Redox synthesis reactions in organic chemistry, Oxidation & Reduction reactions in Organic chemistry, Preparations involving oxidation or reduction, oxidation states of carbon in organic compounds, advanced A level chemistry revision notes

8. Summary of some ORGANIC SYNTHESIS REDUCTION REACTIONS PART 8-11 Redox reactions in Organic Chemistry Summary of page 3 contents Advanced Level Theoretical Organic Chemistry Revision Notes on REDOX reactions involving organic oxidation and reduction synthesis–preparations. Reactions involving the oxidation or reduction of the starting reactant molecules and the chemistry of the appropriate oxidising agents and reducing reagents is described and discussed. How do you work out the oxidation state/number of carbon in organic molecule compounds? A summary of some redox reactions used in organic synthesis is given below. Further details for selected reactions are given below the summary tables. Most of the reactions described are found in one or other of UK based GCE–AS–A2, IB courses and US grades 11-12 K12 AP Honors courses. The application of oxidation states to organic molecules can be tricky, but, (i) use of half–cell equations usually gets round the problem, and (ii) hopefully the oxidation state exemplars in the last section will help illuminate the situation if you are interested, but this knowledge is not required at this level? Important examination note: Unless hydrogen gas or oxygen gas is used directly in the redox synthesis reaction [O] and [H] should be used in simplified equations and examples will be quoted in each section and some syllabuses specifically state so. The concept of functional group level is dealt with in appendix 1. Guide notes: YES/NO denotes whether reaction possible. Lithium tetrahydridoaluminate(III), LiAlH₄ (lithium aluminium hydride), is a more powerful reducing agent than sodium tetrahydridoborate(III), NaBH₄ (sodium borohydride) and accounts for the NO/YES differences in columns (a) and (b). R = H, alkyl or aryl for aldehydes, carboxylic acids and nitriles, R = alkyl or aryl for ketones. Click on 8.1, 8.2, 8.3, 8.4 and 8.5 in the table for more details on (a), (b), (c) or (d) reaction reagents/conditions and the chemical outcome. Important examination note: Unless hydrogen gas or oxygen gas is used directly in the redox synthesis reaction [O] and [H] should be used in simplified equations and examples will be quoted in each section and some syllabuses specifically state so.
homologous series change on reduction	molecular structure change on reduction	reducing agent (a) NaBH₄ in water (lab method)	reducing agent (b) LiAlH₄ in e.g. ether (lab method)	reducing agent (c) reflux with Sn/conc. HCl(aq) (lab method)	reducing agent (d) Ni/H₂ (industry), but many other hydrogenation catalysts based on Pt, Pd etc.
8.1 alkene ====> alkane	>C=C< ====> >CH–CH<	NO	NO	NO	YES, Ni/H₂ used
8.2 aldehyde/ketone ====> primary/secondary alcohol	RCHO ==> RCH₂OH and R₂C=O ==> R₂CHOH	YES	YES	NO but can be reduced with Na/C₂H₅OH or Zn/CH₃COOH mixtures	NO with Ni/H₂ YES with other specialised catalysts
8.3 carboxylic acid ====> primary aliphatic alcohol	RCOOH ====> RCH₂OH	NO	YES	NO	NO with Ni/H₂ YES with other specialised catalysts
8.4 nitrile ====> primary aliphatic amine	RCN ====> RCH₂NH₂	NO	YES	NO	YES with Ni/H₂
8.5 nitro–aromatic ====> primary aromatic amine	e.g. C₆H₅NO₂ ====> C₆H₅NH₂	NO	YES	YES	YES with Ni/H₂
TOP OF PAGE and sub-index 8.1 Reduction of alkenes to alkanes Some further details of the organic reductions tabulated above How can you reduce unsaturated compounds like alkenes? e.g. unsaturated alkenes to saturated alkanes 8.1(a)/(b) Alkenes cannot be reduced by metal hydride complexes (see 8.2/8.3a–b) because the 'attacking' species is nucleophilic e.g. the negative (e^– pair donating) BH₄^– or AlH₄^– ions are repelled too strongly to allow 'disruption' of the electron rich C=C double bond in alkenes. Alkenes tend to react with electron pair seeking electrophilic reagents like H^δ+–Br^δ–. 8.1(c) Reduction reactions by metal occur on the surface of the metal and involves reactants weakly bonding with the metal. 8.1(d) e.g. propene ==> propane: CH₃CH=CH₂ + H₂ == Ni catalyst ==> CH₃CH₂CH₃ This reaction is initiated on the nickel catalyst surface which lowers the activation energy to help break the H–H bonds and 'half' of the C=C bond. (see diagram and explanation of the hydrogenation mechanism and Hydrogenation of alkenes to give saturated alkanes). The catalysed hydrogenation of alkenes is a very important industrial reaction used to convert natural polyunsaturated oils into low melting solid more saturated fats like margarine. 8.2 Reduction of aldehydes to primary alcohols and ketones to secondary alcohols How can you reduce carbonyl compounds like aldehydes and ketones? These reactions are essentially the reduction of the carbonyl group >C=O to >CHOH. 8.2(a) Using sodium tetrahydridoborate(III), NaBH₄ (sodium borohydride) The reaction can be carried out in water. The reduction mechanism is very complicated, but can be considered in a simplistic way as involving the donation of a hydride ion to the aldehyde/ketone. An outline of the nucleophilic addition mechanism is given on the Organic Mechanisms Part 3 page, but the simple equation will suffice here. aldehyde: RCHO + 2[H] ==> RCH₂OH (R = H, alkyl or aryl) e.g. ethanal to ethanol: CH₃CHO + 2[H] ==> CH₃CH₂OH ketone: R₂C=O + 2[H] ==> R₂CHOH (R = alkyl or aryl) e.g. propanone to propan–2–ol: CH₃COCH₃ + 2[H] ==> CH₃CH(OH)CH₃ 8.2(b) Using lithium tetrahydridoaluminate(III), LiAlH₄, (lithium aluminium tetrahydride), LiAlH₄ is a more powerful reducing agent than NaBH₄ and reacts violently with water (and reacts with ethanol too), so the reaction must be carried out in an inert non-aqueous solvent like dry ethoxyethane ('ether'). The initial product is hydrolysed by dilute sulfuric acid. The simplified equations above apply, see 8.2(a) LiAlH₄ is a more powerful reducing agent than NaBH₄ because the Al–H bond is weaker than the B–H bond. This fits in with the aluminium atom having a larger radius than the boron atom giving a longer weaker bond with hydrogen. Covalent radii: B = 0.080 nm (80 pm), Al = 0.125 nm (125 pm). 8.2(c) Aldehydes and ketones can be reduced to alcohols by reacting them with a sodium/ethanol ('alcohol') mixture or a zinc/ethanoic acid mixture, but these reactions are not usually dealt with in UK AS–A2 or IB chemistry. Reduction reactions by metal/acid occur on the surface of the metal and involve electron transfer from the metal to the organic compound. The reduction involves the half–cell reaction: >C=O + 2H⁺ + 2e^– ==> >CHOH Reaction at the zinc metal surface via the half–cell reaction: Zn_(s) ==> Zn²⁺_(aq) + 2e^– 8.2(d) The reduction of aldehydes/ketones with hydrogen and a purely Ni catalyst does NOT work. However, there are other specialised hydrogenation catalysts that will work. aldehyde: RCHO + H₂ ==> RCH₂OH (R = H, alkyl or aryl) ketone: R₂C=O + H₂ ==> R₂CHOH (R = alkyl or aryl) 8.3 Reduction of a carboxylic acid to a primary aliphatic alcohol 8.3(a) NaBH₄, is not a powerful enough reducing agent to reduce carboxylic acids. 