section 2 - alkenes and halogenoalkanes student notes.docx

AS ChemistryUnit 4: Organic Chemistry 1

Part 6: Bonding in methane, ethane and ethene (sigma and pi bonds)

We have seen that cracking leads to the formation of alkenes. These are a family of hydrocarbons that contain only one carbon-carbon double bond. The simplest alkene is ethene, CH2=CH2. The general formula of the homologous series of alkenes is CnH2n.

Task 1

Draw structural formulae for the following alkenes:

a) pent-2-ene

b) hex-3-ene

c) 2,3-dimethylpent-2-ene

d) cyclopenta-1,3-diene

e) 3-ethylhept-1-ene

To understand the origin of the double bond in these molecules, and hence the reactivity of this family, we need an understanding of the bonding in methane and ethane.

Methane, CH4

The simple view of the bonding in methane

Task 2

a) Can you draw a dot-and-cross diagram for methane?

b) Write the electronic configuration for a carbon

atom………………………………………….

c) Write this again using the electrons-in-boxes notation. Only the outer electrons should be shown.

Cambridge A-level Centre


2s 2px 2py

2pz

d) Why does this more sophisticated model not match the dot-and-cross diagram you drew earlier?

……………………………………………………………………………………………………………………………………………

……………………………………………………………………………………………………………………………………………

……………………………………………………………………………………………………………………………………………

Promotion of an electron and hybridisation

When bonds are formed, energy is released and the system becomes more stable. If carbon forms 4 bonds rather than 2, twice as much energy is released and so the resulting molecule becomes even more stable.

There is only a small energy gap between the 2s and 2p orbitals, and so it pays the carbon to provide a small amount of energy to promote an electron from the 2s to

the empty 2p to give 4 unpaired electrons. The extra energy released when the bonds form more

than compensates for the initial input.

The electrons now rearrange themselves again in a process called hybridisation. This reorganises the electrons into four identical hybrid orbitals called sp3 hybrids (because they are made from one s orbital and three p orbitals).

These arrange themselves in space so that they are as far apart as possible (remember VSEPR theory) to form a tetrahedron.

In methane four molecular orbitals are formed with hydrogen. These look rather like the original sp3 hybrids, but with a hydrogen nucleus embedded in each lobe.

The principles involved - promotion of electrons if necessary, then hybridisation, followed by the formation of molecular orbitals - can be applied to any covalently-bonded molecule.

Ethane, C2H6



The formation of molecular orbitals in ethane

Ethane is a simple example of how a carbon-carbon single bond is formed.

Each carbon atom in the ethane promotes an electron and then forms sp3

hybrids exactly as we've described in methane. So just before bonding, the atoms look like this:

The hydrogen atoms bond with the two carbons to produce molecular orbitals just as they did with methane. The two carbon atoms bond by merging their remaining sp3 hybrid orbitals end-to-end to make a new molecular orbital. The bond formed by this end-to-end overlap is called a sigma bond. A σ-bond is symmetrical with respect to rotation about the bond axis. The bonds between the carbons and hydrogens are also therefore sigma bonds.

In any sigma bond, the most likely place to find the pair of electrons is on a line between the two nuclei.

Free rotation about the carbon-carbon single bond

The two ends of this molecule can spin quite freely about the sigma bond so that there are, in a sense, an infinite number of possibilities for the shape of an ethane molecule. Some possible shapes are:

In each case, the left hand CH3 group has been kept in a constant position so that you can see the effect of spinning the right hand one.

Other alkanes



All other alkanes will be bonded in the same way:

The carbon atoms will each promote an electron and then hybridise to give sp3 hybrid orbitals.

The carbon atoms will join to each other by forming sigma bonds by the end-to-end overlap of their sp3 hybrid orbitals.

Hydrogen atoms will join on wherever they are needed by overlapping their 1s1 orbitals with sp3 hybrid orbitals on the carbon atoms.

There will therefore be a tetrahedral arrangement of bonds around a 4-coordinated sp3 carbon atom.

Ethene, C2H4

The simple view of the bonding in ethene

At a simple level, we have drawn ethene showing two bonds between the carbon atoms. Each line in this diagram represents one pair of shared electrons.

Ethene is actually much more interesting than this.

An orbital view of the bonding in ethene

Ethene is built from hydrogen atoms (1s1) and carbon atoms (1s22s22px

12py1).

The carbon atom doesn't have enough unpaired electrons to form the required number of bonds, so it needs to promote one of the 2s2 pair into the empty 2pz orbital. This is exactly the same as happens whenever carbon forms bonds - whatever else it ends up joined to.

Now there's a difference, because each carbon is only joining to three other atoms rather than four - as in methane or ethane. This time, when the carbon atoms hybridise their outer orbitals before forming bonds, they only hybridise three of the orbitals rather than all four. They use the 2s electron and two of the 2p electrons, but leave the other 2p electron unchanged.

