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Alkenes General Akenes and cycloalkenes are aliphatic unsaturated hydrocarbons.

e.g. hex-1-ene , cyclohexene

The alkenes form a homologous series with general formula CnH2n. Each member of the series differs by a -CH2- unit. Benzene (C6H6) is an aromatic arene (rather than the alkene "cyclohexatriene" that you might expect), since the double bonds and single bonds are not alternating around the ring as the displayed formula would suggest, but all equivalent, with the double bond delocalised all around the ring. That's why we use the symbol: rather than: Bonding In addition to the single covalent sigma- bonds which hold the carbon atoms to the hydrogen atoms, in an alkene there is double bond >C=C< between two of the carbon atoms.

A double bond consists of:

• a bonding pair of electrons in a sigma bond between the two carbon atoms, formed by the end-on overlap of two orbitals

• a bonding pair of electrons in a pi bond formed by sideways overlap of adjacent p-orbitals to form a second bond between the two carbon atoms. The electrons in the pi bond are localised – held in place between the two carbon atoms, so this region has high electron density.

The pi-bond is formed parallel to the sigma bond, and it is this bond that restricts rotation of a C=C bond, since the pi bond has to be broken before rotation can take place.

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C C

H

H

H

H

H CH

HCH

HH

Note also the spatial arrangement at each carbon atom: The three bonds from each carbon are trigonal planar with approx 120° bond angle because the bonding pairs repel each other as far as possible. The pi-bond is above and below the plane of the three sigma bonds. e.g. in propene we have CH3CH=CH2, so we have one carbon with 109° bond angles and four sigma bonds, and two carbons with 120° bond angles and 3 sigma bonds plus a pi bond between them:

Physical properties Because of the double bond having different bond angles (120°) to the rest of the chain (109.5°), alkenes can't pack as closely as the corresponding alkanes and therefore have correspondingly lower melting and boiling points. e.g. boiling points in Kelvin: ethene 169K ethane 185K propene 226K propane 231K Reactions of Alkenes Most alkene reactions involve breaking the pi-bond, which has a lower bond enthalpy and is therefore weaker than the sigma- bond. This leaves the sigma bond intact allowing the attacking species to be attached to the chain to form a saturated molecule. These are addition reactions. With lots of negative charge in the area of high electron density between the two C=C carbons, species wanting to accept an electron pair (often positively charged species, but not always) are attracted here where they can attack the double bond. We call species which are electron pair acceptors electrophiles (electron loving). Thus these reactions are termed electrophilic addition reactions. i) reaction with hydrogen, H2 e.g. + H2 !

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C C

H

H

H

H

CH

BrCH

BrH H

C C

H

H

H

H

CH

HCH

ClH H

CH

C CH

HH

H

HCH

BrCH

HH C

H

HH C

H

HCH

BrH C

H

HH

Reagents: Hydrogen Conditions: Nickel catalyst, 150°C ii) reaction with halogens, Cl2, Br2, I2 e.g. + Br2 ! ethene 1,2-dibromoethane Reagents: Bromine Conditions: mix at room temp Use: test for the presence of a C=C – bromine loses its orange colour and is decolourised. Note: F2 is too powerful an oxidising agent and tends to ignite the hydrocarbon ! Chlorine and iodine produce similar addition products. iii) reaction with hydrogen halides, HF, HCl, HBr, HI e.g. + HCl ! ethene chloroethane Reagent: hydrogen chloride Conditions: mix gases at room temperature Note that reactivity increases from HF to HI. HF will react with an alkene only under pressure. Note also that addition of a hydrogen halide to an unsymmetrical alkene can result in two possible isomeric products: CH3CH=CH2 + HBr ! CH3CHBrCH3 or CH3CH2CH2Br propene 2-bromopropane 1-bromopropane iv) reaction with water (steam) Where water is added to a molecule, as here, we call this a hydration reaction. This is not to be confused with a hydrolysis, which is breaking a molecule up due to a reaction with water forming two products.

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CH

C CH

HH

H

HHOH

H CH

CH

CH

HOH

HH H C

HCH

CH

HH

H OH

With an unsymmetrical alkene two isomeric products are possible. e.g. CH3CH=CH2 + H2O(g) ! CH3CH2CH2OH or CH3CH(OH)CH3 propene propan-1-ol propan-2-ol Reagent: water (steam) Conditions: an acid catalyst such as H3PO4 at 300°C and 6MPa pressure Use: this is the industrial route to alcohols not for drinking – methylated spirits,

