Doc Brown's GCE Chemistry  Revising Advanced Level Organic Chemistry

Revision Notes PART 10 Summary of organic reaction mechanisms - A mechanistic introduction to organic chemistry and explanations of different types of organic reactions

10.3 Reaction mechanisms of ALKENES

Part 10.3 ALKENES - introduction to the reaction mechanisms of alkenes. Electrophilic addition of hydrogen bromide [HBr(conc. aq) and HBr(g/non-polar solvent)] to form halogenoalkanes Electrophilic addition of bromine ... with pure bromine or in non-polar solvent (non-aqueous Br2(l/solvent)) to give dibromoalkanes or using bromine water [aqueous Br2(aq)] to give bromo-alcohols Electrophilic addition of sulphuric acid Electrophilic addition of water [acid catalyst] to form alcohols Free radical polymerisation to give poly(alkene) polymers Hydrogenation to give saturated alkanes (in kinetics notes). These revision notes include full diagrams and explanation of the mechanisms of alkenes and the 'molecular' equation and reaction conditions and other con-current reaction pathways and products are also explained.

• Organic Chemistry Part 10 sub-index: and Other advanced level organic chemistry links

• MECHANISMS INDEX

• Part 10.1 Introduction and terms used in organic chemistry

• Part 10.2 ALKANES - introduction

  • Free radical chlorination/bromination to give halogenoalkanes (haloalkanes, alkyl halides)

  • Free radical thermal cracking to give shorter alkanes and alkenes

  • Ionic catalytic cracking to give shorter alkanes and alkenes

• Part 10.3 ALKENES - introduction [this page]

  • Electrophilic addition of hydrogen bromide [HBr_{(conc. aq)} and HBr_{(g/non-polar solvent)}] to form halogenoalkanes

  • Electrophilic addition of bromine ...

    • with pure bromine or in non-polar solvent (non-aqueous Br_2(l/solvent)) to give dibromoalkanes

    • or using bromine water [aqueous Br_2(aq)] to give bromo-alcohols

  • Electrophilic addition of sulphuric acid

  • Electrophilic addition of water [acid catalyst] to form alcohols

  • Free radical polymerisation to give poly(alkene) polymers

  • Hydrogenation to give saturated alkanes (in kinetics notes)

• Part 10.4 HALOGENOALKANES - introduction (haloalkanes, alkyl halides)

  • Nucleophilic substitution by water/hydroxide ion [S_N1 or S_N2, hydrolysis to give alcohols]

    • with extra notes on kinetics, rds, molecularity, activated complex etc.

  • Nucleophilic substitution by cyanide ion to give a nitrile [S_N1 or S_N2]

  • Nucleophilic substitution by ammonia/primary amine to give primary/secondary amines etc. [S_N1 or S_N2]

  • Elimination of hydrogen bromide to form alkenes [E1 and E2]

• Part 10.5 ALCOHOLS - introduction

  • Conversion of an alcohol to a halogenoalkane

  • Elimination of water from an alcohol to give an alkene [acid catalysed, E1 and E2]

• and 10.6 Carbonyl compounds - ALDEHYDES and KETONES - introduction

  • Nucleophilic addition of hydrogen cyanide to form a hydroxy-nitrile

  • Addition of hydrogen - reduction with LiAlH₄ or NaBH₄ to give alcohols

  • The iodination of ketones e.g. a 2-one like propanone (a methyl ketone) to give iodo-ketones

• and 10.7 ACID CHLORIDES - introduction - all nucleophilic addition-elimination reactions

  • Hydrolysis with water to give a carboxylic acid

  • Esterification with alcohols to give an ester

  • Amide formation from reaction with ammonia or primary amines

• Part 10.8 AROMATIC HYDROCARBONS - introduction to arene electrophilic substitutions

  • Nitration to give nitro-aromatics like nitrobenzene

  • Chlorination to chloro-aromatics like chlorobenzene

  • Alkylation to give alkyl-aromatics like methylbenzene [Friedel-Crafts reaction]

  • Acylation to give aromatic ketones [Friedel-Crafts reaction]

  • Sulphonation/sulfonation to give a sulphonic/sulfonic acid like benzenesulphonic acid/benzenesulfonic acid

  • The orientation of products in aromatic substitution (2,4,6 or 3,5 positions, ortho,para/meta substitution products)

• Alkenes are reactive molecules, particularly when compared to alkanes.

