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

10.2 Some reaction mechanisms of ALKANES

The reaction mechanisms of alkanes

Part 10.2 ALKANES - introduction to the reaction mechanisms of alkanes. 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 Revision notes include full diagrams and explanation of the mechanisms of alkanes 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 (this page)

  • 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

  • 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)

• Alkanes are not very reactive molecules. Most reactions require some energy input to initiate a reaction e.g. high temperature and catalyst for cracking, uv light for chlorination or a spark to ignite them (initiating free radical reactions).

• A combination of two main reasons account for this lack of reactivity compared to most other homologous groups of organic molecules.

  1. Bond Strength:

     • The single covalent C-C (bond enthalpy 348 kJ mol^-1) and C-H (bond enthalpy 412 kJ mol^-1) bonds are very strong so bond fission does not readily happen. The carbon atom radius is small, giving a short and strong bond with other small atoms. Therefore the reactions will tend to have high activation energies resulting in slow/no reaction.

  2. Nature of bonding:

     • Carbon and hydrogen have similar electronegativities, so there is no polar bond giving a slightly positive carbon (C^δ+) which can be attacked by electron pair donating nucleophiles. [e.g. see halogenoalkanes (^δ+C-Cl^δ-) or aldehydes/ketones (^δ+C=O^δ-)]

     • All the C-C and C-H bonds are single covalent and no region of particularly high electron density susceptible to attack by electron pair accepting electrophiles. [e.g. like a double C=C bond see alkenes which are highly reactive despite the fact the C=C double bond has]

• What is the reaction mechanism of chlorine reacting with alkanes like methane and ethane etc.

• The basic reaction is: R₃C-H + Cl₂ ==heat/uv==> R₃C-Cl + HCl [mechanism 6]

  • R = alkyl e.g. CH₃, CH₃CH₂ etc. or aryl e.g. C₆H₅, CH₃C₆H₄ etc.

• The reaction is initiated by higher temperatures e.g. 250-400^oC or uv light at room temperature.

• If other hydrogen atoms are available on the original hydrocarbon then polysubstituted chloroalkanes will be formed

  • e.g. methane => chloromethane => dichloromethane => trichloromethane => tetrachloromethane

    • CH₄ ==> CH₃Cl ==> CH₂Cl₂ ==> CHCl₃ ==> CCl₄

  • propane can form initially 1-chloropropane or 2-chloropropane

    • and then 1,1- or 1,2- or 1,3- or 2,2-dichloropropanes etc. etc.!

    • i.e. CH₃CH₂CH₃ ==> CH₃CH₂CH₂Cl or CH₃CHClCH₃ 

    • => CH₃CH₂CHCl₂ or CH₃CHClCH₂Cl or ClCH₂CH₂CH₂Cl or CH₃CCl₂CH₃  

    • etc. and ultimately CCl₃CCl₂CCl₃

• Step (1) is the initiation step when the chlorine molecule is split into two chlorine atoms/radicals by homolytic bond fission by the impact-absorption of the ultraviolet photon. Its quantum of energy, E=hv, must be great enough to break the Cl-Cl bond.

  • Homolytic bond fission means the original pair of (Cl-Cl) bonding electrons is split between the two radicals formed.

  • Step (1) illustrates how to use half-arrows to show a homolytic bond fission step

    • Not all exam boards demand half-arrows, so the 'style' is deliberately varied in the diagram.

  • The red dots represent the unpaired electron on the free radical and the half-arrows show the individual electron 'shifts'.

  • The breaking of the Cl-Cl bond in the chlorine molecules begins the reaction because it is the weakest of the bonds of any reactant molecule involved. Bond enthalpies/kJmol^-1: Cl-Cl = 242, C-C = 348, C-H = 412, and even the new bond formed, C-Cl, is 338.

  • Free radicals are highly reactive species with an unpaired electron and tend to form a new bond as soon as is possible by e.g. in this case by ...

    • abstracting another atom from another molecule e.g. step (2) H abstracted, and step (3) chlorine abstracted or by pairing up with another radical e.g. steps (4) to (6).

• Steps (2) and (3) are chain propagation steps, because as well as producing one of the reaction products, a new free radical is also produced to continue the reaction, which is why such reactions are sometimes referred to as 'chain reactions'.

  • Step (2) Illustrates how to use half-arrows in a chain propagation step where an attacking radical abstracts an atom from a stable and complete molecule and another radical is formed in the process.

• Steps (4) to (6) are three possible chain termination steps which remove the highly reactive free radicals as two unpaired electrons form a new bond, in this case single C-C covalent bonds.

  • Step (5) illustrates how to use half-arrows to indicate a termination step where the unpaired electrons of the two radicals pair up to form a new bond, in this case a C-Cl bond.

• FURTHER COMMENTS

  • The mechanism for bromination is similar.

  • When the alkane is methane, traces of ethane are found in the final mixture of products. This provides evidence for a mechanism involving a methyl radical.

    • It would be formed from combining two methyl radicals: H₃C^. + ^.CH₃ ==> H₃C-CH₃ 

• What is the reaction mechanism of hydrocarbon alkanes being cracked to form alkenes, smaller alkanes and hydrogen.

• e.g. CH₃CH₂CH₃ ==> CH₄, CH₂=CH₂, CH₃CH₃, CH₂=CHCH₃, H₂  [mechanism 31]

  • The equation is not meant to be balanced, but just to show the variety of possible products.

  • Balanced equations e.g.

