Organic synthesis

Class 11 Chemistry, unit 21, book pages 380–395

Practise 48 MCQs Take the unit quiz Review 23 cards

Organic synthesis is the planned making of a required compound from simple starting materials, using the reactions of the functional groups. This unit collects the important reactions of the homologous series into one map, shows how a carbon atom is added through a nitrile and how a multi-step route is analysed, and lists the tests that identify each functional group. It then names the key oxidising, reducing and dehydrating reagents and ends with retrosynthesis, the method of working backwards from a target compound to a route.

Learning outcomes (6)

Explain the concept of organic synthesis and functional group interconversions.
Identify organic functional groups using the reactions in this progression grid.
Predict the properties and the reactions of organic molecules based on functional group presence.
Devise multistep synthetic route for preparing organic molecules.
Analyze a given synthetic route in terms of type of reaction and reagents used for each step of it, and possible by-products.
Explain the concept of retro-synthesis and its applications in organic synthesis.

What is organic synthesis?

Organic synthesis means making a required organic compound from simpler, easily available starting materials by a planned series of reactions. When research chemists want to make a new compound they usually work backwards from the desired compound to a series of reactions that begins with a compound taken from a common raw material. In industry the usual starting materials are hydrocarbons from crude oil and its refining, and compounds taken from plants, such as esters from fats and vegetable oils.

Two skills of a research chemist are needed in this unit. Each one is a type of question you must be able to answer:

Predicting the reactions of a complex molecule you have never seen before that contains more than one functional group.
Suggesting a series of reactions that makes a given compound from a given starting compound.

21.1Important organic reactionsOften examined

To answer synthesis questions you must know the reactions and conditions of every homologous series in the syllabus. Figure 21.1 is a summary of some of the most useful reactions. You could also be asked about others, so the book advises you to make your own summary charts for all the reactions in the organic chapters.

Figure 21.1: Important organic reactions. Each arrow is one reaction; the numbers are explained in the table below.

Reactions in Figure 21.1
Arrow	Change	Reagents and conditions
1	alkene to alkane	$H X_{2} (g)$ , heat under pressure, nickel catalyst
2	alkane to halogenoalkane	halogen ( $Cl X_{2}$ or $Br X_{2}$ ) with ultraviolet light, room temperature and pressure
3	alkene to halogenoalkane	treat with $HX (aq)$ , room temperature and pressure
4	halogenoalkane to amine	heat under pressure with ammonia in alcohol (ethanol)
5	halogenoalkane to alcohol	heat with dilute aqueous alkali
6	alcohol to halogenoalkane	distil from a mixture of the alcohol, conc. $H X_{2} SO X_{4}$ and $KX (s)$
7	alkene to alcohol	pass the alkene and steam over heated phosphoric acid catalyst under pressure
8	alcohol to alkene	hot pumice and $Al X_{2} O X_{3}$
9	alcohol to aldehyde	distil on adding to acidified potassium dichromate(VI)
10	aldehyde to alcohol	warm with $NaBH X_{4}$ in water
11	aldehyde to carboxylic acid	reflux with excess acidified potassium dichromate(VI)
12	carboxylic acid to ester	heat with alcohol and an acid catalyst
13	ester to carboxylic acid	heat with water and an acid catalyst
14	ester to alcohol	heat with water and an acid catalyst (the same hydrolysis gives the acid and the alcohol)

21.1.1Adding carbon atomsOften examined

Sometimes the starting compound from a raw material does not have enough carbon atoms in its molecule to make the desired product. An extra carbon atom can be added by adding the nitrile group, $- CN$ . Nitriles are made from alkyl halides: heating the alkyl halide with ethanolic $KCN$ gives the nitrile.

RBr + KCN RCN + KBr

Here $R$ is an alkyl group, so $RCN$ has one more carbon atom than $RBr$ . The nitrile $RCN$ can then go either of two ways:

Hydrolysed to a carboxylic acid by refluxing with dilute hydrochloric acid.
Reduced to an amine by adding $LiAlH X_{4}$ in dry ether.

RCN dil. HCl, reflux RCOOH

RCN LiAlH X_{4} dry ether RCH X_{2} NH X_{2}

A different way of lengthening a molecule is to add an alkyl or acyl side chain to a benzene ring by a Friedel Crafts reaction. This is another useful reaction when planning synthetic routes.

Example: Friedel Crafts alkylation of benzene

Write the equation for the reaction of benzene with chloroethane in the presence of

AlCl X_{3}

The alkyl chloride is the source of the alkyl group, and anhydrous $AlCl X_{3}$ is the catalyst. The ethyl group replaces one hydrogen atom of the ring.
$C X_{6} H X_{6} + CH X_{3} CH X_{2} Cl AlCl X_{3} C X_{6} H X_{5} CH X_{2} CH X_{3} + HCl$
The product is ethylbenzene, so the side chain has two carbon atoms.

benzene

ethylbenzene

C X_{6} H X_{6} + CH X_{3} CH X_{2} Cl AlCl X_{3} C X_{6} H X_{5} CH X_{2} CH X_{3} + HCl

Concept assessment exercise 21.1

Concept assessment exercise 21.1 (1a)

Name the organic products A to D in the synthetic route:

C X_{2} H X_{5} COOC X_{3} H X_{7}

boiled with

NaOH

gives A; adding excess

HCl

gives B; adding

PCl X_{5}

gives C; adding concentrated

NH X_{3}

gives D.

Boiling the ester (propyl propanoate) with aqueous $NaOH$ hydrolyses it. The acid part is held as its sodium salt and the alcohol part is set free. A is sodium propanoate (the propan-1-ol is the other product).
$C X_{2} H X_{5} COOC X_{3} H X_{7} + NaOH C X_{2} H X_{5} COONa + C X_{3} H X_{7} OH$
Adding excess strong acid protonates the carboxylate ion and frees the carboxylic acid. B is propanoic acid.
$C X_{2} H X_{5} COONa + HCl C X_{2} H X_{5} COOH + NaCl$
$PCl X_{5}$ replaces the $- OH$ of the carboxylic acid by $- Cl$ . C is propanoyl chloride.
$C X_{2} H X_{5} COOH + PCl X_{5} C X_{2} H X_{5} COCl + POCl X_{3} + HCl$
Concentrated ammonia attacks the acyl chloride and replaces $- Cl$ by $- NH X_{2}$ . D is propanamide.
$C X_{2} H X_{5} COCl + 2 NH X_{3} C X_{2} H X_{5} CONH X_{2} + NH X_{4} Cl$

A: propanoate ion (as the sodium salt)

B: propanoic acid

C: propanoyl chloride

D: propanamide

A: sodium propanoate; B: propanoic acid; C: propanoyl chloride; D: propanamide

Concept assessment exercise 21.1 (1b)

(i) Devise a three-stage synthetic route to convert benzene into benzenediazonium chloride. (ii) How would you convert the benzenediazonium chloride into an orange dye?

Stage 1, nitration. Warm benzene with a mixture of concentrated nitric acid and concentrated sulphuric acid (about $5 5^{\circ} C$ ). Nitrobenzene forms.
$C X_{6} H X_{6} + HNO X_{3} conc \cdot H X_{2} SO X_{4} 55 X^{\circ} C C X_{6} H X_{5} NO X_{2} + H X_{2} O$
Stage 2, reduction of the nitro group. Heat nitrobenzene under reflux with tin and concentrated hydrochloric acid, then add excess alkali to free the amine. Phenylamine (aniline) forms. This is the $Sn / HCl$ reduction of $NO X_{2}$ to $NH X_{2}$ from section 21.2.
$C X_{6} H X_{5} NO X_{2} + 6 [H] Sn / conc. HCl C X_{6} H X_{5} NH X_{2} + 2 H X_{2} O$
Stage 3, diazotisation. Treat phenylamine with sodium nitrite and hydrochloric acid (nitrous acid made in the flask) with the temperature kept below $1 0^{\circ} C$ . Benzenediazonium chloride forms.
$C X_{6} H X_{5} NH X_{2} + HNO X_{2} + HCl < 10 X^{\circ} C C X_{6} H X_{5} N X_{2} X^{+} Cl X^{-} + 2 H X_{2} O$
(ii) Couple the cold diazonium salt with an alkaline solution of a phenol (for example phenol or 2-naphthol). An azo compound forms and it is an orange (or orange-red) dye. The temperature must stay below $1 0^{\circ} C$ because the diazonium salt decomposes on warming.

benzene HNO X_{3} / H X_{2} SO X_{4} C X_{6} H X_{5} NO X_{2} Sn / HCl C X_{6} H X_{5} NH X_{2} NaNO X_{2} / HCl, < 1 0^{\circ} C C X_{6} H X_{5} N X_{2} X^{+} Cl X^{-}

21.1.2Multi-step synthetic routesOften examined

Organic compounds can be made from other organic compounds that have a different functional group, by routes of several steps. The book works through one example, the preparation of ethanoic acid from methane.

Preparation of ethanoic acid from methane

Methane has one carbon atom and ethanoic acid has two, so a carbon atom must be added. This is the nitrile route of section 21.1.1: make a halogenoalkane, convert it to a nitrile (one more carbon), then hydrolyse the nitrile to the acid. Each step is described by its type of reaction, reagents and by-product.

The three-step route from methane to ethanoic acid. The nitrile step adds the second carbon atom.

The three-step route from methane to ethanoic acid
Step	Change	Type of reaction	Reagents	By-product
1	methane to chloromethane	free radical substitution	chlorine gas in the presence of ultraviolet light	\ce{HCl}
2	chloromethane to cyanomethane	nucleophilic substitution	\ce{KCN} (ethanolic)	\ce{KCl}
3	cyanomethane to ethanoic acid	hydrolysis	water with \ce{HCl} (dilute acid, reflux)	ammonium chloride (the book writes \ce{NH3})

Equations of the route

Step 1. Chlorine replaces one hydrogen of methane. The reaction needs ultraviolet light to make chlorine free radicals.
\ce{CH4 + Cl2 ->[UV] CH3Cl + HCl}
Step 2. The cyanide ion is a nucleophile and displaces the chloride ion. The carbon skeleton grows from one carbon to two.
\ce{CH3Cl + KCN -> CH3CN + KCl}
Step 3. Water, with hydrochloric acid as catalyst and reactant, hydrolyses the nitrile group to the carboxylic acid group.
\ce{CH3CN + 2H2O ->[HCl] CH3COOH + NH3}

Worked example: ethene to propanoic acid

Plan a three-step synthesis of propanoic acid from ethene. For each step give the reagent, the type of reaction and the equation.

