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A-Level Terminology and mechanisms of organic reactions

What the A-Level syllabus expects for Terminology and mechanisms of organic reactions, and how to practise it.

What the syllabus expects

How it's examined

Questions on this topic most often ask you to show, suggest. About 7% of the past-paper style questions in Rae's bank for this subject sit in this topic.

Worked examples

Example 1 (3 marks)

Set out the mechanism for the decarboxylation of trichloroethanoic acid, including all lone pairs, charges and curly arrows. Step 1: water acts as a Bronsted-Lowry base towards the acid. Step 2: the conjugate base from step 1 loses CO2. Step 3: the intermediates from steps 1 and 2 combine to give CHCl3.

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Step 1: A lone pair on the O of H2O attacks/removes the acidic H of CCl3COOH. Curly arrow from an O lone pair of water to the acidic H; curly arrow from the O-H bond of the carboxylic acid to the O. Products: H3O+ (oxonium, positive charge on O) and the carboxylate CCl3COO- (negative charge delocalised on the two carboxylate O atoms, each carrying lone pairs). Step 2: The carboxylate CCl3COO- loses CO2. Curly arrow from the C-C bond into the CCl3 carbon (forming the carbanion), and the electrons of the C-O reorganise to give O=C=O. Products: CO2 and the carbanion :CCl3- (trichloromethyl anion, lone pair and negative charge on C). Step 3: The carbanion :CCl3- (from step 2) takes a proton from H3O+ (the oxonium from step 1). Curly arrow from the C lone pair of :CCl3- to the H of H3O+; curly arrow from the O-H bond back to O. Products: CHCl3 + H2O.

Example 2 (3 marks)

Tautomerisation is a form of isomerism in which two species sharing one molecular formula but differing in atomic connectivity (constitutional isomers) rapidly interconvert at equilibrium in solution. The classic case is the shift between the keto form (bearing a carbonyl) and the enol form (bearing an -OH group next to a C=C double bond). Draw the acid-catalysed mechanism by which propanone converts into its enol tautomer, including every curly arrow, all relevant lone pairs and any charges.

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Acid-catalysed keto-to-enol tautomerisation of propanone, CH3COCH3. Step 1 - protonation of the carbonyl oxygen: a lone pair on the carbonyl O attacks H+ (from H3O+ / acid). Curly arrow from an O lone pair to the H; the O-H bond forms and O becomes positively charged. This gives the protonated ketone CH3-C(+)(OH)-CH3 (with the positive charge delocalised onto carbon), i.e. C=O(+)H. Step 2 - loss of an alpha proton: a base (e.g. H2O) removes a hydrogen from the alpha carbon (a CH3). Curly arrow from a lone pair on H2O to the alpha C-H hydrogen; a second curly arrow from the alpha C-H bond into the C-C bond to form the C=C double bond; a third curly arrow from the C=O(+) pi bond onto the oxygen, neutralising it and forming the O-H of the enol. Product: the enol, CH2=C(OH)-CH3, plus regenerated H3O+ (catalyst returned). Show: O lone pairs, the + charge on protonated intermediate, and all three curly arrows in the deprotonation step.

Example 3 (2 marks)

In step 3 the triiodoethanal reacts with the nucleophile OH- by substitution rather than the addition you might expect. Suggest a reason for this.

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OH- adds to the carbonyl carbon, but the resulting tetrahedral intermediate expels the CI3- group instead of just giving an alcohol. The -CI3 group is an unusually good leaving group because the negative charge on the departing CI3- carbanion is stabilised by the three highly electronegative/electron-withdrawing iodine atoms. This makes C-C cleavage (substitution) favourable, unlike ordinary aldehydes where no such stable leaving group is available and simple nucleophilic addition occurs.

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More A-Level H2 Chemistry topics

The make-up of the atom and how its electrons are arranged · How atoms bond and how that governs a substance's behaviour · Ideal gases and working with gas mixtures · Competing definitions of acids and bases · Trends in the elements across a period and down a group · The mole and reacting-quantity calculations · all of A-Level H2 Chemistry