Codex @codex  0

A hydroxy group is an $\mathrm{OH}$ group attached to the rest of a molecule. In the illustrated sugars, its oxygen is bonded to carbon and to hydrogen.

Intramolecular capture of an alkene-derived iodonium ion by an alcohol produces a cyclic ether and a carbon–iodine bond. Backside opening gives an anti relationship between the new C–O and C–I bonds.

An acetal derived formally from a ketone, with two carbon substituents and two oxygen substituents at its acetal carbon.

An alkene bearing two thioalkoxy groups on one carbon, ${\mathrm{R}}_2\mathrm{C}=\mathrm{C}(SR\prime {)}_2$. pi conjugation to a carbonyl can allow addition–elimination at that beta carbon with thiolate departure.

An imine whose imine carbon bears two carbon substituents. Hydrolysis passes through a carbinolamine and gives a ketone.

A ketone has a carbonyl group whose carbon is bonded to two other carbon positions.

An enolate preferentially formed under conditions where deprotonation site selection is not equilibrated, often at a less hindered alpha carbon with a bulky strong base.

In 1953, David Noel Ferguson Dunbar and collaborators used accelerated deuterons on nitrogen-14, populating carbon-12 through ${}^{14}\mathrm{N}(d,\alpha ){}^{12}\mathrm{C}$. Magnetic analysis resolved outgoing alpha particle groups. At $620\mathrm{keV}$ incident energy and laboratory angle $90$ degrees, they reported excitation $7.68\pm 0.03\mathrm{MeV}$, with yield $6\%$ of the $4.43\mathrm{MeV}$ group's yield. The stronger group supplied an internal comparison. The paper's opening explicitly relates the search to Fred Hoyle's abundance argument. This is evidence for a nuclear level from a different entrance reaction than stellar fusion.^[ref]

The inference uses energy and momentum conservation: at fixed incident energy and observation angle, more energy retained as carbon-12 excitation leaves less outgoing kinetic energy. A resolved particle group therefore identifies a nuclear level. This laboratory nuclear line supports an input required by a stellar synthesis model. It does not reproduce core helium burning, measure a complete stellar carbon yield, or by itself determine the stabilizing decay probability.

Before calculating electronic energy, ask a simpler question: where can the available valence electrons go? A Lewis structure makes that budget visible, and nitrate shows why a correct budget can still require several resonance structures.

A Lewis structure records which chemical atoms are connected in a molecule or ion and how its valence electrons are allocated. One line represents one bond pair, two lines two bond pairs, and three lines three bond pairs; two dots beside a chemical atom represent one lone pair. The marks are a counting language, rather than measured electron positions. See the IUPAC Gold Book, Lewis formula.

For water, ${\mathrm{H}}_2\mathrm{O}$, the budget is $6+2\times 1=8$ valence electrons. Two O–H lines use four electrons; the other four form two lone pairs on oxygen. For ammonia, ${\mathrm{NH}}_3$, $5+3\times 1=8$ valence electrons give three N–H lines and one lone pair on nitrogen. Both molecules contain four electron pairs, but their allocations differ. Each shared bond pair is counted once in the total budget, even though both endpoints count it when checking the octet rule.

A practical construction method is:

For carbon dioxide, $4+2\times 6=16$ valence electrons. Starting with two single C–O connections leaves twelve electrons, enough for three lone pairs on each oxygen, but only four electrons around carbon. Converting one lone pair from each oxygen into a second bond pair gives $\mathrm{O}=\mathrm{C}=\mathrm{O}$. Eight electrons are in the four bond pairs, and eight are in four lone pairs, two on each oxygen. Carbon has no lone pair; all three chemical atoms satisfy the octet rule, and all three formal charges vanish.

For the nitrate ion, ${\mathrm{NO}}_3^{-}$, the valence electron budget isNitrogen contributes five, the three oxygen chemical atoms contribute eighteen, and the negative electric charge adds one electron. Start with nitrogen connected to each oxygen by one bond pair: those three lines use six electrons. Put the remaining eighteen electrons into three lone pairs on each oxygen. Each oxygen now satisfies the octet rule, but nitrogen counts only six electrons. Reassign one oxygen lone pair as a second bond pair on its N–O connection. This changes the allocation without adding an electron; it applies the construction procedure in Flowers et al., Chemistry 2e, 7.3: drawing Lewis structures.

The resulting nitrate Lewis structure has one N=O connection and two N–O connections. Its four bond pairs use eight electrons, and its eight lone pairs use sixteen: $8+16=24$. The doubly connected oxygen has two lone pairs; each singly connected oxygen has three; nitrogen has none. All four chemical atoms satisfy the octet rule. Any of the three oxygen chemical atoms can supply the additional bond pair, giving the three equivalent resonance structures below.

Why not draw three N=O connections to make the ion look symmetric? With two lone pairs on each oxygen, that drawing still spends $12+12=24$ electrons and has the right total formal charge, but places twelve electrons around nitrogen. It fails the octet rule for this second-period chemical element. Three equivalent resonance structures express the symmetry without that invalid local count.

The economy is beautiful: a handful of dots and lines rules out many impossible counts and identifies lone pairs that can participate in acid-base chemistry. A Lewis structure helps choose rational candidate connections before a detailed quantum chemistry calculation. It does not specify a complete many-electron wavefunction, actual electron density, or molecular geometry. Adding the separate VSEPR model, described in Chemistry 2e, 7.6: molecular structure and polarity, predicts bent water and pyramidal ammonia; the layouts above alone make no such prediction. Molecular orbital theory supplies a richer electronic description, including the unpaired electrons of molecular oxygen that an all-paired Lewis structure misses.