8.3(b) LiAlH₄ is a more powerful reducing agent than NaBH₄, and in ether solvent, readily reduces carboxylic acids to primary alcohols. The reaction can be summarised as: RCOOH + 4[H] ==> RCH₂OH + H₂O (R = H, alkyl or aryl) e.g. propanoic acid to propan–1–ol: CH₃CH₂COOH + 4[H] ==> CH₃CH₂CH₂OH + H₂O or benzoic acid to phenylmethanol: C₆H₅COOH + 4[H] ==> C₆H₅CH₂OH + H₂O 8.3(c) As far as I know, metal/acid reducing agents are not powerful enough to reduce carboxylic acids. 8.3(d) Hydrogen/Ni will NOT reduce carboxylic acids, but there are other specialised catalysts that can effect this reduction. RCOOH + 4[H] ====> RCH₂OH + H₂O (R = H, alkyl or aryl) 8.4 Reduction of nitriles to primary aliphatic amines 8.4(a) NaBH₄, is not a powerful enough reducing agent to effect the change. 8.4(b) LiAlH₄ is a more powerful reducing agent than NaBH₄ and in ether solvent will reduce nitriles to primary aliphatic amines. The reaction can be summarised as: RCN + 4[H] ==> RCH₂NH₂ (R = H, alkyl or aryl) e.g. propanenitrile to propylamine: CH₃CH₂CN + 4[H] ==> CH₃CH₂CH₂NH₂ 8.4(c) Sn/HCl_(aq), is not a powerful enough reducing agent to reduce nitriles to amines. 8.4(d) Hydrogen/150^oC?/Ni catalyst conditions will reduce nitriles to primary aliphatic amines. RCN + 2H₂ ==> RCH₂NH₂ (R = H, alkyl or aryl) 8.5 Reduction of nitro–aromatics to primary aromatic amines 8.5(a) NaBH₄, is not a powerful enough reducing agent to reduce nitro–aromatic compounds. 8.5(b) LiAlH₄ is a more powerful reducing agent than NaBH₄ and in ether solvent readily reduces nitro–aromatics to primary aromatic amines, the simplified equation for nitrobenzene to phenylamine is ... C₆H₅NO₂ + 6[H] ==> C₆H₅NH₂ + 2H₂O and methylnitrobenzenes would be reduced to methylphenylamine primary amines, i.e. CH₃C₆H₄NO₂ + 6[H] ==> CH₃C₆H₄NH₂ + 2H₂O as will any aromatic compound with a nitro group (–NO₂) attached directly to the benzene ring. 8.5(c) The reduction of nitro–aromatics with tin and concentrated hydrochloric acid. In the laboratory, reacting a nitro–aromatic with a mixture of tin and conc. hydrochloric acid by heating under reflux will reduce it to a primary aromatic amine (–NH₂ directly attached to benzene ring). In industry a cheap metal like iron powder and acid are used or a direct reduction in the gas phase with hydrogen/transition metal catalyst (8.5(d). e.g. 2 cm³ of nitrobenzene, 4g of tin and 10 cm³ of conc. hydrochloric acid. In the 'laboratory' preparation, the mixture may need cooling with a beaker of cold water if the reaction is too vigorous, then gentle heating with a beaker of boiled water to complete the reaction (the condenser causes avoids loss of product). The formation of phenylamine (aniline) from nitrobenzene can be summarised as C₆H₅NO₂ + 6[H] ==> C₆H₅NH₂ + 2H₂O but the 'real' equations are rather more complicated, the simplest redox equation I can come up with is 2C₆H₅NO_2(aq) + 14H⁺_(aq) + 3Sn_(s) ==> 2C₆H₅NH₃⁺_(aq) + 3Sn⁴⁺_(aq) + 4H₂O_(l) which shows the formation of the phenylammonium cation because the amine is a base and formed in an acid medium. The tin(IV) ion is actually a chloro-complex ion of tin, SnCl₆^2–, the hexachlorostannate(IV) ion, so the full ionic–redox equation is more correctly written as ... if you really must! 2C₆H₅NO_2(aq) + 14H⁺_(aq) + 18Cl^–_(aq) + 3Sn_(s) ==> 2C₆H₅NH₃⁺_(aq) + 3[SnCl₆]^2–_(aq) + 4H₂O_(l) but you should appreciate that any phenylamine formed, will immediately react with hydrochloric acid and dissolve to form the salt C₆H₅NH_2(l) + HCl_(aq) ==> C₆H₅NH₃⁺_(aq) + Cl^–_(aq) Oxidation state changes: 3Sn (0) inc. to (+4) balanced by 2N decreasing from (+3) to (–3). and then conc. aqueous sodium hydroxide is added to free the amine (immiscible with water) from its arylammonium cation C₆H₅NH₃⁺_(aq) + OH^–_(aq) ==> C₆H₅NH_2(l) + H₂O_(l) The primary aromatic amine is then be extracted by steam distillation. For more on the theory of steam distillation see Equilibria Part 8.5 On addition of conc. sodium hydroxide (a strong soluble base) the amine separates out as an oily layer and the mixture is heated with the steam input via the procedure known as steam distillation. The addition of an alkali is necessary because the phenylamine base is soluble in the excess acid. A mixture of the amine and water 'steam distils' into the condenser and separates into two layers in the collection flask. Steam distillation is the most efficient way of extracting the phenylamine from the reaction mixture, leaving behind all the inorganic residues. The inorganic residues are soluble, so can't be filtered of. It would prove difficult to extract the phenylamine from the reaction mixture by direct distillation OR trying to use an extracting solvent directly on the reaction mixture. Its best leave as much of the reaction residues behind before attempting the final purification procedures. 2g of salt can be added to the immiscible liquid mixture to help the two layers separate out. The phenylamine layer is separated out with a separating funnel Some methods use ether solvent to extract the phenylamine from the steam distilled mixture. Can't say I like this idea since a fractional distillation is done later, ether is very volatile and very flammable! Either way, the 'damp' phenylamine or mixture with ether, can be dried with anhydrous solid sodium hydroxide or anhydrous potassium carbonate. The drying agent is filtered off and the dried liquid fractionally distilled to obtain pure phenylamine liquid. If ether isn't used, the phenylamine is distilled with an air condenser (just a tube, not Liebig style), though some texts say distil under reduced pressure because phenylamine can decompose at its boiling point. Phenylamine distils over between 180-185^oC at normal pressure, quite a high boiling point, and a water condenser can crack of very hot phenylamine vapour. 8.5(d) In the chemical industry aromatic nitro–compounds are more efficiently reduced with hydrogen gas using a Ni or Cu catalyst at elevated temperatures, rather than a 'laboratory style' preparation. The resulting primary aromatic amines are very important intermediate compounds in dye and drug manufacture e.g. C₆H₅NO₂ + 3H₂ ==> C₆H₅NH₂ + 2H₂O (nitrobenzene ==> phenylamine) CH₃C₆H₄NO₂ + 3H₂ ==> CH₃C₆H₄NH₂ + 2H₂O (nitromethylbenzenes ==> aminomethylbenzenes/methylphenylamines) CH₃C₆H₃(NO₂)₂ + 6H₂ ==> CH₃C₆H₃(NH₂)₂ + 4H₂O (dinitromethylbenzenes ==> diaminomethylbenzenes)