The three-step route from ethene to propanoic acid. The nitrile step adds the third carbon atom.

The product has one more carbon than ethene, so the nitrile route must be used. First turn the alkene into a halogenoalkane, by electrophilic addition of \ce{HBr} at room temperature.
\ce{CH2=CH2 + HBr -> CH3CH2Br}
Step 2: heat the halogenoalkane with ethanolic \ce{KCN}. The cyanide ion is a nucleophile and displaces bromide (nucleophilic substitution). The nitrile has three carbons.
\ce{CH3CH2Br + KCN -> CH3CH2CN + KBr}
Step 3: hydrolyse the nitrile by refluxing with dilute hydrochloric acid.
\ce{CH3CH2CN + 2H2O ->[HCl] CH3CH2COOH + NH3}
Check step 3: \ce{C} 3 = 3, \ce{H} 5 + 4 = 9 = 6 + 3, \ce{N} 1 = 1, \ce{O} 2 = 2.

\ce{CH2=CH2 ->[HBr] CH3CH2Br ->[KCN] CH3CH2CN ->[H2O / HCl] CH3CH2COOH}

21.1.3Tests used for the identification of functional groupsOften examined

A functional group is identified by a simple test: a reagent is added and a clear change (colour, precipitate, gas) is observed. The book lists eight tests.

Test 1: Bromine water test

This test identifies unsaturation (\ce{C=C} or \ce{C#C}) in an organic compound.

Reagent: aqueous bromine.
Observation with an alkene or alkyne: the reddish-brown colour of bromine fades away (the solution is decolourised).

\ce{>C=C< + Br2 -> >CBr-CBr<}

Test 2: Silver nitrate test

This test identifies the halogen present in a halogenoalkane. The halogen leaves as a halide ion, which forms a silver halide precipitate.

Reagent: aqueous silver nitrate in ethanol.

Silver nitrate test for halogenoalkanes
Halogenoalkane	Observation	Equation (as in the book)
chloroalkane \ce{R-Cl}	white precipitate of \ce{AgCl}	\ce{R-Cl + AgNO3 + CH3CH2OH -> ROCH2CH3 + AgCl + HNO3}
bromoalkane \ce{R-Br}	cream precipitate of \ce{AgBr}	\ce{R-Br + AgNO3 + CH3CH2OH -> ROCH2CH3 + AgBr + HNO3}
iodoalkane \ce{R-I}	yellow precipitate of \ce{AgI}	\ce{R-I + AgNO3 + CH3CH2OH -> ROCH2CH3 + AgI + HNO3}

Test 3: Iodoform test

This test is given by certain alcohols, aldehydes and ketones, those that contain the group \ce{CH3-CO-} or \ce{CH3-CH(OH)-} (which is oxidised to it by the iodine reagent).

In primary alcohols only ethanol gives a positive test.
In secondary alcohols only those with the \ce{O-H} group on carbon number 2 give the test, for example \ce{CH3CH(OH)-}.
The test is not given by tertiary alcohols.
In aldehydes only ethanal gives a positive test.
In ketones only those whose carbonyl group is at position 2, \ce{CH3-CO-R}, give the test.

propanone (CH3-CO-CH3) a methyl ketone, which gives the test

Reagent: iodine and aqueous sodium hydroxide.
Observation: a yellow precipitate of triiodomethane (\ce{CHI3}).

\ce{CH3COR + 3I2 + 4NaOH -> CHI3 v + RCOONa + 3NaI + 3H2O}

The book gives the skeleton \ce{organic compound + I2 + NaOH -> CHI3 + RCOONa + NaI}; the balanced form is shown here.

Test 4: 2,4-DNPH test

This is the identifying test for the carbonyl group in aldehydes and ketones.

Reagent: 2,4-dinitrophenylhydrazine (2,4-DNPH).
Observation: a yellow or orange precipitate of the 2,4-dinitrophenylhydrazone.

The carbonyl oxygen and two hydrogen atoms of the \ce{NH2} group leave as water, and the carbon and nitrogen are joined by a double bond. The book shows ethanal.

\ce{CH3CHO + H2N-NH-C6H3(NO2)2 -> CH3CH=N-NH-C6H3(NO2)2 + H2O}

ethanal

2,4-dinitrophenylhydrazine

ethanal 2,4-dinitrophenylhydrazone

Test 5: Fehling solution test

This test is given by aldehydes only, so it is used to differentiate aldehydes from ketones.

Reagent: Fehling's solution, a mild oxidising agent. It is an alkaline solution of \ce{Cu^2+} ions with sodium or potassium tartrate.
Observation: a red (brown-red) precipitate of \ce{Cu2O} forms with an aldehyde, not with a ketone. The blue solution loses its colour as \ce{Cu^2+} is reduced.

\ce{CH3CHO + 2Cu^2+ + 5OH- -> CH3COO- + Cu2O v + 3H2O}

Ionic equation; the tartrate keeps \ce{Cu^2+} in solution in alkali.

Test 6: Silver mirror test (Tollens' test)

This test is also given by aldehydes only and differentiates aldehydes from ketones.

Reagent: ammoniacal silver nitrate (Tollens' reagent), which contains \ce{[Ag(NH3)2]+}.
Observation: a silver mirror of \ce{Ag} forms with an aldehyde, not with a ketone.

\ce{CH3CHO + 2[Ag(NH3)2]+ + OH- -> CH3COO- + 2NH4+ + 2Ag v + 2NH3}

Telling an aldehyde from a ketone
Test	Aldehyde	Ketone
Fehling's solution	red precipitate of \ce{Cu2O}	no change
Tollens' reagent	silver mirror	no change
2,4-DNPH	yellow or orange precipitate	yellow or orange precipitate

Test 7: Test to distinguish primary, secondary and tertiary alcohols

Reagent: potassium dichromate(VI) in the presence of concentrated sulphuric acid.
Primary alcohol: the orange solution turns green. The alcohol is oxidised first to an aldehyde. The aldehyde can be oxidised further to a carboxylic acid, especially if heated or if excess oxidising agent is used. To obtain only the aldehyde, heat the mixture and distil it off as it forms.
Secondary alcohol: the orange solution turns green. The secondary alcohol is oxidised to a ketone, which gives no colour with Fehling's solution.
Tertiary alcohol: there is no colour change of the potassium dichromate. Tertiary alcohols resist oxidation.

\ce{RCH2OH + [O] -> RCHO}

\ce{RCHO + [O] -> RCOOH}

\ce{RCH(OH)R + [O] -> RCOR}

Test 8: Test of a carboxylic acid

Reagent: sodium carbonate (\ce{Na2CO3}).
Observation: effervescence (bubbles of carbon dioxide).

\ce{2CH3COOH + Na2CO3 -> 2CH3COONa + H2O + CO2 ^}

The book shows the unbalanced equation \ce{CH3COOH + Na2CO3 -> CH3COONa + H2O + CO2}.

Summary of the eight tests
Functional group	Reagent	Positive observation
\ce{C=C} or \ce{C#C}	aqueous bromine	reddish-brown colour fades
halogen in \ce{R-X}	aqueous \ce{AgNO3} in ethanol	white (\ce{Cl}), cream (\ce{Br}), yellow (\ce{I}) precipitate
\ce{CH3CO-} or \ce{CH3CH(OH)-}	\ce{I2} and \ce{NaOH(aq)}	yellow precipitate of \ce{CHI3}
\ce{C=O} (aldehyde, ketone)	2,4-DNPH	yellow or orange precipitate
aldehyde	Fehling's solution	red precipitate of \ce{Cu2O}
aldehyde	Tollens' reagent	silver mirror
primary or secondary alcohol	acidified \ce{K2Cr2O7}	orange turns green; tertiary: no change
carboxylic acid	\ce{Na2CO3}	effervescence of \ce{CO2}

Worked example: choosing a test to tell two compounds apart

Name one test that distinguishes each pair, and give the observation for each member: (a) propene and propane, (b) chloroethane and bromoethane, (c) ethanol and ethanoic acid, (d) propan-2-ol and 2-methylpropan-2-ol.

Pair	Reagent	First compound	Second compound
(a) propene, propane	aqueous bromine	Propene: reddish-brown colour fades	Propane: no change
(b) chloroethane, bromoethane	aqueous \ce{AgNO3} in ethanol	Chloroethane: white precipitate of \ce{AgCl}	Bromoethane: cream precipitate of \ce{AgBr}
(c) ethanol, ethanoic acid	sodium carbonate	Ethanol: no effervescence	Ethanoic acid: effervescence of \ce{CO2}
(d) propan-2-ol, 2-methylpropan-2-ol	acidified \ce{K2Cr2O7}, warm	Propan-2-ol (secondary): orange turns green	2-Methylpropan-2-ol (tertiary): stays orange

Choose the functional group that differs, then the reagent that reacts with that group only. In (a) it is the \ce{C=C} bond, in (b) the halogen, in (c) the \ce{-COOH} group and in (d) the type of alcohol.
Common mistake in (d): an orange to green change happens for primary and secondary alcohols, so it cannot separate those two. Here the second compound is tertiary and is not oxidised.

Worked example: deducing a compound from its reactions

Compound Y, \ce{C3H8O}, turns acidified dichromate from orange to green and gives a yellow precipitate with iodine and sodium hydroxide. Its oxidation product gives an orange precipitate with 2,4-DNPH but no silver mirror with Tollens' reagent. Identify Y.

Orange to green shows that Y is a primary or secondary alcohol (a tertiary alcohol is not oxidised).
The iodoform test is given only by alcohols with \ce{CH3CH(OH)-}. Of the two alcohols \ce{C3H8O}, propan-1-ol does not have this group but propan-2-ol does. So Y is propan-2-ol.
Oxidation of a secondary alcohol gives a ketone.
\ce{CH3CH(OH)CH3 + [O] ->[K2Cr2O7 + H2SO4] CH3COCH3 + H2O}
Propanone gives the 2,4-DNPH precipitate (carbonyl group) but no silver mirror (not an aldehyde). This agrees with the data. Propan-1-ol would give propanal, which would give a silver mirror.

