Organic Synthesis and Reaction Pathways

AS · 12 min

Organic synthesis is where all of AS organic chemistry comes together. Given a starting material and a target, can you plan the steps in between, with the right reagent and conditions for each? Given a molecule with three different functional groups, can you predict what each reagent will do to it? Given someone else's route, can you name each type of reaction and spot the by-products? The syllabus asks for exactly these three skills, and they carry a large share of the marks in the longer Paper 2 questions. This note gathers every AS reaction into one map and one table, then sets out methods for planning and analysing routes, with graded examples.

The map of AS reactions

Map of the AS organic interconversions. Each numbered arrow is a reaction in the table below; double-headed arrow 21 means esterification one way and hydrolysis the other.

Key result

no.	conversion	reagents and conditions	type of reaction
1	alkane → halogenoalkane	$\ce{Cl2}$ or $\ce{Br2}$ , UV light	free-radical substitution
2	alkene → alkane	$\ce{H2}$ , Ni catalyst, heat (or Pt, room temperature)	addition (hydrogenation)
3	alkene → halogenoalkane	HX(g), room temperature (or $\ce{X2}$ , room temperature, giving a dihalogenoalkane)	electrophilic addition
4	halogenoalkane → alkene	NaOH in ethanol, heat under reflux	elimination
5	alkene → alcohol	steam, $\ce{H3PO4}$ catalyst, about $300\ ^\circ\text{C}$ , $6\ \text{MPa}$	electrophilic addition (hydration)
6	alcohol → alkene	heated $\ce{Al2O3}$ , or conc. $\ce{H2SO4}$ (or $\ce{H3PO4}$ ), heat	elimination (dehydration)
7	alkene → diol	cold, dilute, acidified $\ce{KMnO4}$	oxidation
8	alkene → poly(alkene)	heat and high pressure, or a catalyst	addition polymerisation
9	halogenoalkane → alcohol	NaOH(aq), heat under reflux	nucleophilic substitution (hydrolysis)
10	alcohol → halogenoalkane	HX; KBr + conc. $\ce{H2SO4}$ ; KI + conc. $\ce{H3PO4}$ ; $\ce{PCl3}$ + heat; $\ce{PCl5}$ ; $\ce{SOCl2}$	substitution
11	halogenoalkane → primary amine	excess $\ce{NH3}$ in ethanol, heated under pressure	nucleophilic substitution
12	halogenoalkane → nitrile (+1 C)	KCN in ethanol, heat under reflux	nucleophilic substitution
13	nitrile → carboxylic acid	dilute HCl, heat under reflux (or NaOH(aq), reflux, then acidify)	hydrolysis
14	primary alcohol → aldehyde	acidified $\ce{K2Cr2O7}$ , distil off the aldehyde as it forms	oxidation
15	aldehyde → primary alcohol	$\ce{NaBH4}$ (or $\ce{LiAlH4}$ )	reduction
16	secondary alcohol → ketone	acidified $\ce{K2Cr2O7}$ , heat	oxidation
17	ketone → secondary alcohol	$\ce{NaBH4}$ (or $\ce{LiAlH4}$ )	reduction
18	aldehyde (or primary alcohol) → carboxylic acid	acidified $\ce{K2Cr2O7}$ (or $\ce{KMnO4}$ ), excess, heat under reflux	oxidation
19	aldehyde or ketone → hydroxynitrile (+1 C)	HCN with KCN catalyst, heat	nucleophilic addition
20	hydroxynitrile → 2-hydroxycarboxylic acid	dilute HCl, heat under reflux	hydrolysis
21	carboxylic acid + alcohol ⇌ ester	forward: conc. $\ce{H2SO4}$ catalyst, heat; back: dilute acid or dilute NaOH(aq), heat under reflux	condensation (esterification); hydrolysis
22	alcohol → ester	carboxylic acid, conc. $\ce{H2SO4}$ , heat	condensation
23	carboxylic acid → primary alcohol	$\ce{LiAlH4}$ in dry ether	reduction

Not on the map but also needed: complete and incomplete combustion; cracking of alkanes; hot concentrated acidified $\ce{KMnO4}$ splitting alkenes into ketones, carboxylic acids and $\ce{CO2}$ ; and the reactions of carboxylic acids as acids (metals, alkalis, carbonates).