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9. Summary of some ORGANIC SYNTHESIS OXIDATION REACTIONS Guide notes Some direct catalytic oxidations at higher temperatures used in industry are included, but most are 'school laboratory' reactions. YES/NO – for a laboratory synthesis. R = alkyl or aryl for primary/secondary alcohols. Click on 9.1, 9.2, 9.3, 9.4 and 9.5 in the table for more details on (a), (b) and (c) Important examination note: Unless hydrogen gas or oxygen gas is used directly in the redox synthesis reaction [O] and [H] should be used in simplified equations and examples will be quoted in each section and some syllabuses specifically state so.
homologous series change on oxidation	molecular structure change	(a) heat with mod conc. H₂SO₄ and K₂Cr₂O_7(aq) (lab method)	(b) reflux with KMnO₄/NaOH_(aq) (lab method)	(c) oxygen + catalyst or thermal decomposition (industrial methods)
9.1 primary alcohol ==> aldehyde ==> carboxylic acid	RCH₂OH ===> RCHO ===> RCOOH	YES, can separate intermediate aldehyde, or allow complete oxidation to carboxylic acid	YES but only get RCOOH and of little synthetic use	e.g. CH₃CH₂OH ==> CH₃CHO (Cu/500^oC)
9.2 secondary alcohol ==> ketone	R₂CHOH ===> R₂C=O	YES	YES but of little synthetic use	(CH₃)₂CHOH ==> CH₃COCH₃ (Cu/500^oC)
9.3 tertiary alcohol ==> ?	R₃C–OH fairly stable (if oxidised C–C bonds broken ==> lower RCOOH, CO₂, H₂O products)	not readily oxidised – no synthetic use	not readily oxidised – no synthetic use	not readily oxidised – no synthetic use
9.4 alkyl groups on benzene ring	e.g. C₆H₅CH₃ ==> C₆H₅COOH	YES	YES	YES air/150^oC/Co salt
9.5 alkene ==> ?	details in appropriate box	NO	ethene ==> ethane–1,2–diol (at room temp.)	e.g. ethene ==> epoxyethane (Ag/250^oC)
Some further details of the organic oxidations tabulated above 9.1 Oxidation of primary alcohols to aldehydes and carboxylic acids (includes oxidation of aldehydes to carboxylic acids) 9.1(a) It is possible using the same reagent of aqueous sodium/potassium dichromate(VI)–sulfuric acid to oxidise a primary alcohol to either the aldehyde, or the carboxylic acid, depending on the reaction conditions. In order to selectively isolate the aldehyde this initial oxidation products must be removed from the reaction mixture as quickly as possible, otherwise oxidation proceeds to the carboxylic acid (diagram PD1 below). The 25% sulfuric acid is placed in the flask and gently simmered. The alcohol and aqueous sodium/potassium dichromate(VI) solution is dripped onto the hot acid. Immediately, the orange dichromate(VI) is reduced by the alcohol to the green chromium(III) ion and the alcohol is oxidised to the aldehyde or ketone. The diagram shows a bunsen burner being used to supply the heat ('my days'), these days its more likely, and safer, to use an electrical heater that the round bottomed flask fits in snugly. A spot of theory to explain the separation of the aldehyde/ketone from the reaction mixture. For the same carbon number, the boiling point of the polar aldehyde/ketone (^δ+C=O^δ–, but no H bonding) is lower than the original more polar alcohol (^δ–O–H^δ+, hydrogen bonding) whose bpt. is higher. Therefore, as long as the bpt. of the aldehyde/ketone is not too high, in the set–up shown above, the aldehyde rapidly distils over and condenses in the collection tube/flask with some water. This rapid in situ extraction ensures that most of the aldehyde (or ketone), 9.1(a) is not oxidised further. If the carboxylic acid of the same carbon number is required from a primary alcohol, the mixture is refluxed using the set–up illustrated in diagram PD2. (i) primary alcohol ==> aldehyde Cr₂O₇^2–_(aq) + 3RCH₂OH_(aq) + 8H⁺_(aq) ==> 3RCHO_(aq) + 2Cr³⁺_(aq) + 7H₂O_(l) reduction half reaction: Cr₂O₇^2–_(aq) + 14H⁺_(aq) + 6e^– ==> 2Cr³⁺_(aq) + 7H₂O_(l) oxidation half reaction: RCH₂OH_(aq) ==> RCHO_(aq) + 2H⁺_(aq) + 2e^–_(aq) (R = alkyl or aryl) Examples using simplified symbol equations e.g. ethanol ==> ethanal: CH₃CH₂OH + [O] ==> CH₃CHO + H₂O propan–1–ol (1–propanol, n–propyl alcohol, n–propanol) ==> propanal CH₃CH₂CH₂OH + [O] ==> CH₃CH₂CHO + H₂O The technique illustrated on the right (diagram PD5) is called heating under reflux, a method which enables a reaction to be carried out at a higher temperature than room temperature to speed up the reaction AND retain the solvent (reaction medium e.g. water) and any volatile reactant or product (e.g. an alcohol/aldehyde/ketone). As the mixture boils, the vapours of the solvent or volatile reactant/product are condensed back into the flask in the vertical condenser, so any volatile reactant is used up and no volatile product lost (at least at this stage in a preparation!). then under reflux conditions the further oxidation ... (ii) aldehyde ==> carboxylic acid Cr₂O₇^2–_(aq) + 3RCHO_(aq) + 8H⁺_(aq) ==> 3RCOOH_(aq) + 2Cr³⁺_(aq) + 4H₂O_(l) (R = alkyl or aryl) reduction half reaction: Cr₂O₇^2–_(aq) + 14H⁺_(aq) + 6e^– ==> 2Cr³⁺_(aq) + 7H₂O_(l) oxidation half–reaction: RCHO_(aq) + H₂O_(l) ==> RCOOH_(aq) + 2H⁺_(aq) + 2e^–_(aq) Examples using simplified symbol equations: ethanal ==> ethanoic acid, CH₃CHO + [O] ==> CH₃COOH propanal (propionaldehyde) ==> propanoic acid (propionic acid) CH₃CH₂CHO + [O] ==> CH₃CH₂COOH butanal (butyraldehyde) ==> butanoic acid (butyric acid) CH₃CH₂CH₂CHO + [O] ==> CH₃CH₂CH₂COOH so overall for reflux conditions (i) + (ii) gives (iii) primary alcohol ==> carboxylic acid 2Cr₂O₇^2–_(aq) + 3RCH₂OH_(aq) + 16H⁺_(aq) ==> 3RCOOH_(aq) + 4Cr³⁺_(aq) + 11H₂O_(l) the oxidation half–reaction: RCH₂OH_(aq) + H₂O_(l) ==> RCOOH_(aq) + 