Propan-2-ol (Y)

Propanone (oxidation product)

\text{Y is propan-2-ol, }\ce{CH3CH(OH)CH3}

21.1.4Types of reaction given by functional groupsOften examined

Each functional group has its own typical reaction types. If you can name the group, you can name the reaction. The book takes the groups in turn.

1. Reactions of alkanes

Free radical substitution

Alkanes undergo free radical substitution. A hydrogen atom of the alkane is displaced by a free radical, for example \ce{Cl.}. The general reaction needs diffused (diffuse) sunlight or ultraviolet light.

\ce{Alkane + Cl2 ->[\text{sunlight}] \text{Chloroalkane} + HCl}

2. Reactions of unsaturated hydrocarbons (alkenes and alkynes)

Unsaturated hydrocarbons undergo two types of reaction: electrophilic addition and oxidation.

i. Electrophilic addition reactions

In these reactions one of the two bonds in the carbon-carbon double bond is broken and a new single bond is formed from each of the two carbon atoms.

Figure 21.2: General equations for addition reactions. (a) With \ce{HBr}, \ce{H2O} or \ce{HOX} (\ce{X-Y}). (b) With \ce{H2} or \ce{Br2} (\ce{Z-Z}). In an addition reaction a single product is always formed.

\ce{>C=C< + X-Y -> -C(X)-C(Y)-}

ii. Oxidation reactions

Alkenes can be oxidised to diols by cold dilute alkaline \ce{KMnO4}. For example propene gives propane-1,2-diol.

\ce{CH3CH=CH2 + KMnO4 ->[OH-][cold] CH3CH(OH)CH2OH}

propene

propane-1,2-diol

Alkenes also undergo ozonolysis to form aldehydes and ketones. This is also an oxidation of the alkene. The \ce{C=C} pi bond of the alkene is broken and two new carbonyl compounds are formed. In the book's example the alkene is but-1-ene, which gives propanal and methanal.

\ce{CH3CH2CH=CH2 ->[1) O3][2) Zn, HCl] CH3CH2CHO + HCHO}

but-1-ene

propanal

methanal

An alkyne on ozonolysis gives carboxylic acids (one from each end of the triple bond).

\ce{R-C#C-R' ->[1) O3][2) H2O] RCOOH + R'COOH}

3. Reactions of alkyl halides

Alkyl halides undergo two types of reaction: nucleophilic substitution and elimination.

i. Nucleophilic substitution reaction

The halogen of the alkyl halide (\ce{R-X}) is displaced by an attacking nucleophile such as \ce{OH-}, \ce{NH3} or \ce{CN-}.

\ce{R-X ->[OH- / NH3 / CN-] R-OH / R-NH2 / RCN + X-}

ii. Elimination reactions

In an elimination reaction hydrogen and halogen are removed from adjacent carbons of the alkyl halide, leaving an alkene as the organic product. The reagent is ethanolic \ce{NaOH} or \ce{KOH}, with heat.

\ce{-CH-CH(X)- + KOH(alc.) ->[\Delta] >C=C< + KX + H2O}

4. Reactions of alcohols

Alcohols undergo oxidation, substitution, dehydration and esterification. Figure 21.3 shows the types of reaction of an alcohol.

Figure 21.3: Different types of reactions of an alcohol: oxidation, dehydration and esterification.

Reactions of an alcohol (Figure 21.3)
Type	Reagent and condition	Product
oxidation of primary alcohol	acidified \ce{K2Cr2O7}	aldehyde, then on further oxidation a carboxylic acid
oxidation of secondary alcohol	acidified \ce{K2Cr2O7}	ketone
oxidation of tertiary alcohol	acidified \ce{K2Cr2O7}	no reaction
dehydration	excess alcohol at 140^\circ\mathrm{C} with conc. \ce{H2SO4}	ether
dehydration	excess conc. \ce{H2SO4} at 170^\circ\mathrm{C}	alkene
esterification	carboxylic acid, acid catalyst	ester

5. Reactions of carbonyl compounds

Carbonyl compounds undergo nucleophilic addition, oxidation and reduction.

In nucleophilic addition a hydroxynitrile (cyanohydrin) is formed with hydrogen cyanide.
Aldehydes are oxidised to carboxylic acids, while ketones resist oxidation.
Aldehydes are reduced to primary alcohols and ketones are reduced to secondary alcohols.

\ce{CH3CHO + HCN -> CH3CH(OH)CN}

6. Reactions of carboxylic acids

Carboxylic acids undergo acid-base reactions and esterification.

Carboxylic acids react with bases, metals and metal carbonates. These are acid-base reactions.
The reaction of a carboxylic acid with an alcohol is called esterification.

\ce{RCOOH + R'OH <=>[{conc.} H2SO4] RCOOR' + H2O}

Reaction type for each functional group
Functional group	Main reaction types
alkane	free radical substitution
alkene, alkyne	electrophilic addition, oxidation
alkyl halide	nucleophilic substitution, elimination
alcohol	oxidation, substitution, dehydration, esterification
aldehyde, ketone	nucleophilic addition, oxidation, reduction
carboxylic acid	acid-base reaction, esterification

Worked example: naming the type of reaction

Name the type of reaction in each case: (a) ethene with \ce{HBr}, (b) ethane with chlorine in sunlight, (c) bromoethane with aqueous KOH, (d) bromoethane with ethanolic KOH and heat, (e) ethanal with \ce{HCN}.

Case	Equation	Type of reaction
(a)	\ce{CH2=CH2 + HBr -> CH3CH2Br}	Electrophilic addition
(b)	\ce{C2H6 + Cl2 -> C2H5Cl + HCl}	Free radical substitution
(c)	\ce{C2H5Br + KOH(aq) -> C2H5OH + KBr}	Nucleophilic substitution
(d)	\ce{C2H5Br + KOH(alc.) -> C2H4 + KBr + H2O}	Elimination
(e)	\ce{CH3CHO + HCN -> CH3CH(OH)CN}	Nucleophilic addition

Look at what happens to the organic molecule. If a double bond is used up and one product forms, it is addition. If one atom or group is replaced by another, it is substitution. If a small molecule is removed and a double bond appears, it is elimination.
Common mistake: (c) and (d) use the same reagents in different solvents. Aqueous KOH replaces Br by OH (substitution). Ethanolic KOH removes H and Br from neighbouring carbons (elimination).

21.2Key reagents used for synthesisOften examined

A synthetic route is built from a small group of reagents. They fall into three classes: oxidising agents, reducing agents and dehydrating agents.

Oxidising agents

The oxidising agents are acidified potassium dichromate (\ce{K2Cr2O7} with \ce{H2SO4}) and acidified potassium permanganate (\ce{KMnO4}). They oxidise primary alcohols to aldehydes and then to carboxylic acids. They oxidise secondary alcohols to ketones.

Reducing agents

Reducing agents named in the book
Reducing agent	What it reduces
\ce{NaBH4}	a carbonyl group to an alcohol
\ce{LiAlH4}	a stronger reducing agent than \ce{NaBH4}; it reduces a carboxylic acid to an alcohol
\ce{H2} with nickel catalyst	a \ce{C=C} double bond
tin and hydrochloric acid (\ce{Sn/HCl})	\ce{NO2} to \ce{NH2}

Dehydrating agents

A dehydrating agent is one that removes water from a molecule. The common dehydrating agents are \ce{Al2O3} and concentrated \ce{H2SO4}.

Key reagents at a glance
Reagent	Class	Typical use
acidified \ce{K2Cr2O7} or \ce{KMnO4}	oxidising	alcohol to aldehyde, ketone or carboxylic acid
\ce{NaBH4}	reducing	\ce{C=O} to alcohol
\ce{LiAlH4}	reducing (stronger)	carboxylic acid to alcohol
\ce{H2/Ni}	reducing	\ce{C=C} to \ce{C-C}
\ce{Sn/HCl}	reducing	\ce{NO2} to \ce{NH2}
\ce{Al2O3}, conc. \ce{H2SO4}	dehydrating	alcohol to alkene (or ether)

Flow chart of the different reactions of functional groups

Figure 21.4 joins the functional groups through the reactions you have met: addition, substitution, oxidation, reduction, elimination, hydrolysis, acylation and esterification.

Figure 21.4: Organic functional group interconversion. Each arrow is a reaction; the numbers are explained in the table below.

Reactions in Figure 21.4
Arrow	Change	Reagents and conditions
1	alkane to alkene	cracking (heat, catalyst)
2	alkene to alkane	\ce{H2}, nickel catalyst, about 180^\circ\mathrm{C}
3	alkane to halogenoalkane	\ce{Cl2} or \ce{Br2} with ultraviolet light
4	alkene to halogenoalkane	\ce{HBr} (or \ce{HX}) at room temperature
5	halogenoalkane to alcohol	\ce{NaOH(aq)}, heat
6	halogenoalkane to nitrile	\ce{KCN} in ethanol, heat
7	halogenoalkane to amine	concentrated ammonia in ethanol, heat under pressure
8	nitrile to amine	reduction with \ce{LiAlH4} in dry ether
9	nitrile to carboxylic acid	hydrolysis, heat with dilute acid (\ce{HCl} or \ce{H2SO4})
10	alkene to alcohol	steam with phosphoric acid catalyst (hydration)
11	alcohol to alkene	\ce{Al2O3} or excess conc. \ce{H2SO4} at about 170^\circ\mathrm{C} (dehydration)
12	alcohol to ether	excess alcohol with conc. \ce{H2SO4} at 140^\circ\mathrm{C}
13	primary alcohol and aldehyde	oxidation: acidified \ce{K2Cr2O7}, distil; reduction: \ce{NaBH4}
14	secondary alcohol and ketone	oxidation: acidified \ce{K2Cr2O7}; reduction: \ce{NaBH4} or \ce{LiAlH4} in ether
15	aldehyde to carboxylic acid	acidified \ce{K2Cr2O7}, heat under reflux
16	carboxylic acid and ester	forwards: alcohol with conc. \ce{H2SO4}; backwards: water with acid catalyst (hydrolysis)

Concept assessment exercise 21.3

Concept assessment exercise 21.3 (1)

Compound X, \ce{C4H8}, undergoes these reactions: X with \ce{Br2} gives Y (\ce{C4H8Br2}); Y with aqueous \ce{NaOH} and heat gives Z (\ce{C4H10O2}); and X with \ce{H2/Ni} gives methylpropane. (a) Suggest a synthesis of the starting material \ce{C4H8}. (b) What is the formula of compound Z?

