topic guide 13.4: the chemistry of the industrial · pdf file• understand the chemistry...

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Unit 13: Industrial chemistry

In the previous two topic guides, you have looked at some of the factors that are important in a viable chemical process and related these to the underlying chemical ideas. Now you will put these ideas together and examine them in detail in relation to your chosen industrial process.

You will bring together ideas used in previous portfolio activities in order to produce a study of a single chemical process, including consideration of alternative pathways for the manufacture of the product.

The case studies in this topic guide relate to two different methods for the manufacture of propanone.

On successful completion of this topic you will: • understand the chemistry of the industrial process (LO4).

To achieve a Pass in this unit you need to show that you can: • justify the choice of reaction and identify the mechanism of each stage

and its relation to the reaction conditions (4.1) • explain the factors affecting yields including those of alternative pathways (4.2) • describe the processes of separation and purification and their influence on

the overall yield of product available (4.3) • explain the potential uses and commercial values of the principal and

co-products (4.4).

The chemistry of the industrial process13.4

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13.4: The chemistry of the industrial process

1 Choice of reactionEvaluating process factorsYou saw in Topic guide 13.2 how industrial reactions can be described in terms of a range of process factors:

• the starting materials (raw materials) needed in the reaction • the yield and rate of the reaction • the operating conditions • the carbon footprint of the process.

These process factors can be used to evaluate a reaction in order to select the most industrially viable process:

• Which reaction uses the cheapest and most readily available raw materials? • Which reaction produces the greatest yield and/or the fastest rate? • Which reaction uses conditions that are cheaper to produce and maintain? • Which reaction has the lowest carbon footprint and is most sustainable?

Case study: Manufacture of propanonePropanone (acetone) is a simple ketone containing three carbon atoms, as shown in Figure 13.4.1.

C

CH

H

H

H

H O

CH

Figure 13.4.1: The structure of propanone.

Propanone is widely used as a solvent and is converted into methyl isobutyl ketone, which has important uses as a solvent for plastics and adhesives. The UK annual production of propanone is about 150 000 tonnes.

Two processes for propanone production will be considered: • the oxidation of cumene (see Figure 13.4.2), and • the catalytic dehydrogenation of propan-2-ol (see Figure 13.4.3).

H3C CH3 H3CH3C

OH OH

+

CH3

H3C

O

Figure 13.4.2: Oxidation of cumene.

H2+

CH3

H3C

O

CH3

H3C

OH

Figure 13.4.3: Dehydrogenation of propan-2-ol.

In process 1, air oxidation of cumene occurs at temperatures of 90–130 °C and a slightly raised pressure (1–10 bar), which is used to keep the organic reactants and products in a liquid state. This oxidation produces an unstable hydroperoxide, which then breaks down at 60–70 °C by the action of a sulfuric acid catalyst to form propanone and phenol. The yield is approximately 85–87%.

In process 2, propan-2-ol vapour is passed over a copper catalyst at 400–500 °C to form propanone. The yield is approximately 95%. continued

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Questions:1 Identify the starting materials needed for each of these processes. Using ideas from Topic

guide 13.2, classify these as raw materials, feedstocks or commodity chemicals.2 Find a current price for these starting materials in $ (or £) per tonne. Which process uses the

cheapest starting materials? 3 Comment on the yield obtained in each process. Use your answers to 2 and 3 to comment on

the most cost-effective process in terms of the cost of starting materials needed to produce 1 tonne of product.

4 Compare the conditions used in the two processes and the yields obtained. Which process appears to be most cost-efficient in terms of energy use?

5 Comment on the carbon footprint of the two processes.6 On the basis of this information, which process is likely to be most cost-effective and sustainable?

Take it furtherDetails of the manufacturing processes described in the case studies in this topic guide can be found in The Essential Chemical Industry (Allan Clements et al., 2010) on p147 and p154.

ActivityYou should have collected information about the starting materials, conditions used and yield of your chosen process in completing your portfolio for LO2 and LO3.

