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Incoming links: Sodium hydride
Ionic approximation of bonds Created 2026-10-14 Updated 2026-10-20
For the routine examples here, assign all bonding electrons between different chemical elements to the more electronegative partner; divide bonding electrons between identical chemical elements equally. Retain each lone pair on its original chemical atom. Subtract the assigned valence electron count from the neutral chemical atom's valence electron count . The IUPAC convention uses Allen electronegativity, based on free-chemical atom valence electron energies. Close electronegativity comparisons and some reversibly bound chemical compounds require care beyond this simple rule.[ref]
Comprehensive definition of oxidation state (IUPAC Recommendations 2016) by Pavel Karen and others. §3 and §4.1, pp. 833–834: Allen electronegativity, electron allegiance and the reversible Lewis-acid-ligand caveat; §5.5, pp. 836–837: ambiguous assignments
The examples explain exceptions to memorized rules:
- Water, : oxygen receives both O–H bond pairs plus its two lone pairs, giving eight assigned valence electrons and . Each hydrogen receives none, giving .
- Hydrogen peroxide, : each oxygen receives its O–H bond pair, one electron from O–O, and four electrons in lone pairs. Thus each has , while each hydrogen remains . The O–O connection is why the usual oxygen shortcut fails.
- Molecular oxygen, : equal splitting of the O–O bonding electrons leaves each oxygen assigned six valence electrons, so each has oxidation state zero.
- Sodium hydride, : hydrogen is more electronegative than sodium. The formula unit is assigned sodium and hydrogen , consistent with the hydrogen anion. The usual hydrogen shortcut fails here.
These drawings record allocations, not molecular geometry or a physical process of breaking the chemical bonds. In particular, the O–O line in the molecular oxygen panel only marks the connection; it does not assert a complete Lewis structure or spin state.
Past exam of the chemistry B course of the University of Cambridge 2022 Paper 1 B2 d Solution 2026-10-20
Let and number the carbon substituent from oxygen: . sodium hydride removes a proton from C3, alpha to the nitrile, forming hydrogen and a nitrile-stabilized carbon anion. That carbon attacks the nitroaryl ipso carbon bearing O. The tether closes a five-membered cyclic Meisenheimer intermediate with sequence C3–C2–C1–O–aryl ipso carbon. The para nitro group stabilizes the negative charge generated by temporary loss of aromaticity.
Rearomatization breaks the aryl–O bond, releasing the tether oxygen as an alkoxide while retaining the new aryl–C3 bond. protonation gives . This is a Truce–Smiles rearrangement: the aryl group migrates from oxygen to carbon. The nitrile remains attached to its original side-chain carbon; the original ether oxygen becomes the terminal alcohol oxygen. The substrate and product have identical atomic composition. Ordinary substitution at a saturated chain carbon would not produce this aryl–carbon bond or restore the terminal OH.
Past exam of the chemistry B course of the University of Cambridge 2023 Paper 1 B1 b i Solution 2026-10-20
The malonate carbon receives two allyl groups. With , attached through its first carbon, the product issodium hydride removes a methylene proton from diethyl malonate; molecular hydrogen formation accompanies production of an enolate stabilized by resonance in chemistry. Its carbon attacks the bromomethylene carbon of allyl bromide in an SN2 reaction, expelling bromide. The monoalkylated malonate still has one hydrogen between the carbonyls. Excess base removes that hydrogen, and a second equivalent of allyl bromide produces a second carbon–carbon bond. The final malonate carbon has no remaining hydrogen and cannot undergo a third analogous deprotonation there.
The two alkenes are preserved because the substitution occurs at the primary carbon bearing bromine. Counting the introduced groups gives for the diallylated diethyl ester. Its two identical allyl groups also mean that the central carbon is not a stereocenter. This stage is the alkylation portion of the malonic ester synthesis.
