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$\mathrm{C}{\mathrm{H}}_2=\mathrm{C}(\mathrm{Me})\mathrm{C}{\mathrm{O}}_2\mathrm{Et}$, an alpha-methyl-substituted acrylate. Conjugate addition installs a chain whose carbon next to the ester retains the methyl substituent.

The body-centred cubic iron-rich solid phase of ordinary low-temperature steel, with low equilibrium carbon solubility.

In a Fischer projection, each cross represents a carbon position: horizontal chemical bonds point toward the viewer and vertical chemical bonds point away. For the open-chain aldehyde sugars illustrated here, place $\mathrm{CHO}$ at the top and ${\mathrm{CH}}_2\mathrm{OH}$ at the bottom. This is a convention for stereochemistry, not a measured molecular shape. See Moss, IUPAC Recommendations 1996: Fischer projection.

In these Fischer projections, D specifies that the hydroxy group at the lowest stereocenter lies on the right: C4 for ribose and deoxyribose, C5 for glucose. D does not specify every other stereocenter, and it does not specify the direction of rotation of polarized light. See McMurry: the meaning of D and L.

The formal charge of a chemical atom in a Lewis structure assigns it all of its nonbonding electrons and half of the electrons in its connections:Here $v$ is the neutral chemical atom's valence electron count, $n$ is the number of nonbonding electrons assigned to it, and $b$ is the number of bonding electrons incident on it. Count electrons, not electron pairs: two lone pairs mean $n=4$, and two double connections mean $b=8$. The number $\mathrm{FC}$ gives electric charge in units of the positive elementary charge. Equal splitting is a bookkeeping convention, regardless of the actual electron density. See the IUPAC Gold Book, formal charge.

For $\mathrm{O}=\mathrm{C}=\mathrm{O}$, each oxygen has $v=6$, $n=4$, $b=4$, giving formal charge $6-4-4/2=0$. Carbon has $v=4$, $n=0$, $b=8$, giving formal charge $4-0-8/2=0$. For ${\mathrm{O}}^{-}-\mathrm{C}\equiv {\mathrm{O}}^{+}$, the left oxygen has formal charge $6-6-2/2=-1$, carbon still has formal charge zero, and the right oxygen has formal charge $6-2-6/2=+1$. The net electric charge is zero in either drawing.

The Lewis structure with no separated formal charges is the dominant contributor in this elementary comparison; the charge-separated drawing is a less important contributor, rather than an equally dominant alternative. This is a qualitative selection rule, not a calculation of energy or numerical chemical resonance weights. See Flowers et al., Chemistry 2e, 7.4: comparing carbon dioxide structures. A zero formal charge does not imply zero partial atomic charge: the latter depends on partitioning the actual electron density, which a Lewis structure does not supply.

As a short consistency derivation, summing $n+b/2$ over all chemical atoms counts every drawn electron once: a nonbonding electron is assigned once, and a bonding electron contributes half at each of its two endpoints. Thuswhere $N_{\mathrm{valence}}$ is the total drawn valence electron count. The right side is the net electric charge in units of elementary charge. This catches a wrong budget even when individual octet rule checks look plausible.

Glucose is a monosaccharide with molecular formula ${\mathrm{C}}_6{\mathrm{H}}_{12}{\mathrm{O}}_6$. Its familiar biological form, D-glucose, is an aldohexose. In the open-chain Fischer projection, $\mathrm{CHO}$ is at the top and ${\mathrm{CH}}_2\mathrm{OH}$ at the bottom; the hydroxy groups at C2, C3, C4, and C5 point right, left, right, right. This full pattern distinguishes D-glucose from other six-carbon aldehyde sugars. See McMurry, Organic Chemistry, 25.4: configurations of the aldoses.

The molecular formula can be checked directly: one $\mathrm{CHO}$ group, four $\mathrm{CHOH}$ groups, and one ${\mathrm{CH}}_2\mathrm{OH}$ group give six carbon, twelve hydrogen, and six oxygen chemical atoms. All four interior carbon positions are stereocenters in this open-chain structure. Neither their orientations nor the arrangement of chemical bonds can be recovered from the molecular formula alone.