    • propene ===> methane + ethene

      • CH₃CH₂CH₃ ===> CH₄ + CH₂=CH₂

    • propane ==> propene + hydrogen

      • CH₃CH₂CH₃ ===> CH₂=CHCH₃ + H₂

• When alkane hydrocarbons are heated to a high temperature (450-900^oC, with/without superheated steam) they are thermally decomposed or 'cracked' to form mainly alkanes of lower C number, alkenes of equal or smaller C number and hydrogen.

mechanism 31 - free radical thermal cracking of alkanes

• When the temperature is high enough, the kinetic energy of the particles is sufficient to cause bond fission on collision, and this initiates a free radical chain reaction.

• Step (1) is the initiation step when a C-C bond in an alkane molecule is split into two alkyl free radicals by homolytic bond fission. This means the original C-C bonding electron pair is split between the two alkyl radicals formed. 

  • The weakest bonds will break first in the initiation step, so the C-C bond (bond enthalpy 348 kJmol^-1) will tend to break first as the C-H bond is stronger (bond enthalpy is 412 kJmol^-1).

  • The red dots represent the unpaired electron on the free radical.

  • Free radicals are highly reactive species with an unpaired electron and tend to form a new bond as soon as is possible, in this case by ...

    • abstracting a hydrogen from another molecule e.g. steps (2) to (6) or pairing up with another radical e.g. steps (7) and (8).

• Steps (2) to (6) are chain propagation steps, because as well as a product, a free radical is also produced to continue the chain reaction and lead to, in this case, a variety of other products.

• Steps (7) and (8) are two possible chain termination steps which remove the highly reactive alkyl free radicals. The unpaired electrons from the two radicals 'pair up' to form a new bond.

  • You can also have ethyl radical termination by simultaneous alkene and alkane formation (not shown in mechanism 31).

    • e.g. 2CH₃CH₂^. ==> CH₂=CH₂ + CH₃CH₃ 

    • or even isomerise on combination to form methylpropane (CH₃)₂CHCH₃ as well as the expected butane.

• FURTHER COMMENTS

  • Not all possible steps are shown, and not necessarily in that order all the time, but those that are shown illustrate the possibilities of how a wide variety of products can be formed.

10.2.4 An ionic mechanism for catalytic cracking

• (a) Catalytic cracking occurs at lower temperatures over a ceramic/zeolite catalytic material based on silica and/or aluminium oxide (I've called it Z in the mechanisms). The mechanism is believed to be ionic and some examples of reaction steps are given below. Z represents a matrix element of the alumino-silicate catalyst and can act in a variety ways e.g. Lewis acid or base, Bronsted-Lowry acid or base. R₁, R₂, R₃  = alkyl groups.

  • Step 1 Initiation: This can occur via an alkane or an alkene previously formed from an alkane.

    • (i) R₁-CH₂-CH₂-R₂ + Z⁺ ==> R₁-CH₂-CH⁺-R₂ + HL 

    • (ii) R₁-CH=CH-R₂ + HZ ==> R₁-CH₂-CH⁺-R₂ + Z^- 

  • Step 2 Propagation: This example shows chain scission forming a 'smaller' alkene and a 'smaller' carbocation to continue the chain.

    • R₁-CH₂-CH⁺-R₂ ==> R₁⁺ + CH₂=CH-R₂ 

  • Step 3 Termination: A lower alkene or alkane is formed.

    • R₁⁺ + Z^- ==> R₃CH=CH₂  + ZH 

      • (R₃ < R₁ in alkene, but no scission, so same chain length overall)

    • R₁⁺ + ZH ==> R₁H  + Z⁺ 

      • (no bond scission, so R₁ same length in carbocation or alkane formed)

• (b) A catalytic 'ionic' cracking cycle via alkenes:

  • I found this example in the literature, and the aluminosilicate/zeolite matrix (Z) acts alternately as an acid in step 1 and a base in step 4. Unfortunately it does involve starting with a rather a 'big' alkene called 2,4,4-trimethylpent-1-ene.

  • Step 1 initiation

    • (CH₃)₃C-CH₂-C(CH₃)=CH₂ + HZ ==> (CH₃)₃C-CH₂-C⁺(CH₃)₂ + Z^- 

    • The 'big' alkene is protonated by the catalyst and forms a 'big' carbocation (tertiary, most stable).

  • Step 2 propagation:

    • (CH₃)₃C-CH₂-C⁺(CH₃)₂ ==> (CH₃)₃C⁺ + CH₂=C(CH₃)₂ 

    • The 'big' carbocation from step 1 splits into 2-methylpropene (a cracking product) and a 'smaller' carbocation that continues the chain in step 3.

  • Step 3 propagation:

    • (CH₃)₃C-CH₂-C(CH₃)=CH₂ + (CH₃)₃C⁺ ==> (CH₃)₃C-CH₂-C⁺(CH₃)₂ + CH₂=C(CH₃)₂ 

    • From another 'big' starter alkene, and the carbocation from step 2, another molecule of 2-methylpropene is formed and the 'big' carbocation that was also formed in step 1, so allowing the 'ionic chain' reaction to continue via steps 2 and 3.

  • Step 4 termination:

    • e.g. (CH₃)₃C⁺ + Z^- ==> CH₂=C(CH₃)₂ + HZ 

    • The 'smaller' carbocation from step 2 is deprotonated to form another molecule of 2-methylpropene, and HZ is ready for step 1 again.

• (c) FURTHER COMMENTS

  • I found little help for ionic cracking mechanisms in either textbooks or the internet, I'm quite happy with the catalytic cycle in (b), but frankly I found the descriptions for (a) not easy to follow and cannot vouch for their absolute authenticity. I just my best to make sense of it and illustrate how the ionic mechanism might operate. (I wonder what does the IB syllabus require? Can anyone help me on this one?)