X adds \ce{H2} to give methylpropane, \ce{(CH3)2CHCH3}. So X has the carbon skeleton of methylpropane with one \ce{C=C}. The only such alkene is 2-methylpropene.
\ce{(CH3)2C=CH2 + H2 ->[Ni] (CH3)2CHCH3}
Bromine adds across the double bond (electrophilic addition). Y is 1,2-dibromo-2-methylpropane, \ce{C4H8Br2}.
\ce{(CH3)2C=CH2 + Br2 -> (CH3)2CBrCH2Br}
Aqueous \ce{NaOH} with heat replaces both bromine atoms by \ce{-OH} (nucleophilic substitution). Z is 2-methylpropane-1,2-diol.
\ce{(CH3)2CBrCH2Br + 2NaOH ->[aq., heat] (CH3)2C(OH)CH2OH + 2NaBr}
(a) Synthesis of X (2-methylpropene) by elimination. Either heat 2-bromo-2-methylpropane with ethanolic \ce{KOH} (elimination from a halogenoalkane), or dehydrate 2-methylpropan-2-ol by passing it over hot \ce{Al2O3} (or heating with excess conc. \ce{H2SO4}).
\ce{(CH3)3CBr + KOH ->[ethanol, \Delta] (CH3)2C=CH2 + KBr + H2O}
(b) Count atoms in \ce{(CH3)2C(OH)CH2OH}: four C, ten H (six + one + two + one) and two O.
\ce{Z = C4H10O2}

X: 2-methylpropene

Y: 1,2-dibromo-2-methylpropane

Z: 2-methylpropane-1,2-diol

\ce{X = (CH3)2C=CH2},\quad \ce{Y = (CH3)2CBrCH2Br},\quad \ce{Z = (CH3)2C(OH)CH2OH = C4H10O2}

Concept assessment exercise 21.3 (2)

Prepare ethanal and propanone starting from 2-methylbut-2-ene.

2-Methylbut-2-ene is \ce{CH3CH=C(CH3)2}. A route that breaks a \ce{C=C} into two carbonyl compounds is ozonolysis (section 21.1.4). The carbon on one side of the double bond has \ce{CH3} and \ce{H}, so it becomes an aldehyde (ethanal). The other carbon has two \ce{CH3} groups, so it becomes a ketone (propanone).
Treat with ozone, then reduce the ozonide with zinc and hydrochloric acid (or zinc and water). One step gives both products.
\ce{CH3CH=C(CH3)2 ->[1) O3][2) Zn, HCl] CH3CHO + (CH3)2C=O}

2-methylbut-2-ene

ethanal

propanone

\ce{CH3CH=C(CH3)2 ->[1) O3][2) Zn / HCl] CH3CHO + CH3COCH3}

Concept assessment exercise 21.3 (3)

Devise the best synthetic route to prepare 2-hydroxypropanedioic acid starting from ethanal.

The book's name 2-hydroxypropanedioic acid describes \ce{HOOC-CH(OH)-COOH} (two acid groups). That cannot be made from ethanal by the reactions of this unit: ethanal has two carbons, the nitrile route adds only one, and there is no step that turns the \ce{CH3} group into a second \ce{COOH}. The compound that ethanal does give, in two steps, is 2-hydroxypropanoic acid (lactic acid), \ce{CH3CH(OH)COOH}, and this is almost certainly the compound meant. Its route is below.
Step 1, nucleophilic addition. Ethanal reacts with hydrogen cyanide (with a little \ce{KCN} as catalyst) to give 2-hydroxypropanenitrile. One carbon is added.
\ce{CH3CHO + HCN ->[KCN] CH3CH(OH)CN}
Step 2, hydrolysis. Reflux the hydroxynitrile with dilute hydrochloric acid. The \ce{-CN} group becomes \ce{-COOH}.
\ce{CH3CH(OH)CN + 2H2O + HCl -> CH3CH(OH)COOH + NH4Cl}

ethanal

2-hydroxypropanenitrile

2-hydroxypropanoic acid

\ce{CH3CHO ->[HCN / KCN] CH3CH(OH)CN ->[{dil.} HCl, reflux] CH3CH(OH)COOH}

21.3Retrosynthetic analysis

Retrosynthesis is a method of chemical synthesis in which a target molecule is deconstructed into readily available, simple starting materials to assess the best synthetic route. The main difference between the two ideas is the direction of thinking:

Synthesis and retrosynthesis
	Synthesis	Retrosynthesis
direction	forward: from simple starting materials to the target	backward: from the target to simple precursors
what is done	combines simple reactions to form an organic compound	works backwards from the target to devise a suitable route
starting point	a simple precursor molecule	the target organic compound

Retrosynthesis for ethyl propanoate

Start with the target and work backwards. Ethyl propanoate has the structure \ce{CH3CH2COOCH2CH3}. Break it into simpler fragments that come from known materials. The two fragments are \ce{CH3CH2CO-} and \ce{-O-CH2CH3}.

For \ce{CH3CH2CO-} use \ce{CH3CH2COOH} (propanoic acid).
For \ce{-O-CH2CH3} use \ce{HO-CH2CH3} (ethanol).

These two precursors are combined to form ethyl propanoate by esterification: the ester is made by the reaction of propanoic acid with ethanol in the presence of sulphuric acid as catalyst.

Retrosynthesis of ethyl propanoate: the ester is broken at the \ce{C-O} bond into propanoic acid and ethanol, which esterification joins again.

\ce{CH3CH2COOH + HOCH2CH3 <=>[{conc.} H2SO4] CH3CH2COOCH2CH3 + H2O}

The retrosynthetic pathway is summarised as: ethyl propanoate \Longleftarrow propanoic acid + ethanol.

ethyl propanoate

propanoic acid

ethanol

Retrosynthesis of cyanopropane

Cyanopropane has an alkyl chain (propane) with a cyano group (\ce{-CN}) attached to one of its carbons: \ce{CH3CH2CH2-CN}. Break it into two simpler fragments, the propane backbone \ce{CH3CH2CH2-} and the cyano group \ce{-CN}. Then find simpler molecules that can supply them.

For the propane backbone use \ce{CH3CH2CH2-Cl}.
For the cyano group use \ce{KCN}.

\ce{CH3CH2CH2-Cl + KCN -> CH3CH2CH2CN + KCl}

The pathway is cyanopropane \Longleftarrow 1-chloropropane + \ce{KCN}. The forward step is nucleophilic substitution (heat with ethanolic \ce{KCN}).

cyanopropane (butanenitrile)

1-chloropropane

Worked example: retrosynthesis of an amide

Use retrosynthesis to plan a synthesis of N-ethylethanamide, \ce{CH3CONHCH2CH3}.

Retrosynthesis of N-ethylethanamide: the amide is broken at the C–N bond next to the carbonyl group into an acid chloride and a primary amine.

Find the functional group in the target: it is an amide, \ce{-CO-NH-}.
Break the bond next to the functional group, the \ce{C-N} bond. The two fragments are \ce{CH3CO-} and \ce{-NHCH2CH3}.
Replace each fragment by a real compound: \ce{CH3COCl} (ethanoyl chloride) for the first and \ce{CH3CH2NH2} (ethylamine) for the second.
Write the forward reaction (an amine reacts with an acid halide to give an amide and HCl).
\ce{CH3COCl + CH3CH2NH2 -> CH3CONHCH2CH3 + HCl}
Check: \ce{C} 2 + 2 = 4, \ce{H} 3 + 7 = 10 = 9 + 1, \ce{N} 1, \ce{O} 1, \ce{Cl} 1.

\text{amide}\Longleftarrow\ce{CH3COCl}+\ce{CH3CH2NH2}

Application of retrosynthesis

Retrosynthesis enables the development of multiple reaction pathways for a given target compound. These pathways can be more cost-effective, more environmentally transparent (friendly), and need fewer reaction steps.

Concept assessment exercise 21.4

Discuss the retrosynthesis of each compound. In each answer the target is broken at the bond next to its functional group, and the precursors are named with the forward reaction.

Concept assessment exercise 21.4 (a)

Discuss retrosynthesis of \ce{CH3CH2CH2Cl}.

Functional group: a halogenoalkane. Break the \ce{C-Cl} bond: fragments \ce{CH3CH2CH2-} and \ce{Cl-}.
Precursor: propan-1-ol supplies the propyl group, and a chlorinating reagent supplies chlorine. Here the alcohol is converted to the halogenoalkane by distilling it with \ce{KCl} and conc. \ce{H2SO4} (arrow 6 of Figure 21.1), or with \ce{PCl5}.
\ce{CH3CH2CH2OH + PCl5 -> CH3CH2CH2Cl + POCl3 + HCl}
Alternatively propane with \ce{Cl2} and ultraviolet light gives it, but this gives a mixture of 1-chloro and 2-chloropropane, so it is a poor route.
Pathway:
\ce{CH3CH2CH2Cl} \Longleftarrow \ce{CH3CH2CH2OH}

\ce{CH3CH2CH2Cl} \Longleftarrow \ce{CH3CH2CH2OH} + \ce{KCl} / \ce{{conc.} H2SO4}

Concept assessment exercise 21.4 (b)

Discuss retrosynthesis of \ce{CH3CH2CH2OH}.

Functional group: a primary alcohol. Break the \ce{C-OH} bond: fragments \ce{CH3CH2CH2-} and \ce{OH-}.
Precursor: 1-chloropropane (or 1-bromopropane) supplies the propyl group and aqueous alkali supplies \ce{OH-}. The forward reaction is nucleophilic substitution (hydrolysis).
\ce{CH3CH2CH2Cl + NaOH ->[aq., heat] CH3CH2CH2OH + NaCl}
Another disconnection works too: reduction of propanal with \ce{NaBH4} gives propan-1-ol (and \ce{LiAlH4} reduces propanoic acid to it).
\ce{CH3CH2CHO ->[NaBH4] CH3CH2CH2OH}
Pathway:
\ce{CH3CH2CH2OH} \Longleftarrow \ce{CH3CH2CH2Cl} + \ce{NaOH(aq)}

\ce{CH3CH2CH2OH} \Longleftarrow \ce{CH3CH2CH2Cl} + \ce{NaOH(aq)}

Concept assessment exercise 21.4 (c)

Discuss retrosynthesis of \ce{CH3CH2CONHCH3}.