Task 3

What do you think these hybrid orbitals might be called?.................................................



These new orbitals look rather like sp3 orbitals except that they are shorter and fatter (they have more s character and less p character). The orbitals arrange themselves as far apart as possible - which is at 120° to each other in a plane (i.e. a trigonal planar arrangement). The remaining p orbital is at right angles to them.

The two carbon atoms and four hydrogen atoms would look like this before they joined together:

The various atomic orbitals which are pointing towards each other now merge to give molecular orbitals, each containing a bonding pair of electrons. These are sigma bonds - just like those formed by end-to-end overlap of atomic orbitals in, say, ethane.

Notice that the p orbitals are so close that they are overlapping sideways.

This sideways overlap also creates a molecular orbital, but of a different kind. In this one the electrons aren't held on the line between the two nuclei, but above and below the plane of the molecule. A bond formed in this way is called a pi bond.

Be clear about what a pi bond is. It is a region of space in which you can find the two electrons which make up the bond. Those two electrons can live anywhere within that space. It would be quite misleading to think of one living in the top and the other in the bottom.

The pi bond dominates the chemistry of ethene. It is very vulnerable to attack - a very negative region of space above and below the plane of the molecule. It is also somewhat distant from the control of the nuclei and so is a weaker bond than the sigma bond joining the two carbons.

All double bonds (whatever atoms they might be joining) will consist of a sigma bond and a pi bond.

Task 4

Dehydration of pentan-1-ol produces pent-1-ene. Sketch on the diagram below the orbital overlap between the two carbon atoms. Label the bonds.



ReferencesA-level Chemistry pages 300-303Chemistry in Context pages 419

Learning ObjectivesCandidates should be able to

describe covalent bonding in terms of orbital overlap, giving and bonds.

explain the shape of, and bond angles in, ethane and ethane molecules in terms of and bonds.



Part 7: What is stereoisomerism? Part 1: Geometric Isomerism

In stereoisomerism, the atoms making up the isomers are joined up in the same order, but still manage to have a different arrangement in space

Geometric (cis / trans) isomerism

Geometric isomerism, also known as cis-trans isomerism, is one type of stereoisomerism.

Task 1

Look at the two displayed formula below. Do these diagrams represent isomers? Explain your answer.

………………………………………………………………………………………………………………………………………………………..

………………………………………………………………………………………………………………………………………………………..

………………………………………………………………………………………………………………………………………………………..

………………………………………………………………………………………………………………………………………………………..

………………………………………………………………………………………………………………………………………………………..


ISOMERISM

STRUCTURAL ISOMERISM STEREOISOMERISM

GEOMETRIC ISOMERISM OPTICAL ISOMERISM


But what happens if you have a carbon-carbon double bond - as in 1,2-dichloroethene?

These two molecules aren't the same. The carbon-carbon double bond won't rotate and so you would have to take the models to pieces in order to convert one structure into the other one. That is a simple test for isomers. If you have to take a model to pieces to convert it into another one, then you've got isomers. If you merely have to twist it a bit, then you haven't!

Drawing structural formulae for the last pair of models gives two possible isomers.

In one, the two chlorine atoms are locked on opposite sides of the double bond. This is known as the trans isomer. (trans : from latin meaning "across" - as in transatlantic).

In the other, the two chlorine atoms are locked on the same side of the double bond. This is know as the cis isomer. (cis : from latin meaning "on this side")

Task 2

The most likely example of geometric isomerism you will meet at A-level is but-2-ene. Can you draw and name its two geometric isomers?

The importance of drawing geometric isomers properly



It's very easy to miss geometric isomers in exams if you take short-cuts in drawing the structural formulae. For example, it is very tempting to draw but-2-ene as:

CH3CH=CHCH3

If you write it like this, you will almost certainly miss the fact that there are geometric isomers. For many situations the formulae given above is fine. However, if there is even the slightest hint in a question that isomers might be involved, always draw compounds containing carbon-carbon double bonds showing the correct bond angles (120 °) around the carbon atoms at the ends of the bond. In other words, use the format shown for 1,2-dichloroethene earlier.

What needs to be attached to the carbon-carbon double bond?

Think about this case:

Although we've swapped the right-hand groups around, these are still the same molecule. To get from one to the other, all you would have to do is to turn the whole model over.

You won't have geometric isomers if there are two identical groups on one end of the bond. So . . . there must be two different groups on the left-hand carbon and two different groups on the right-hand one. The cases we've been exploring earlier are like this:

Here, the two different groups are either on the same side of the bond or the opposite side.