Two features of the map are worth fixing in your mind:

The alcohol is the hub. Almost everything can be made from it or converted into it.
Only two reactions make a new C–C bond, increasing the chain length: 12 (KCN with a halogenoalkane) and 19 (HCN with an aldehyde or ketone). If the target has more carbons than the starting material, one of these must be in the route.

Which reagent reacts with which group?

When a molecule has several functional groups, each reagent attacks only some of them. This table is the key to predicting products.

reagent	reacts with	does not react with
$\ce{Br2(aq)}$ (room temp.)	C=C	alcohols, carbonyls, acids, esters, halogenoalkanes
Na metal	any O–H (alcohols, carboxylic acids, water)	C=C, C=O of aldehydes/ketones, esters
$\ce{Na2CO3}$ or $\ce{NaHCO3}$	carboxylic acids only	alcohols
NaOH(aq)	carboxylic acids (neutralisation); esters and halogenoalkanes on heating (hydrolysis/substitution)	alcohols, aldehydes, ketones, alkenes
acidified $\ce{K2Cr2O7}$	primary and secondary alcohols, aldehydes	tertiary alcohols, ketones, carboxylic acids, C=C (at AS)
$\ce{NaBH4}$	aldehydes and ketones (C=O)	C=C, carboxylic acids, esters
$\ce{LiAlH4}$	aldehydes, ketones, carboxylic acids	C=C
$\ce{H2}$ / Ni, heat	C=C (and C=O, C≡N under more forcing conditions)
$\ce{PCl5}$	any O–H (alcohols, acids), giving steamy HCl fumes	aldehydes, ketones, esters
2,4-DNPH	aldehydes and ketones	acids, esters, alcohols
Tollens' or Fehling's	aldehydes	ketones, alcohols, acids
alkaline $\ce{I2}$	$\ce{CH3CO-}$ and $\ce{CH3CH(OH)-}$ groups	other groups
HCN / KCN	aldehydes and ketones	C=C, acids, esters

Planning a synthetic route

Method

Devising a multi-step synthesis

Compare the start and the target. Write both as structural formulae. Note (a) the functional groups in each, (b) the number of carbon atoms, and (c) the positions of the groups.
Carbon count. If the target has one more carbon, include step 12 (KCN) or step 19 (HCN). At AS, no other reaction makes C–C bonds.
Work backwards from the target. Ask "what could this be made from in one step?", using the map. Repeat until you reach something you can make from the starting material. This is called retrosynthesis.
Check positions. Markovnikov addition puts a halogen or OH on the more substituted carbon; oxidation of a secondary alcohol gives a ketone, not an aldehyde; elimination from an unsymmetrical halogenoalkane gives a mixture.
Write each step with: the intermediate's structure and name, the reagent(s), the conditions (solvent, heat, reflux, distil, catalyst), and the type of reaction.
Look for problems: by-products (isomers, polysubstitution), reagents that would also attack another group in the molecule, and the number of steps (each step loses yield, so fewer is better).

Overall yield

Each step has its own percentage yield, and the overall yield is the product of them. A four-step synthesis with yields of 80%, 75%, 90% and 60% has an overall yield of $0.80 \times 0.75 \times 0.90 \times 0.60 = 0.324$ , about 32%. This is why chemists prefer short routes.

Analysing a given route

Method

Analysing a synthetic route

For each step, compare the functional groups before and after.
Name the type of reaction (use the two-word mechanism name where there is one: nucleophilic substitution, electrophilic addition and so on; otherwise oxidation, reduction, hydrolysis, condensation, elimination).
Give the reagents and conditions, including solvent and whether heating under reflux or distillation is needed.
Identify by-products: the inorganic products (water, HBr, $\ce{NH4Br}$ ) and any organic side products (a positional isomer from Markovnikov addition, a second alkene from elimination, secondary amines, polysubstituted halogenoalkanes).