industrial alcohol etc. Check your understanding: Write equations (displayed or skeletal formulae) and show all the possible products produced when: i) chlorine reacts with pent-2-ene ii) limonene (right) reacts with bromine water iii) hydrogen chloride reacts with 1,2-dimethylcyclohex-2-ene iv) hydrogen iodide reacts with buta-1,3-diene v) steam reacts with methylpropene vi) pent-1-ene is burnt in excess oxygen Mechanism of electrophilic addition The reaction begins with the approach of a molecule with a dipole - either permanent, e.g. with HCl - or induced by the presence of the double bond e.g. with Br2. The area of high electron density in the C=C repels the electrons in Br-Br, pushing them towards the end of the molecule furthest away from the C=C which therefore becomes δ- while the end of the molecule nearest the C=C becomes δ+, and is therefore attracted towards the negative charge of the C=C. There are two steps. This example shows the reaction of bromine with ethane, but these can be replaced by any other halogen or alkene in this mechanism. Step 1: Approach of electrophile and movement of an electron pair from the pi-system to form a bond to the δ+ end, followed by heterolytic fission of the bond in the electrophile. A curly arrow show where the electron pair comes from and where they go. They always come FROM a bond or a lone pair on an atom, and go TO an atom where the new bond is being formed.

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Step 2: The intermediate carbocation (ion containing a carbon atom with a + charge) is very reactive and immediately reacts with the negative ion from step 1, forming a covalent bond

Note: when drawing mechanisms, there are marks for - relevant dipoles - lone pairs - curly arrows (showing the movement of a PAIR of electrons) Check your understanding: vii) Draw the mechanism for the electrophilic addition of HCl to propene to form

2-chloropropane. viii) Draw the mechanism for the addition of H2O(g) to cyclohexa-1,3-diene to form cyclohexane-1,4-diol Markownikoff’s Rule Reactions of hydrogen halides with unsymmetrical alkenes produce two different possible products, but these are not equally likely to be formed. One will be the major product, and one the minor product. Markownikoff’s rule states that: the hydrogen atom of the H-X is added to the carbon of the C=C which has the greatest number of hydrogen atoms/least number of carbon atoms bonded to it. To understand why this happens, we need to look again at the mechanism – in particular the intermediate carbocation that is formed when the H atom from the H-X has been added in the first step. The most stable carbocation will be most likely to be formed. Stability of carbocations increases with the number of alkyl groups bonded to the carbon with the positive charge:

à

> >

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Application If we add HCl to propene, CH3CH=CH2, then we can see that the carbon in the middle of the propene has one H and one C attached, while the one on the end has two H bonded to it. Markownikoff’s rule says the H of the HCl will be added here so the major product will be 2-chloropropane and the minor product 1-chloropropane. To explain this, in step 1 we could form a primary carbocation, CH3CH2CH2

+, by adding the hydrogen to the middle carbon, or we could form a secondary carbocation CH3CH+CH3 by adding the hydrogen to the end carbon. The secondary carbocation is more stable, so this is formed preferentially so that in the second step the chloride ion is added to the middle carbon to form 2-chloropropane as the major product. Check your Understanding: ix) Draw the structure of the major product formed when HBr reacts in an electrophilic

addition with 1-methylcyclohex-1-ene. More about stereoisomerism in alkenes It is easy to determine with is the cis- and which is the trans- isomer, because this definition is based on there being two identical groups on either carbon of the C=C which are either on the same side (cis-) or opposite sides (trans-). Of more general use is the notation of E- and Z- isomers because it can be applied to any alkene having different groups on each carbon of the C=C. The rules for determining which isomer is E- and which is Z- are known as the CIP rules (Cahn-Ingold-Prelog). The essence of the rules is that an E- isomer has the highest priority groups on either side, and a Z- isomer has the highest priority groups on the same (zame?) side. The priority of of the groups increases with the atomic number of the bonded atoms, stepping outwards from the carbon atoms of the C=C to the first point of difference. Application e.g. isomers of 1-chloropropene: The right C=C carbon is bonded to a C atom (of the -CH3) and an H atom. This is the first point of difference, where the higher atomic number of the C makes the -CH3 the priority group. The left C=C carbon is bonded to a Cl atom and an H atom. The higher atomic number of the Cl atom makes the –Cl the priority group. The E- isomer therefore has the –CH3 and -Cl on either side of the C=C, and the Z- isomer has them on the same side. e.g. isomers of 1-chloro-2-methylbut-2-ene: The right C of the C=C has a C (of the CH3) and an H atom bonded to it, so the highest priority group is the –CH3. The left C of the C=C has two carbon atoms bonded to it,