• Find out whether there are other possible methods of manufacturing your chosen chemical and (where possible) research the starting materials, conditions used and yield of any other method(s).

• Use this information to suggest reasons why your chosen process is used to manufacture this chemical.

2 Mechanism of reactionIn this section, you will be looking in detail at the mechanism of the two stages in the production of propanone by cumene oxidation.

The mechanism of a reaction is a description of what happens in a chemical reaction at a molecular level. Reactions that are represented by a single overall equation may, in reality, occur by a series of identifiable steps involving bond breaking, bond forming, electron movement, and so on. Examining what happens in these steps helps chemists to understand the role of reactants and catalysts in the reaction and, by identifying intermediates formed during the process, can make it easier to understand (and hence minimise) the formation of unwanted side-products.

Reactions involving oxygen, or air, often occur by an autoxidation mechanism involving free radicals.

Key terms

Key termsIntermediate: A short-lived substance formed during the mechanism of a chemical reaction.

Autoxidation: An oxidation process that involves oxygen or air, and that proceeds by a free-radical mechanism.

Free radical: A reactive species (molecule, ion or atom) that contains one or more unpaired electrons. This is often shown in the structural formula using the convention of a dot (•) to represent the position of the unpaired electron (for example, H–O–O•).

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Case study: The mechanism for cumene oxidationCumene oxidation is a free-radical process. The presence of oxygen at slightly raised temperature creates conditions ideal for autoxidation.

The initiation step is the loss of an H atom from the cumene molecule to create a free radical, by the action of atmospheric oxygen.

H3C CH3

O O

H3CC

CH3

OHO+ +

Figure 13.4.4: Equation for initiation step of cumene oxidation.

Once cumene radicals are produced, a rapid free-radical chain reaction occurs. In the first stage the cumene radical reacts with an oxygen molecule. This reacts with another cumene molecule to form cumene hydroperoxide (CHP).

H3CO O+

H3CC

CH3O

H3C O

+ +

H3CC

CH3OH

H3CH3C

OH3C CH3O

H3CH3C

O

Figure 13.4.5: The free radical chain reaction that forms cumene hydroperoxide.

Questions:1 Explain why the free-radical reaction of the cumene radical is described as a ’chain reaction’.2 Reactions involving free radicals commonly form significant amounts of side products.

Suggest why.3 Combine the two equations of the free-radical chain reaction to deduce an overall equation for

these two steps.

The mechanism of many organic reactions (such as those shown in the case study) can be represented by so-called ’curly arrow‘ diagrams, in which an arrow is used to represent the movement of a pair of electrons. The base of the arrow shows the original location of the electrons (such as a covalent bond or a lone pair) and the arrow represents the final location of the electron pair. In mechanisms involving free radicals, ’half-arrows‘ are used to show the movement of a single electron.

Take it furtherThe use of curly arrows in representing mechanisms is described further at http://www.chemguide.co.uk/basicorg/conventions/curlies.html.

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Case study: Formation of propanone and phenolPropanone, and the co-product phenol, are formed by an acid-catalysed hydrolysis reaction.

In the first stage, an H+ ion (proton) attaches itself to the peroxide group (OOH) of the cumene hydroperoxide, forming an OH

2+ group. This is easily lost, causing a shift in a pair of electrons that

results in the formation of a new bond between the phenyl group and the remaining oxygen atom.

The carbon atom that was originally bonded to the phenyl group is now left as a carbocation.

H3O+

+

+0OH2+

OOOH

H2O

Figure 13.4.6: The first stage in the hydrolysis of cumene hydroperoxide.

In the second stage, the carbocation is attacked by a water molecule. An H+ from this attached water molecule is then transferred to the other oxygen in the molecule (proton transfer) in a reversible process, and finally removal of another H+ from the molecule results in the break up of the molecule into the products phenol and propanone. Propanone is separated and purified by distillation (b.p. = 56 °C).

H3O+

OH2

OH2

OH O

H

HO

+ +

+O O

+HO HO

+

Figure 13.4.7: The mechanism of the second stage in the hydrolysis reaction.