Summary

To get geometric isomers you must have:



restricted rotation (involving a carbon-carbon double bond for A-level purposes);

two different groups on the left-hand end of the bond and two different groups on the right-hand end. It doesn't matter whether the left-hand groups are the same as the right-hand ones or not.

Geometrical isomers normally have similar chemical properties but often their physical properties are markedly different. For example the table below shows the melting point and boiling point of the cis and trans isomers of 1,2-dichloroethene.

isomer

melting point (°C)

boiling point (°C)

cis -80 60

trans -50 48

In each case, the higher melting or boiling point is shown in bold.

You will notice that:

the trans isomer has the higher melting point; the cis isomer has the higher boiling point.

This is common. You can see the same effect with other cis and trans isomers. The differences are due to two factors:

the relative strengths of the intermolecular forces; how well the molecules can pack together in the solid.

Task 3

Complete the worksheet ‘Geometrical Isomerism’.

References

A-level Chemistry page 281Chemistry in Context pages 418-419

Learning ObjectivesCandidates should be able to

describe cis-trans isomerism in alkenes, and explain its origin in terms of restricted rotation due to the presence of π bonds.

deduce the possible isomers for an organic molecule of known molecular formula.

identify cis-trans isomerism in a molecule of given structural formula.



Part 8: What is stereoisomerism? Part 2: Optical Isomerism

Optical isomerism arises because of the different ways in which you can arrange four groups around a carbon atom. It is a type of stereoisomerism.

Four single bonds around a carbon atom are arranged tetrahedrally. When four different atoms or groups are attached to these four bonds, the molecules can exist in two isomeric forms.

Look at the diagram below. This illustrates the two optical isomers of the amino acid alanine (2-aminopropanoic acid).

They are different because they are mirror images. All molecules have mirror images, but they don’t all exist as two isomers. What makes alanine exist in two forms is that the mirror image and the original molecule are non-superimposable. If you move the mirror image across to the left, the H atom, C atom and CH3 group will coincide, but NH2 and COOH will be in the wrong places. No amount of twisting and turning will put things right! The only way you can make them superimpose is to break bonds and swap the two groups around.

Key Terms

Optical isomers are named according to their effect on plane polarised light. There are a number of different ways of naming the enantiomers, e.g. (+) or (-), L- and D-, or R- and S-. At A-level you will not be expected to name them.

Molecules such as the two forms of alanine are called optical isomers (or enantiomers). You don’t have to build models to find them. Whenever a molecule contains a carbon atom that is surrounded by four different atoms or groups of atoms there will be optical isomerism.

Molecules that are not superimposable on their mirror images are called chiral molecules. A carbon atom that is surrounded by four different groups is called a chiral centre.

E.g. Butan-2-ol



The asymmetric carbon atom in a compound (the one with four different groups attached) is often shown by a star.

It's extremely important to draw the isomers correctly. Draw one of them using standard bond notation to show the 3-dimensional arrangement around the chiral carbon atom. Then draw the mirror to show the examiner that you know what you are doing, and then the mirror image.

Task 1

Can you draw the two optical isomers of butan-2-ol

Notice that you don't literally draw the mirror images of all the letters and numbers! It is, however, quite useful to reverse large groups.

It doesn't matter in the least in what order you draw the four groups around the central carbon. As long as your mirror image is drawn accurately, you will automatically have drawn the two isomers.

How do enantiomers differ?

Enantiomers behave identically in ordinary test-tube chemical reactions. Most of their physical properties, such as melting point, density and solubility, are also the same. But enantiomers behave differently in the presence of other chiral molecules.

The proteins in our body are built up from only one enantiomer of each acid.

Enantiomers can interact differently with the chiral ‘taste buds’ on your tongue. One form of amino acids all taste sweet, while the other is often tasteless or bitter.

Enantiomers can smell different. For example the different smells of oranges and lemons are caused by enantiomers.

Many beneficial medicines have enantiomers which have little or no pharmacological effect. In the case of one medicine, thalidomide, the apparently non-active enantiomer was found to damage unborn children when the drug was taken during pregnancy. A similar situation arises with L-dopa, used in the treatment of Parkinson’s Disease.



Task 2

Question 15.11 on page 282 of A-level Chemistry.

Task 3

Complete question from worksheet ‘Optical Isomers’.

References

A-level Chemistry pages 282-285Chemistry in Context pages 392-393

Learning Objectives

Candidates should be able to: explain what is meant by a chiral centre and that such a centre gives

rise to optical isomerism. deduce the possible isomers for an organic molecule of known

molecular formula. identify chiral centres in a molecule of given structural formula.



Part 9: Electrophilic Addition?

Task 1

Can you complete the gaps in the passage below?

Most alkene reactions involve __________the bond. This is __________ than the C-C __________bond and reacts with a variety of reagents. The characteristic __________ of an alkene involves a simple __________ (such as hydrogen, water or bromine) joining across the __________ bond to form a __________ product. Such reactions are called __________ reactions.

