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Synthesis Explorer
Teachers Guide
Part 1: Carbon Chains and Functional Groups
1.1

The molecular, displayed and skeletal formulas of propane can be seen on the
structures tab − right hand screen of Synthesis Explorer.

After fractional distillation of crude oil, propane would be a component of the refinery
gas fraction.

Ethene could be obtained from propane by a cracking reaction, in which the propane
is heated in the presence of a suitable catalyst.

In the mass spectrum of propane, the molecular ion (parent ion) peak can be seen at
a mass/charge ratio of 44; after cracking, the molecular ion of ethane (a smaller
molecule) can be seen at 28.

Ethene can be oxidised to ethane-1,2-diol using alkaline potassium manganate(VII)
solution.

Commercially ethene and oxygen are passed over a silver catalyst at 250 °C to form
the three-membered ring of epoxyethane which is then hydrolysed with water or dilute
acid to form ethane-1,2-diol.

The C=C double bond is responsible for the reactivity of ethene.

Ethene and other alkenes will immediately decolourise an aqueous solution of
bromine; alkanes will not react in this way.

Ethene can be easily converted to bromoethane by mixing it with gaseous HBr. The
reaction occurs at room temperature. The electrophilic addition mechanism can be
found in any organic chemistry textbook.

Bromoethane can be converted to ethanol by gentle warming with an aqueous alkali.
The reaction happens by a nucleophilic substitution mechanism.

It is more efficient to produce ethanol directly from ethene by passing a mixture of
ethene and steam over a catalyst of phosphoric acid on silica at 300 °C and a
pressure of ~60 atm.

In an older method ethene is reacted with concentrated sulphuric acid followed by
water. The oldest known method of producing ethanol is by fermentation of starch- or
sugar-containing plant material.

By placing methane on the Synthesis Explorer reaction canvas, its molecular formula
and structure can be seen in the right hand panel.
1.2
1.3
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
Right-clicking on methane on the reaction canvas will reveal its reactions, including
that with bromine in the presence of UV to form bromomethane.

The molar mass of bromomethane is revealed on the ‘physical’ tab in the right hand
panel. The mass spectrum shows two peaks at ~94 and 96, resulting from
approximately equal amounts of two isotopes of bromine of masses ~79 and 81.

The reaction canvas will show the conversion of bromomethane to methanol by the
action of aqueous alkali.

The mass spectrum of methanol shows a molecular ion at a mass/charge ratio of 32;
the twin peaks of the bromine isotopes have disappeared.

The molecular, skeletal and displayed (2D and 3D) formulas of butane are shown on
the structure tab on the reaction canvas.

The formula mass is shown on the physical tab and a molecular ion peak appears at
m/z 58 on the mass spectrum. The cracking reaction occurs at a high temperature in
the presence of a catalyst.

Ethane and ethene are possible products, together with methane and propene.

Low density poly(ethene) (LDP) is produced by heating ethene at high pressure in the
presence of a little oxygen as initiator. High density poly(ethene) (HDP) requires
much lower pressure in the presence of a Ziegler Natta, or more recently,
metallocene, catalyst.

Poly(ethene) is saturated and therefore will not decolourise aqueous bromine.

The given % composition does not add up to 100% so the remaining 35% must be
oxygen. The empirical formula is C2H6O, relative mass 46 g mol−1.

Structures corresponding to ethanol CH3CH2OH (alcohol) or methoxymethane
CH3OCH3 (ether) can be drawn.

Ethers do not (currently) feature in Synthesis Explorer, but ethanol could be identified
by its boiling point of 78.5 °C.

Its mass spectrum shows a molecular ion at m/z 46 and a base peak at m/z 31,
corresponding to the removal of a methyl group CH3.

The RMM of 1-bromopropane is 123 g mol−1.

The pair of molecular ion peaks at m/z 122 and 124 arise from the presence of two
isotopes of bromine.

The base peak at m/z 43 corresponds to the fragmentation of a bromine atom from
the molecule. (The pair of isotope peaks are not so obvious here – there is a clearer
version of the spectrum on SpectraSchool).

The structural formula of 1-bromopropane is CH3CH2CH2Br. Its displayed formula is
shown on the Synthesis Explorer reaction canvas.
1.4
1.5
1.6
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
2-bromopropane is a position isomer of 1-bromopropane. Its structural formula is
CH3CHBrCH3.