Functional group: an N-substituted amide (N-methylpropanamide). Break the \ce{C-N} bond of the amide: fragments \ce{CH3CH2CO-} and \ce{-NHCH3}.
Precursors: propanoyl chloride supplies the acyl part and methylamine supplies the amine part.
\ce{CH3CH2COCl + 2CH3NH2 -> CH3CH2CONHCH3 + CH3NH3+ Cl-}
Go back one more step. Propanoyl chloride comes from propanoic acid with \ce{PCl5} (as in Concept assessment exercise 21.1). Methylamine comes from chloromethane with excess ammonia.
\ce{CH3CH2COOH + PCl5 -> CH3CH2COCl + POCl3 + HCl}
Pathway:
\ce{CH3CH2CONHCH3} \Longleftarrow \ce{CH3CH2COCl} + \ce{CH3NH2}

\ce{CH3CH2CONHCH3} \Longleftarrow \ce{CH3CH2COCl} + \ce{CH3NH2}

Concept assessment exercise 21.4 (d)

Discuss retrosynthesis of \ce{CH3CH2COOH}.

Functional group: a carboxylic acid. Two sensible disconnections:
(i) Oxidation. The acid is made by oxidising propan-1-ol with excess acidified \ce{K2Cr2O7} under reflux.
\ce{CH3CH2CH2OH ->[K2Cr2O7 / H+][reflux] CH3CH2COOH}
(ii) Hydrolysis of a nitrile. The acid comes from \ce{CH3CH2CN}, which comes from chloroethane (or bromoethane) and \ce{KCN} (the carbon chain grows by one carbon; section 21.1.1).
\ce{CH3CH2CN + 2H2O + HCl -> CH3CH2COOH + NH4Cl}
Pathway:
\ce{CH3CH2COOH} \Longleftarrow \ce{CH3CH2CN} \Longleftarrow \ce{CH3CH2Cl} + \ce{KCN}

\ce{CH3CH2COOH} \Longleftarrow \ce{CH3CH2CH2OH}\ \text{or}\ \ce{CH3CH2CN}

Concept assessment exercise 21.4 (e)

Discuss retrosynthesis of \ce{CH3CH2NHCH3}.

Functional group: a secondary amine. Break the \ce{C-N} bond: fragments \ce{CH3CH2-} and \ce{-NHCH3}.
Precursors: chloroethane supplies the ethyl group and methylamine the nitrogen part. The forward reaction is nucleophilic substitution, heating under pressure with the amine in ethanol.
\ce{CH3CH2Cl + CH3NH2 -> CH3CH2NHCH3 + HCl}
Methylamine itself comes from chloromethane and excess ammonia. Chloroethane comes from ethanol (arrow 6 of Figure 21.1).
\ce{CH3Cl + NH3 -> CH3NH2 + HCl}
Pathway:
\ce{CH3CH2NHCH3} \Longleftarrow \ce{CH3CH2Cl} + \ce{CH3NH2}

\ce{CH3CH2NHCH3} \Longleftarrow \ce{CH3CH2Cl} + \ce{CH3NH2}

Concept assessment exercise 21.4 (f)

Discuss retrosynthesis of \ce{HCOOC2H5}.

Functional group: an ester (ethyl methanoate). Break the \ce{C-O} bond next to the carbonyl: fragments \ce{HCO-} and \ce{-OC2H5}.
Precursors: methanoic acid and ethanol, joined by esterification (conc. \ce{H2SO4} catalyst, heat).
\ce{HCOOH + C2H5OH <=>[{conc.} H2SO4] HCOOC2H5 + H2O}
Going back further: methanoic acid comes from the oxidation of methanol, and ethanol from the hydrolysis of chloroethane or the hydration of ethene.
Pathway:
\ce{HCOOC2H5} \Longleftarrow \ce{HCOOH} + \ce{C2H5OH}

\ce{HCOOC2H5} \Longleftarrow \ce{HCOOH} + \ce{C2H5OH}

Formula sheet

Nitrile formation (adds one carbon)

\ce{RBr + KCN -> RCN + KBr}

\ce{R} is an alkyl group; the nitrile has one more carbon than the halogenoalkane. Hydrolysis then gives \ce{RCOOH} and reduction with \ce{LiAlH4} gives \ce{RCH2NH2}.

no unit

Friedel Crafts alkylation

\ce{C6H6 + RCl ->[\ce{AlCl3}] C6H5R + HCl}

\ce{R} is an alkyl group added to the benzene ring; \ce{AlCl3} is the catalyst.

no unit

Oxidation of alcohols

\ce{RCH2OH ->[O] RCHO ->[O] RCOOH},\qquad \ce{R2CHOH ->[O] R2C=O}

\ce{[O]} is acidified \ce{K2Cr2O7}; tertiary alcohols are not oxidised.

no unit

Esterification

\ce{RCOOH + R'OH <=>[{conc.} H2SO4] RCOOR' + H2O}

A carboxylic acid and an alcohol give an ester and water; the reverse is hydrolysis.

no unit

Ozonolysis of an alkene

\ce{>C=C< ->[1) O3][2) Zn, HCl] >C=O + O=C<}

The double bond is split into two carbonyl compounds (aldehydes and ketones); an alkyne gives carboxylic acids.

no unit

Key terms

Organic synthesis: Making a required organic compound from simpler starting materials by a planned series of reactions.
Retrosynthesis: A method of working backwards from the target compound to simple, readily available precursors in order to devise a synthetic route.
Nitrile: A compound containing the \ce{-CN} group, made from a halogenoalkane and \ce{KCN}; used to add one carbon atom.
Friedel Crafts reaction: A reaction of benzene with an alkyl or acyl halide and \ce{AlCl3} that adds an alkyl or acyl side chain to the ring.
Free radical substitution: The reaction of an alkane with a halogen in sunlight or ultraviolet light, in which a hydrogen atom is replaced by a halogen atom.
Electrophilic addition: The reaction of an alkene or alkyne in which one bond of the multiple bond breaks and a new single bond forms to each of the two carbon atoms.
Nucleophilic substitution: The reaction of an alkyl halide in which the halogen is displaced by a nucleophile such as \ce{OH-}, \ce{NH3} or \ce{CN-}.
Elimination reaction: The removal of hydrogen and halogen from an alkyl halide, by heating with ethanolic \ce{KOH} or \ce{NaOH}, to leave an alkene.
Ozonolysis: The oxidation of an alkene with ozone, then \ce{Zn/HCl}, which breaks the \ce{C=C} bond to give aldehydes and ketones; an alkyne gives carboxylic acids.
Esterification: The reaction of a carboxylic acid with an alcohol, with an acid catalyst, to form an ester and water.
Bromine water test: Test for unsaturation: the reddish-brown colour of bromine fades with an alkene or alkyne.
Iodoform test: Test with iodine and aqueous \ce{NaOH} that gives yellow \ce{CHI3} with ethanol, ethanal, \ce{CH3CH(OH)-} alcohols and \ce{CH3CO-} ketones.
2,4-DNPH test: Test for the carbonyl group of aldehydes and ketones, which gives a yellow or orange precipitate of the hydrazone.
Fehling's solution: A mild oxidising agent, alkaline \ce{Cu^2+} with tartrate, that gives a red precipitate of \ce{Cu2O} with aldehydes only.
Tollens' reagent: Ammoniacal silver nitrate, which gives a silver mirror with aldehydes only.
Dehydrating agent: A substance that removes water from a molecule, such as \ce{Al2O3} or concentrated \ce{H2SO4}.
Oxidising agent (organic): Acidified potassium dichromate(VI) or acidified potassium permanganate, used to oxidise primary alcohols to aldehydes and acids and secondary alcohols to ketones.
\ce{NaBH4} and \ce{LiAlH4}: Reducing agents: \ce{NaBH4} reduces a carbonyl group to an alcohol and the stronger \ce{LiAlH4} also reduces a carboxylic acid to an alcohol.

Multiple choice questionsp. 390–392

The product of the reaction between propanone and hydrogen cyanide is hydrolysed under acidic conditions. What is the formula of the final product?

A\ce{CH3CH(OH)CO2H}
B\ce{CH3CH2CH2CO2H}
C\ce{(CH3)2CHCONH2}
D\ce{(CH3)2C(OH)CO2H}Answer

Why

Propanone is a ketone, so HCN adds across its C=O group (nucleophilic addition) to give the cyanohydrin \ce{(CH3)2C(OH)CN}, 2-hydroxy-2-methylpropanenitrile. Acidic hydrolysis then turns the \ce{-CN} group into \ce{-CO2H}, so the final product is \ce{(CH3)2C(OH)CO2H} (2-hydroxy-2-methylpropanoic acid).

Option A has the same functional groups but only three carbons; it is the product from ethanal, not propanone. Option B has four carbons and no OH group. Option C is an amide, which is an intermediate of the hydrolysis, not the final acid.

Note. Printed as 'ydrolysed' (the h is missing); read as 'hydrolysed'. The book prints no answer key for this question; the key was worked out.

Compound X undergoes the following reactions. \ce{C4H9Br ->[NaOH(aq)/heat] C4H10O ->[Cr2O7^2-, H+ (aq)/heat] C4H8O only}. What is X?

A1-bromobutane
B2-bromobutaneAnswer
C1-bromo-2-methylpropane
D2-bromo-2-methylpropane

Why

Hot aqueous NaOH replaces the Br by OH, so \ce{C4H10O} is a butanol. Acidified dichromate then oxidizes it to \ce{C4H8O}, which is the formula of an aldehyde or ketone (the alcohol has lost \ce{H2}) and not of a carboxylic acid (\ce{C4H8O2}). Only a secondary alcohol gives a ketone that stops at \ce{C4H8O}: butan-2-ol gives butanone, \ce{CH3COCH2CH3}.

So X is 2-bromobutane. A primary alcohol (from options A or C) would be oxidized further to a carboxylic acid, \ce{C4H8O2}, under reflux. The tertiary alcohol from option D cannot be oxidized at all.

