Carbonyl Compounds: Identification and Reactions of Aldehydes and Ketones, Study notes of Chemistry

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OCR Chemistry A H432 Carbonyl Compounds Page 1 Carbonyl Chemistry Carbonyl compounds are those which contain >C=O - aldehydes - ketones - carboxylic acids - esters You should recall how to name aldehydes and ketones: - longest chain is 5 carbons, so "-pentan-" stem - ketones have "-one" ending - C=O is located on 3rd carbon in chain - has a methyl group - methyl group on 2nd carbon (numbering for smallest) 2-methylpentan-3-one - longest chain is 5 carbons so "-pentan-" stem - aldehydes have an "-al" ending - numbering starts from the end with the aldehyde - methyl group is therefore on 4th carbon 4-methylpentanal The simplest aromatic aldehyde and ketone are: benzaldehyde - colourless liquid (at RT) with almond-like aroma - almond essence, gives flavour to marzipan phenylethanone - used to create fragrances which resemble cherry, strawberry, honeysuckle or jasmine Chemical tests to distinguish carbonyl compounds 1: Detecting an aldehyde or ketone Aldehydes and ketones react with 2,4-dinitrophenylhydrazine (2,4-DNP or 2,4-DNPH) to form an orange or yellow precipitate. No precipitate is formed with other carbonyl compounds such as carboxylic acids or esters. Brady's Reagent is a solution of 2,4-DNPH in methanol and sulphuric acid. A few drops of the carbonyl compound are put in a test tube with about 5cm3 of Brady's reagent. The precipitate formed, referred to as a 2,4-dinitrophenylhydrazone derivative, can be used to help identify the specific aldehyde or ketone, after purifying, by CH3 CH2 C CH O CH3 CH3 CH3 CH CH2 CH2 C O HCH3 C OH CH3 C O OCR Chemistry A H432 Carbonyl Compounds Page 2 measuring its melting point. This works well because the different derivatives have melting points that are many degrees apart. e.g. heptan-2-one b.p. = 151°C m.p of 2,4-DNP derivative = 90°C cyclohexanone b.p. = 156°C m.p of 2,4-DNP derivative = 162°C octan-2-one b.p. = 173°C m.p of 2,4-DNP derivative = 58°C Identifying an aldehyde/ketone from the 2,4-DNPH derivative: ! The orange/yellow solid substance is purified by recrystallisation ! Melting point is determined ! Melting point is compared to a database of published melting points for these derivatives You will not be asked to write an equation for the formation of a 2,4-DNP derivative or to recall who to draw the structure of a 2,4-dinitrophenylhydrazone derivative. 2: Telling an aldehyde from a ketone A further test is necessary to distinguish an aldehyde from a ketone. Aldehydes can be further oxidised to carboxylic acids, but ketones cannot. Tollens' reagent is a weak oxidising agent containing silver nitrate in ammonia. Aldehydes can be oxidised further whereas ketones are not oxidised. The oxidising agent is the aqueous silver (I) ion, Ag+(aq). When warmed with Tollens reagent, the aldehyde is oxidised to a carboxylic acid, and the silver ions in solution are reduced to silver metal. A "silver mirror" is formed on the walls of the test tube (or sometimes just a silver-grey solid is formed). Oxidation of the aldehyde: R-CHO + [O] " R-COOH Reduction of the silver ions: Ag+(aq) + e- " Ag(s) How the carbonyl group reacts Firstly we need to understand a little more about the carbonyl group: 1. Like a C=C double bond, it is comprised of a sigma bond and a pi bond formed by the overlap of p-orbitals on the C and O atoms. C O C C O C à π-bond above and below plane of other bonds σ-bond OCR Chemistry A H432 Carbonyl Compounds Page 5 Reagent: A suitable reagent is a solution of NaBH4(aq), sodium borohydride (sodium tetrahydridoborate III). This is a source of hydride ions, H-, which are the actual reducing agent. We can represent these in equations by [H]. Conditions: • Usually carried out by warming the carbonyl compound with the reducing agent. • Water is used as the solvent. Equations: • use [H] for the reducing agent • never write as ambiguous molecular formulae: structural, displayed or skeletal! Aldehydes are reduced to primary alcohols, e.g. CH3CH2CHO + 2[H] " CH3CH2CH2OH Ketones are reduced to secondary alcohols, e.g. CH3COCH3 + 2[H] " CH3CH2(OH)CH3 Mechanism of nucleophilic addition: The H- ion is a hydrogen atom with an extra electron in its shell, meaning it has a lone pair, and is negatively charged. It can donate this lone pair to form a bond, so it is a nucleophile. Nucleophiles attack δ+ centres – in this case the δ+ carbon of the C=O. In the first stage • the hydride ion attacks the δ+ carbon, using its lone pair to form a new bond. • the pi-bond between C and O atoms breaks, leaving an intermediate C-O- with a single bond and a negative charge on the O atom. In the second stage • the O atom of the intermediate donates a lone pair to a hydrogen atom of a water molecule forming a dative bond. • the O-H bond in water breaks heterolytically leaving hydroxide ion. e.g. reduction of propanal Don’t represent the second stage as O-: forming a bond to an H+ ion – there aren’t H+ ions available, this isn’t done under acidic conditions, unlike with NaCN/H+ when there are, and H+ can be used. H C H H C H H C O H :H- δ+ δ- " H C H H C H H C O H H intermediate .. H O H " H C H H C H H C OH H H + OH- OCR Chemistry A H432 Carbonyl Compounds Page 6 ii) reaction with HCN to form hydroxynitriles Hydrogen cyanide adds across the C=O to produce an –OH group and add a –CN (nitrile) group to the molecule. This is very useful in synthesis as it extends the carbon skeleton. Reagent: Hydrogen cyanide is colourless, extremely poisonous, and volatile boiling just above room temperature – it cannot be used safely in the laboratory so sodium cyanide and sulphuric acid are used to produce HCN in-situ. The reaction is still very hazardous. Conditions: Not required. Equations: H2SO4/NaCN e.g. CH3CH2CHO + HCN " CH3CH2CH(OH)CN propanal 2-hydroxybutanenitrile H2SO4/NaCN CH3COCH3 + HCN " CH3C(OH)(CN)CH3 propanone 2-hydroxy(-2-)methylpropanenitrile Mechanism of nucleophilic addition: The initial attack on the δ+ carbon of the C=O is by the cyanide ion, often written as CN-. This is a little misleading, as the negative charge is on the carbon atom, and it is a lone pair on the carbon that attacks. It is better to represent as [:CN]- or -:CN. In the first stage • the -:CN ion attacks the δ+ carbon, using its lone pair to form a new bond. • the pi-bond between C and O atoms breaks, leaving an intermediate C-O-. In the second stage • the O atom of the intermediate donates a lone pair to a hydrogen atom of a water molecule forming a dative bond (or an H+ ion in solution, OK for this reaction) • the O-H bond in water breaks heterolytically leaving hydroxide ion.