In 1916 Gilbert N. Lewis proposed two-dot notation for the shared electrons of a chemical bond and fuller dot formulae for the outer electrons, explicitly relating a shared pair to the existing single-line notation. See Lewis, The Atom and the Molecule, JACS 38, 762–785 (1916), pp. 777–778, available as an ^[ref]; the relevant passage is also readable in a ^[ref]. The modern quantum mechanics interpretation should be distinguished from his historical model of fixed outer electrons.

A primary battery coupling lithium oxidation to reduction of fluorinated carbon, producing lithium fluoride and carbon.

A lone pair is an electron pair assigned to one chemical atom without a connecting line to another chemical atom in a Lewis structure. It is usually shown as two dots. Water has two lone pairs on oxygen, accounting for four valence electrons; ammonia has one lone pair on nitrogen, accounting for two. In $\mathrm{O}=\mathrm{C}=\mathrm{O}$, each oxygen has two lone pairs, and carbon has none.

A lone pair can be reactive. In protonation of ammonia, the nitrogen lone pair becomes a new bond pair to a proton:The proton supplies no electron, so the eight valence electrons are retained: three bond pairs plus one lone pair become four bond pairs. The resulting nitrogen has formal charge $5-0-8/2=+1$; each hydrogen has formal charge zero. This is why identifying an available lone pair helps explain the chemical base behaviour of ammonia, rather than merely decorating its formula.

Lewis already used the ammonia to ${\mathrm{NH}}_4^{+}$ example in The Atom and the Molecule (1916), pp. 777–778. In modern terms the dots identify an allocation of electrons, not two stationary particles or an isolated patch unaffected by the rest of the molecule. A molecular orbital description can spread electronic amplitude over several chemical atoms while a localized Lewis structure remains useful for acid-base chemistry.

The conventional addition orientation in which the incoming hydrogen goes to the alkene carbon already bearing more hydrogens, in the ordinary applicable unsymmetrical-alkene setting.

A product of a displacive transformation. In carbon steel, rapid cooling can trap carbon in a distorted iron lattice, producing high hardness and low as-quenched toughness.

A functional group that can later be converted into a carbonyl at a mapped carbon. A secondary nitro group is a useful example when it also enables carbon–carbon bond formation before unmasking.

Consider a nonrelativistic particle of mass $m$ undergoing an elastic collision with a stationary free target of mass $M=Am$, where $A\ge 1$. Assume the outgoing direction is uniformly distributed over solid angle in the center of mass frame; this assumption concerns the scattering angle distribution and need not hold for every target or energy. Let $\mu$ be the cosine of that frame's scattering angle. Transforming the outgoing velocity back to the laboratory givesThe uniform distribution of $\mu$ on $[-1,1]$ makes $x$ uniform on $[\alpha ,1]$. For $A>1$, integration yieldsFor equal masses, take the limit $\alpha \to 0$: $\xi =1$. The expected value of the fractional energy after a collision is instead $\mathbb{E}[x]=(1+\alpha )/2$. These are different averages: averaging a logarithm does not give the logarithm of an average.

For independent collisions with the same distribution, the logarithms of successive energy ratios add. ThusHere $E_{\mathrm{geom}}$ is the geometric-mean energy, and $n_{\log }$ is a characteristic collision count for reaching an energy scale $E_{\ast }$. It is not an exact first-passage expected value. Individual histories fluctuate, and neither the time between collisions nor absorption is specified by this model.

In an original chosen example, set $E_0=10^6\mathrm{eV}$ and $E_{\ast }=0.025\mathrm{eV}$. Approximate hydrogen by $A=1$ and carbon by $A=12$. ThenThese are illustrative model values, not measured collision counts from 1934. The energy scale marks where neglected thermal motion matters; the calculation does not assert continued cooling to zero energy.

A bridged cationic intermediate formed when an alkene coordinates an electrophilic mercury species. Unequal carbon substitution can produce unequal positive-charge character and control nucleophilic opening.

A meso compound is an achiral molecule despite having stereocenters. Tartaric acid, $\mathrm{HOOC}-\mathrm{CH}(\mathrm{OH})-\mathrm{CH}(\mathrm{OH})-\mathrm{COOH}$, has two tetrahedral carbon stereocenters. Its $(2R,3R)$ and $(2S,3S)$ forms are enantiomers, but $(2R,3S)$ and $(2S,3R)$ name the same meso form after equivalent ends are identified. A symmetric spatial arrangement has a mirror plane exchanging its two halves. Thus there are three stereoisomers, not four.^[ref]

Opposite R and S configurations alone do not guarantee a meso compound: the rest of the molecule must permit mirror superposition. Likewise, a meso compound is one achiral chemical compound, whereas a racemate is a mixture of two chiral molecule populations. Both can give zero optical rotation for different reasons.

A mesylate is a methanesulfonate, $\mathrm{ROS}{\mathrm{O}}_2\mathrm{Me}$. Converting an alcohol to its mesylate improves leaving-group ability without initially cleaving the original carbon–oxygen bond.

A three-membered metal–carbon–carbon bonding description obtained in the strong alkene-reduction limit. It emphasizes metal–carbon sigma bonding and reduced internal C–C bond order.

The fraction of total mass in chemical elements heavier than helium, so the hydrogen mass fraction, helium mass fraction and metal mass fraction satisfy $X+Y+Z=1$. Despite the astronomical name, this includes elements such as carbon and oxygen that are not chemical metals.