On gentle warming with aqueous alkali, a nucleophilic substitution reaction ccurs
forming propan-1-ol. The mechanism can be found in any A level textbook. But on
heating with an ethanolic solution of sodium hydroxide, an elimination reaction will
produce propene.
Part 2: Chains and Rings with Polar Groups
2.1

Esters can be produced by refluxing an alcohol with a carboxylic acid (or derivative)
in the presence of concentrated sulphuric acid as a catalyst.

Oxidise ethanol to ethanoic acid by refluxing with excess acidified potassium
dichromate solution. Then react with more ethanol in the presence of concentrated
sulphuric acid to produce ethyl ethanoate.

In the IR spectrum of ethanol, the broad O-H absorption is prominent at ~3300 cm−1.
There are no peaks between 1500 and 2500 cm−1. For ethyl ethanoate, the broad OH peak has disappeared but a sharp peak has appeared at ~1750 cm−1 confirming
the presence of a carbonyl group C=O.

In the first step ethanol should be converted to bromoethane by reacting with HBr.
This must be generated in situ from potassium (or sodium) bromide and concentrated
sulphuric acid.

The second step is achieved by heating the bromoethane (in ethanol solution) with
ammonia under pressure (in a sealed tube).

Both of these steps occur by a nucleophilic substitution mechanism. (This last section
is not examined in most A level specifications).

The nucleophile attacks the back of the halogenoalkane molecule (i.e. the side
opposite the halogen). So the configuration of the product is inverted. In this case it
would not be noticeable. But butan-2-ol contains a chiral carbon atom. If the starting
material was a single enantiomer, the product of each nucleophilic substitution
reaction would be the mirror image.

The first step is to dehydrate butan-1-ol to but-1-ene, by passing the alcohol vapour
over heated Al2O3. This is an elimination reaction.

In the second step bromine adds across the alkene double bond to form 1,2dibromobutane. This is an electrophilic addition mechanism which will be described in
any textbook.

The empirical formula is C3H8O.
2.2
2.3
2.4
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
The reaction with acidified potassium dichromate suggests that the compounds could
be alcohols, including the functional group O-H.

The two compounds could be propan-1-ol CH3CH2CH2OH and propan-2-ol
CH3CHOHCH3.

As they are isomers, both compounds exhibit a molecular ion at m/z 60. But the base
peak for propan-1-ol is at 31, representing the removal of an ethyl group, while the
base peak for propan-2-ol is at 45, representing the removal of a methyl group.

But their IR spectra are very similar, both showing the broad O-H absorption at
~3300 cm−1 and C-H at ~3000 cm−1.

On oxidation with acidified potassium dichromate solution, propan-1-ol forms the
aldehyde propanal, CH3CH2CHO and propan-2-ol forms the ketone propanone,
CH3COCH3.

In both IR spectra, the broad O-H absorption disappears, to be replaced by the sharp
C=O peak at ~1750 cm−1.

This could be done in three steps: oxidise the ethanol to ethanoic acid by refluxing
with acidified potassium dichromate; convert ethanoic acid to ethanoyl chloride by
reacting with phosphorus pentachloride; react the ethanoyl chloride with more ethanol
in the presence of an acid catalyst to produce ethyl ethanoate.

Full balanced equations would not be required for the first two steps. For example, in
the first step the oxidising agent could be represented by [O].

In the reaction between ethanol and ethanoyl chloride, a lone pair on the hydroxyl
group of the alcohol acts as the nucleophile, attacking the electrophilic carbon atom
of the acyl chloride. This is followed by elimination of HCl.

In the IR spectra of both ethanol and ethanoic acid a broad O-H peak will be visible,
although a good student should be able to distinguish the alcohol from the acid by its
position and shape. The characteristic carbonyl group peak will be visible in the acid,
acyl chloride and ester, but not, of course, in the alcohol.

Ethanoic anhydride could be used to replace ethanoyl chloride.

As Y can be both oxidised and reduced it is likely to be an aldehyde. Propanal has
the structure CH3CH2CHO.

Z can be reduced but cannot be oxidised so it is likely to be a ketone. Propanone has
the structure CH3COCH3.