The structure of ethene

We are going to start by looking at ethene, because it is the simplest molecule containing a carbon-carbon double bond. What is true of C=C in ethene will be equally true of C=C in more complicated alkenes.

Ethene, C2H4, is often modelled as shown on the right. The double bond between the carbon atoms is, of course, two pairs of shared electrons.

What the diagram doesn't show is that the two pairs aren't the same as each other.

One of the pairs of electrons is held on the line between the two carbon nuclei, a sigma bond, but the other is held in a molecular orbital above and below the plane of the molecule. This bond is called a pi bond. The electrons in the pi bond are free to move around anywhere in this shaded region and can move freely from one half to the other.

The pi electrons are not as fully under the control of the carbon nuclei as the electrons in the sigma bond and,

because they lie exposed above and below the rest of the molecule, they are relatively open to attack by other things.

Electrophiles

An electrophile is something which is attracted to electron-rich regions in other molecules or ions. Because it is attracted to a negative region, an electrophile must be something which carries either a full positive charge, or has a slight positive charge on it somewhere. Ethene and the other alkenes are attacked by electrophiles.

The electrophilic addition of bromine to ethene

The facts



Alkenes react in the cold with both pure liquid bromine, or with a bromine solution. The double bond breaks, and a bromine atom becomes attached to each carbon atom. The bromine loses its original red-brown colour to give a colourless liquid. In the case of the reaction with ethene, 1,2-dibromoethane is formed.

Task 2

Write an equation for the reaction above using structural formulae

This decolourisation of bromine is often used as a test for a carbon-carbon double bond. The other halogens, apart from fluorine, behave similarly. (Fluorine reacts explosively with all hydrocarbons - including alkenes - to give carbon and hydrogen fluoride.)

The mechanism for the reaction between ethene and bromine

The reaction is an example of electrophilic addition.

Bromine as an electrophile

The bromine is a very "polarisable" molecule and the approaching bond in the ethene induces a dipole in the bromine molecule. If you draw this mechanism in an exam, write the words "induced dipole" next to the bromine molecule - to show that you understand what's going on.

The simplified version of the mechanism

Summary notes on Electrophilic additions

Alkenes can become saturated by the addition of small molecules across the double bond.

The most common reactions of the alkenes are described as electrophilic additions.

You should be able to write a mechanism for the reaction of ethene with bromine water (a test for alkenes).

You should also know the reactions (no mechanisms required) and conditions for the reaction of alkenes with both hydrogen, steam and hydrogen halides, and other halides.

ReferencesCambridge A-level Centre


A-level Chemistry page 301-303Chemistry in Context pages 421-424

Learning ObjectivesCandidates should be able to:

describe the mechanism of electrophilic addition in alkenes, using bromine/ethene as an example.

describe the chemistry of alkenes as exemplified, where relevant, by the following reactions of ethene: addition of hydrogen, steam, hydrogen halides and halogens.



Part 10: Electrophilic addition to unsymmetrical alkenes

Task 1So far we have only considered electrophilic additions involving ethene. In the space below write a mechanism for the reaction of propene with hydrogen bromide.

You should have found that two products are possible. This situation occurs when an asymmetrical alkene reacts with another asymmetrical molecule. When this reaction is actually carried out, it is found that one product is more abundant, and therefore more stable, than the other.

The major product is the one formed via the more stable carbocation.

The order of stability of carbocations is:tertiary (3o) > secondary (2o) > primary (1o).

A primary carbocation has one alkyl group attached to C+. A secondary carbocation has two alkyl groups attached to C+. A tertiary carbocation has three alkyl groups attached to C+.

The increased stability of the tertiary carbocation is due to the electron-releasing (inductive) effect of the attached alkyl groups. Alkyl groups tend to push electrons slightly towards any carbon atoms to which they are attached.

Task 2Draw the four possible carbocations of C4H9

+ and label each as 1o, 2o and 3o.

Task 3Use the information above to deduce the major product of the reaction between propene and hydrogen bromide.



Task 4Questions 3-5 on page 307 of A-level Chemistry

References

Chemistry in Context pages 423-424A-level Chemistry page 303

Learning Objectives

Candidates should be able to:

describe the chemistry of alkenes as exemplified, where relevant, by the following reactions of ethene: addition of hydrogen, steam, hydrogen halides and halogens.



Part 11: The Polymerisation of alkenes

Here we look at the polymerisation of alkenes to produce polymers like poly(ethene) (usually known as polythene) and PVC.

Poly(ethene) (polythene or polyethylene)

Manufacture

During polymerisation, an alkene molecule undergoes an addition reaction to itself. As we have seen an addition reaction is one in which two or more molecules join together to give a single product. A polymer is a long molecule (or macromolecule) made up from lots of small molecules called monomers.

