Hydrocarbons (13) Geometric (EZ) Isomers in Alkenes

What are Geometric Isomers in alkenes and how do they arise?

Let's go back the the basic alkene structure of ethene C2H4.

The carbon carbon double bond composed of the σ and the π bond cannot rotate easily.  

To rotate the σ bond, the π bond would have to break requiring some input of energy.

So if there are different groups attached to the double bond carbons, different structures fit the same molecular formula.  

We have what are called geometric isomers.

A common example quoted is the literature is but-2-ene C4H8.

You can find a short (less than 2mins) tutorial here on geometric isomerism though you might be a tad frustrated with the fact that its a preview version — still better than paying I say.

Here are the two geometric isomers of but-2-ene:

So the condition for geometric isomerism is that there are two different groups attached to the Carbon atoms of the double bond.

In this case a methyl group (CH3) and a hydrogen atom.  

Trans refers to the isomer where the groups are on different sides of the double bond.

Trans is Latin for "across", i.e. the groups are across from one another.

Cis refers to the isomer where the groups are on the same side of the double bond.  

Other examples are:

This last example shows how the geometric isomerism can exist around other structures not just the carbon carbon double bond.

So what happens to cis and trans designations when you have this kind of a substituted alkene?

With more complex alkenes, the designations E (for the German Entgegen) and Z (for the German Zusamen) are used to distinguish the isomers.

This is how it works:

This compound is (E) 1-bromo-2-chloro-2-fluoroethene.

The different ends of the molecule are treated separately.

The two groups on the end of the double bond are compared to see which has the higher atomic number.

In the example above, on the left bromine has the higher atomic number and on the right chlorine has the higher atomic number.

Atoms with the higher atomic number have what is called priority.

If higher priority atoms (or groups) are cis to each other that is the Z isomer.

In the above example the higher priority atoms are trans to each other so that is the E isomer.

If there are just alkyl groups on the ends of the double bond then as a general rule the group with the higher number of carbons has priority.

See the example below where on the left the CH3— has priority over H–— and on the right (CH3)2CH— has priority over CH3CH2— putting higher priority groups on the same side therefore it is the Z isomer:

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Hydrocarbons (12) Propene and Hydrogen Bromide (HBr): Markovnikov addition

It certainly does get very interesting when propene is used instead of ethene to illustrate the electrophilic addition across the double bond using bromine in water.

What difference could propene possibly make?

Let's recap for a minute.

What happens when bromine reacts with water?

Br2   +    H2O      =        HBr     +     HOBr

Two compounds form: hydrobromic acid and bromic(I)acid.

Both these compounds are polar and could therefore engage in the electrophilic addition mechanism for the bromination of propene.

With bromine and with water the example in the previous post (Hydrocarbons(11)) of ethene and bromine water is followed.

Here is the equation:

1,2-dibromopropene CH3CHBrCH2Br is formed together with the bromohydrin CH3CHBrCH2OH

The interesting case is that of hydrobromic acid (HBr) or hydrogen bromide.

HBr is already a polar molecule due to the differences in electronegativity between hydrogen (2.2 on the Pauling Scale) and bromine (2.96).
                                                            δ+   δ-
We can write hydrobromic acid as   H—Br.

When HBr adds across the propene double bond, it adds via the more stable carbocation.

This is referred to as Markovnikov's Rule.

To quote ROC Norman: "The underlying principle is that the electron releasing alkyl groups stabilise carbocations more effectively when they are bound directly to the positively charged carbon than when they are further removed from it."

So the primary carbocation:

is less stable than the secondary carbocation:

The electron releasing alkyl groups are the CH3– methyl groups and ethyl group (CH3CH2–) attached to the C+ atom.

There is only one of these in the primary propene carbocation but two alkyl groups in the secondary carbocation.

Therefore the secondary carbocation will be formed faster by the addition of a proton to propene.

Here is the electrophilic addition of the hydrogen bromide to propene:

Here then is summary of the energetic stability of carbocations (once called carbonium ions)

R— stands for an alkyl group and R' or R'' are different types of alkyl group.

The energetic stability increases left to right in the table.

Here is another diagram to illustrate that the increased inductive or
electron pushing effect of the alkyl groups stabilises the carbocation:










In this diagram above energetic stability increases right to left.

The arrow on the bond is a way of showing the inductive or electron pushing effect of the alkyl groups.

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Hydrocarbons (11) Electrophilic addition of bromine to ethene in the presence of water

So in the previous post (Hydrocarbons(10)) I explained how electrophilic addition of bromine to an alkene like propene or ethene could take place.