Aldehydes and ketones can be distinguished by using either Fehling’s solution or
Tollens’ reagent. Aldehydes give a red precipitate of copper(I) oxide when warmed
with Fehling’s solution, while ketones do not react. Similarly aldehydes produce a
silver mirror on the inside of the test tube when warmed with Tollens’ reagent while
ketones do not react. Both of these reactions show that aldehydes are reducing
agents, themselves being oxidised to carboxylic acids.
2.5
2.6
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
In the mass spectrum of Y, the base peak can be seen at m/z value 29. The propanal
molecule has fragmented into two equal halves, as both the CH3CH2 and the CHO
fragments have a relative mass of 29.

In the mass spectrum of Z, the base peak can be seen at m/z value 43, a methyl
group having fragmented off the propanone molecule.

Both IR spectra show the characteristic peak of the carbonyl group C=O at around
1730 cm−1.

This synthesis initially proceeds from ethanol by two separate tracks.

In the first track ethanol is converted to bromoethane by a nucleophilic substitution
reaction using KBr and concentrated sulphuric acid to generate HBr. Then by another
nucleophilic substitution bromoethane is converted to ethylamine by heating with
ammonia in a sealed tube to increase the pressure.

In the second track ethanol can be oxidised to ethanoic acid by refluxing with acidified
potassium dichromate solution. Then ethanoic acid is converted to ethanoyl chloride
by a substitution reaction using PCl5 (or SOCl2).

Finally the products form each track, ethylamine and ethanoyl chloride, can be
reacted to achieve the desired secondary amide. In this reaction the nucleophilic
amine attacks the electrophilic acyl chloride.
2.7
Part 3: Pulling it All Together
3.1

Step 1 – convert ethanol to bromoethane using NaBr and concentrated sulphuric acid
to generate HBr.

Step 2 – in another nucleophilic substitution reaction convert bromoethane to
propanenitrile by warming with an ethanolic solution of KCN. This step adds a carbon
atom to the chain.

Finally propanenitrile can be hydrolysed by warming with dilute acid, producing
propanoic acid.

The main difference in the IR spectra is the presence of the C=O peak at about
1720 cm−1 in the spectrum of propanoic acid. A shift in the O-H peak from about
3400 cm−1 in the alcohol to 3100 cm−1 in the acid may also be observed.

The α carbon atom in lactic acid is a chiral centre and therefore two mirror image
molecules (enantiomers) can be drawn.

Ethanal reacts with KCN in ethanolic solution by a nucleophilic addition reaction to
produce 2-hydroxypropanenitrile. This can be hydrolysed with dilute acid to produce
2-hydroxypropanoic acid (lactic acid).
3.2
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
2-hydroxypropanoic acid contains both a secondary alcohol and a carboxylic acid
group. These two functional groups can react to produce an ester.

Two molecules of lactic acid can condense to form a cyclic ester with the elimination
of two molecules of water.

Lactic acid can also form a condensation polymer PLA, which has the repeated unit
-O-CH(CH3)-CO-.

Methylbenzene can be oxidised to benzoic acid using an alkaline solution of
potassium manganate (VII). Then the benzene ring can be nitrated by an electrophilic
substitution reaction, using a nitrating mixture of concentrated nitric and sulphuric
acids and keeping the temperature low to prevent multiple substitution. (At this level it
may be assumed that the nitro group is substituted in the 3-position). Then the nitro
compound can be reduced using Sn/HCl.

3-aminobenzoic acid can undergo condensation polymerisation forming a polymeric
aromatic amide (aramid). Fire-resistant and able to be made into fibres, this was a
pre-cursor of Kevlar.

In step 1, benzene is reacted with ethanoyl chloride in the presence of aluminium
chloride (Friedel Crafts acylation) to form phenylethanone. This happens by an
electrophilic substitution mechanism.

In step 2, phenylethanone can be reduced by NaBH4 to produce 1-phenylethanol
(ketone reduced to secondary alcohol).

If phenylethanol is dehydrated by passing its heated vapour over aluminium oxide,
the alkene phenylethene C6H5CH=CH2 is produced. Its common name is styrene and
the polymer polystyrene is used for mouldings, or, in its expanded form, for ceiling
tiles and insulation.

1-phenylethanol could be reacted with a carboxylic acid or derivative (acyl chloride,
anhydride) to produce a sweet smelling ester.