• They are reactive towards electron pair accepting electrophiles because of the high density of negative electron charge associated with the ∏ electrons of the double bond.

• However they can also readily undergo free radical reactions  e.g. their peroxide catalysed polymerisation to form a poly(alkane) and these reactions also involve the interaction of free radicals with the ∏ electrons.

• The electrophilic addition reactions of alkenes are compared with the nucleophilic addition to carbonyl compounds in the aldehydes and ketones section.

• What is the mechanism for the addition of hydrogen bromide to an alkene?

• Does the mechanism change if the solvent is changed?

• Do the products of the reaction depend on the solvent used?

• Can isomeric products be formed in the addition of hydrogen bromide to an alkene?

• AQUEOUS media: [see mechanism 39 below]

  • If a liquid alkene is mixed with, or a gaseous alkene is bubbled through, concentrated hydrobromic acid, HBr_(aq) (hydrogen bromide dissolved in water) a bromoalkane is formed. 

  • overall reaction: R₂C=CR₂ + HBr ==> R₂CH-CBrR₂ 

  • In water, hydrogen bromide is a strong acid i.e. completely ionises to give the oxonium ion and bromide ion.

    • HBr_(g/aq) + H₂O_(l) ==> H₃O⁺_(aq) + Br^-_(aq) 

mechanism 39 - electrophilic addition of hydrogen bromide to an alkene in aqueous media

• In the acid solution via step (1) the H₃O⁺ or oxonium ion (hydrated proton) is the 'attacking electrophile' and protonates the alkene to form the intermediate positive carbocation R₂CHCR₂⁺. The oxonium ion is an electrophile because it accepts a pair of electrons from the alkene ∏ bond to form the new C-H bond.

• In step (2) the (already present) negative bromide ion rapidly combines with the carbocation to form the bromoalkane product. The bromide ion donates a pair of electrons to form the new C-Br bond.

  • With the high concentration of water present, a water molecule could also interact with the carbocation to eventually form a small amount of the alcohol R₂CHCR₂OH, this again provides evidence of an ionic mechanism. 

• NON-AQUEOUS media: [mechanism 3 below] Addition will also occur if the alkene is mixed with hydrogen bromide gas, or the HBr gas is dissolved in a non-polar organic solvent and mixed with the alkene.

mechanism 3 - electrophilic addition of hydrogen bromide to an alkene in non-aqueous media

• In this case, for step (1), the attacking electrophile is the already polarised hydrogen bromide molecule, H^d+Br^d-, which splits heterolytically to protonate the alkene, forming the carbocation and a bromide ion. The HBr molecule is an electrophile because it accepts a pair of electrons from the alkene ∏ bond to form the new C-H bond.

• In step (2) the bromide ion formed in step (1) rapidly combines with the carbocation to form the bromoalkane. The bromide ion donates a pair of electrons to form the new C-Br bond.

• FURTHER COMMENTS

  • EVIDENCE for an IONIC MECHANISM

    • Below is a general comment for all the electrophilic addition reactions of alkenes.

    • If the reaction is carried out in the presence of other negative ions e.g. chloride ion from adding sodium chloride salt to an aqueous reaction mixture, then some chloroalkane is produced via step (2).

      • R₂CH-CR₂⁺ + Cl^- ==> R₂CH-CR₂Cl 

        • Without a carbocation intermediate formed it is difficult to explain the formation of such products.

        • In fact any anion present, X^-, produces some R₂CH-CR₂X

  • A symmetrical alkene is when the atoms/groups are the same on either side of the C=C double bond.

    • e.g. ethene H₂C=CH₂ or but-2-ene CH₃-CH=CH-CH₃ 

    • This means which ever way round the HX addition takes place onto the double bond, you always get the same product.

  • An non-symmetrical alkene is when the atoms/groups are NOT the same on either side of the C=C double bond e.g.

    • propene CH₃-CH=CH₂, methylpropene (CH₃)₂C=CH₂ or but-1-ene CH₂=CH-CH₂-CH₃ 

    • This means that when addition to the double bond with a non-symmetrical reagent itself, e.g. like H-X, you have the possibility of two different isomeric addition products.