During the polymerisation of ethene, thousands of ethene molecules join together to make poly(ethene) - commonly called polythene.

Task 1

Can you write an equation for the polymerisation of ethene?

The number of molecules joining up is very variable, but is in the region of 2000 to 20000.

ConditionsTemperature: about 200°CPressure: about 2000 atmospheres

Initiator:often a small amount of oxygen as an impurity

The mechanism

The over-all process is known as a free radical addition.

Chain initiation

The chain is initiated by free radicals, Ra , produced by reaction between some of the ethene and the oxygen initiator. You are not expected to know any details.

Chain propagation



Each time a free radical hits an ethene molecule a new longer free radical is formed.

etc.

Chain termination

Eventually two free radicals hit each other producing a final molecule. The process stops here because no new free radicals are formed.

Because chain termination is a random process, poly(ethene) will be made up of chains of all sorts of different lengths.

Properties and uses

Poly(ethene) made in this way is called low density poly(ethene). One polymer chain is held to its neighbours in the solid structure by weak van der Waals dispersion forces (temporary dipole – induced dipole interactions). Poly(ethene) has quite a lot of branching along the hydrocarbon chains, and this prevents the chains from lying very close to each other. As a result the material is soft and flexible. Low density poly(ethene) is used for familiar things like plastic carrier bags and other similar low strength and flexible sheet materials.

Another method of making polythene was developed by Ziegler in the 1950s. This process uses catalysts at lower temperatures and pressure. The molecules produced have few branches and can pack closely together. This is known as high-density polythene (HDPE). It is more rigid and has a higher melting point. It is used for moulding rigid articles such as crates.

Poly(chloroethene) (polyvinyl chloride): PVC

Poly(chloroethene) is commonly known by the initials of its old name, PVC.

Structure

Poly(chloroethene) is made by polymerising chloroethene, CH2=CHCl. Working out its structure is no different from working out the structure of poly(ethene) (see above). As long as you draw the chloroethene molecule in the right way, the structure is pretty obvious.



The equation is usually written:

It doesn't matter which carbon you attach the chlorine to in the original molecule. Just be consistent on both sides of the equation.

Properties and uses

Pure poly(chloroethene) tends to be rather hard and rigid. This is because of the presence of additional dipole-dipole interactions due to the polarity of the carbon-chlorine bonds. Chlorine is more electronegative than carbon, and so attracts the electrons in the bond towards itself. That makes the chlorine atoms slightly negative and the carbons slightly positive.

Plasticisers are added to the poly(chloroethene) to reduce the effectiveness of these attractions and make the plastic more flexible. The more plasticiser you add, the more flexible it becomes.

Poly(chloroethene) is used to make a wide range of things including guttering, plastic window frames, electrical cable insulation, sheet materials for flooring and other uses, footwear, clothing, and so on and so on.

Task 2

Can you write balanced equations for the polymerisation of propene and tetrafluoroethene?

Disposal of polymers

Polymers are strong and lightweight. As a result they have found many uses in our modern world. However, there are problems associated with the widespread use of polymers because they are difficult to dispose of. They are resistant to chemical attack and to bacteria (non-biodegradable).

Task 3



Using pages 305-307 of your textbook can you summarise some of the advantages and disadvantages of the various methods used to dispose of polymers.

Method Comments

Landfill

Incineration

Recycling

Feedstock recycling

References

Chemistry in Context pages A-level Chemistry page

Learning Objectives

Candidates should be able to: describe the chemistry of alkenes including polymerisation. describe the characteristics of addition polymerisation as exemplified

by poly(ethene) and PVC. Recognize the difficulty of the disposal of poly(alkene)s, i.e. non-

biodegradability and harmful combustion products.



Part 12: Oxidation of alkenes

Reactions of alkenes with manganate (VII) ions, MnO4-

Manganate (VII) ions are oxidizing agents. Alkenes undergo different reactions with manganate (VII) ions, depending on the conditions.

1. In the presence of dilute (acidified or alkaline) potassium manganate (VII).

Alkenes react readily at room temperature (i.e. in the cold). The purple colour disappears and a diol is formed.

CH2=CH2 + H2O + [O] HOCH2CH2OHethane – 1,2-diol

2. In the presence of a hot, concentrated solution of acidified potassium manganate (VII), any diol formed is split into two fragments which are oxidized further to carbon dioxide, a ketone or a carboxylic acid. Analysis of the products can indicate where the carbon-carbon double bond was in the alkene.

Fragment Product=CH2 CO2

R-CH=

Aldehyde

→ carboxylic

acid

R2C=Ketone

Task

Some alkenes were heated in the presence of an acidified solution of potassium manganate (VII). The products obtained in each reaction are listed below. In each case identify the initial alkene and give its structural formula.