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so this is not a point of difference, ands we have to step outwards, to where there is a difference – one carbon has an H and one has a Cl bonded to it, so the one with the Cl becomes the priority group. The illustrated stereoisomer has the priority groups on either side, so is the E- isomer. Check your Understanding: x) Draw and label the E- and Z- isomers of 2-bromo-1-chloropropene xi) Draw and label the E- and Z- isomers of 3-methylpent-2-ene xii) A molecule has structural formula CH3CH=C(CH2CH2Cl)CH2CH2OH. Draw the

skeletal formula of the Z-isomer. Addition Polymers Alkenes can undergo addition reactions in which alkene molecules join to each other to form long chains – often tens of thousands of carbon atoms long. The individual alkene molecules are referred to as monomers, and the resulting chain as a polymer. Note that (ignoring the ends of the chain) the empirical formula of the monomer and of the polymer are the same. We ignore the ends of the chain because they are insignificant compared with the length of the polymer chain. Examples: But-1-ene and but-2-ene both react to form polymers. We may be required to draw the monomers, sections of the polymers, the repeat units, or write a balanced equation for the reactions: but-1-ene but-2-ene It may be helpful to re-draw the monomer molecules starting with the C=C and adding to each carbon the atoms/groups which are connected to it above and below:

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We then draw adjacent monomers, replacing the C=C with –C-C– and using these ‘dangling’ bonds to join the C=C carbons into a section of the polymer chain.

Drawing a repeat unit: A repeat unit can be found by taking any two linked carbons in the polymer chain and drawing a ring around them and their attached side groups. If starting with the monomer, simply draw the monomer then turn the double bond into a single bond and add connecting bonds to the two carbons which were originally double-bonded. Drawing the monomer from repeat unit or a section of chain: Take two adjacent carbons in the chain, plus their attached side groups. Remove the connecting bonds and replace them with the double bond. For example, here's a repeat unit of poly(pent-2-ene), from which we can construct the monomer:

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Writing a balanced equation: We don't know how many monomers form the chain, so we use 'n' to be the number of monomers. This will produce a chain 'n' repeat units long. Note the connecting bonds sticking through the brackets, and don't forget the 'n' after the brackets to show the number of times the repeat unit repeats.

Disposal of Waste Polymers While addition polymers from crude-oil derived raw materials tend to be cheap, they are also non-biodegradable (not broken down by micro-organisms). Having strong covalent bonds which are mostly non-polar makes them resistant to chemical attack too. This makes polymer waste hard to dispose of because it endures in the environment without decomposing. It is unsuitable for disposal in landfill sites (although that is where much polymer waste currently turns up) because more and more land has to be found. More sustainable use of polymer waste can be achieved by:

• combustion for energy generation, although it is necessary to remove toxic waste products such as HCl formed during the combustion of halogenated plastics such as PVC (polychloroethene). This reduces our dependence on finite fossil fuel resources.

• use as an organic feedstock for the production of plastics and other organic chemicals (e.g. polymer waste can be cracked to produce shorter chain alkanes plus alkenes)

• reuse and repurposing of polymer waste materials Research chemists can reduce environmental damage and improve sustainability in our use of polymers by the development of photodegradable polymers or biodegradable polymers, making use of sustainable sources of raw materials such as plant starches. This reduces our dependence on finite crude-oil resources. Check your understanding: xiii) Write a balanced equation for the addition polymerisation used to form polyethenol xiv) Draw a section of the polymer formed by addition polymerisation of methypropene xv) Draw the monomer from which this polymer is made:

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OH

OH

CH

C CH

HH

H

H H CH

HCH

CH

HH+

CCl

C CHH

H

H

HH

H

Answers to Check your Understanding questions: Write equations (displayed or skeletal formulae) and show the product(s) produced when: i) chlorine reacts with pent-2-ene Cl2 + ! ii) 2 Br2 + ! iii) HCl + !

iv) 2 HI + ! v) H2O(g) + or ! vi) + 7½ O2 ! 5 CO2 + 5 H2O vii) H δ+ | Cl δ-

à

:Cl-

à

Cl

Cl

Br

BrBr

Br

Cl

Cl

or

I

I

I I

I

I

or or

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viii)

N.B. mechanism could be drawn with both H2O attacking simultaneously rather than sequentially

ix)

x) xi) xii) The first point of difference on the right C of the

C=C is at the Cl and O atoms. The Cl has the higher atomic number, so is on the same side as the higher priority –CH3 group on the other carbon in the Z-isomer.

HOH OH

HOH

OHOH

OH

δ+

δ-..OH

δ+δ-

..OH

HBr

Br

Br

Cl

Br

Cl

Z- isomer E- isomer

E- isomer Z- isomer

Cl

OH

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C CH

OH

H

H

C CH

OH

H

H

C CH

H

CH3

CH3

C CH

H

CH3

CH3

C CH

H

CH3

CH3

C CH

H

CH3

CH3

xiii) xiv) methypropene can be drawn as so the section of the polymer is xv) the monomer is phenylethene:

n à

n