      • e.g. CH₃-CH=CH₂ + H-X ==> CH₃-CHX-CH₃ or CH₃-CH₂-CH₂-X

      • Which begs the questions, which isomer predominates? and why?

  • The Markownikoff rule predicts which isomer is likely to predominate for adding a non-symmetrical reagent to a non-symmetrical alkene and the rule can be stated in various ways but the IUPAC definition of 1997 states: For the heterolytic addition of a polar molecule to an alkene (or alkyne), the more electronegative (nucleophilic like OH^- or Br^- etc.) atom (or part) of the polar molecule becomes attached to the carbon atom bearing the smaller number of hydrogen atoms [or you can say the least electronegative (most electrophilic like Br⁺ or H⁺ etc.) will attach to the carbon atom bonded with the most H atoms). BUT the 'rule' only applies to the ionic mechanism, you can get the opposite effect in free radical addition in the presence of peroxides!

    • The orientation of the products from non-symmetrical addition (HX or Br_2(aq) see later) is governed by the stability of the carbocation intermediate formed by the protonation of the alkene by the attacking H-X electrophile, and explains the Markownikoff rule.

    • The order of carbocation stability is tertiary > secondary > primary, because alkyl groups give a slight electron donating inductive effect (+I) via the attraction of the positively charged carbon atom. This spreads the positive charge of the carbocation and gives the carbocation more stability by lowering its potential energy. It is a general rule of physics that spreading out electric charge lowers the potential energy and increases the stability of a situation.

    • The most stable carbocation will be the one most likely to exist with a sufficient life-time to be hit by the electron pair donating ion (e.g. X^-) or any other electron pair donor, including water (see addition of bromine water). NOTE: The positive carbon of the most stable carbocation, has attached to it the most alkyl groups and the least hydrogen atoms.

      • e.g. from protonating propene CH₃CH=CH₂ you expect the carbocation stability to be ...

        •  CH₃CH⁺CH₃ (sec) > CH₃CH₂CH₂⁺ (prim)

      • or from protonating 2-methylbut-2-ene (CH₃)₂C=CHCH₃ you expect the carbocation stability to be ...

        • (CH₃)₂C⁺CH₂CH₃ (tert) > (CH₃)₂CHC⁺HCH₃ (sec)

    • So for adding HX to a non-symmetrical alkene you would expect the major isomer to be e.g.

      • from propene, CH₃CH=CH₂ you expect mainly CH₃CHX-CH₃

        • and some CH₃CH₂-CH₂X

      • from methylpropene, (CH₃)₂C=CH₂ you expect mainly (CH₃)₂CX-CH₃  

        • and some (CH₃)₂CH-CH₂X

      • from 2-methylbut-2-ene you expect mainly (CH₃)₂CXCH₂CH₃ > and some (CH₃)₂CHCHXCH₃

      • from but-1-ene, CH₂=CHCH₂CH₃ you expect mainly CH₃-CHXCH₂CH₃  

        • with some XCH₂-CH₂CH₂CH₃ 

  • What happens in terms of optical isomers/activity if the product has a chiral carbon^*?

    • An example of a  chiral carbon results from when four different atoms or groups (a to d) is bonded to the same carbocation i.e. ^*Cabcd, so the carbocation has a plane of symmetry. This symmetrical arrangement means that if the product is potentially optically active, a racemic mixture will be formed because the e.g. bromide ion, can add with equal probability on both sides of the carbocation. This will result in equal quantities of the optical isomers (enantiomers), giving an optically inactive racemic mixture.

    • e.g. but-1-ene, CH₂=CHCH₂CH₃ on adding HX, will give a racemic mixture of the optical isomers of CH₃^*CHXCH₂CH₃ with some XCH₂-CH₂CH₂CH₃ which is incapable of optical isomerism because it does not have a chiral carbon.

  • Free radical addition of hydrogen bromide

    • The addition of hydrogen bromide using a peroxide catalyst produces 'anti-Markownikoff' addition.