4H⁺_(aq) + 4e^–_(aq) (R = alkyl or aryl) Examples using simplified symbol equations: ethanol ==> ethanoic acid CH₃CH₂OH + 2[O] ==> CH₃COOH + H₂O propan–1–ol (1–propanol, n–propyl alcohol, n–propanol) ==> propanoic acid CH₃CH₂CH₂OH + 2[O] ==> CH₃CH₂COOH + H₂O 9.1(b) Heating a primary alcohol with a aqueous sodium hydroxide/potassium manganate(VII) mixture under reflux (diagram PD2) will give the sodium salt of the carboxylic acid and it is not possible to isolate the intermediate aldehyde. However, the acid/dichromate(VI) method 9.1(a) under reflux is better, and the carboxylic acid is less liable to further degradative oxidation. The complex reaction can be summarised as: RCH₂OH_(aq) + NaOH_(aq) + 2[O] ==> RCOO^–Na⁺_(aq) + 2H₂O_(l) (R = alkyl or aryl) After removing the excess KMnO₄/MnO₂ the weak acid is freed from its sodium salt by adding strong dilute hydrochloric acid. RCOO^–_(aq) + H⁺_(aq) ==> RCOOH_(aq/s) 9.1(c) Many chemical feedstocks are oxidised directly with molecular oxygen/transition metal catalyst to produce useful products in industry. e.g. 2CH₃OH + O₂ ==> 2HCHO + 2H₂O (Ag/500^oC, methanol ==> methanal) or 2CH₃CH₂OH + O₂ ==> 2CH₃CHO + 2H₂O (Ag/500^oC, ethanol ==> ethanal) and the latter reaction can also be achieved via a thermal decomposition using a different catalyst, e.g. CH₃CH₂OH ==> CH₃CHO + H₂ (Cu/500^oC), which is still an oxidation, right carbon (–1) to (+1) and 2 x hydrogen (+1) to (0). 9.2 Oxidation of secondary alcohols to ketones 9.2(a) In the case of secondary alcohols you only get the ketone, unless you reflux the alcohol/K₂Cr₂O₇/H₂SO_4(aq) for a long time, in which case the ketone can be oxidised to lower carbon number carboxylic acids, carbon dioxide and water etc. if the carbon chain is broken Ketones are quite stable to further oxidation due to the strong carbon–carbon (C–C) bonds that have to be broken. As with the aldehyde, you can distil off the ketone, having a lower boiling point than the parent secondary alcohol. To be on the safe side it is better to make the ketone under the same restricted reaction conditions used to produce the aldehyde (details above with diagram PD1). Cr₂O₇^2–_(aq) + 3R₂CHOH_(aq) + 8H⁺_(aq) ==> 3R₂C=O_(aq) + 2Cr³⁺_(aq) + 7H₂O_(l) oxidation half–reaction: R₂CHOH_(aq) ==> R₂C=O_(aq) + 2H⁺_(aq) + 2e^–_(aq) (R = alkyl or aryl) 9.2(b) Ketones are produced by refluxing secondary alcohols with NaOH/KMnO_4(aq), but further oxidation is likely to take place because this reagent is a stronger oxidising agent than acidified potassium dichromate(VI). (CH₃)₂CHOH + [O] ==> (CH₃)₂C=O + H₂O (propan–2–ol ==> propanone) 9.2(c) Many chemical feedstocks are oxidised directly with molecular oxygen/transition metal catalyst to produce useful products in industry. 2(CH₃)₂CHOH_(g) + O_2(g) ==> 2(CH₃)₂C=O_(g) + 2H₂O_(g) (Ag/500^oC) and this reaction can also be achieved via a thermal decomposition using a different catalyst. (CH₃)₂CHOH_(g) ==> (CH₃)₂C=O_(g) + H_2(g) (Cu/500^oC) which is still an oxidation, right carbon (–1) to (+1) and 2 x hydrogen (+1) to (0). 9.3 Oxidation of tertiary alcohols 9.3(a)–(d) Tertiary alcohols, R₃COH (R = alkyl or aryl), are not readily oxidised because strong carbon–carbon bonds have to be broken. The products would be lower chain carboxylic acids, carbon dioxide and water and therefore of no synthetic use. 9.4 Oxidation of alkyl–aromatic hydrocarbons to aromatic carboxylic acids 9.4(a) Acidified potassium dichromate(VI) will oxidise alkyl benzene compounds to benzoic acid, but I think it is slower than with the alkaline manganate(VII) method described below. Overall change is represented by the equations ... C₆H₅CH₃ + 3[O] ==> C₆H₅COOH + H₂O 9.4(b) Aromatic are not easily oxidised and longish reflux times are necessary (illustrated, fig. PD2 right). Hydrocarbons are difficult to oxidise with typical organic oxidising agents compared to compounds like alcohols. However, aromatic hydrocarbons with an alkyl side chain can be oxidised with strong reagents such as aqueous potassium manganate(VII)/sodium hydroxide. Whatever the length of the alkyl group on a benzene ring it gets whittled down to a carbon atom of the carboxylic acid group e.g. propyl benzene ends up as benzoic acid. The more stable aromatic benzene ring is left intact. The overall process for producing benzoic acid from methylbenzene can be summarised .. C₆H₅CH₃(l) + NaOH(aq) + 3[O] ====> C₆H₅COO^–Na⁺(aq) + 2H₂O(l) After removing the excess KMnO₄/MnO₂ the weak benzoic acid is freed from its sodium salt by adding strong dilute hydrochloric acid. C₆H₅COO^–_(aq) + H⁺_(aq) ==> C₆H₅COOH or in principle you eventually get benzene–1,2–dicarboxylic acid from 1,2–dimethylbenzene + 6[O] ==> + 2H₂O 9.4(c) Aromatic hydrocarbons with alkyl groups can be directly oxidised with air/oxygen at elevated temperatures and pressures. C₆H₅CH_3(g) + ³/₂O_2(g) ==> C₆H₅COOH_(g) + H₂O_(g) (e.g. air/150^oC/Co salt catalyst) 9.5 Oxidation of alkenes 9.5(a) There are no useful oxidations of alkenes with acidified potassium dichromate(VI) as far as I know. 9.5(b) At room temperature alkenes react with alkaline potassium manganate(VII), KMnO₄/NaOH_(aq), to form diols e.g. ethene ==> ethane–1,2–diol (at room temp.) CH₂=CH₂ + H₂O + [O] ==> HO-CH₂-CH₂-OH 9.5(c) In industry alkenes can be oxidised directly with molecular oxygen to useful products e.g. epoxyalkanes at elevated temperatures and transition metal catalysts. ethene + oxygen ==> epoxyethane (Ag catalyst/250^oC) 2CH₂=CH₂ + O₂ ==> 2 propene + oxygen ==> epoxypropane (Ag catalyst/250^oC ?) 2CH₃–CH=CH₂ + O₂ ==> 2