Note. Option C is 1-bromo-2-methylpropane, which gives a primary alcohol. The book prints no answer key for this question; the key was worked out.

iii

The reaction scheme outlines the production of one of the monomers of nylon 66 from compound X. \ce{compound X ->[KCN, in ethanol] NCCH2CH2CH2CH2CN ->[reduction] H2N(CH2)6NH2}. Which compound could be X?

A\ce{BrCH2CH2CH2CH2Br}Answer
B\ce{CH2=CHCH=CH2}
C\ce{HOCH2CH2CH2CH2OH}
D\ce{HO2CCH2CH2CH2CH2CO2H}

Why

KCN in ethanol replaces both bromine atoms of \ce{BrCH2CH2CH2CH2Br} (1,4-dibromobutane) by \ce{-CN} groups, giving \ce{NC(CH2)4CN}. The two nitrile groups add two carbon atoms, so the chain now has six carbons. Reduction turns each \ce{-CN} into \ce{-CH2NH2}, giving hexane-1,6-diamine, \ce{H2N(CH2)6NH2}, the nylon 66 monomer.

Option B has no leaving group for KCN to replace. Options C and D are not halides, so KCN cannot substitute them; option D is already a dicarboxylic acid with six carbons.

Note. The book prints no answer key for this question; the key was worked out.

Burnt sugar has two functional groups indicated by Q and R. When this compound is tested in a laboratory with 2,4-dinitrophenylhydrazine and Fehling's reagent, which functional groups are responsible for positive tests?

A2,4-DNPH: Q and R; Fehling's: Q and R
B2,4-DNPH: R only; Fehling's: Q and R
C2,4-DNPH: Q and R; Fehling's: R onlyAnswer
D2,4-DNPH: Q only; Fehling's: R only

Why

The compound is \ce{CH3COCHO}. Q is the ketone C=O group and R is the aldehyde \ce{-CHO} group. Both carbonyl groups react with 2,4-dinitrophenylhydrazine to give an orange precipitate, so the first column is 'Q and R'.

Fehling's reagent is reduced only by aldehydes, so only R (the \ce{-CHO} group) gives the red precipitate of \ce{Cu2O}. Ketone Q does not react, so the second column is 'R only'. Option C is therefore correct.

Note. The book prints no answer key for this question; the key was worked out.

Ethanal may be converted into a three-carbon acid in a two-step process. Which compound is the intermediate?

A\ce{CH3CO2H}
B\ce{CH3CN}
C\ce{CH3CH2CN}
D\ce{CH3CH(OH)CN}Answer

Why

Step 1 is the addition of HCN to ethanal (nucleophilic addition), which gives the cyanohydrin \ce{CH3CH(OH)CN}, 2-hydroxypropanenitrile. This adds one carbon atom, so the intermediate has three carbons.

Step 2 is acidic hydrolysis of the nitrile to \ce{CH3CH(OH)CO2H}, a three-carbon acid. Options A and B have only two carbons; option C has three carbons but cannot be made from ethanal in this way.

Note. The book prints no answer key for this question; the key was worked out.

Consider the sequence of reactions: \ce{CH3CH2OH ->[step I] CH3CHO ->[step II] CH3CH(OH)CH2CHO ->[step III] CH2=CHCH=CH2} (buta-1,3-diene). Which term could be used to describe step I?

Acondensation
Bdehydration
CdehydrogenationAnswer
Dhydrogenation

Why

In step I, ethanol becomes ethanal: \ce{CH3CH2OH -> CH3CHO + H2}. Hydrogen is removed from the molecule, which is a dehydrogenation (and is also an oxidation).

Dehydration would remove water, which does not happen here. Hydrogenation adds hydrogen, the reverse of this step. Condensation joins two molecules, which is what step II (the aldol reaction) does.

Note. The book prints no answer key for this question; the key was worked out.

vii

Which compound reacts with its own oxidation product (an oxidation which involves no loss of carbon) to give a sweet-smelling liquid?

Apropanal
Bpropanoic acid
Cpropanone
Dpropan-1-olAnswer

Why

Propan-1-ol is a primary alcohol. Oxidation keeps all three carbons and gives propanal and then propanoic acid, \ce{CH3CH2CO2H}. Propan-1-ol reacts with this acid by esterification to give propyl propanoate, a sweet-smelling ester.

Propanal is oxidized to propanoic acid too, but an aldehyde does not form an ester with its acid, so option A is wrong. Propanone is a ketone that cannot be oxidized without breaking the chain, so option C is wrong.

Note. The book prints no answer key for this question; the key was worked out.

viii

Butanedioic acid may be synthesised in two steps from 1,2-dibromoethane. \ce{BrCH2CH2Br ->[step 1] X ->[step 2] HO2CCH2CH2CO2H}. Which reagents could be used for this synthesis?

Astep 1: \ce{HCN(g)}; step 2: \ce{HCl(aq)}
Bstep 1: \ce{HCO2Na(aq)}; step 2: \ce{HCl(aq)}
Cstep 1: \ce{KCN(aq/alcoholic)}; step 2: \ce{H2SO4(aq)}Answer
Dstep 1: \ce{NaOH(aq)}; step 2: \ce{K2Cr2O7/H2SO4(aq)}

Why

Step 1 is the substitution of both bromine atoms by cyanide, \ce{BrCH2CH2Br + 2KCN -> NCCH2CH2CN + 2KBr}. This adds two carbons, so \ce{X} is butanedinitrile with four carbons. Step 2 is acidic hydrolysis of both nitrile groups with aqueous \ce{H2SO4}, giving butanedioic acid.

Option A uses HCN gas, which does not supply the cyanide ion needed for the substitution. Option B gives a formate ester, not a nitrile. Option D gives ethanedioic acid (two carbons), because the chain is broken and not extended.

Note. The book prints no answer key for this question; the key was worked out.

Y and Z are two widely-used selective weed killers. Which reagent will distinguish Y from Z?

Y Y

Z Z

Aacidified \ce{AgNO3(aq)}
BFehling's solution
CNa
D\ce{Na2CO3(aq)}Answer

Why

Y contains a carboxylic acid group, which reacts with sodium carbonate to release \ce{CO2} (effervescence). Z has only an alcohol group, which does not react with \ce{Na2CO3}. So \ce{Na2CO3(aq)} distinguishes them.

Sodium metal reacts with both compounds, because both have an O-H group, so option C does not separate them. Fehling's solution needs an aldehyde, which neither has. Silver nitrate gives no precipitate because the chlorine on the benzene ring is not ionized.

Note. The book prints no answer key for this question; the key was worked out.

Tartaric acid is present in some wines. It may be synthesised in the laboratory in two steps. \ce{OHCCHO ->[step 1] intermediate ->[step 2] HO2CCH(OH)CH(OH)CO2H} (tartaric acid). Which reagents could be used for this synthesis?

Astep 1: \ce{HCl(aq)}; step 2: \ce{HCN(g)}
Bstep 1: \ce{HCN}, \ce{NaCN(aq/alcoholic)}; step 2: \ce{H2SO4(aq)}Answer
Cstep 1: \ce{H2SO4(aq)}; step 2: \ce{K2Cr2O7/H2SO4(aq)}
Dstep 1: \ce{KCN(aq/alcoholic)}; step 2: \ce{K2Cr2O7/H2SO4(aq)}

Why

Step 1 adds cyanide to both aldehyde groups of ethanedial (glyoxal), giving \ce{NCCH(OH)CH(OH)CN}, the bis-cyanohydrin. Step 2 is acidic hydrolysis of both nitrile groups with \ce{H2SO4(aq)}, which gives tartaric acid, \ce{HO2CCH(OH)CH(OH)CO2H}.

Option D is wrong in step 2, because dichromate oxidizes alcohols, not nitriles, so it cannot make the diacid. Option C does not add carbon or form the cyanohydrin. Option A uses HCl in step 1, which does not form a C-C bond.

Note. Option B lists both HCN and NaCN(aq/alcoholic) in step 1. The book prints no answer key for this question; the key was worked out.

Short questionsp. 393–395

24 marks

Four organic compounds have been identified in the mandibular glands of ants, and are related to this behaviour. The structures of these molecules are as follows. Outline a series of chemical tests, with the observations, to distinguish each compound (A to D) from the others.

hexan-1-ol A

undecan-3-one B

2-butyloct-2-enal C

hexanal D

Solution

Use three tests in this order. Each one removes a compound, and a fourth test confirms the carbonyl groups. Bromine water tests for the C=C bond, Fehling's solution tests for an aldehyde, and acidified potassium dichromate tests for an alcohol that can be oxidized (a ketone does not change).

Test	A (alcohol)	B (ketone)	C (unsaturated aldehyde)	D (aldehyde)
Bromine water, shake	No change (redish brown stays)	No change	Redish brown colour fades (C=C)	No change
Fehling's solution, boil	No change	No change	Red precipitate of \ce{Cu2O}	Red precipitate of \ce{Cu2O}
Acidified \ce{K2Cr2O7}, warm	Orange to green (primary alcohol oxidised)	No change (ketone resists oxidation)	Orange to green	Orange to green
2,4-dinitrophenylhydrazine	No change	Yellow/orange precipitate	Yellow/orange precipitate	Yellow/orange precipitate

Identification: C decolourises bromine water, so it is the only alkene. D gives the red precipitate with Fehling's solution, so it is the aldehyde that is not C. Of the rest, A turns the dichromate green (primary alcohol) and B does not change (ketone), so A and B are separated. The 2,4-DNPH test shows that B, C and D are carbonyl compounds and A is not.

Note. Book gives no marks; 4 marks is the typical allocation for this question.