And we were supported in our thinking by my organic chemistry hero ROC Norman!!

So what happens when we do that simple alkene test using bromine water?

How does the water interfere?

If you've done any simple halogen chemistry at this point you should be thinking yea I remember what happens between water and a halogen.

So what does happen say between water and bromine?

The reaction produces two acids:

Br2  +   H2O     =    HBr      +      HOBr

The two acids are hydrobromic acid (HBr), an analogue of hydrochloric acid, and bromic (I) acid (HOBr)

Both these acids are ionic molecules and that means they can be involved in an ionic mechanism like electrophilic addition.

Let's look at what happens then with ethene and water:

Here's the first stage:

But the bromide ion (Br-) isn't the only species in the reaction that could add to the carbocation.

Water is just as able to add to the carbocation.

When it does a bromohydrin forms.

See the equation below:

CH2=CH2  +  Br—Br(aq)  =   +H2O—CH2—CH2Br     =      HOCH2—CH2Br     +      H+

To quote ROC Norman: The intermediate carbocation may be attacked by any nucleophile which is present.  For example, the addition of bromine to ethene is aqueous solution gives both the dibromide and the bromohydrin since water is nucleophilic.

And the bromohydrin is also colourless.

The action of hydrobromic acid is similar in ethene the bromoetheae is formed.

CH2=CH2  +  H—Br(aq)    =     +CH2—CH3    +  Br-    =      BrCH2—CH3

But it gets interesting when propene is used instead of ethene as I will discuss in the next post (Hydrocarbons (12)).

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Hydrocarbons (10) Electrophilic Addition in Alkenes

This post is about a particular addition reaction mechanism found in alkene chemistry.

We've all seen the test for the alkene double bond.

Nice bit of chemistry as the alkene decolorises the bromine water.

What's exactly going on in that reaction?

Is it a straightforward as it seems ?

Well, as you might have guessed, I wouldn't be asking the question if it wasn't exactly a straight forward reaction,

The mechanisms are complex and there are two of them if you use bromine water (Br2) and propene (CH3CHCH2).

So in this post let's
just look at the reaction between bromine itself and propene and then in the next post on hydrocarbons (Hydrocarbons (11)) look at the influence of the water.

Overall the reaction goes like this:

CH3CH=CH2    +       Br2         =           CH3CHBrCH2Br

propene                 bromine               1,2-dibromopropane

Now the reaction has to go via collision between the bromine molecule and the propene molecule.

The discussion lies around how that collision happens and what sort of entity results from it.

In our example then we have a double bond with π and σ bonds.

You'll see from the diagram that the π bond is quite exposed as it is above and below the plane of the alkene molecule.

We also know that the π bond is weaker than the σ bond.

Collision between the bromine molecule and the alkene at the π bond is likely to result in a change especially if the collision is energetic enough.

The reaction happens at room temperature even in the dark so the collision is probably energetic enough.

But what will happen if the bromine molecule approaches the alkene double bond?

The π electrons could affect the bromine molecule and in fact we say they induce a temporary dipole in the bromine molecule.

You can see in the diagram to the left that a pair of electrons moves towards the bromine molecule.

The Bromine molecule is polarised in the presence of the π bond as shown by the δ+ and δ- charges.

The effect of the approach of the π bond is to push a pair of bonding electrons onto the δ- bromine atom as shown by the other double headed curly arrow.


Let's see how that looks in propene:

You can see that the initial product is an unstable cation because the triangular ring structure is very strained.

As ROC Norman (my organic chemistry hero by the way!!) states, and you'll hear more from him if you stick with this blog, "the intermediate carbocation is not able to undergo free rotation about the new C-C bond but is held rigid by interaction with the electrophile which has been added."

Hence the triangular structure on the right is rigid and the C-C bond in it cannot rotate.

It is this carbocation that is then attacked (very rapidly) from the opposite side to the bromine atom already attached.

Here is the overall mechanism of electrophilic addition of bromine in propene:

You'll see in the representation that I have left the intermediate as a simple carbocation and ignored the triangular structure (its called an epoxide ring structure`)

But note too how the curly arrow from the bromide ion moves towards the positive charge on the intermediate.

So this is called an electrophilic addition mechanism.

Electrophilic because the bromine molecule acts as in electrophile once a dipole is induced in it.

Addition because the two reactants from one product.

Check out the next post (Hydrocarbons(11)) for what happens when this reaction is carried out in the presence of water.

