organic reaction mechanismsDoc Brown's GCE Chemistry  Revising Advanced Level Organic Chemistry

A Level 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

Electrophilic addition of bromine and hydrogen bromide

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. These revision notes include full diagrams and explanation of the ionic electrophilic addition reaction mechanisms of alkenes and the 'molecular' equation and reaction conditions and other con-current reaction pathways and products are also explained.

Part 10.3 ALKENES

10.3.1 Introduction

  • 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 pi 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 ∏ pi electrons.

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


10.3.2 The electrophilic addition of hydrogen bromide to alkene

The organic synthesis of bromoalkanes by reacting hydrogen bromide with alkenes

  • Examples of the addition of hydrogen bromide to alkenes

    • (i) ethene + hydrogen bromide ==> bromoethane

    • alkenes structure and naming (c) doc b + HBr ==> (c) doc b

    • (ii) propene + hydrogen bromide ==> 2-bromopropane

    • alkenes structure and naming (c) doc b+ HBr ==> (c) doc b

      • some of the isomer 1-bromopropane is also formed in this reaction

      • minority product (c) doc b from a con-current reaction.

  • What is the reaction 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) 

organic reaction mechanisms

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.

organic reaction mechanisms

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, Hd+Brd-, 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 Br2(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.

  • FURTHER COMMENTS

    • 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, Br2(aq) and H2SO4 etc. all broadly follow the Markownikoff addition rule.

    • 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 Iδ+-Clδ-.

 


keywords phrases: electrophile mechanism steps reagents reaction conditions formula intermediates organic chemistry reaction mechanisms steps electrophilic addition of hydrogen bromide hydrobromic acid to alkenes ethene propene butene R2C=CR2 + HBr ==> R2CH-CBrR2 HBr(g/aq) + H2O(l) ==> H3O+(aq) + Br-(aq) R2CH-CR2+ + Cl- ==> R2CH-CR2Cl ethene H2C=CH2 or but-2-ene CH3-CH=CH-CH3 propene CH3-CH=CH2, methylpropene (CH3)2C=CH2 or but-1-ene CH2=CH-CH2-CH3 CH3-CH=CH2 + H-X ==> CH3-CHX-CH3 or CH3-CH2-CH2-X CH3CH+CH3 (sec) > CH3CH2CH2+ (prim) (CH3)2C+CH2CH3 (tert) > (CH3)2CHC+HCH3 (sec) CH3CHX-CH3 CH3CH2-CH2X (CH3)2CH-CH2X (CH3)2CXCH2CH3 > and some (CH3)2CHCHXCH3 from but-1-ene, CH2=CHCH2CH3 CH3-CHXCH2CH3 XCH2-CH2CH2CH3

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