The entire sequence of reactions occurs in aqueous solution, using dilute sulfuric acid (0.1%–1.0% w/v) as a solvent.

Questions:1 Explain why the H+ ion can be regarded as a catalyst in this sequence of reactions.2 Increasing the pressure will have no effect on the rate of this reaction. Explain why.3 Why is it not possible to carry out this reaction at high temperature (for example, above 100 °C)?4 Suggest why this sequence of reactions occurs rapidly enough at such low temperatures.

Activity • Find out what you can about the mechanism for all or part of your chosen process. Describe

the mechanism using equations to represent the different steps. Include a curly arrow representation of the reaction step(s), if appropriate.

• If possible, use ideas about the reactivity of the molecules to make any relevant comments on the conditions used in the reaction.

3 Reaction conditionsIn this section, you will look in detail at the conditions used in the production of propanone from cumene.

As you saw in Topic guide 13.3 reaction conditions can include such factors as: • temperature of the reactants • pressure inside the reaction vessel • presence of a catalyst

Key termsHydrolysis: The breaking apart of a molecule by the reaction with a molecule of water (usually involving H+ or OH− ions as a catalyst).

Carbocation: A positively charged carbon atom; such a carbon atom possesses only six electrons in its outer shell and is thus very reactive, undergoing rapid attack by nucleophiles such as water.

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• surface area of the catalyst or other solid reactant, if present • solvent (if used) • concentration of reactants.

The conditions are chosen to ensure that the process is as cost-effective as possible, while also taking into account the need for safety and sustainability.

Factors to consider in selecting these conditions will include: • the rate of the reaction • the yield of the reaction (which may depend on the position of equilibrium of

any reversible reactions) • the cost of fuel required to maintain conditions of high pressure and high

temperature • the need to keep reactants in a particular state (e.g. liquid) • the carbon footprint of the process • the potential for leaks, fires or explosions during the operation of the process.

Case study: Temperature conditions in the manufacture of propanoneThe temperature of the first stage in the reaction (cumene oxidation) is approximately 90–130 °C.

The second stage of the reaction uses a temperature of around 60–70 °C.

Questions:1 Comment on these temperatures in terms of fuel costs and carbon footprint. 2 Use ideas about the mechanisms of the reactions to suggest why acceptable rates for these two

stages can be achieved at low temperatures.3 Write a short paragraph justifying the choice of temperature in the manufacturing process.

Case study: Pressure conditions in the manufacture of propanoneA slightly raised pressure (1–10 bar) is used in the first stage (cumene oxidation). The second stage is carried out at approximately atmospheric pressure.

Questions:1 Identify which, if any, of the reactions involve gaseous reagents. How will increasing pressure

affect the rate (if at all) in the two stages of the reaction?2 Identify which, if any, of the reactions involving gaseous reagents are reversible processes. Use

your answer to comment on whether changing the pressure will have any effect on the yield of propanone.

3 Are there any other reasons mentioned in earlier case studies for using raised pressure in the first stage?

4 Write a short paragraph justifying the choice of pressure in the manufacturing process.

Portfolio activity (4.1)Describe the process used to form your chosen product in industry and explain the choice of this process. In your answer:

• comment on the existence of any other methods for manufacturing this product • compare the starting materials, conditions and yield of alternative methods with the method

used in your chosen process • describe the mechanism of your chosen process, and relate this mechanism to the conditions

used in the process.

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4 Yield of reactionIn this section the case studies will illustrate the factors affecting yield by (once again) comparing two alternative pathways for manufacturing propanone – the cumene oxidation and catalytic hydrogenation of propan-2-ol, which you studied in Section 1.

As discussed above, the factors that affect yield can include the position of equilibrium of a reversible reaction.

Other factors can include: • losses due to the formation of unwanted by-products • losses during separation (see Section 5 of this topic guide) • the number of stages in the process.