Worked examples

Routine: a two-step route

Describe how ethanoic acid can be made from ethene in two steps. For each step give the reagents, conditions, an equation and the type of reaction.

Solution

Step 1: ethene to ethanol. Steam with phosphoric acid catalyst, about $300\ ^\circ\text{C}$ and $6\ \text{MPa}$ . Electrophilic addition (hydration).

\ce{CH2=CH2 + H2O -> CH3CH2OH}

Step 2: ethanol to ethanoic acid. Excess acidified potassium dichromate(VI), heat under reflux. Oxidation.

\ce{CH3CH2OH + 2[O] -> CH3COOH + H2O}

Routine: predicting products in a molecule with two groups

Prop-2-en-1-ol, $\ce{CH2=CHCH2OH}$ , contains a C=C and a primary alcohol group. Give the organic product with each reagent: (a) bromine water; (b) sodium; (c) acidified potassium dichromate(VI), distilling the product as it forms; (d) hydrogen with a nickel catalyst, heated.

Solution

(a) Addition to C=C: $\ce{CH2BrCHBrCH2OH}$ , 2,3-dibromopropan-1-ol. (The OH is unaffected.)

(b) Reaction at O–H: $\ce{CH2=CHCH2O^-Na+}$ and hydrogen gas. (The C=C is unaffected.)

(d) Addition of hydrogen to C=C: $\ce{CH3CH2CH2OH}$ , propan-1-ol.

Standard: a branched acid from propene

Plan a three-step synthesis of 2-methylpropanoic acid, $\ce{(CH3)2CHCOOH}$ , from propene. Give reagents, conditions and intermediates.

Solution

Carbon count: propene has 3 C; the target has 4 C. A cyanide step is needed. The COOH carbon is attached to the middle carbon of the original three-carbon chain, so the cyanide must go onto C2.

Step 1: HBr(g), room temperature (electrophilic addition). Markovnikov addition puts Br on C2: $\ce{CH3CHBrCH3}$ , 2-bromopropane.

Step 2: KCN in ethanol, heat under reflux (nucleophilic substitution): $\ce{(CH3)2CHCN}$ , 2-methylpropanenitrile.

Step 3: dilute HCl, heat under reflux (hydrolysis): $\ce{(CH3)2CHCOOH}$ , 2-methylpropanoic acid.

By-product: some 1-bromopropane forms in step 1, which would lead to butanoic acid. Markovnikov's rule makes 2-bromopropane the major product.

Standard: making lactic acid from ethanol

2-hydroxypropanoic acid (lactic acid), $\ce{CH3CH(OH)COOH}$ , is to be made from ethanol in three steps. Give each step.

Solution

Carbon count: 2 → 3, so a cyanide step is needed. The target has an OH and a COOH on adjacent carbons: the signature of a hydroxynitrile hydrolysed (steps 19 then 20). The hydroxynitrile $\ce{CH3CH(OH)CN}$ comes from ethanal.

Step 1: acidified $\ce{K2Cr2O7}$ , warm, distilling off the ethanal as it forms (oxidation): $\ce{CH3CH2OH + [O] -> CH3CHO + H2O}$ .

Step 2: HCN with KCN catalyst, heat (nucleophilic addition): $\ce{CH3CHO + HCN -> CH3CH(OH)CN}$ .

Step 3: dilute HCl, heat under reflux (hydrolysis): $\ce{CH3CH(OH)CN + 2H2O + HCl -> CH3CH(OH)COOH + NH4Cl}$ .

The product is a racemic mixture of the two optical isomers, because the cyanide ion attacks the planar carbonyl group equally from either side.

Exam-hard: a molecule with three functional groups

4-hydroxybutan-2-one, $\ce{CH3COCH2CH2OH}$ , is treated separately with each reagent. Predict the organic product, or the observation, in each case.