In water, glucose exists mainly in six-membered rings in chemical equilibrium with a small amount of open-chain glucose. Ring closure connects the hydroxy group at C5 to the aldehyde carbon at C1; the ring contains five carbon chemical atoms and one oxygen, with C6 outside it. No chemical atoms are lost, so the molecular formula remains ${\mathrm{C}}_6{\mathrm{H}}_{12}{\mathrm{O}}_6$. See McMurry, Organic Chemistry, 25.5: cyclic structures of monosaccharides.

Glucose is useful both as a substrate in cellular metabolism and as a building block of polysaccharides. Cellulose chains assembled from glucose units help support plant cell walls. The same small sugar thus connects chemical fuel to a structural material; its role depends on how it is transformed or linked. See Cooper, The Cell: A Molecular Approach, the molecular composition of cells.

Glycerol has three carbon chemical atoms and three $\mathrm{OH}$ groups. It supplies the scaffold of triglycerides and glycerophospholipids, but is not itself a long hydrocarbon chain. See Biology 2e: glycerol.

Glycolysis converts one glucose into two pyruvates, yielding two net ATP and reducing two NAD+ to two NADH in the canonical pathway. It first spends two ATP, then produces four by substrate-level phosphorylation. The pathway itself consumes no molecular oxygen; sustained operation nevertheless requires regeneration of NAD+.^[ref]

The useful organizing idea is three separate ledgers: carbon chemical atoms, transfers between ADP and ATP, and redox transfers between NAD+ and NADH. Splitting the carbon skeleton doubles the number of downstream conversions; recycling a cofactor permits repeated operation. These counts explain the yield without memorizing a drawing of every enzyme. The two net ATP belong to glycolysis alone, not to complete cellular respiration.

Follow one glucose through canonical glycolysis, with no diversion of intermediates into other metabolic pathways. The investment stage uses two ATP to prepare the six-carbon substrate. Cleavage and rearrangement provide two glyceraldehyde 3-phosphate molecules, each containing three carbon chemical atoms. Each then yields one three-carbon pyruvate, one NADH, and two ATP in the payoff stage.^[ref]

The material check is $6=2\times 3$: all six carbon chemical atoms remain in the two pyruvates. No carbon dioxide has been released within this boundary. In the ATP ledger, each three-carbon route contains two ATP-forming transfers, soThe first pair of payoff transfers only restores the initial ATP inventory; the second pair creates the net gain. Four newly formed ATP is therefore compatible with only two additional ATP remaining.

For an original scaled example, stipulate complete conversion of $5\mu \mathrm{mol}$ of glucose. The initial $30\mu \mathrm{mol}$ of carbon chemical atoms becomes $10\mu \mathrm{mol}$ of pyruvate, still containing $30\mu \mathrm{mol}$ of carbon chemical atoms. The pathway spends $10\mu \mathrm{mol}$ of ATP, forms $20\mu \mathrm{mol}$ of ATP, and adds $10\mu \mathrm{mol}$ to the ATP inventory before any other ATP use. It also reduces $10\mu \mathrm{mol}$ of NAD+. These are calculated ideal amounts, not observed yields; side routes or incomplete conversion change the outcome.

A common biochemical net chemical equation isHere ${\mathrm{P}}_i$ is inorganic phosphate. The names group biochemical protonation and metal-binding states; the displayed proton and water terms use a common shorthand rather than a unique specification of every chemical species. The robust ledger records two net ADP-to-ATP conversions and two NAD+-to-NADH conversions. Summing the individual chemical equations cancels two invested ATP against four formed ATP and cancels the internal substrates, giving those coefficients.

An oxocarbenium ion at the anomeric carbon of a cyclic glycosyl fragment. Positive charge is delocalized between that carbon and the ring oxygen. A counterion can remain nearby and influence which face a nucleophile approaches.