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10. Other miscellaneous Organic Redox Reactions

This is a 'collection' of reactions not dealt with in sections 8. and 9.

They may/may not be useful reactions.

Section 10. reaction sub–index

10.1 Cannizzaro reaction

10.2 aldehydes/ketones tests

10.3 Combustion

10.4 Fuel cells

10.5 Use of 1,4–dihydroxybenzene (quinol, hydroquinol) in photography

10.1 The Cannizzaro reaction
- Aldehydes which do not have a hydrogen atom on the carbon next to the carbon of the carbonyl group (C=O) undergo the Cannizzaro reaction with concentrated aqueous sodium hydroxide in which one molecule of the aldehyde is reduced to a primary alcohol and another is oxidised to the sodium salt of a carboxylic acid.
- This is an organic example of disproportionation in which the same carbon atoms of the reactant molecule simultaneously increase and decrease their oxidation state e.g.
  - methanal changes to methanol and sodium methanoate
    - 2HCHO + Na⁺OH^– ==> CH₃OH + HCOO^–Na⁺
  - benzaldehyde changes to phenylmethanol (benzyl alcohol) and sodium benzoate
    - 2C₆H₅CHO + Na⁺OH^– ==> C₆H₅CH₂OH + C₆H₅COO^–Na⁺
10.2 Simple chemical tests to distinguish aldehydes from ketones
- The tests depend on the relative redox stability of ketones compared to the much more readily oxidised aldehydes. These tests also give positive results with many reducing sugars and some rather more stable aromatic aldehydes e.g. benzaldehyde, may not give a positive result at all.
- In these tests, because aldehydes are stronger reducing agents than ketones, they reduce the metal ion and are oxidised in the process i.e. RCHO + [O] ==> RCOOH, a change which is equivalent to a 2 electron loss by the RCHO.
- Tollens reagent is a colourless solution of silver nitrate in aqueous ammonia.
- When an aldehyde is warmed with Tollens reagent it is oxidised to a carboxylic acid and the silver ion (in ammine complex form) is reduced to silver, forming a silver mirror on the side of the test tube.
2[Ag(NH₃)₂]⁺_(aq) + R–CHO_(aq) + H₂O_(l) ==> 2Ag_(s) + 4NH_3(aq) + R–COOH_(aq) + 2H⁺_(aq)
simplified: 2Ag⁺_(aq) + R–CHO_(aq) + H₂O_(l) ==> 2Ag_(s) + R–COOH_(aq) + 2H⁺_(aq)
or 2[Ag(NH₃)₂]⁺_(aq) + R–CHO_(aq) + 2OH^–_(aq) ==> 2Ag_(s) + 4NH_3(aq) + R–COOH_(aq) + H₂O_(l)
simplified: 2Ag⁺_(aq) + R–CHO_(aq) + 2OH^–_(aq) ==> 2Ag_(s) + R–COOH_(aq) + H₂O_(l)
- Ketones show no reaction because of their greater stability to oxidation.
- Fehlings or Benedict's solution consists of a copper(II) ion complexed with an organic carboxylic acid.
- The deep blue copper(II) ion in the complex is reduced to a red–brown precipitate of copper(I) oxide.
- Ketones show no reaction due to their greater oxidation stability.
- 2Cu²⁺_(complex/aq) + R–CHO_(aq) + 2H₂O_(l) ==> Cu₂O_(s) + R–COOH_(aq) + 4H⁺_(aq)
- or
- 2Cu²⁺_(complex/aq) + R–CHO_(aq) + 4OH^–_(aq) ==> Cu₂O_(s) + R–COOH_(aq) + 2H₂O_(l)
- Ketones show no reaction because of their greater reluctance to oxidation.
- There are lots more organic chemical tests described – Chemical Identification Tests (with alphabetical index).
10.3 All organic compound air/oxygen combustion reactions are oxidations
- in terms of the carbon atoms of the organic molecule e.g. carbon's oxidation state is increased from (–4) in methane to (+4) in CO₂ for complete combustion and (+2) if carbon monoxide formed or (0) in carbon (soot) if the combustion is inefficient/incomplete. In each case oxygen's oxidation state changes from (0) to (–2) to offset the increase in carbon's oxidation state. e.g.
  - CH_4(g) + 2O_2(g) ==> CO_2(g) + 2H₂O_(l)
  - 2CH_4(g) + 3O_2(g) ==> 2CO_(g) + 4H₂O_(l)
  - CH_4(g) + O_2(g) ==> C_(s) + 2H₂O_(l)
10.4 Organic Fuel Cells
- Many hydrocarbon molecules are burned as fuels, but since it is a redox reaction, theoretically it can be done as a combined half–cell oxidation/reduction electron transfer reaction.
- In practice organic compounds can be oxidised by oxygen in a way that can be used to generate electricity directly in a fuel cell (right diagram) rather than release the energy as heat e.g.
- In reaction 9.1(a) ethanol was oxidised to ethanoic acid by acidified potassium dichromate(VI). The same result can be obtained by reaction of ethanol with oxygen in a fuel cell.
- The reaction is very exothermic, so ethanol is a good source of chemical potential energy.
- Ethanol is used directly as a fuel in a direct ethanol fuel cell (a DEFC fuel cell).
- (i) CH₃CH₂OH_(aq) + O_2(g) ==> CH₃COOH_(aq) + H₂O_(l) (ΔH^Ø = –494 kJ mol^–1)
- The half–cell reactions are:
  - (ii) Reduction, +ve electrode:
  - O_2(g) + 4H⁺_(aq) + 4e^– ==> 2H₂O_(l) (E^Ø = +1.23V)
  - (iii) Oxidation, –ve electrode:
  - CH₃CH₂OH_(l) + H₂O_(l) ==> CH₃COOH_(aq) + 4H⁺_(aq) + 4e^– (E^Ø = +0.06V)
  - Adding (ii) + (iii) = equation (i) so ...
    - E^Ø_cell = E^Ø_+/red'n – E^Ø_–/ox'n = 1.23 – 0.06 = +1.17V
  - The reactions must take place on catalytic electrodes made of platinum and other transition metals and here the inner 'electrolyte' is as a polymer proton exchange membrane of the fuel cell (a PEFC fuel cell), though it can be a concentrated phosphoric acid solution (in a PAFC fuel cell).