33 marks

Samples of the following compounds were labelled A to E as shown: A \ce{CH3CH2CH2OH}; B \ce{CH3CH2CH(OH)CH3}; C \ce{CH3CHO}; D \ce{CH3COCH3}; E \ce{CH3CH=CHCH3}. Complete the table below by inserting the letter (or letters) of the compounds that correspond to each test.

reagent	observation	letter(s)
acidified potassium dichromate(VI)	green colour obtained on boiling
acidified potassium manganate(VII)	ethanoic acid obtained on boiling
hydrogen in the presence of a platinum catalyst	hydrogen absorbed
Fehling's reagent	brown-red precipitate obtained on boiling
2,4-dinitrophenylhydrazine	orange precipitate
bromine in an inert solvent	solution decolourised

Solution

Each test is matched to the functional group it detects. Primary and secondary alcohols and aldehydes are oxidized by dichromate, a C=C or C=O group absorbs hydrogen over platinum, and only aldehydes reduce Fehling's reagent.

reagent	observation	letter(s)
acidified potassium dichromate(VI)	green colour obtained on boiling	A, B, C
acidified potassium manganate(VII)	ethanoic acid obtained on boiling	C, E
hydrogen in the presence of a platinum catalyst	hydrogen absorbed	C, D, E
Fehling's reagent	brown-red precipitate obtained on boiling	C
2,4-dinitrophenylhydrazine	orange precipitate	C, D
bromine in an inert solvent	solution decolourised	E

Notes: propanone (D) gives no green colour with dichromate, and propan-1-ol (A) gives propanoic acid, not ethanoic acid, so neither is in the second row. Hot acidified permanganate cleaves the C=C of but-2-ene (E) into two molecules of ethanoic acid. Ethanal (C) is the only aldehyde, so it is the only positive Fehling's test.

Note. The book gives no marks or answer; the table is worked out.

94 marks

Consider the following conversion. Write down the name of reagents used in the following synthesis also write type of reaction in each case. (a) \ce{CH3COOH -> CH3CH(OH)CN} (b) \ce{CH3CH2CH2CH3 -> CH3CH2CH(Br)CH3} (c) \ce{CH3CH(OH)CH3 -> CH3CH=CH2} (d) \ce{CH3CH=CH3 -> CH3CH(OH)CH2OH}

Solution

Each part is answered separately below. The reagent and the type of reaction are given for each step.

Part	Reagent	Type of reaction
(a)	HCN (or KCN then dilute acid)	Nucleophilic addition
(b)	\ce{Br2} with UV light	Free radical substitution
(c)	Conc. \ce{H2SO4}, heat at about 170^\circ\mathrm{C}	Dehydration (elimination)
(d)	Cold dilute alkaline \ce{KMnO4}	Oxidation (hydroxylation of C=C)

(a) \ce{CH3COOH -> CH3CH(OH)CN}

The product \ce{CH3CH(OH)CN} is the cyanohydrin of ethanal, so the printed starting material must be ethanal, \ce{CH3CHO}. Ethanoic acid has one carbon fewer and cannot give this product in one step.

Reagent: hydrogen cyanide, \ce{HCN} (or \ce{KCN} followed by dilute acid, to make HCN in the mixture). Type of reaction: nucleophilic addition to the C=O group of the aldehyde, \ce{CH3CHO + HCN -> CH3CH(OH)CN}.

(b) \ce{CH3CH2CH2CH3 -> CH3CH2CH(Br)CH3}

Reagent: bromine, \ce{Br2}, in the presence of ultraviolet light (or sunlight). Type of reaction: free radical substitution, in which an H atom of butane is replaced by Br. The secondary H is replaced to give 2-bromobutane, \ce{CH3CH2CHBrCH3}.

Reagent: concentrated sulphuric acid, heated at about 170^\circ\mathrm{C} (or heated \ce{Al2O3}). Type of reaction: dehydration (elimination), in which the water of the alcohol is removed to give propene, \ce{CH3CH(OH)CH3 -> CH3CH=CH2 + H2O}.

(d) \ce{CH3CH=CH3 -> CH3CH(OH)CH2OH}

Reagent: cold, dilute, alkaline potassium manganate(VII), \ce{KMnO4/OH-}. Type of reaction: oxidation, in which the C=C bond becomes two C-OH groups, giving propane-1,2-diol. The product is \ce{CH3CH(OH)CH2OH}, so the starting material is propene, \ce{CH3CH=CH2}.

Note. Two misprints: part (a) starting material printed as CH3COOH (read as CH3CHO) and part (d) printed as CH3CH=CH3 (read as CH3CH=CH2).

Long answer questionsp. 393–395

46 marks

Compound G, in which R- represents the rest of the molecule, was made for use as a tear gas in World War 2. Compound G was made by the following sequence of reactions: \ce{R-CH3 ->[stage I] R-CH2Cl ->[stage II] R-CH2CN ->[stage III] R-CHBrCN} (compound G). (a) Suggest reagents for each step. (b) Consider the following sequence of reactions: \ce{R-CHBrCN ->[stage IV] R-CHBrCO2H ->[stage V] R-CH(OH)CO2H ->[stage VI] R-CH2CO2H}. Suggest reagents for each step from iv to vi.

Solution

Compound G is an \alpha-bromo nitrile. Each stage either adds one carbon (stage II, cyanide), or changes one functional group into another. Stages IV to VI turn the nitrile into an acid, then replace Br by OH and finally remove that OH, so the net result is \ce{R-CH2CO2H}.

(a) Reagents for stages I to III

Stage I: \ce{R-CH3 -> R-CH2Cl}

Free radical substitution. Chlorine with ultraviolet light, \ce{Cl2} in the presence of UV light, replaces one H atom of the alkyl group by Cl. \ce{R-CH3 + Cl2 -> R-CH2Cl + HCl}

Stage II: \ce{R-CH2Cl -> R-CH2CN}

Nucleophilic substitution. Heat with potassium cyanide in ethanol under reflux. The \ce{CN-} ion replaces Cl and adds one carbon atom, \ce{R-CH2Cl + KCN -> R-CH2CN + KCl}.

Stage III: \ce{R-CH2CN -> R-CHBrCN}

Free radical substitution. Bromine with ultraviolet light, \ce{Br2}, replaces one H atom on the carbon next to the CN group by Br, \ce{R-CH2CN + Br2 -> R-CHBrCN + HBr}. This gives compound G.

(b) Reagents for stages IV to VI

Stage IV: nitrile to carboxylic acid

Hydrolysis. Heat with dilute hydrochloric acid (or dilute sulphuric acid) under reflux. The \ce{-CN} group becomes \ce{-CO2H} and the Br stays on the carbon, \ce{R-CHBrCN + 2H2O + H+ -> R-CHBrCO2H + NH4+}.

Stage V: replacement of Br by OH

Nucleophilic substitution. Warm with aqueous sodium hydroxide, which replaces Br by OH, \ce{R-CHBrCO2H + NaOH -> R-CH(OH)CO2Na + NaBr}. Then add dilute acid to give the free acid \ce{R-CH(OH)CO2H}.

Stage VI: removal of the OH group

Reduction. Heat with hydroiodic acid and red phosphorus. The OH group is replaced by H, \ce{R-CH(OH)CO2H + 2HI -> R-CH2CO2H + I2 + H2O}, giving the acid with one CH2 group.

Note. Stage VI (hydroiodic acid with red phosphorus) is not in the book's list of reducing agents in 21.2. It is the standard reagent for removing an OH group from the carbon next to CO2H, so it is given here.

52025 Annual I, Set D, Q.2 (vii) OR5 marks

(a) Ethanol may be converted into propanoic acid in a three-stage process which uses ethanol as the only organic compound. \ce{C2H5OH ->[step I] K ->[step II] L ->[step III] C2H5CO2H}. (i) Give the structural formulae of the intermediate compounds K and L. (ii) Suggest reagents for each step.

Solution

The three stages change OH to Br, Br to CN (adding a carbon) and CN to CO2H. Each stage is a standard reaction of the homologous series.

(i) Structural formulae of K and L

Each step changes one group. Step I replaces OH by Br (a halogenoalkane). Step II replaces Br by \ce{CN}, which adds one carbon atom to give a nitrile. Step III hydrolyses the nitrile to the acid \ce{CH3CH2CO2H}, which has three carbons, one more than ethanol.

Working

K is bromoethane, formed by substituting OH by Br.
\ce{CH3CH2OH -> CH3CH2Br}
L is propanenitrile, formed by substituting Br by CN (this adds one carbon).
\ce{CH3CH2Br + KCN -> CH3CH2CN + KBr}
Propanoic acid is formed by hydrolysis of the nitrile.
\ce{CH3CH2CN + 2H2O + H+ -> CH3CH2CO2H + NH4+}

K: bromoethane

L: propanenitrile

(ii) Reagents for steps I to III

Step I: distil ethanol with concentrated sulphuric acid and potassium bromide (or with HBr), giving bromoethane. If \ce{PCl5} is used instead, K is chloroethane, \ce{CH3CH2Cl}, and the later steps are the same.

Step II: potassium cyanide in ethanol, heated under reflux (nucleophilic substitution).

Step III: dilute hydrochloric acid (or dilute sulphuric acid), heated under reflux (hydrolysis of the nitrile).

65 marks

Lactic acid, 2-hydroxypropanoic acid, \ce{CH3CH(OH)CO2H}, occurs in sour milk. Glycolic acid, 2-hydroxyethanoic acid, \ce{HOCH2CO2H}, occurs in sugar cane. (a) Lactic acid may be synthesized from propene. Write down the steps involved in this synthesis. (b) Convert glycollic acid to ethanoic acid.

Solution

Propene gives lactic acid by oxidation, oxidation and reduction of a keto acid. Glycolic acid loses its OH group by reduction.

(a) Steps from propene to lactic acid

The OH group must go on C-2 and the acid group on C-1. Oxidation of a diol converts the primary OH to \ce{CO2H} and the secondary OH to a ketone, so the route is: alkene to diol, diol to keto acid, and keto acid to hydroxy acid.

Step 1: propene to propane-1,2-diol

Reagent: cold, dilute, alkaline potassium manganate(VII), \ce{KMnO4/OH-}. Both C atoms of the C=C bond gain an OH group (oxidation). \ce{CH3CH=CH2 -> CH3CH(OH)CH2OH}

Step 2: propane-1,2-diol to 2-oxopropanoic acid

Reagent: acidified potassium dichromate(VI), heated under reflux. The primary OH becomes \ce{CO2H} and the secondary OH becomes a C=O group. \ce{CH3CH(OH)CH2OH -> CH3COCO2H}

Step 3: 2-oxopropanoic acid to lactic acid

Reagent: sodium tetrahydridoborate(III), \ce{NaBH4}, in water. It reduces the ketone to a secondary alcohol but does not reduce the \ce{CO2H} group. \ce{CH3COCO2H -> CH3CH(OH)CO2H}

propane-1,2-diol

2-oxopropanoic acid

lactic acid

(b) Glycollic acid to ethanoic acid

The OH group on the carbon next to \ce{CO2H} must be replaced by H. Heating an \alpha-hydroxy acid with hydroiodic acid and red phosphorus removes the OH group and leaves the acid, \ce{HOCH2CO2H + 2HI -> CH3CO2H + I2 + H2O}.

glycollic acid

ethanoic acid

Note. The book gives no route for this question. The route in part (a) follows the book's own method for converting an alkene to a diol and then to a keto acid (see question 10).

