Sunday 18 August 2013

Electrophilic addition reactions of alkenes


  • 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 of hydrogen bromide to alkene
  • 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) 
electrophilic addition of hydrogen bromide to an alkene in aqueous mediaorganic reaction mechanisms
  • In the acid solution via step (1) the H3O+ or oxonium ion (hydrated proton) is the 'attacking electrophile' andprotonates 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
    • 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 Bretc.) 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). NOTEThe 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 ofCH3*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, Br2(aq) and H2SO4 etc. all broadly follow the Markownikoff addition rule.



10.3.3 The electrophilic addition of bromine to alkene (non-aqueous media)
  • 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.
organic reaction mechanisms
mechanism 4 - electrophilic addition of bromine to an alkene in non-aqueous media
  • 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.
    • organic reaction mechanisms
    • 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 Iδ+-Clδ-.
    • -


10.3.4 The electrophilic addition of bromine to alkene (aqueous media)
  • 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.
organic reaction mechanisms
mechanism 5 - electrophilic addition of bromine to an alkene in aqueous media
  • 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 acarbocation. 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.



10.3.5 The electrophilic addition of conc. sulphuric acid to alkene
  • 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.
organic reaction mechanisms
mechanism 28 - electrophilic addition of sulphuric acid (conc.) to an alkene
  • 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 + H2==> 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 + 2H2==> 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).



10.3.6 The acid catalysed electrophilic addition of water to alkene
  • 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 + H2==>  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!
organic reaction mechanisms
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.
organic reaction mechanisms
mechanism 32 - the peroxide catalysed free radical polymerisation of ethene
  • Step (1) For peroxide catalysed polymerization, the initial free radicals are formed by the homolytic bond fission of the O-O bond in a peroxide molecule.
    • The O-O bond (bond enthalpy 146 kJmol-1) is the weakest and will break before any others e.g. the C-C bond enthalpy is 348 kJmol-1 and the C-H bond enthalpy is even stronger at 412 kJmol-1. (see similar arguments in alkane chlorination)
      • Step (1) illustrates how to use half-arrows to show a homolytic bond fission step.
  • Step (2) is a chain propagation step, in which the 'alkoxy radical' adds onto an alkene molecule.
    • If you split open half the alkene double bond (i.e. one bond pair), one electron pairs up with the unpaired electron of the 'alkoxy radical' to form a C-O bond. The other electron remains unpaired at the end of the radical, so that one active free radical is replaced by another.
      • Step (2) illustrates how to use half-arrows in a polymerisation chain propagation step, where the radical adds onto the polymer molecule to make an even longer radical to continue the chain reaction.
  • Step (3) illustrates all the further subsequent chain propagation steps in which the polymer free radical adds onto another monomer alkene molecule to make the chain longer and longer.
  • Step (4) represents one possible chain termination, were the electrons of two free radicals pair up to form a covalent C-C bond. 
  • FURTHER COMMENTS
    • At GCSE level the presence of a very tiny % of oxygen atom in poly(ethene) is unlikely to have been mentioned, but has no influence on the properties of the polymer, since the extremely long chain of -CH2-CH2-'s predominates!
    • The use of Ziegler-Natta catalysts to make more stereo-specific high density poly(ethene) and high density poly(propene) involves an ionic mechanism.
  • -

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