  - The electrons will flow through the external circuit from the –ve electrode to the +ve electrode.
  - Free energy change: ΔG^Ø = –nE^ØF
  - = – 4 x 1.17 x 96500 = –451620 J mol^–1 = –451.6 kJ mol^–1
  - n = number of electrons transferred, E^Ø = cell voltage, F = Faraday constant in C mol^–1 coulombs/mole
  - This sort of chemistry is being developed to make reasonably efficient portable fuel cells
- Commercially developed fuel cells will hopefully convert the ethanol completely into water and carbon dioxide to give a greater and more efficient energy output, but its still generates a voltage of 1.0–1.2V per cell, in this case ...
  - (i) Oxidation, –ve electrode: C₂H₅OH_(l) + 3H₂O_(l) ==> 12H⁺_(aq) + 2CO_2(aq/g) + 12e^–
  - (iii) Reduction, +ve electrode: 3O_2(aq/g) + 12H⁺_(aq) + 12e^– ==> 6H₂O_(l)
  - adding (i) + (iii) gives: C₂H₅OH_(l) + 3O_2(aq/g) ==> 3H₂O_(l) + 2CO_2(aq/g)
  - + other organic products as the reaction is not completely efficient.
  - The cells can obviously be connected in series to give larger voltages.
  - The ethanol ('alcohol', C₂H₅OH) can be bio–sourced from sugar beet, potatoes and cereal crops and other plant material that can fermented with enzymes. The ethanol is fractionally distilled from the fermented mixture and constitutes a renewable fuel.
- However there seems to have been more work done? on the Direct Methanol Fuel Cell (a DMFC fuel cell), though there are concerns over methanol's toxicity and the very costly platinum catalytic electrodes required, but the DMFC is essentially like the DEFC described above and the principles illustrated in the diagram.
- Methanol can be synthesised in the reforming reaction: CO_(g) + 2H_2(g) ==> CH₃OH_(l)
  - half–cell reactions for the DMFC cell:
  - (i) Reduction, +ve electrode:
    - ³/₂O_2(g) + 6H⁺_(aq) + 6e^– ==> 3H₂O_(l) (E^Ø = +1.23V)
  - (ii) Oxidation, –ve electrode:
    - CH₃OH_(aq) + H₂O_(l) ==> CO_2(aq/g) + 6H⁺_(aq) + 6e^– (E^Ø = ?V)
  - so overall the cell reaction is ... with a maximum output voltage of about 1.0V.
  - CH₃OH_(aq) + ³/₂O_2(g) ==> CO_2(aq/g) + 3H₂O_(l)
- Hopefully, efficient cells using ethanol can be developed because ethanol compared to methanol has a higher energy density (e.g. kJ/kg), is less toxic and can be bio–resourced as a renewable fuel.
- There is a huge amount of research going on in fuel cell development to try to use cheaper, but equally effective tiny particle metal catalysts of Fe, Co and Ni instead of costly platinum, but the most efficient metals are still the most costly?
10.5 The use of 1,4–dihydroxybenzene (p–quinol, hydroquinol) in photography
- 1,4–dihydroxybenzene (Quinol) is an ingredient in photographic developing solutions. In the transparent plastic emulsion film of the exposed film it will reduce the silver ions in silver halide salts, (which have not already been decomposed by light to silver and bromine), to silver as well. It would appear that these extra silver atoms produced by the developer, cluster around the already formed silver atoms (acting as 'nuclei') from the action of light, to enhance the image in 'shades' and 'contrast' to give a more clearly defined negative.
  - The relevant half–cell reactions and their standard redox electrode potentials are ...
  - AgBr_(s) + e^– ==> Ag_(s) + Br^–_(aq) (E^θ_Ag/AgBr = +0.073V at 298K)
  - _(aq) + 2H⁺_(aq) + 2e^– ==> _(aq) (E^θ_Q/QH2 = +0.059V (at 298K, pH 8.5)
  - These two organic molecules are referred to by a variety of systematic and trivial names, confusing when searching the web! e.g.
  - left (ref Q): 2,5–cyclohexadiene–1,4–dione, p–quinone, ('one' higher ranking than 'ene')
  - right (ref QH2): 1,4–dihydroxybenzene, 1,4–benzenediol, benzene–1,4–diol, p–quinol, p–hydroquinone, p–dihydroxybenzene
- It is convention to show the half–cell reactions as reductions when quoting them with the half–cell potential.
- In the reactions the silver ions in the silver bromide salt are reduced to silver atoms and 1,4–dihydroxybenzene (quinol) is oxidised to 2,5–cyclohexadiene–1,4–dione (quinone). The bromide ions 'dissolve' and two hydrogen ions are released from the quinol. Therefore the basic reaction via the alkaline developer is ....
  - _(aq) + 2AgBr_(s) ==> _(aq) + 2Ag_(s) + 2Br^–_(aq) + 2H⁺_(aq)
  - Sometimes more simply written as ...
  - C₆H₄(OH)_2(aq) + 2AgBr_(s) ==> C₆H₄O_2(aq) + 2Ag_(s) + 2HBr_(aq)
  - E^θ_reaction = E^θ_red – E^θ_ox = E^θ_AgBr/Ag – E^θ_Q/QH2
  - E^θ_reaction = (+0.073) – (+0.059) = +0.014V
  - The positive value for E^θ_reaction shows the reaction is feasible. In fact the E^θ_Q/QH2 becomes less positive as the pH is increased making the reaction more feasible, but the E_AgBr/Ag is theoretically independent of pH. The hydrogen ions would be neutralised by the alkaline media.
- Oxidation number analysis for the molecules in question is shown below ...
  - using graphic structural formulae and
  - Oxidation states in organic compounds is explained in section 11. below.