(a) 2,4-DNPH (b) Tollens' reagent (c) alkaline aqueous iodine (d) $\ce{NaBH4}$ (e) excess acidified $\ce{K2Cr2O7}$ , heat under reflux (f) concentrated $\ce{H2SO4}$ , heat

Solution

Groups: ketone ( $\ce{CH3CO-}$ , a methyl ketone) and a primary alcohol ( $\ce{-CH2OH}$ ).

(a) Orange precipitate (the ketone).

(b) No silver mirror: a ketone does not reduce Tollens' reagent, and the primary alcohol does not react either.

(d) The ketone is reduced: $\ce{CH3CH(OH)CH2CH2OH}$ , butane-1,3-diol. (It has a chiral centre at C3.)

(e) The primary alcohol is oxidised to a carboxylic acid; the ketone is unaffected: $\ce{CH3COCH2COOH}$ , 3-oxobutanoic acid.

(f) Dehydration of the alcohol (OH from C4, H from C3): $\ce{CH3COCH=CH2}$ , but-3-en-2-one.

Exam-hard: analysing a route and its yield

A student converts propan-1-ol into propan-2-ol in two steps: (1) heat with concentrated phosphoric acid; (2) react the product with steam over a phosphoric acid catalyst.

(a) Identify the intermediate and name each type of reaction. (b) Explain why the final product contains some propan-1-ol. (c) The student now wants propanone. Give one further step, with reagents and conditions, and explain why propan-1-ol cannot simply be oxidised to propanone. (d) The two steps have yields of 85% and 60%. What mass of propan-2-ol is obtained from $15.0\ \text{g}$ of propan-1-ol? ( $A_r$ : H 1.0, C 12.0, O 16.0)

Solution

(a) Intermediate: propene, $\ce{CH3CH=CH2}$ . Step 1: elimination (dehydration). Step 2: electrophilic addition (hydration).

(b) In step 2 the $\ce{H+}$ can add to either carbon of the C=C. Adding to C1 gives the more stable secondary carbocation and hence propan-2-ol (major, Markovnikov); adding to C2 gives a primary carbocation and some propan-1-ol (minor).

(c) Oxidise the propan-2-ol with acidified potassium dichromate(VI), heating: $\ce{CH3CH(OH)CH3 + [O] -> CH3COCH3 + H2O}$ . Propan-1-ol is a primary alcohol; its OH is on the end carbon, so oxidation gives propanal and then propanoic acid. A ketone needs the C=O within the chain, which only comes from a secondary alcohol; that is why the OH must first be moved to C2.

(d) $n(\ce{C3H7OH}) = 15.0 / 60.0 = 0.250\ \text{mol}$ . Overall yield $= 0.85 \times 0.60 = 0.51$ . $n(\text{propan-2-ol}) = 0.250 \times 0.51 = 0.1275\ \text{mol}$ ; mass $= 0.1275 \times 60.0 = 7.65\ \text{g}$ .

Watch out

Missing the carbon-count clue. If the target has one more carbon than the starting material and your route has no cyanide step, it is wrong.
Incomplete conditions. " $\ce{K2Cr2O7}$ " alone is not enough: write "acidified potassium dichromate(VI), heat under reflux" (or "distil"). "NaOH" needs "aqueous" or "in ethanol".
Reagents that attack two groups. $\ce{LiAlH4}$ reduces a carboxylic acid as well as a ketone; use $\ce{NaBH4}$ if only the ketone should change. Acidified dichromate oxidises every primary and secondary alcohol in the molecule.
Ignoring Markovnikov's rule. Adding HBr or steam to an unsymmetrical alkene puts the Br or OH on the carbon with fewer hydrogens.
Oxidising a primary alcohol to a ketone. Primary alcohols give aldehydes and acids. A ketone needs a secondary alcohol.
Too many steps. Each extra step lowers the overall yield. Look for the shortest route that works.