Graphene is a single layer of carbon chemical atoms arranged in a honeycomb lattice. A nearby gate can select predominantly electron or electron hole conduction, as demonstrated by the graphene exfoliation and field-effect experiments.^[ref]

A useful question is how geometry changes motion. The nearest-neighbour model of graphene starts with three equivalent bonds from each site and two sites per primitive cell. Interference between those three hopping paths makes two energy bands meet; near their meeting, energy grows linearly with distance in wave vector. The resulting Dirac cones explain why an atomic sheet can imitate one aspect of a relativistic particle while remaining an ordinary material with contacts, defects and boundaries.

A layered solid form of carbon, with covalent bonds joining chemical atoms within each layer.

A helium-burning shell converts helium into carbon and oxygen outside a helium-exhausted stellar core. Thin-shell burning on the Asymptotic giant branch can develop a thermal pulse of an asymptotic-giant-branch star.

The helium mass fraction is the fraction of a material's mass carried by helium nuclei. Hydrogen burning raises it, whereas core helium burning lowers it while producing carbon and oxygen.

A hemiacetal has an hydroxy group and an $\mathrm{OR}$ group on the same saturated carbon; $R$ is attached through carbon.

A hemiketal is the addition product of an alcohol and a ketone, with hydroxy group and alkoxy substituents on the former carbonyl carbon. Intramolecular formation can temporarily tether an alcohol to a ketone before another chemical reaction changes the carbon skeleton.

The hydrocarbon $\mathrm{HC}\equiv \mathrm{C}-\mathrm{C}{\mathrm{H}}_2-\mathrm{C}{\mathrm{H}}_2-\mathrm{C}\equiv \mathrm{CH}$. Its triple-bond fragments are linear, but the entire carbon skeleton need not be. The extended anti conformation has an inversion centre at the central bond midpoint and belongs to the C2h molecular point group.

The unbranched six-carbon acid $\mathrm{C}{\mathrm{H}}_3(\mathrm{C}{\mathrm{H}}_2{)}_4\mathrm{C}{\mathrm{O}}_2\mathrm{H}$. Its carbonyl carbon is included in the six-carbon total.

When the bottom of an Asymptotic giant branch star's convective envelope becomes hot enough for hydrogen burning, the CNO cycle can convert newly dredged-up carbon into nitrogen. Its onset and strength depend on mass, composition and envelope structure; it is not a universal property of all stars at one initial mass.

A hydrocarbon contains only carbon and hydrogen. A fatty acid also contains oxygen; calling its long chain a hydrocarbon region does not make the whole fatty acid a hydrocarbon. See IUPAC Gold Book: hydrocarbons.

A hydrogen bond donor is the $X-\mathrm{H}$ group supplying the attached hydrogen in a hydrogen bond $X-\mathrm{H}\cdots Y$. The donor atom $X$, or the whole molecule containing it, may also be called the donor. The terminology does not imply transfer of a free proton: in the ordinary water pair, the O–H covalent bond remains. Familiar donors include O–H groups in water and N–H groups in ammonia. See Andrew Glasfeld, Intermolecular Attraction, pp. 12–13: donor and acceptor roles.

The attached hydrogen is essential to this role. At the nitrogen site, an unprotonated amine $R_3\mathrm{N}$ can be a hydrogen bond acceptor, but has no N–H hydrogen bond donor. Protonation gives $R_3{\mathrm{NH}}^{+}$: the new N–H group can donate, while that nitrogen no longer has the accepting lone pair. This is a change of chemical form, not a permanent label attached to a chemical element. Here $R$ denotes a group attached through carbon; the statement concerns the nitrogen site and does not rule out weaker C–H associations elsewhere. See McMurry, Organic Chemistry, 24.3: amine protonation and Glasfeld: protonated amines as donors.

An amide with an N–H covalent bond can donate even though its nitrogen is ordinarily not an accepting site. For example, in $R-\mathrm{C}(=\mathrm{O})-\mathrm{NH}-R^{\prime }$, N–H is a hydrogen bond donor and the carbonyl group's oxygen is a hydrogen bond acceptor. Replacing the remaining N–H by another group attached through carbon removes this N–H donor without removing the accepting oxygen. The analogous division of roles in a peptide bond helps explain protein structure. See Glasfeld: amide roles and biological interactions. The useful question is therefore which site supplies the hydrogen, rather than whether the whole substance is simply “hydrogen bonding”.