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Hydrocarbons (9) Designer polymers GORE-TEX

One of the beauties of polymers is that they can be designed to fit particular uses.

So some ingenious guy decided that combining nylon and PTFE would produce a breathable AND waterproof material.

Nylon being tough, lightweight and hydrophobic (that is water repelling) means that using it for a waterproof jacket, sweat is going to build up inside it.

Some of us can remember this happening in the kind of outdoor kit we bought in the 60's and 70's!!

But if the nylon could be made to breathe, that is: be both water repellent and allow sweat to escape, then that would help those of us who engage in active outdoor activities because we'd be better able to cope with perspiration wetness.

GORE-TEX is a laminated membrane, i.e. it is a two layer material, that claims to do just that help us cope with perspiration wetness.

Below is a microphotograph of a GORE-TEX membrane.

GORE-TEX is a specially engineered hydrophobic material encasing a microporous expanded polytetrafluoroethene membrane.

This membrane has pores, as you can see in the microphoto, that are engineered small enough to keep soil particles and water droplets from entering but molecules of water, that are about 1000 times smaller than the pores, are unimpeded in their path out to the air through the material.

This expended PTFE membrane is too fragile to exist on its own without support so it is laminated to a nylon support.

Here is a GORE-TEX promotional diagram showing how it works at keeping people dry from perspiration wetness in wet conditions.

Thing is, this fragile PTFE membrane is just that - fragile, and collapses readily and loses its performance characteristics because the pores clog up with dirt and close in on themselves.

PTFE stands for polytetrafluoroethene.

It is an addition polymer.

Here is its structure:

Fluorine atoms give the polymer its inert/unreactive properties because the C-F bond is so strong and difficult to break.

Nylon is of a different class of polymer called condensation polyamides. 

Nylons have several different structures so here is a typical example:

The differences between the types of nylon arise because the group in the oblong in the diagram can be a hydrocarbon chain of different lengths. 

For example, Nylon 6:6 has a (CH2)6 group in each box whereas nylon 6:10 has a (CH2)10 group in the diamine.

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Hydrocarbons (8) Thermoplastics and Thermosets

Polymers are everywhere.

You find them used in virtually every electronic device and from the ubiquitous (nice word there!!) polythene supermarket bag to the bearings of the decking of the Humber Bridge.

Polymer clothing is common too.

The point being of course is that the properties of polymers can be tuned to fit the use required.

And there has been massive investment in the development of polymers for sports and outdoor use.

So we are thinking about the versatility of polymers like nylon and materials like GORE-TEX.

But first a simple interesting thing about polythene bags.

Now every tried to rip a polythene bag apart?

If you stop now and find an old bag and try to rip it up you will find it easier to do this in one direction rather than the other.

Which is the stronger direction and why?

This is the reason why:

The polymer fibres and to a large extent the polymer molecules themselves are aligned one way in the bag that is from the handle to the bottom of the bag.

So it is stronger in the direction that it needs to be to carry a large weight of shopping.

But at right angles to this direction the plastic is much weaker.

The strength comes from the forces within as opposed to the forces between the polymer fibres and molecules.

There are comparatively weaker forces between the polymer chains in polythene.

But there are really strong forces within the polymer chains themselves because the carbon atoms are held together with strong covalent bonds.

Polymer like polythene and other addition polymers like PVC, polystyrene and polypropene are called thermo softs or thermoplastics.

The test comes when they are gently heated because they can be deformed and be remoulded at relatively low temperatures into a different shaped object.

Thermosplastic polymers can be easily stretched because the polymer chains can easily slide over each other.

Now chemists can strengthen the links between polymer chains adding what are called crosslinks and create a a stronger plastic but sacrifice a great deal of flexibility like this:

The strong covalent bonds between the polymer chains now add rigidity to the plastic.

It will char on heating rather than softening and being remoulded.

It cannot be stretched.

It will have a high melting point

Examples of thermosets are in this table:

Thermosoftening plastics are in the table below:

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Hydrocarbons (7) Free Radical Chlorination of Methane

So the alkanes are not supposed to be that reactive.

Actually that's not so really, just another one of those chemistry myths you meet in school courses, across the world not just in the UK or the US.

Given the quantity of hydrocarbon alkanes separated from crude oil per annum they had better be reactive in some way or other or we are going to waste so much bonded carbon atom and turn them into greenhouse causing CO2 through using them as fuels - a not very creative approach to say the least.