      • e.g. propene, CH₃CH=CH₂ produces mainly CH₃CH₂-CH₂Br

      • but only hydrogen bromide gives this peroxide effect, the addition of H-Cl, Br_2(aq) and H₂SO₄ etc. all broadly follow the Markownikoff addition rule.

• What is the mechanism for the addition of bromine to an alkene?

• Does the mechanism change if the solvent is changed?

• Do the products of the reaction depend on the solvent used?

• Can isomeric products be formed in the addition of bromine to an alkene?

• R₂C=CR₂ + Br₂ ==> R₂CBr-CBrR₂ [see mechanism 4 below]

• The alkene is mixed with bromine liquid or a solution of bromine in an organic (non-aqueous, non-polar) solvent.

• In step (1) The non-polar bromine molecule is the electrophile, and becomes polarised on collision with the traces of water or ions on the reaction vessel surface^*. The collision causes the bromine molecule to split heterolytically so that that the equivalent of a Br⁺ bonds to one of the double bond carbons to give a carbocation. The electrophilic Br⁺ accepts the pair of ∏ electrons from the C=C double bond to form the 1st new C-Br bond.

  • ^*Apparently completely dry bromine and alkene, and a paraffin coated reaction vessel, produce zero reaction!

• In step (2) the bromide ion formed in step (1) rapidly combines with the carbocation to form the dibromoalkane, by donating a pair of electrons to make the new 2nd C-Br bond.

• FURTHER COMMENTS

  • There is considerably evidence (beyond the academic scope of the page) to show that the 1st stage in the mechanism of bromine addition (non-aqueous or aqueous) actually goes via a triangular bromonium ion shown in mechanism 43 below. However many exam boards and older textbooks seem happy with the carbocation mechanism 4 shown above. Chlorine reacts similarly via a chloronium ion.

  • 

  • The reaction mechanism is similar for non-aqueous chlorine.

  • The Markownikoff rule does NOT apply to this reaction, whatever the mechanistic details, because the reagent itself is symmetrical i.e. Br-Br, so different isomeric products are NOT expected. However the rule does apply when using aqueous bromine (see mechanism 5 below) or using a mixed halogen reagent (see next point)

  • Addition of mixed halogen compounds (inter-halogen compounds), such as iodine(I) chloride ICl, will also add to the alkene double bond. 

    • e.g. CH₃CH=CH₂ + ICl ==> CH₃CHI-CH₂Cl or CH₃CHCl-CH₂I 

    • From the Markownikoff rule 2-chloro-1-iodopropane should be the principal product because chlorine is more electronegative than iodine, so think of it as the addition of I^δ+-Cl^δ-.

  • -

• R₂C=CR₂ + 2H₂O + Br₂ ==> R₂C(OH)-CBrR₂ + H₃O⁺ + Br^- [see mech 5 below]

  • If the alkene is not symmetrical about the C=C bond, isomeric products can be formed.

  • with the chance of a little R₂CBr-CBrR₂ too via the small concentration of bromide ion.

• In step (1) The bromine molecule is the electrophile, and becomes polarised on collision with water. It splits heterolytically so that that the equivalent of Br⁺ bonds to one of the double bond carbons to give a carbocation. The electrophilic Br⁺ accepts the pair of ∏ electrons from the C=C double bond to form a new C-Br bond. So step (1) is the same for non-aqueous bromine, however step (2) is different!

• In step (2), unlike with non-aqueous bromine, the much greater concentration of water, compared to the bromide ion, ensures the most probable addition to the carbocation is a water molecule. Water acts as an electron pair donor and on rapid combination with the carbocation, a protonated alcohol is formed.

• In step (3) a 2nd water molecule then removes a proton to leave the bromo-alcohol product.

• FURTHER COMMENTS

  • The triangular bromonium ion mechanism described above for non-aqueous bromine also applies here and the reaction is similar with chlorine water.

  • There is of course a small chance that a bromide will combine with the carbocation, so a little of the dibromoalkane is formed to.

  • The Markownikoff rule for a non-symmetrical alkenes does apply here, so the initially added 'Br⁺' will end up combined with the carbon atom of the double bond with the most hydrogen atoms and the H₂O/OH ends up bonded to the C atom with the least number of H atoms e.g.