1. CO2

2.

+ CO2


http://images.google.com/imgres?imgurl=http://upload.wikimedia.org/wikipedia/commons/thumb/f/f6/Aldehyde-skeletal.png/636px-Aldehyde-skeletal.png&imgrefurl=http://commons.wikimedia.org/wiki/Image:Aldehyde-skeletal.png&h=600&w=636&sz=18&hl=en&start=1&um=1&tbnid=UKi4zMEau3STCM:&tbnh=129&tbnw=137&prev=/images?q=aldehyde&um=1&hl=en&newwindow=1&rlz=1T4ADBS_enCN238CN238&sa=N

http://images.google.com/imgres?imgurl=http://upload.wikimedia.org/wikipedia/commons/thumb/4/48/Ketone-general.png/664px-Ketone-general.png&imgrefurl=http://commons.wikimedia.org/wiki/Image:Ketone-general.png&h=599&w=664&sz=24&hl=en&start=1&um=1&tbnid=IMF0prdyN3PVZM:&tbnh=124&tbnw=138&prev=/images?q=ketone&um=1&hl=en&newwindow=1&sa=N

http://www.uoregon.edu/~ch111/images/ethanoic.gif


3.

+ CO2

4.

References

A-level Chemistry page 304

Learning Objectives

Candidates should be able to describe the oxidation of alkenes by: cold, dilute, acidified manganate(VII) ions to form the diol, and hot, concentrated, acidified manganate(VII) ions leading to the

rupture of the carbon-to-carbon double bond in order to determine the position of alkene linkages in larger molecules.


http://images.google.com/imgres?imgurl=http://www.uoregon.edu/~ch111/images/propanone.gif&imgrefurl=http://www.uoregon.edu/~ch111/L31.htm&h=95&w=119&sz=1&hl=en&start=2&um=1&tbnid=r8RaHTF2QW2EwM:&tbnh=70&tbnw=88&prev=/images?q=propanone&um=1&hl=en&newwindow=1


Part 13: Halogenoalkanes (haloalkanes)

The simple halogenoalkanes are a homologous series of compounds in which one or more hydrogen atoms in an alkane have been replaced by halogen atoms (fluorine, chlorine, bromine or iodine). They have the general formula CnH2n+1X. They do not occur much in nature, but they are useful for all sorts of human purposes, so chemists make and use them a lot. As with all functional groups, the halogen atom modifies the properties of the relatively unreactive hydrocarbon chain.

They are named after the parent alkanes, with the halogen atom added as the prefix fluoro-, chloro-, bromo- or iodo-.

Task 1

Use systematic nomenclature to name the following halogenoalkanes. Draw the displayed formula of each.

a. CHCl3b. CH3CHClCH3

c. CF3CCl3

Typically, halogenoalkanes are volatile liquids that do not mix with water.

Task 2

a) Explain why 1-chloropropane, C3H7Cl, is a liquid at room temperature whereas butane is a gas.

b) Why is it that halogen compounds such as 1-chloropropane do not mix with water?

Task 3

Can you solve the anagrams (words in bold) to complete the notes below?

Nucleophilic substitution

Halogen atoms are negotiate clever. As a result, the carbon-halogen bond is alp or. The electrons in the C-X bond are cadet tart towards the halogen



atom which gains a slight eat given charge, leaving the carbon atom electron enticed if.

The d+ carbon is susceptible to attack by chenille soup, i.e. ions or molecules with a lone pair of electrons. The C-X bond is broken releasing the had lie ion. A nucleophilic stubs tuition reaction has occurred.

Task 4

Using the notes above as a guide can you write a general reaction mechanism for this type of reaction. You should represent the nucleophile asN u -. Please note, nucleophiles do not have to carry a negative charge.

This is known as an SN2 reaction. S stands for substitution, N for nucleophilic, and 2 because the initial stage of the reaction involves two species.

All reactions of the haloalkanes involve breaking the C-X bond. The stronger the bond, the more difficult it is to break, and the lower the rate of reaction.

Task 5

Look at the table of bond strengths and electronegativity values below. Which halogenoalkane would you expect to be the most reactive and why?

Halogen F Cl Br IElectronegativity 4.0 3.0 2.8 2.5

Bond strength (C-X) kJ mol-1

484 338 276 238

______________________________________________________________

______________________________________________________________

______________________________________________________________

______________________________________________________________

______________________________________________________________



Task 6

How could you measure the relative rates of these reactions?

Nucleophilic substitution reactions

Task 7

Using the general equation as your guide, can you write a mechanism for the following reaction of bromoethane.

1. Hydroxide ions

When haloalkanes are warmed with NaOH (aq) or KOH(aq), alcohols are formed. N.B. halogenoalkanes do not mix with water, so they are first mixed with a little ethanol.