The first product is , with the original alkene geometry retained. methoxide attacks the primary saturated carbon bearing bromine from the backside. Its oxygen lone pair forms the new C–O bond as the C–Br bonding pair leaves with bromide. The other bromine is vinylic, directly bonded to an sp carbon: ordinary SN2 reaction is unfavourable there, and ionization would require an unstable vinyl cation. Thus the primary alkyl bromide is replaced selectively, rather than replacing the vinylic halide or moving the double bond. There is no observable inversion label at the attacked carbon because it is a methylene.
The second product is sodium 4-propoxybutane-1-sulfonate, . sodium hydride first removes the alcohol proton, giving propoxide and hydrogen. The propoxide oxygen then attacks the primary ring carbon bonded to the ring oxygen; simultaneous cleavage of that carbon–oxygen bond opens the sultone. The departing ring oxygen remains attached to sulfur as part of the resonance-stabilized sulfonate anion. The carbon attached directly to sulfur is not the substitution site, because cleavage there would not expel the same stable sulfonate leaving group. This is carbon-centred ring opening of a cyclic sulfonate ester, not addition across an S=O bond.
The left ring system is spiro, and the right-hand substituent is , not OMe directly attached to the cyclohexene ring. Let denote the right cyclohexenyl group bearing that methoxyethyl substituent cis to its linker. The connection to the left alkene carbon is . These observations suggest two Diels–Alder constructions. A left-hand dienehas the original cyclopentane carbon as one terminal diene atom; cycloaddition with an ethylene equivalent gives the spiro six-membered ring. Its residual double bond is between the two internal diene atoms, putting the linker on the required alkene carbon.
Build and its four-carbon alcohol appendage first. Protect the OH group of the indicated 4-Bromobutan-1-ol, for example as a THP ether. Convert the primary bromide to its grignard reagent and formylate with DMF, giving . olefination with an allylidenetriphenylphosphorane, , supplies a conjugated diene ; select the required diene geometry rather than assume a single E/Z outcome.
Cycloaddition with methyl vinyl ketone produces a cyclohexene with a protected four-carbon alcohol chain at one saturated ring carbon and an acetyl group at its adjacent ring carbon. Choose and isolate the adjacent-substituent cis regioisomer, matching the two dashed substituent bonds in the target. Endo preference and diene geometry can aid the selection, but an unsymmetrical diene is not guaranteed to give a unique regioisomer. Reduce the acetyl ketone to , then methylate that alcohol with sodium hydride/methyl iodide while the chain-end OH is protected. Remove the THP group under mild acid conditions to reveal , preserving the secondary methyl ether and the ring alkene.
Oxidize this primary alcohol to . Add vinylmagnesium bromide to give , then oxidize the allylic alcohol to the enone . This carbonyl carbon is the original alcohol-end carbon of 4-bromobutanol; after olefination it will be the linker-bearing carbon of the left ring alkene.
In the other branch, reduce the specified cyclopentanone to cyclopentanol, convert to bromocyclopentane, and react with triphenylphosphine to obtain the cyclopentylphosphonium salt. Deprotonation gives the cyclopentylidene phosphorane, whose ylidic carbon is part of the five-membered ring. Wittig olefination of the just-prepared enone gives the substituted diene displayed above. This carbonyl–ylide pairing keeps the ring carbon terminal in the diene and avoids accidentally constructing a fused ring instead of a spiro ring.
Finally, cycloadd an ethylene equivalent. Ethylene itself requires suitably forcing/high-pressure conditions; a more reactive alternative is phenyl vinyl sulfone, followed by reductive removal of the sulfone substituent using a method compatible with the methyl ether and retained C=C bonds. The latter installs the same two otherwise unsubstituted saturated carbons. Control or separate any additional cycloaddition isomers, and retain the previously selected cis configuration on .
The carbon count provides a strong check: cyclopentanone supplies five, bromobutanol four, DMF formylation one, the first allylidene ylide three, methyl vinyl ketone four, O-methylation one, vinylmagnesium two, and the final ethylene equivalent two, totalling 22 target carbons. Three original bromobutanol carbons form the linker; its fourth becomes the substituted carbon of the spiro ring's alkene. The proposal is a multistep constitutional/relative-stereochemical route with explicit regioisomer and diastereomer selection, not an unsupported claim of a single quantitative cycloaddition product.
