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Concept 04 / 08Organic Chemistry - Basic Principles and Techniques

Organic Compounds: Tetravalence of Carbon, Structural Representations and Classification

Tetravalence and hybridisation of carbon in organic molecules, sigma and pi bonds, complete, condensed and bond-line formulas, three-dimensional wedge-dash representation, classification of organic compounds into acyclic, alicyclic, aromatic and heterocyclic types, homologous series, functional groups and degree of unsaturation.

Figure 1
Organic chemistry is the chemistry of carbon compounds (other than simple ones like CO, CO2\mathrm{CO_{2}} and carbonates). Carbon's ability to catenate and to form strong single and multiple bonds with itself and with H, O, N, S and halogens gives millions of compounds. Tetravalence and hybridisation. Carbon forms four covalent bonds. sp3sp^{3}: four single bonds, tetrahedral (109.5∘109.5^{\circ}), as in alkanes. sp2sp^{2}: one double bond, trigonal planar (120∘120^{\circ}), as in alkenes and carbonyl compounds. spsp: a triple bond or two cumulated double bonds, linear (180∘180^{\circ}), as in alkynes and nitriles. More s character means a shorter bond, a more electronegative carbon (sp>sp2>sp3sp > sp^{2} > sp^{3}) and, for C-H, a more acidic hydrogen (ethyne is far more acidic than ethane). A π\pi bond forms by sideways overlap of parallel p orbitals; it is weaker than a σ\sigma bond, prevents free rotation, and its electrons are exposed to attack by electrophiles. Structural representations (see figure). Complete (expanded) formula: every bond shown by a line. Condensed formula: bonds omitted, identical groups grouped: CH3CH(OH)CH2CH3\mathrm{CH_{3}CH(OH)CH_{2}CH_{3}}, or CH3(CH2)4CH3\mathrm{CH_{3}(CH_{2})_{4}CH_{3}} for hexane. Bond-line formula: carbons are the corners and line ends of a zigzag; hydrogens on carbon are not drawn; heteroatoms and their hydrogens are shown. Three-dimensional: wedges for bonds towards the viewer, dashes for bonds away, plain lines in the plane of the paper. Classification. Acyclic (open chain, aliphatic): straight or branched chains. Cyclic: alicyclic (carbocyclic, non-aromatic: cyclohexane), aromatic (benzenoid: benzene, naphthalene; non-benzenoid: tropolone), and heterocyclic (ring containing N, O or S: pyridine, furan, thiophene). Homologous series: a family with the same functional group, successive members differing by CH2\mathrm{CH_{2}} (mass 1414), with a general formula and gradually changing physical properties (alkanes CnH2n+2\mathrm{C_{n}H_{2n+2}}, alkenes CnH2n\mathrm{C_{n}H_{2n}}, alcohols CnH2n+1OH\mathrm{C_{n}H_{2n+1}OH}). Functional group: the atom or group that determines characteristic reactions: −OH\mathrm{-OH}, −CHO\mathrm{-CHO}, >C=O\mathrm{>C=O}, −COOH\mathrm{-COOH}, −NH2\mathrm{-NH_{2}}, −X\mathrm{-X}, −NO2\mathrm{-NO_{2}}, −C≡N\mathrm{-C \equiv N}, and so on. Degree of unsaturation (double bond equivalents). For CcHhNnXxOo\mathrm{C_{c}H_{h}N_{n}X_{x}O_{o}}: DBE=c+1−h+x−n2\text{DBE} = c + 1 - \frac{h + x - n}{2} (oxygen does not count). Each ring or double bond counts 11; a triple bond counts 22; benzene has DBE 44. Key Takeaways: - sp3sp^{3} 109.5∘109.5^{\circ}, sp2sp^{2} 120∘120^{\circ}, spsp 180∘180^{\circ}; more s character, more electronegative carbon. - Bond-line formulas omit C and the H on C. - Homologues differ by CH2\mathrm{CH_{2}}; the functional group decides the chemistry. - DBE=c+1−h+x−n2\text{DBE} = c + 1 - \frac{h + x - n}{2}. - Pitfall: in bond-line formulas, the hydrogens on heteroatoms (OH, NH2\mathrm{NH_{2}}) must be shown. Brain-Teaser 1: Find the hybridisation of each carbon in CH2=CH−C≡N\mathrm{CH_{2}=CH-C \equiv N} and count its σ\sigma and π\pi bonds. Solution: C1 (CH2\mathrm{CH_{2}}) sp2sp^{2}, C2 (CH) sp2sp^{2}, C3 (nitrile carbon) spsp. Bonds: three C-H, one C-C, one C=C, one C≡N\mathrm{C \equiv N}. σ\sigma: 3+1+1+1=63 + 1 + 1 + 1 = 6; π\pi: 1+2=31 + 2 = 3. Brain-Teaser 2: Calculate the degree of unsaturation of C6H5NO2\mathrm{C_{6}H_{5}NO_{2}} (nitrobenzene) and of C4H6\mathrm{C_{4}H_{6}}, and suggest structures for the latter. Solution: Nitrobenzene: 6+1−5−12=56 + 1 - \frac{5 - 1}{2} = 5 (benzene ring 44 plus one N=O). C4H6\mathrm{C_{4}H_{6}}: 4+1−3=24 + 1 - 3 = 2: a diene (buta-1,3-diene), an alkyne (but-1-yne, but-2-yne), a cycloalkene (cyclobutene), or a bicyclic compound. Brain-Teaser 3: Why is the C-H bond in ethyne more acidic than that in ethene or ethane? Solution: The spsp carbon has 50%50\% s character, so it holds the electron pair of the carbanion closer to the nucleus, stabilising HC≡C−\mathrm{HC \equiv C^{-}}. Ethene (33%33\%) and ethane (25%25\%) carbanions are less stable. Ethyne reacts with sodium or sodamide to form sodium acetylide; ethane does not.