This process is sometimes called hydrolysis.

A similar reaction occurs more slowly with water.

CH3CH2Br + H2O CH3CH2OH + HBr

2. Cyanide ions

Nucleophilic substitutions with cyanide ions add an extra carbon atom to the chain. This reaction can, therefore, be useful in organic synthesis. The compounds formed are known as nitriles, RCN. The reaction is carried out in a warm ethanolic solution.

Task 8

Can you write an equation for the reaction of bromoethane with KCN?



3. Ammonia

Primary amines are formed when an ethanolic solution of the haloalkanes are warmed with an excess of ammonia in a sealed container (i.e. under pressure).

Task 9

Can you write equations for the two reactions below?

For example, bromoethane forms ethylamine:

Since the acid HBr immediately reacts with the base NH3, this equation is more correctly written as:

Task 10

Past paper question



References

A-level Chemistry pages 328 - 331Learning Objectives

Candidates should be able to recall the chemistry of halogenoalkanes as exemplified by the following nucleophilic substitution reactions of bromoethane: hydrolysis; formation of nitriles; and the formation of primary amines by reaction with ammonia.



Part 14: Substitution vs. Elimination

The halogenoalkanes can be divided into three groups depending on the position of the halogeno- group along the carbon chain.

Task 1

Can you draw appropriate structures to illustrate the three classes of halogenoalkanes in the table below?

Type of halogenoalkane

Position of halogeno- group

Example

primary at end of chain: bromoethane

secondary in middle of chain: 2-bromopropane

tertiary attached to a carbon atom which carries no H atoms:

2-bromo-2-methylpropane

Nucleophilic substitution

The mechanism of nucleophilic substitution in a tertiary halogenoalkane is different from that discussed above for a primary halogenoalkane.

Task 2

Read the description below carefully and use the information to draw a reaction mechanism for this hydrolysis reaction.Cambridge A-level Centre


The hydrolysis of 2-bromo-2-methylpropane, a tertiary halogenoalkane, involves the breaking of the C-Br bond followed by attack by OH - ions on the resulting carbocation.

Explanation: The electron-releasing methyl groups enhance the stability of the carbocation, thus changing the mechanism of the hydrolysis. As the first step involves only 1 species this mechanism is described as SN1.

The nucleophilic substitution of secondary halogenoalkanes involves both mechanisms.

Elimination

We have already seen how the hydroxide ion can act as a nucleophile in the reaction with a haloalkane to form an alcohol. You need to be aware however, that this ion can also act as a strong base. An alternative reaction can take place in which HBr is removed and an alkene is formed. This is known as elimination.

E.g. CH3CH2Br + NaOH CH2=CH2 + NaBr + H2O

Task 3

Can you copy this mechanism into the space below?

Substitution vs. Elimination

Which mechanism is most prevalent depends on a number of factors:

The structure of the haloalkane1o haloalkanes - predominantly substitution



2o haloalkanes – both reactions occur3o haloalkanes – predominantly elimination

The reaction conditionsHigher temperatures favour elimination

The choice of solvent Elimination is favoured by hot ethanolic conditions Substitution is favoured by warm aqueous conditions.

References

A-level Chemistry pages 328 - 331

Learning Objectives

Candidates should be able to

recall the chemistry of halogenoalkanes as exemplified by the elimination of hydrogen bromide from 2-bromopropane.

describe the mechanism of nucleophilic substitution (by both SN1 and SN2mechanisms) in halogenoalkanes.

Part 15: Pros and cons of the halogenoalkanes

Task 1

Read the passage below on the uses of halogenoalkanes (and some related compounds). See if you can find a replacement for each of the words/phrases in bold which explains their meaning.

CFCs and their replacements

CFCs are chlorofluorocarbons. Two common CFCs are:

CFC-11

CCl3F

CFC-12

CCl2F2

How to work out the formula from the CFC number: You add 90 to the CFC number to give a new number. For example, CFC-11 generates the number 101 (11+90)



The first digit of the new number tells you how many carbon atoms there are. The second number tells you the number of hydrogens, and the third the number of fluorines. You calculate the number of chlorines from the formula Cl = 2(C+1)-H-F.

Uses of CFCs

CFCs are non-flammable and not very toxic. They therefore had a large number of uses.

They were used as refrigerants, propellants for aerosols, for generating foamed plastics like expanded polystyrene or polyurethane foam, and as solvents for dry cleaning and for general degreasing purposes.

Unfortunately, CFCs are largely responsible for destroying the ozone layer. In the high atmosphere, the carbon-chlorine bonds break to give chlorine free radicals. It is these radicals which destroy ozone. CFCs are now being replaced by less environmentally harmful compounds.

