy difference. ΔE is the energy gap between the two forms, and nothing else. It is not a free energy, so the result below is not a free-energy ratio either. Divide 3179.84 J/mol by RT, 2478.957 J/mol, and the units cancel, so the exponent is the pure number 1.28273. Take e to the minus that power, and the factor is 0.27728 per gauche form. The sign is minus because gauche sits above anti, so each gauche form gets a share smaller than anti's.

Here is the error. Treat gauche as one state, and anti : gauche comes out as 78.3 : 21.7. That is wrong. There are two gauche forms, one twisted left and one twisted right. So multiply 0.27728 by 2, which gives 0.55456. Anti's own factor is 1, because it is the reference state. Add 1 and 0.55456, and the total is 1.55456. Anti is 1 divided by 1.55456, which is 64.3 per cent. Gauche is 0.55456 divided by 1.55456, which is 35.7 per cent. So anti : gauche is 64.3 : 35.7, and that is the right figure. The 2 is called a degeneracy, the number of states sharing one energy. These figures are at 298.15 K, the standard reference temperature every table in this course uses. The bench's slider moves in steps of ten kelvin, so its nearest stop is 300 K. At 300 K the bench draws 64.1 : 35.9, not 64.3 : 35.7. Ten kelvin moved the anti share by 0.2 points, about two parts in a thousand. That is how sensitive a Boltzmann population is to temperature near room temperature. Slide to 200 K and anti : gauche is 77.2 : 22.8. Slide to 400 K and it is 56.5 : 43.5. The colder the gas, the more of it sits in the lowest well.

Methylcyclohexane is the contrast. Its methyl group sits either equatorial, out from the ring, or axial, up from it. Axial costs 7.32 kJ/mol more, from Booth and Everett, 1980. Here there is one state on each side, so the degeneracy is 1 : 1. Divide 7320 J/mol by 2478.957 J/mol, and the exponent is 2.9529. This time take e to the plus that power, because the ratio is written low over high, equatorial over axial. It is the same Boltzmann factor turned upside down, and it comes to 19.16. Add 1 for the axial state, and the total is 20.16. Equatorial is 19.16 divided by 20.16, which is 95.0 per cent, and axial is 5.0. At the bench's 300 K that still rounds to 95.0 : 5.0. At 200 K it is 98.8 : 1.2, and at 400 K it is 90.0 : 10.0. Each extra RT of gap multiplies that ratio by e, about 2.7, again. So a gap of nearly three RT swings the split much further than a gap near one RT. That is why a gap only about twice butane's gives a far bigger split.

Here is a distinction the sentence you keep depends on. Stable has two meanings, and they are different. One is thermodynamic. It asks whether the products sit lower than the reactants. For methane and oxygen they do, by a long way. So carbon compounds are not sitting at the bottom of anything. Towards oxygen, almost all of them are unstable in this first sense.

That sounds like the opposite of §2, so settle it now. §2's table compared carbon's self-bond against a single bond to oxygen, 350 kJ/mol. Burning does not make single bonds to oxygen. It makes double ones. C=O is 741 kJ/mol, from the same OpenStax table. Divide 741 by 350, and the double bond is 2.12 times the single. Carbon does not lose to C–O. It loses to C=O. Both sections are right, about two different bonds.

The other meaning of stable is kinetic. It asks whether the path to the products is open at this temperature. Module 22 built exactly this distinction. ΔG is Module 14's referee, the free energy change that says whether a change runs by itself. Written ΔG°, it is that change under standard conditions. A favourable ΔG° did not by itself mean a reaction happened. The furnace still had to be hot. So those bonds last means three things at once. The self-bond is strong in absolute terms, 345 kJ/mol. It does not lose badly to a comparable bond, the single C–O. And the barrier to the reaction that would destroy it is high. §2 supplied the first two. This section supplies the third. Read the sentence with all three, and §2 and §6 are halves of one claim. Carbon compounds persist under ordinary conditions. They are not forbidden from reacting. A spark ends the persistence at once, and the drop is then paid out as heat.

Here is the arithmetic, and it cross-checks two routes to one number. Start with the measured value. NIST's WebBook gives methane's enthalpy of combustion as −890.35 kJ/mol, by calorimetry, with liquid water as the product. The sign is negative, and negative means heat is given out, so the reaction is exothermic. The reaction is CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(l).

Now the same number by Module 12's other tool, Hess's law. Hess's law says the heat of a reaction is the same whichever route of steps you add up. The steps here are standard enthalpies of formation, the heat of making one mole of a compound from its elements. Methane gas is −74.6 kJ/mol, carbon dioxide gas is −393.5 kJ/mol, and liquid water is −285.8 kJ/mol. Oxygen is an element, so its formation enthalpy is zero by definition. Add the products, and −393.5 plus 2 × (−285.8) is −393.5 minus 571.6, which is −965.1 kJ/mol. Subtract the reactant, and −965.1 minus (−74.6) is −965.1 plus 74.6, which is −890.5 kJ/mol. The sign is negative again. The two routes differ by 890.5 minus 890.35, which is 0.15 kJ/mol. Divide 0.15 by 890.35 and multiply by 100, and the difference is 0.017 per cent. A thermochemical cycle and a calorimeter agree to better than two parts in ten thousand. So the drop is real, measured two ways. And methane in a pipe still sits there. Nothing about the size of that drop says when the reaction starts.

What holds the reaction back is a barrier. Module 16 called it the activation energy. Before better bonds can form, existing bonds must be stretched and broken. That costs energy first, before any is paid back. Paper is a good case, because paper is cellulose. Cellulose is a long chain of sugar units, all carbon, hydrogen and oxygen. Antal and Várhegyi measured a substantial barrier that cellulose must climb before it breaks down, in a primary study. That study's figure comes with its own kinetic model and conditions, so this page prints no value for it. What it can do is show what a barrier of that size does, whatever the exact number.

RT at 298.15 K is 2.478957 kJ/mol, from §5. Take a barrier of 50 kJ/mol, and divide 50 by 2.478957, which is 20.17 times RT. Module 15's tail puts the fraction of molecules above that at e to the minus 20.17. That is 1.7 × 10⁻⁹, about one in 580 million. Take a barrier of 100 kJ/mol instead, and 100 divided by 2.478957 is 40.34 times RT. The fraction above it is e to the minus 40.34. That is 3.0 × 10⁻¹⁸, about one in 3 × 10¹⁷. Both are energies per mole, so both exponents are pure numbers. Doubling the barrier did not halve the fraction. It divided it by a billion. A page of paper lasts for centuries in a library. It is not stable against oxygen. It is slow. This page prints no ignition temperatures. The ones found came from an uncited table, and an uncited number does not reach a page.