Exam tip

Synthesis questions usually award a mark for each correct intermediate and a mark for each correct set of reagents and conditions. Give all three (structure, reagent, conditions) for every step.
"Name the type of reaction" in route analysis: use the precise term (nucleophilic substitution, electrophilic addition, nucleophilic addition, elimination, oxidation, reduction, hydrolysis, condensation, free-radical substitution).
In "identify the functional groups and predict the reactions" questions, go through the molecule group by group, and for each reagent ask which group(s) it attacks. The selectivity table above is the tool.
When a question shows a reaction scheme with letters (A, B, C...), use molecular formulae and test results to fix each compound, and write a reason for each deduction.

Summary

Learn the 23 numbered conversions with reagents, conditions and reaction type: the alcohol is the hub of the map.
C–C bonds are made only by KCN with halogenoalkanes and HCN with carbonyl compounds (each adds one carbon).
Plan routes by comparing functional groups, carbon count and positions, and working backwards from the target.
Reagents are selective: $\ce{NaHCO3}$ (acids only), Na (all O–H), $\ce{NaBH4}$ (aldehydes and ketones only), $\ce{LiAlH4}$ (also acids), $\ce{Br2(aq)}$ (C=C), acidified dichromate (primary and secondary alcohols, aldehydes).
Analyse routes by naming each reaction type, its reagents and conditions, and the possible by-products.
Overall yield is the product of the yields of the individual steps.

Practice

Question

Give the reagents and conditions for converting ethene into ethylamine in two steps, naming the intermediate.
Describe how 1-bromobutane can be converted into butanal in two steps.
Describe how but-1-ene can be converted into butanone in two steps, explaining why the route works.
Plan a three-step synthesis of ethanoic acid from methane.
Three compounds have the formula $\ce{C3H6O}$ : propanal, propanone and prop-2-en-1-ol. Describe two tests that together identify all three.
4-hydroxybut-2-enal, $\ce{HOCH2CH=CHCHO}$ , is treated with (a) excess hydrogen and a nickel catalyst; (b) $\ce{NaBH4}$ ; (c) bromine water; (d) Tollens' reagent. Give the organic product in each case.
In the scheme propene → $\ce{CH3CHBrCH3}$ → $\ce{CH3CH(OH)CH3}$ → $\ce{CH3COCH3}$ → $\ce{(CH3)2C(OH)CN}$ , give the reagents and conditions and the type of reaction for each step, and name one organic by-product of the first step.
A four-step synthesis has yields of 80%, 75%, 90% and 60%. Starting from $4.20\ \text{g}$ of propene, calculate the mass of 2-hydroxy-2-methylpropanenitrile obtained by the route in question 7. ( $A_r$ : H 1.0, C 12.0, N 14.0, O 16.0)
Alkene A, $\ce{C4H8}$ , reacts with HBr to give mainly B, which is hydrolysed by NaOH(aq) to C. C is also formed directly from A and steam, and is not oxidised by acidified dichromate(VI). A reacts with cold dilute acidified $\ce{KMnO4}$ to give D, and with hot concentrated acidified $\ce{KMnO4}$ to give propanone and carbon dioxide. Identify A, B, C and D.
Using bromoethane as the only organic starting material, plan a synthesis of ethyl propanoate. Give every step with reagents, conditions and intermediates.