Note: Not all the silver bromide reacts in this way, so in the fixing process the remaining silver bromide is removed by sodium thiosulfate to produce the 'light stable' negative.
The transition metal complex ion [Ag(S₂O₃)₂]^3–_(aq) is formed when sodium thiosulfate (Na₂S₂O₃) is used to remove unreacted silver bromide (AgBr) crystals in developing photographic films.
- AgBr_(s) + 2S₂O₃^2–_(aq) ==> [Ag(S₂O₃)₂]^3–_(aq) + Br^–_(aq)
- This is NOT a redox reaction, Ag is +1 and Br is –1 throughout the reaction. The thiosulfate ion is here acting as a ligand and not a reducing agent e.g. like its reaction with iodine.

TOP OF PAGE and sub-index

11. Oxidation state of carbon in organic compounds

Usually the oxidation state of hydrogen is +1, and oxygen –2 in organic compounds, but carbon's oxidation state is variable.

This section should enable you to answer questions such as 'what is the oxidation state of carbon in methane?', 'what is the oxidation state of carbon in methanol?', ''what is the oxidation state of carbon in methanol?', 'what is the oxidation state of carbon in methanoic acid?', ''what is the oxidation state of carbon in ethane?', ''what are the oxidation states of carbon in ethanol?', ''what are the oxidation states of carbon in ethanal?', ''what are the oxidation states of carbon in ethanoic acid?', ''what is the oxidation state of carbon in ethene?', ''what are the oxidation states of carbon in propene?',

(The quoted Pauling electronegativities are C = 2.5, H = 2.2 and O = 3.5 which gives the lead in assigning oxidation numbers in this section i.e. the highest oxidation state is assigned to the least electronegative atom and vice versa for hydrogen and oxygen BUT beware for carbon, logical deduction can give some surprising, but correct results!)

On this basis you can achieve a useful oxidation number analysis of simple organic compounds in an oxidation sequence e.g. the oxidation sequence below, with oxidation state of carbon in () with hydrogen (+1) and oxygen (-2).

CH₄ (–4) == ox'n ==> CH₃OH (–2) == ox'n ==> HCHO (0) == ox'n ==> HCOOH (+2) ==> CO₂ (+4)

The above sequence can described in terms of the 'level of the functional group' which is equal the number of bonds the carbon atom forms with more electronegative atoms like oxygen.

Therefore in the above sequence:

alkane hydrocarbons are level zero (carbon and hydrogen have virtually the same electronegativity

alcohols are level 1 (C–OH, as are halogenoalkanes/haloalkanes e.g. C–Cl)

carbonyl is level 2 (e.g. C=O in aldehydes/ketones)

carboxylic acids are level 3 (C=O and C–OH)

and finally the fully oxidised carbon in carbon dioxide is level 4.

Note that these levels equate with a 'functional group level' and applies to a single carbon atom e.g. the carbon of the functional group, and not the full molecule. (See Appendix 1.)

Similarly for the oxidation sequence from ethane to ethanoic acid ...