If there are two stages in a process, each with a yield of 95%, the overall yield will be:

95100

×

95100

×

100 = 90.25%

ActivityConsider a process that occurs in four stages. These are the percentage yields for each stage:

Stage 1: 95%Stage 2: 90%Stage 3: 90%Stage 4: 98%

Calculate the overall yield of this process.

Case study: Yield of the two pathwaysPathway one: Oxidation of cumene

H3C CH3 H3CH3C

OH OH

+

CH3

H3C

O

Figure 13.4.8: Reaction sequence for the oxidation of cumene.Yield: 85–87%

Pathway two: Dehydrogenation of propan-2-ol

+

CH3

H3C

Cu

H2

O

CH3

H3C

OH

Figure 13.4.9: Equation for the dehydrogenation of propan-2-ol.Yield: 95%

Question:1 Suggest reasons in each case why the yield of these processes is not 100%.

Activity • For your chosen process, find out the overall yield. If the process occurs in a number of stages,

try to obtain data for these individual steps. • Identify possible reasons why the overall yield is not 100%.

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5 Separation and purificationIn Topic guide 13.3 you looked at a number of possible techniques for separation and purification, such as:

• distillation (simple and fractional) • solvent extraction • chromatography • crystallisation.

Purification processes and yieldIn each of these techniques, the separation or purification process can result in loss of product, and therefore contribute to a reduction in percentage yield.

• If the boiling point of the product is similar to that of other components then a very large number of boiling/condensation cycles may be needed for complete separation. If these do not occur then small amounts of the product may be left in the bulk liquid rather than in the condensate.

• Solvent extraction makes use of the partition coefficient between two solvents. Unless this is very large, small amounts of the product will not be extracted across into the solvent layer.

• During chromatography, if the migration rates of the components are similar, then other components may be mixed with the collected product.

• In crystallisation, some product will always be left in solution uncrystallised.

While many of these problems can be solved by repeated purification, this increases the costs of the process and reduces the overall rate at which pure product can be formed.

Case study: Separation and purification of cumene hydroperoxideIn the oxidation of cumene, the desired product is cumene hydroperoxide. The reaction mixture may contain unreacted cumene and small amounts of by-products formed as a result of the free-radical reaction.

cumene hydroperoxide: b.p. = 125 °C (but decomposes explosively)

cumene: b.p. = 152 °C

Other possible products: acetophenone: b.p. = 202 °C, methyl styrene: b.p. = 165 °C.

The cumene and other by-products that are not miscible with water can be removed from the mixture by a process known as steam distillation. Contact with steam reduces the boiling point of these organic compounds and allows them to be distilled off at a lower temperature.

Questions:1 Suggest why conventional distillation cannot be used to remove cumene hydroperoxide from

the reaction mixture.2 What other methods might also be investigated to separate and purify the liquid cumene

hydroperoxide from the reaction mixture?

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Case study: Separation and purification of propanoneThe propanone (b.p.: 55 °C) is produced in the reactor at a temperature of 60–70 °C. It is removed from the reaction mixture by distillation.

Phenol (b.p.: 182 °C) is an important co-product and is also collected and purified by distillation.

The other main substances present in the reaction mixture are water (b.p.: 100 °C) and the sulfuric acid dissolved in the water (b.p.: 337 °C).

Questions:1 Comment on how easy it will be to separate propanone from the other substances present.2 The propanone has a lower boiling point than the temperature of the reactor. This means it

will be rapidly removed as it forms. Suggest why it can be an advantage, in terms of yield, to remove products as they are formed. (Hint: consider the possibility that the reaction that forms the product may be reversible.)

3 Is it necessary to separate the pure sulfuric acid from the water? Explain why or why not.

Activity • For your chosen process, find out what methods of separation and purification are used to

remove and purify the products from other components of the reaction mixture. • Suggest what effect these methods may have on the yield of the process.