Answers

Step 1: HBr(g) at room temperature (electrophilic addition), giving bromoethane, $\ce{CH3CH2Br}$ . Step 2: excess ammonia in ethanol, heated under pressure in a sealed tube (nucleophilic substitution), giving ethylamine, $\ce{CH3CH2NH2}$ .
Step 1: NaOH(aq), heat under reflux (nucleophilic substitution) to butan-1-ol, $\ce{CH3CH2CH2CH2OH}$ . Step 2: acidified potassium dichromate(VI), warm, distilling the butanal off as it forms (oxidation), giving $\ce{CH3CH2CH2CHO}$ .
Step 1: steam with $\ce{H3PO4}$ catalyst, about $300\ ^\circ\text{C}$ , $6\ \text{MPa}$ (electrophilic addition); by Markovnikov's rule the OH goes mainly onto C2, giving butan-2-ol, $\ce{CH3CH(OH)CH2CH3}$ . Step 2: acidified potassium dichromate(VI), heat (oxidation): a secondary alcohol gives the ketone butanone, $\ce{CH3COCH2CH3}$ . (Alternatively HBr then NaOH(aq), via 2-bromobutane, then oxidation: three steps.)
Step 1: $\ce{Cl2}$ (or $\ce{Br2}$ ), UV light (free-radical substitution): $\ce{CH3Cl}$ , using excess methane. Step 2: KCN in ethanol, heat under reflux (nucleophilic substitution): $\ce{CH3CN}$ , ethanenitrile. Step 3: dilute HCl, heat under reflux (hydrolysis): $\ce{CH3COOH}$ .
Bromine water: only prop-2-en-1-ol decolourises it (orange to colourless). Tollens' reagent, warm, on the other two: propanal gives a silver mirror; propanone does not. (Or alkaline iodine: only propanone gives a pale yellow precipitate.)
(a) Both C=C and C=O are reduced: $\ce{HOCH2CH2CH2CH2OH}$ , butane-1,4-diol. (b) Only the aldehyde is reduced: $\ce{HOCH2CH=CHCH2OH}$ , but-2-ene-1,4-diol. (c) Bromine adds across C=C: $\ce{HOCH2CHBrCHBrCHO}$ . (d) The aldehyde is oxidised: $\ce{HOCH2CH=CHCOOH}$ (present as its carboxylate in the alkaline reagent), and a silver mirror forms.
Step 1: HBr(g), room temperature; electrophilic addition. By-product: 1-bromopropane, $\ce{CH3CH2CH2Br}$ . Step 2: NaOH(aq), heat under reflux; nucleophilic substitution. Step 3: acidified $\ce{K2Cr2O7}$ , heat; oxidation. Step 4: HCN with KCN catalyst, heat; nucleophilic addition.
$n(\ce{C3H6}) = 4.20 / 42.0 = 0.100\ \text{mol}$ . Overall yield $= 0.80 \times 0.75 \times 0.90 \times 0.60 = 0.324$ . $n(\text{product}) = 0.0324\ \text{mol}$ . $M_r(\ce{C4H7NO}) = 85.0$ . Mass $= 0.0324 \times 85.0 = 2.75\ \text{g}$ .
Hot $\ce{KMnO4}$ gives propanone (from a $\ce{=C(CH3)2}$ end) and $\ce{CO2}$ (from a $\ce{=CH2}$ end), so A is methylpropene, $\ce{CH2=C(CH3)2}$ . HBr adds by Markovnikov's rule via the tertiary carbocation: B is 2-bromo-2-methylpropane, $\ce{(CH3)3CBr}$ . C is 2-methylpropan-2-ol, $\ce{(CH3)3COH}$ , a tertiary alcohol (consistent with no oxidation by dichromate; steam also adds by Markovnikov's rule). D is 2-methylpropane-1,2-diol, $\ce{(CH3)2C(OH)CH2OH}$ .
Ethanol: bromoethane + NaOH(aq), heat under reflux (nucleophilic substitution): $\ce{CH3CH2OH}$ . Propanoic acid: bromoethane + KCN in ethanol, heat under reflux (nucleophilic substitution): $\ce{CH3CH2CN}$ ; then dilute HCl, heat under reflux (hydrolysis): $\ce{CH3CH2COOH}$ . Ester: propanoic acid + ethanol, a few drops of concentrated $\ce{H2SO4}$ , heat (condensation): $\ce{CH3CH2COOCH2CH3}$ , ethyl propanoate.

The map of AS reactions

Which reagent reacts with which group?

Planning a synthetic route

Overall yield

Analysing a given route

Worked examples

Practice

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