The destruction of atmospheric ozone

Ozone, O3, is constantly being formed and broken up again in the high atmosphere by the action of ultraviolet light. Ordinary oxygen molecules absorb ultraviolet light and break into individual oxygen atoms. These have unpaired electrons, and are known as free radicals. They are very reactive.

The oxygen radicals can then combine with ordinary oxygen molecules to make ozone.

Ozone can also be split up again into ordinary oxygen and an oxygen radical by absorbing ultraviolet light.

This formation and breaking up of ozone is going on all the time. Taken together, these reactions stop a lot of harmful ultraviolet radiation penetrating the atmosphere to reach the surface of the Earth.

The catalytic reaction we are interested in destroys the ozone and so stops it absorbing UV in this way.



The slow breakdown of chlorofluorocarbons (CFCs) like CF2Cl2 in the atmosphere produces chlorine atoms - chlorine free radicals. These catalyse the destruction of the ozone.

This happens in two stages. In the first, the ozone is broken up and a new free radical is produced.

The chlorine radical catalyst is regenerated by a second reaction. This can happen in two ways depending on whether the ClO radical hits an ozone molecule or an oxygen radical.

If it hits an oxygen radical (produced from one of the reactions we've looked at previously):

Or if it hits an ozone molecule:

Because the chlorine radical keeps on being regenerated, each one can destroy thousands of ozone molecules.

CFCs can also cause global warming. One molecule of CFC-11, for example, has a global warming potential about 5000 times greater than a molecule of carbon dioxide.

On the other hand, there is far more carbon dioxide in the atmosphere than CFCs, so global warming isn't the major problem associated with them.

Replacements for CFCs

These are still mainly halogenoalkanes, although simple alkanes such as butane can be used for some applications (for example, as aerosol propellants).

Hydrochlorofluorocarbons, HCFCs

These are carbon compounds which contain hydrogen as well as halogen atoms. For example:



HCFC-22 CHClF2

The formula can be worked out from the number in the name in exactly the same way as for CFCs.

These have a shorter life in the atmosphere than CFCs, and much of them is destroyed in the low atmosphere and so doesn't reach the ozone layer. HCFC-22 has only about one-twentieth of the effect on the ozone layer as a typical CFC.

Hydrofluorocarbons, HFCs

These are compounds containing only hydrogen and fluorine attached to carbon. For example:

HFC-134a

CH2F-CF3

Because these HCFCs don't contain any chlorine, they have zero effect on the ozone layer. HFC-134a is now widely used in refrigerants, for blowing foamed plastics and as a propellant in aerosols.

Hydrocarbons

Again, these have no effect on the ozone layer, but they do have a down-side. They are highly flammable and are involved in environmental problems such as the formation of photochemical smog.

Other uses of organic halogeno compounds

In making plastics

Strictly speaking, the compounds we are talking about here are halogenoalkenes, not halogenoalkanes.

Chloroethene, CH2=CHCl, is used to make poly(chloroethene) - usually known as PVC.

Tetrafluoroethene, CF2=CF2, is used to make poly(tetrafluoroethene) - PTFE.

Lab uses of halogenoalkanes

Halogenoalkanes are useful in the lab as intermediates in making other organic chemicals.



Additional uses of halogenoalkanes

Task 2

Use page 452 of ‘Chemistry in Context’ and pages 331-332 of ‘AS Level Chemistry’ to answer the following questions.

1. Anaesthetics

The substance 2-bromo-2-chloro-1,1,1-trifluoroethane (or Halothane) has been widely used as an anaesthetic in hospitals since its discovery in 1956.

a. Draw the full structural formula of halothane in the space below.

b. Halothane contains fluorine, chlorine and bromine atoms. What properties do these atoms lend to the molecule:

Fluorine:

Chlorine:

Bromine:

2. Flame retardants

BCF was widely used as a fire-fighting agent.

a. Which compound is BCF?____________________________________

b. Draw the full structural formula of BCF below:

c. Why is BCF good at extinguishing fires?



_______________________________________________________

_______________________________________________________

_______________________________________________________

_______________________________________________________

_______________________________________________________

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_______________________________________________________

d. Why is BCF no longer in general use?

_______________________________________________________

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References

A-level Chemistry pages 331-333

Learning Objectives

Candidates should be able to

interpret the different reactivities of halogenoalkanes e.g. CFCs; anaesthetics; flame retardants; plastics with particular reference to hydrolysis and to the relative strengths of the C-Hal bonds;

explain the uses of fluoroalkanes and hydrofluorooalkanes in terms of their relative chemical inertness;

recognise the concern about the effect of chlorofluoroalkanes on the ozone layer.


section 2 - alkenes and halogenoalkanes student notes.docx

Documents

learning objectives

higher melting

sp3 hybrid

full structural

alkyl groups

carbon double

carbon single

nucleophilic