Portfolio activity (4.2 and 4.3)Discuss the factors that affect the yield of the product in your chosen process. In your answer:

• give a figure for the percentage yield which is achieved by the process • identify reasons why the yield is not 100% • explain how the product is separated and purified and discuss the effect that these processes

have on the figure for the yield • if alternative pathways are available to produce the same product, compare the yield that could

be achieved by these pathways and suggest reasons for the difference.

6 Use and recycling of co-productsAs you saw in Topic guide 13.2, co-products are products that are formed alongside the main product in a chemical process. Co-products appear in the equation for the reaction that occurs in the process.

Although co-products increase the separation costs of a process, they frequently help to increase the cost-effectiveness of a process as they can be purified and sold, or alternatively recycled back into earlier stages of the process to decrease the amount of raw material required.

However, in some cases, the co-product may have little economic value and it must be separated and disposed of safely because it may be environmentally damaging.

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Case study: Co-products in reactionsHere are three processes that you have studied in this unit. In each case there are co-products. Use your knowledge of these processes to decide which of the following options should happen to the co-product:

A: be separated and sold onB: be recycled back into the processC: be separated and disposed of safely.

1 Formation of propanone by dehydrogenation of propan-2-ol:

CH3CH(OH)CH

3 ➝ CH

3COCH

3 + H

2

2 Formation of silicon by reduction of silicon dioxide: SiO

2 + 2C ➝ Si + 2CO

3 Hydrolysis of chlorosilanes to form disilanol: (CH

3)

2SiCl

2 + 2H

2O ➝ (CH

3)

2Si(OH)

2 + 2HCl

Case study: Co-products in propanone formationThe overall equation for the production of propanone by cumene oxidation is:

C6 H

5CH(CH

3)

2 + O

2 ➝ C

6H

5OH + CH

3 COCH

3

The co-product of this reaction is phenol, which has a considerable market value as a commodity chemical.

1 kg of cumene feedstock produces 0.46 kg of propanone and 0.75 kg of phenol.

Questions:1 Find out some of the main uses of phenol in the chemical industry.2 Find an up-to-date figure for the price (in $ or £ per tonne) for propanone and phenol.3 Calculate the total market value of the products formed from 1 tonne of cumene.

Portfolio activity (4.4)Find out whether any co-products are produced in your chosen process and how the co-products are used. In your answer:

• use a suitable chemical equation to identify the co-products formed • comment on the market value of these co-products, or any other use to which they can be put • describe whether these co-products are sold, recycled or just disposed of safely.

Further readingDetails of the manufacturing processes described in the case studies in this topic guide can be found in The Essential Chemical Industry (Allan Clements et al., 2010) on p147 and p154.

A classic text which introduces many of the key concepts underpinning organic mechanisms is A Guidebook to Mechanism in Organic Chemistry (Sykes, 1986, Longman). Written for undergraduate chemistry students it would be appropriate for a level 4 or 5 student interested in how organic reactions occur. At a more introductory level, the Chemguide website has a description of some of the more basic organic reactions (http://www.chemguide.co.uk/mechmenu.html#top).

Since Topic guide 13.4 brings together many of the principles introduced in Topic guides 13.1–13.3, many of the resources mentioned in the Further reading sections of previous guides will also be appropriate for this guide.

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AcknowledgementsThe publisher would like to thank the following for their kind permission to reproduce their photographs:

Imagestate Media: John Foxx Collection

All other images © Pearson Education

Every effort has been made to trace the copyright holders and we apologise in advance for any unintentional omissions. We would be pleased to insert the appropriate acknowledgement in any subsequent edition of this publication.

About the authorDavid Goodfellow studied Natural Sciences at Cambridge and spent 20 years teaching A-level Chemistry in a sixth-form college. He was lead developer for the OCR AS Science in 2008 and for several years was chief examiner for the course. He now works as a freelance writer and examiner alongside part-time work as a teacher. Publications include a textbook for the AS Science course, teaching materials to accompany Chemistry GCSE courses and contributions to textbooks for BTEC First Applied Science.

topic guide 13.4: the chemistry of the industrial · pdf file• understand the chemistry...

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