CH₃CH₃ (–3,–3) = ox'n => CH₃CH₂OH (–3,–1) = ox'n => CH₃CHO (–3,+1) = ox'n => CH₃COOH (–3,+3)

In terms of oxidation states, as far as I can tell, in most organic compounds, hydrogen always seems to be +1 and oxygen –2 (except in e.g. organic peroxides R-O-O-R, R = alkyl or aryl)

Note the rise of carbon's oxidation state in increments of 2, see oxidation equations for acidified potassium dichromate(VI) reaction with alcohols and aldehydes in section 9.1(a) where the half–cell oxidation equations involve a 2 electron loss from the organic molecule.

Other organic molecules and redox sequences can be similarly 'analysed'

ethene H₂C=CH₂ (–2,–2) + H₂ (0) == reduction/Ni ==> ethane CH₃–CH₃ (–3,–3), (+1)

propene CH₃–CH=CH₂ (–3,–1,–2) + H₂ == reduction/Ni ==> CH₃–CH₂–CH₃ (–3,–2,–3)

ethanol CH₃–CH₂–OH (–3,–1) == ox'n ==> ethanal CH₃CHO (–3,+1) == ox'n ==> CH₃COOH (–3,+3)

TOP OF PAGE and sub-index

Appendix 1. The Concept of Functional Group Level

This concept of functional group level has been introduced into some UK A level pre–university courses.
It is related to an increase in the oxidation state of carbon.
As the oxidation state of a carbon atom is increased by combining with a more electronegative atom, so the functional group level is increased.
In the explanation with examples below assume R = a H, alkyl or aryl grouping.
The concept of functional group level of a carbon atom is derived from counting the number of bonds of an individual carbon atom to electronegative atoms X, i.e. C-X
- (For C–X, where I assume X is more electronegative than carbon i.e. oxygen, nitrogen or halogens such as chlorine)
If no such C–X bonds exist the carbon is described as being at the hydrocarbon level.
With one C–X bond the carbon atom is at the alcohol level e.g. alcohols (C–OH), monohaloalkanes (R₃C–Cl) and ethers (R₃C–O–CR₃) where R is alkyl.
With two C–X bonds the carbon atom is at the carbonyl level e.g. aldehydes and ketones (R₂C=O) and dihydroxy compounds with the >C(OH)₂ grouping.
With three C–X bonds the carbon atom is at the carboxylic acid level RCOOH (RO–C=O grouping), esters RCOOR, amides RCONH₂, nitrile RCN (RC≡N), X is O or N.
Finally with four C–X bonds we reach the carbon dioxide level e.g. carbon dioxide itself CO₂ or an ether such as C(OR)₄
The concept of functional group level often applies to a single carbon atom, not the whole molecule e.g.
- alcohol COH, chloroalkane C–Cl, aldehyde CHO,
- carboxylic acid level: carboxylic acid COOH, amide CONH₂,
- but not exclusively e.g. C=C in alkenes, but would this would be considered at the hydrocarbon level despite being at a higher oxidation state than a saturated alkane? (see section 11).
If a reaction takes place within a level you are swapping one heteroatom (non–carbon atom) for another
- e.g. hydrolysis of a halogenoalkane: R₃C–Cl + NaOH ==> R₃COH + NaCl
- which is occurring at the alcohol level
In order to move a carbon atom up a level requires an oxidizing agent (dealt with on this page in section 9).
- e.g. the oxidation of alcohols using acidified potassium dichromate(VI) solution
- R–CH₂OH == Cr₂O₇^2–/H⁺ ==> RCHO == Cr₂O₇^2–/H⁺ ==> RCOOH
- alcohol level ==> carbonyl level ==> carboxylic acid level
To move a carbon atom down a level requires a reducing agent (dealt with in section 8) or carbanion equivalent (not dealt with yet)
- e.g. the reduction of carboxylic acids to primary alcohols with lithium tetrahydridoaluminate(III)
- RCOOH + 4[H] ==> RCH₂OH + H₂O
- In this case the carbon atom of the functional group is reduced two levels i.e. from the carboxylic level to the alcohol level.

Learning objections for redox reactions involving organic molecules

Know that organic synthesis can involve redox reactions, that is synthesis routes employing oxidising agents and reducing agents, often to change or form a different organic functional group.

Know that reducing agents can be selective - don't assume they can reduce any molecule capable of being reduced.

Know which reducing agents can be used for a particular reduction synthesis reactions.

Know and be able to describe and explain the reduction of alkenes to alkanes with hydrogen and nickel catalyst.

Know and be able to describe and explain that aldehydes can be reduced to primary alcohols and ketones reduced to secondary alcohols using sodium tetrahydridoborate(III), NaBH₄ (sodium borohydride)

Know and be able to describe and explain that carboxylic acids can be reduced to primary alcohols using lithium tetrahydridoaluminate(III), LiAlH₄, (lithium aluminium tetrahydride).

Know and be able to describe and explain that nitriles can be reduced to primary aliphatic amines using lithium tetrahydridoaluminate(III), LiAlH₄, (lithium aluminium tetrahydride).

Know and be able to describe and explain that nitroaromatic compounds can be reduced to primary aromatic amines using tin/hydrochloric acid or lithium tetrahydridoaluminate(III), LiAlH₄, (lithium aluminium tetrahydride).

Know which oxidising agents can be used for a particular oxidation synthesis reactions.

Know and be able to describe and explain the oxidation of primary alcohols to aldehydes and carboxylic acids and the oxidation of secondary alcohols to ketones and know that tertiary alcohols resist oxidation until the carbon chain is broken.

Know and be able to describe and explain that alkyl groups directly attached to a benzene ring are completely oxidised to leave a carboxylic acid group i.e. an aromatic carboxylic acid is formed.

Know and be able to describe and explain the tests for aldehydes and ketones.

Know and be able to describe and explain the use of organic molecules in fuel cells and appreciate the redox chemistry involved.

Know and be able to appreciate that carbon exhibits different oxidation states in organic compounds e.g. -4, -2, 0 and +2.

Know and be able to describe, understand and explain the concept of functional group level e.g.

hydrocarbon level ==> alcohol level ==> carbonyl level (aldehydes/ketones) ==> carboxylic acid level

Know and be able to describe and explain the

Know and be able to describe and explain the

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