Step 1 ethene + HBr  bromoethane

Step 2 bromoethane + KCN  propanenitrile (nucleophilic substitution)

Step 3 propanenitrile + LiAlH4 (in ether)  1-aminopropane (reduction)
3.3
3.4
3.5
(electrophilic addition)
Mechanisms can be found in a textbook or try www.chemtube3d.com
Part 4: Building Useful Molecules
4.1

Step 1 – nitrate the phenol by reacting with a nitrating mixture of concentrated nitric
and sulphuric acids, keeping the mixture cool to minimise multiple substitution. (At
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this level you should assume that substitution occurs at the required position).
Multiple substituted derivatives such as 2,4,6-trinitrophenol could be detected by TLC.

Step 2 – reduce the nitro group using Sn/HCl to produce 4-aminophenol.

Step 3 – Acylate the amine with ethanoyl chloride to produce paracetamol.

The phenol O-H group is obvious in both IR spectra – very broad in phenol, showing
hydrogen bonding. The paracetamol spectrum shows a C=O peak at ~1650 cm−1.

The proton NMR spectrum of phenol shows two peaks – the phenol proton and the
aromatic protons, while paracetamol shows five peaks – methyl, amide and phenol
protons and two for the aromatic protons. Some simple discussion of splitting could
be expected.

The reactions described suggest that compounds A and B both contain an alkene
double bond and that carboxylic acid groups are present.

The number of oxygen atoms in the molecular formula suggests the presence of two
carboxylic acid groups per molecule.

Compounds A and B could be Z (cis) and E (trans) isomers of butendioic acid

HO2C-CH=CHCO2H, commonly known as maleic and fumaric acids respectively.

Addition of bromine across the double bond results in two chiral centres and therefore
four optical isomers.

The higher melting point of the E-isomer results from inter-molecular hydrogen
bonding. In the Z-isomer, intra-molecular hydrogen bonding is possible between the
two acid groups, so there are fewer intermolecular links.

On heating, the molecular mass of compound A decreases by 18, which suggests
that dehydration has happened, producing Z-butendioic (maleic) anhydride. This
cyclic compound contains a five-membered ring (named a furan as one of the atoms
is oxygen) and is a versatile reagent. The reaction can be reversed by hydrolysis.

Credit would be given for the following suggested steps:
4.2
4.3
o
Reaction of phenylamine with chloromethane (molar ratio 1:2) to produce N,Ndimethylphenylamine;
o
Sulphonation of phenylamine to produce 4-aminobenzenesulphonic acid.
o
Diazotisation of 4-aminobenzenesulphonic acid.
o
Coupling the diazonium salt with N,N-dimethylphenylamine to produce methyl
orange.
4.4

During acid hydrolysis, the amide and ester bonds in aspartame would break to leave
two amino acids – aspartic acid and phenylalanine (both with amine groups
protonated), together with methanol.
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
The amino acids can be separated by paper chromatography and located using
ninhydrin.

If the hydrolysis had been carried out under alkaline conditions, the amino acids
would be obtained as their sodium salts.

Recognise that the 5 carbon atoms can be divided between the acid and alcohol
precursors of each ester, and that both straight and branched chains are possible.
Straight chain examples should be easy to name.
4.5
o
Derivatives of methanoic acid:
H COO CH2 CH2 CH2 CH3
butyl methanoate
H COO CH2 CH (CH3)2
H COO CH (CH3) CH2 CH3
H COO C (CH3)3
o
Derivatives of ethanoic acid:
CH3 COO CH2 CH2 CH3
propyl ethanoate
CH3 COO CH (CH3)2
o
Derivative of propanoic acid:
CH3 CH2 COO CH2 CH3
o
ethyl propanoate
Derivatives of butanoic acid:
CH3 CH2 CH2 COO CH3
methyl butanoate
(CH3)2 CH COO CH3

The tertiary butyl ester of methanoic acid shown in bold above would exhibit only two
singlet peaks (integration ratio 1:9) in its proton NMR spectrum.

To obtain and identify the alcohol and carboxylic acid, the ester should be hydrolysed
by refluxing with moderately concentrated alkali. (It might then be possible to
crystallise the sodium salt of the acid.) After acidification to release the free acid,
fractional distillation could be used to identify the boiling point of each component.