    • from propene CH₃CH=CH₂, the majority product is 1-bromopropan-2-ol, CH₃CHOHCH₂Br and the 

    • minority products are 2-bromopropan-1-ol, CH₃CHBrCH₂OH and 1,2-dibromopropane, CH₃CHBrCH₂Br 

  • The reaction is similar with chlorine water.

• What is the mechanism for the addition of concentrated sulfuric acid to an alkene?

• Can isomeric products be formed in the addition of sulphuric acid to an alkene?

• R₂C=CR₂ + H₂SO₄ ==>  R₂CH-C(OSO₂OH)R₂    [see mechanism 28 below]

  • Isomeric products can be formed if the alkene not symmetrical.

• In step (1) The sulphuric acid molecule is the electrophile by nature of the highly polar O-H bond which splits heterolytically to protonate the alkene molecule to form the carbocation.

• In step (2) the hydrogensulphate ion formed in step (1) combines with the carbocation to give the alkyl hydrogensulphate product.

• FURTHER COMMENTS

  • The Markownikoff rule predicts for a non-symmetrical alkene, the proton from the sulphuric acid will mainly combine with the carbon atom with the most number of hydrogen atoms.

    • So for propene the majority product is CH₃-CH(OSO₂OH)CH₃ with a little HOSO₂OCH₂-CH₂CH₃ 

  • On warming the product with water, the alkyl hydrogensulphate is hydrolysed to an alcohol is. (see also mechanism 29 below).

    • R₂CH-C(OSO₂OH)R₂ + H₂O ==> R₂CH-C(OH)R₂ + H₃O⁺ + HSO₄^- 

    • In terms of organic synthesis, for example, propene eventually forms mainly propan-2-ol.

      • stage 1: CH₂=CHCH₃ + H₂SO₄ ==> CH₃-CH(OSO₂OH)CH₃ 

        • stage 2: CH₃-CH(OSO₂OH)CH₃ + 2H₂O ==> CH₃-CH(OH)CH₃ + H₃O⁺ + HSO₄^- 

  • If the product of the addition can exhibit optical isomerism, an optically inactive racemic mixture will be formed (see above for fuller discussion).

• What is the mechanism for the catalysis reaction on addition of water to an alkene?

• Can isomeric products be formed in the addition of water to an alkene?

• R₂C=CR₂ + H₂O ==>  R₂CH-C(OH)R₂    [see mechanism 29 below]

  • Isomeric products can be formed if the alkene is not symmetrical about the C=C bond.

  • This one stage synthesis is equivalent to the two stage synthesis of alcohols from alkenes using concentrated sulphuric acid and hydrolysing the product (see above), but the catalysis mechanism has three steps!

mechanism 29 - the acid catalysed electrophilic addition of water to an alkene

• In step (1) the oxonium ion (the electrophile from the acid) protonates the alkene to form the carbocation.

• In step (2) a water molecule, acting as an electron lone pair donor, bonds with the carbocation to form the protonated alcohol.

• A proton transfer occurs in step (3) as a water molecule accepts a proton from the protonated alcohol to leave the free alcohol product and reforming the catalytic oxonium ion H₃O⁺.

• FURTHER COMMENTS

  • The Markownikoff rule predicts for a non-symmetrical alkene, the proton from the sulphuric acid will mainly combine with the carbon atom with the most number of hydrogen atoms.

    • So for propene CH₃CH=CH₂ the majority product is propan-2-ol CH₃-CH(OH)CH₃ 

      • with a little propan-1-ol CH₃CH₂-CH₂OH

  • In the industrial process, catalysed by phosphoric(V) acid, the proton can come from H₃PO₄, in step (1)

    • and the H₂PO₄^- ion formed in step (1) will remove the H⁺ from the protonated alcohol in step (3).

  • If the product of the addition can exhibit optical isomerism, an optically inactive racemic mixture will be formed (see above for fuller discussion).

10.3.7 The free radical addition polymerisation of an alkene

• What is the mechanism of polymerising ethene to form poly(ethene)?

• e.g. nCH₂=CH₂ ==> -(-CH₂-CH₂-)-_n      [see mechanism 32 below]

• Polyalkenes such as low density poly(ethene) can be made by polymerising the alkene monomer with oxygen or peroxide catalysts.