One very imaginative reaction of the alkanes puts a halogen atom in the molecule and that halogen can be pretty labile, replaceable with other groups and deliver on a variety of petrochemical precursors.

The halogen atoms usually used to illustrate this reaction are chlorine atoms in the form of molecular chlorine.

And the simplest example alkane usually used in the illustration is methane CH4.

But what we say below about chlorine and methane applies equally to say bromine and  hexane.

Now if you are thinking like a chemist you will be asking yourself how are we going to have two gases react together and in what proportions?

Even more so what will the result be if the proportions of gases reacting are different?

If methane is in excess what will the product(s) be?

If chlorine is in excess how will the products be different if at all?

So let's think first about how this reaction can occur.

I'm going to assume you have some familiarity with organic reaction mechanisms.

Reaction mechanisms reveal the possible stages a reaction goes through to produce products from reactants.

Here is the overall reaction of excess methane with chlorine:

methane       +      chlorine       =         chloromethane    +    hydrogen chloride

This is a substitution reaction because a chlorine atom takes the place of a hydrogen atom in methane.

The conditions used are symbolised with hν which refers to high energy uv light.

A Rescue Box explains the hν symbol here:

Now all the compounds involved in the reaction are gases.

When the reaction takes place four key observations can be made:

1. Molecular hydrogen (H2) is absent during the course of the reaction, it is never detected in this reaction.
2. The reaction can go at about 300oC but it proceeds at much higher rate in sunlight at room temperature.
3. Light of a wavelength of around 400nm or less, with an energy of around 300kJ/mol of photons or more, is every effective in increasing the reaction rate.
4. In the right light, for each photon of light absorbed, many thousands of chloromethane molecules are produced. 

Let's see how these observations help us to work out or elucidate the reaction mechanism.

The light seems very important especially its energy value per photon why is this?

Let's look at some bond energies for the relevant molecules:

Cl2     =     Cl•     +     Cl•        +242kJ/mol     homolytic fission of chlorine molecule

Cl2     =     Cl+     +    Cl-        +1147kJ/mol   heterolytic fission of chlorine molecule

CH4   =     CH3•    +    H•       +435kJ/mol    homolytic fission of C-H bond in methane

CH4   =     CH3-     +    H+     +1306kJ/mol   heterolytic fission of C-H bond in methane

[You should realise that I have selected a relevant set of reactions here from a list of over 10 possible reactions that the uv light could cause to happen to chlorine or methane molecules in the mixture.]

If the gas mix is exposed to photons with energies around 300kJ/mol then absorption of light of wavelength around 400nm would provide sufficient energy for the homolytic fission of chlorine molecules to occur.

As you can see, the other bond fissions require much more energy per mole of bonds.

[What's more if the reaction is carried out in an inert non-polar solvent then the effect of the solvent will be to suppress the formation of ions.]

So we can suggest that the initial change in this mixture of gases exposed to uv light will be the production of chlorine free radicals like this:

Note the use of one half arrow to show the movement of each bond electron to each chlorine atom.

This is called the Initiation Step in the reaction mechanism.

What could happen next?

Well chlorine free radicals have great energy of movement and will collide with molecules nearby almost immediately.

Some will collide with other nearby chlorine radicals and reverse the initiation step.

But others will collide with methane molecules.

So which of the two possible reactions below are likely to occur?

1.    CH4     +     Cl•     =      CH3Cl    +     H•           ΔH  =  +85kJ/mol

2.   CH4     +     Cl•     =       CH3•     +     HCl         ΔH  =   +4kJ/mol

Now one of our initial observations was that no molecular hydrogen is detected in the final products.

But if reaction 1. occurred then some molecular hydrogen would be produced by the combination of hydrogen radicals.

Furthermore, it is more likely that a less endothermic reaction would occur more readily.

So reaction 2. occurs after the initiation step.

But this reaction still leaves a very reactive methyl free radical.

What could happen to this free radical?

There is still enough chlorine available for the reactive methyl free radical to collide with a chlorine molecule like this:

3.  Cl2    +     CH3•       =        CH3Cl     +    Cl•

If it did, the reaction would regenerate the initial chlorine free radical that came from the initiation step.

And yes that is what happens many times over.

Reactions 2. and 3. form what is called the Propagation Step.

This is how we said "In the right light, for each photon of light absorbed, many thousands of chloromethane molecules are produced."

What's more combining the energy profiles of reactions 2 and 3 shows that the overall propagation process is exothermic even if the reaction 2 is endothermic as you can see below:

We call the combination of reaction 2 and reaction 3 a chain reaction.