74 marks

Outline steps of sequence used to synthesise ethanoic acid from methanol.

Solution

The carbon count rises from one (methanol) to two (ethanoic acid), so a cyanide step is needed. The route is the same as the book's route from methane (Section 21.1.2).

Step 1: methanol to chloromethane

Reagent: concentrated hydrochloric acid with anhydrous zinc chloride (or PCl5). Nucleophilic substitution of OH by Cl, \ce{CH3OH -> CH3Cl + H2O}.

Step 2: chloromethane to ethanenitrile

Reagent: potassium cyanide in ethanol, heated under reflux. The \ce{CN-} ion replaces Cl and adds one carbon, \ce{CH3Cl + KCN -> CH3CN + KCl}.

Step 3: ethanenitrile to ethanoic acid

Reagent: dilute hydrochloric acid, heated under reflux. The nitrile is hydrolysed to the acid, \ce{CH3CN + 2H2O + H+ -> CH3CO2H + NH4+}.

methanol

chloromethane

ethanenitrile

ethanoic acid

\ce{CH3OH ->[HCl, ZnCl2] CH3Cl ->[KCN, ethanol, heat] CH3CN ->[dilute HCl, heat] CH3CO2H}

84 marks

(a) Complete the following reaction which starts with 1-bromobutane. In each empty box, write the structural formula of the organic compound that would be formed. \ce{CH3CH2CH2CH2Br ->[NaOH(aq), heat under reflux] W ->[K2Cr2O7/H+, heat under reflux] Y} and \ce{CH3CH2CH2CH2Br ->[NaOH in ethanol][heat under reflux] X ->[{conc.} KMnO4/H+][heat under reflux] Z}. (b) One of the compounds W, X, Y or Z can be polymerized. Identify this compound by its letter.

Solution

The four compounds come from two routes: substitution, then oxidation (W to Y), and elimination, then oxidative cleavage (X to Z).

(a) Structures of W, X, Y and Z

W is formed by substitution. Hot aqueous NaOH replaces Br by OH, giving butan-1-ol.

X is formed by elimination. NaOH in ethanol removes HBr from 1-bromobutane, giving but-1-ene.

Y is formed by oxidation of W. Acidified dichromate oxidizes the primary alcohol through butanal to butanoic acid.

Z is formed by oxidation of X. Hot, concentrated, acidified KMnO4 breaks the C=C bond. The CH2= end becomes CO2 (given off), and the rest of the chain becomes propanoic acid, \ce{CH3CH2CH=CH2 -> CH3CH2CO2H + CO2}.

W W: butan-1-ol

X X: but-1-ene

Y Y: butanoic acid

Z Z: propanoic acid

(b) Compound that can be polymerized

Answer: X. But-1-ene has a C=C double bond, so its molecules can join by addition polymerization to give poly(but-1-ene). W, Y and Z have no C=C bond and cannot polymerize in this way.

Note. The book does not show the CO2 produced in the oxidation of X. It is included in the answer to explain why Z is propanoic acid.

105 marks

Starting from But-1-ene, \ce{CH3CH2CH=CH2}, how would you synthesis 2-oxo butanoic acid? Outline steps of sequence in this conversion.

Solution

2-oxobutanoic acid, \ce{CH3CH2COCO2H}, has a ketone at C-2 and an acid at C-1. The oxidation of a diol gives the same pattern (acid at the primary carbon, ketone at the secondary carbon), so the route is alkene to diol and then diol to keto acid.

Step 1: but-1-ene to butane-1,2-diol

Reagent: cold, dilute, alkaline potassium manganate(VII), \ce{KMnO4/OH-}. Each carbon of the C=C bond gains one OH group, \ce{CH3CH2CH=CH2 -> CH3CH2CH(OH)CH2OH}.

Step 2: butane-1,2-diol to 2-oxobutanoic acid

Reagent: acidified potassium dichromate(VI), heated under reflux. The primary OH is oxidized to \ce{CO2H} and the secondary OH to a C=O group, \ce{CH3CH2CH(OH)CH2OH -> CH3CH2COCO2H}.

but-1-ene

butane-1,2-diol

2-oxobutanoic acid

\ce{CH3CH2CH=CH2 ->[KMnO4/OH-, cold] CH3CH2CH(OH)CH2OH ->[K2Cr2O7/H+, heat] CH3CH2COCO2H}

115 marks

Compare and contrast between synthesis and retrosynthesis.

Solution

Definitions

Synthesis is the forward process of combining simple starting materials (building blocks) by a series of reactions to make the required organic compound. Retrosynthesis is the reverse process: the target compound is broken down into simpler precursors, which can be obtained from known starting materials, to find the best route.

Similarities

Both use the same reactions of the homologous series (substitution, addition, oxidation, reduction, esterification and so on) and the same reagents.
Both aim at a practical route with as few steps as possible.
Both are based on functional group interconversion: a functional group is changed into another one, step by step.

Differences

Point	Synthesis	Retrosynthesis
Direction	Forward: from the starting material to the target	Backward: from the target back to the starting material
Starting point	Simple, readily available compounds	The target compound that is to be made
Method	Combine reactions to build the molecule	Break the molecule into simpler fragments and precursors
Example	Ethanol and propanoic acid, with conc. \ce{H2SO4}, give ethyl propanoate	Ethyl propanoate is broken into propanoic acid and ethanol
Purpose	To make the compound	To plan and compare possible routes before making it

Application

Retrosynthesis gives more than one pathway for a target compound. A pathway may be more cost-effective, more transparent to the environment and need fewer steps. Synthesis then carries out the chosen pathway in the laboratory.

FBISE 2025 questions

Questions from the FBISE HSSC-I 2025 (1st Annual) papers that belong to this unit, answered the way an examiner expects.

2025 Annual I, Set D, Q.2 (vii) OR3 marks

Ethanol can be converted into propanoic acid in three stage process: \ce{C2H5OH ->[I] A ->[II] B ->[III] C2H5-COOH}. Write down the balanced chemical equation to show the conversion in each step.

Solution

The three stages must end with propanoic acid, which has one more carbon than ethanol. So one stage adds a carbon through the nitrile group: convert the alcohol to a halogenoalkane (A), then to a nitrile (B), then hydrolyse.

Step I: replace \ce{OH} by \ce{Br}. Distil ethanol with potassium bromide and concentrated sulphuric acid. A is bromoethane.
\ce{C2H5OH + HBr -> C2H5Br + H2O}
Step II: heat bromoethane under reflux with potassium cyanide in ethanol. \ce{CN-} replaces \ce{Br-} and adds a carbon. B is propanenitrile.
\ce{C2H5Br + KCN -> C2H5CN + KBr}
Step III: hydrolyse the nitrile by heating under reflux with dilute hydrochloric acid.
\ce{C2H5CN + 2H2O + HCl -> C2H5COOH + NH4Cl}

A: bromoethane

B: propanenitrile

\ce{C2H5OH ->[\text{KBr, {conc.} }H2SO4] C2H5Br ->[KCN, \text{ethanol}] C2H5CN ->[\text{dil. HCl, reflux}] C2H5COOH}

Note. This is part (a) of the book's long question v (long-q2).

Exam focus

Convert ethanol into propanoic acid in three steps (ethanol, halogenoalkane, nitrile, acid) with reagents and equations. (long)
Asked in 2025 (Set D Q.2 (vii) OR) and is part (a) of the book's long question v.
Go to answer
Lengthen a carbon chain: halogenoalkane to nitrile with KCN, then hydrolysis to an acid or reduction to an amine. (long)
The nitrile step is the core of 2025 Set D Q.2 (vii) OR; the book's MCQ iii and long question iv use it too. Carboxylic acid preparation from nitriles was also asked in 2026 (Q.4) and 2025 (Q.4, MCQ 15), which are solved in unit 19.
Go to answer
Chemical tests to distinguish two compounds: iodoform, Fehling's, Tollens', 2,4-DNPH, sodium carbonate and acidified dichromate. (short)
Asked in 2026 (Set B Q.2 (ix) propan-1-ol and propan-2-ol, solved in unit 18; Set D Q.2 (ix) propanoic acid and ethanal, solved in unit 19) and in the book's short questions i and ii.
Go to answer
Name the reagent and the type of reaction for each step (oxidation, reduction, substitution, elimination, addition, dehydration). (short)
2026 Set B MCQ 14 (oxidation of ethanol, solved in unit 18), 2025 Set D Q.2 (iv) OR (diol, ester hydrolysis, reduction; solved in unit 18) and the book's short question iii and MCQ vi.
Go to answer
Reducing agents and what each reduces: NaBH4 (carbonyl to alcohol), LiAlH4 (acid or nitrile), H2/Ni (C=C), Sn/HCl (NO2 to NH2). (short)
2026 Set B Q.2 (ix) OR (Sn/HCl and Fe/H2SO4 on nitro compounds, solved in unit 20) and 2025 Set D Q.2 (iv) OR (iii) (NaBH4 on benzaldehyde, solved in unit 18).
Go to answer
Oxidise alcohols and aldehydes with acidified dichromate (reflux for the acid, distil for the aldehyde) and use the products to tell alcohols apart. (short)
Behind 2026 Set B MCQ 14 and Q.2 (ix) (both solved in unit 18); the 2026 and 2025 Q.4 acid preparations (solved in unit 19) rest on the same reaction.
Go to answer
Plan a multi-step synthesis with intermediates, for example lactic acid from ethanal or 2-oxobutanoic acid from but-1-ene. (long)
Not asked as a separate question in the 2025 or 2026 paper, but it is the unit's main skill and the book's long questions vi and vii.
Go to answer
Compare and contrast synthesis and retrosynthesis, and work backwards from a target molecule. (long)
Not in the 2025 or 2026 paper; it is the book's long question 11 and the subject of the last section.
Go to answer