But this isn't the end of the process.

You should see by now that reactive free radicals can react quite easily with each other and when they do that terminates the chain reaction.

So here are three possible termination reactions:

Cl•    +     Cl•      =       Cl2

Cl•    +      CH3•     =      CH3Cl

and also not surprisingly:

CH3•     +      CH3•      =      C2H6

These are all called Termination Steps

The overall mechanism looks like this:

This mechanism fits the conditions in which there is excess methane but what if there were excess chlorine instead what would change?

Well here let me suggest that other chlorination products form like CH2Cl2, CHCl3 and CCl4.

Can you satisfy yourself that you could write equations for the formation of these other products and show a mechanism for the reactions too?

And lastly, why is it that lead tetramethyl Pb(CH3)4 introduced into the reaction mixture increases the rate of formation of chloromethane?

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Hydrocarbons (6) Free Radical Addition Polymerisation

What are the catalysts used in addition polymerisation?

 

A typical catalyst is di-benzoyl peroxide (C₆H₅CO₂)₂.  

Its mode of action in the high pressure conditions of addition polymerisation is to cleave at the O-O bond into two free radicals.

The breaking of a covalent bond in this way is called homolytic fission.

This is the initiation step in the polymerisation process

The resulting particles are very reactive and soon break up again into carbon dioxide and a benzyl free radical C6H5•

The outline equations below picture this initiation process:

The benzyl free radical is also very reactive and when it collides with an alkene molecule it has sufficient energy to break the π bond and attach itself to the alkene creating a larger free radical like this:

In this example R• stands for the phenyl radical and it attacks the styrene (or phenylethene ) molecule at its double bond.

A bond forms between the styrene and phenyl species from two electrons: the odd electron in the radical and a π bond electron. 

The other π bond electron is available as the odd electron to form a new bond with another styrene molecule so:

This is the propagation stage. 

Clearly, if two growing chains each with a free radical end approach each other the chains will terminate.

You ought to satisfy yourself that you can construct equations of this type for simple alkenes like propene or ethane using benzoyl peroxide as the free radical initiator.   

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Hydrocarbons (5) Making Addition Polymers

We've seen how hydrocarbons split into three groups: alkanes, alkenes and alkynes.

These groups are formally called homologous series.

They each have a general formula, a gradation in physical properties and similar chemical properties.

For example the general formulae for these homologous series are:

Alkanes     Cn H2n+2
Alkenes     Cn H2n
Alkynes     Cn H2n-2

We've also seen that all alkanes are saturated molecules.

All alkanes have single covalent bonds between their atoms of hydrogen and carbon.

Alkenes by contrast have at least one double carbon carbon bond.

The genius of this double bond is however that the two bonds are not identical, one is weaker than the other.

Why is this genius?

The weaker bond allows the alkenes to be much more reactive than the alkanes.

The alkenes also allow chemists to be much more creative in their chemistry.

So alkenes decolorise orange or brown bromine water whereas alkanes do not.

The bromine water reaction is a test for the double bond:

The reaction with bromine is an addition reaction because two reactants add together to make one product.

You can catch a video of this test here

Now if the molecule of bromine can add across an alkene double bond, what's to stop other molecules adding across the double bond - all they need is a weak covalent bond.

Alkenes themselves have a weak covalent bond so couldn't they add to each other?

When this happens many thousands of times the product is an addition polymer

Polymer is this peculiar word for a molecule composed of many other molecules poly - many and mer - molecule.  

How can we represent this on paper?

Here is the example of polythene:  

n is a very large number say 1500 and there are four monomer units shown in the polythene molecule.

We can simplify the structure of a polymer like polythene in this way:

Here the polymer repeating unit is placed in brackets with the n outside; note also that the bonds, joining the polymer repeat units together, extend outside the brackets.

The reaction takes place using a catalyst and under high pressure.  

Under high pressure the monomer molecules of ethene are more likely to collide with each other.   

Here are a few other example polymerisation reactions:

Other groups of atoms can replace the chlorine atom in chloroethene and give rise to other addition polymers.  

This is polystyrene:

This is polypropene:

n

propene                                             polypropene  

You ought to be able to draw the displayed formula of an addition polymer from the displayed formula of its monomer.

You also ought to be able to draw the displayed formula of the monomer from the displayed formula of its polymer.  

In a subsequent post, I will discuss free radical polymerisation, the reason why the catalyst is used in polymerisation.

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