10.4 HALOGENOALKANES (old names 'haloalkanes' or 'alkyl halides')

mechanism 26 elimination of HBr from a halogenoalkane by hydroxide ion (E2 'bimolecular' mechanism)

• E2 mechanism via single step bimolecular collision process. [mechanism diagram 26 above]

  • The hydroxide ion acts both as a strong base (proton accepting) and as a nucleophile (electron pair donor), by removing a proton from the halogenoalkane to form water.

  • As the C-H bond pair of electrons (bottom left) shifts to complete the C=C double bond of the alkene product, simultaneously, the C-Br bond pair (bottom right) leaves with the bromine atom to give the bromide ion.

mechanism 27 - elimination of HBr from a halogenoalkane by hydroxide ion (E1 'unimolecular')

• E1 mechanism, a two step process via carbocation formation. [mechanism 27 above]

• In step (1) the polar and weakest bond, C^δ+-Br^δ-, breaks heterolytically to form a carbocation and bromide ion.

• In step (2) the strongly basic and nucleophilic hydroxide ion abstracts a proton from the carbocation to form water and simultaneously the C-H bond pair (bottom left) shifts to complete the C=C double bond of the alkene.

• FURTHER COMMENTS

• Con-current nucleophilic substitution AND elimination reactions for RX

  • RX = halogenoalkane/haloalkane/alkyl halide/halogenated alkanes.

  • i.e. nucleophilic substitution (SN) to give an alcohol versus elimination (E) to give an alkene.

  • With halogenoalkanes of at least C₂, the use of strong alkali like sodium hydroxide can also produce alkene products by an elimination whose mechanism is discussed above.

    • elimination reaction: R₂CH-CBrR₂ + NaOH ==> R₂C=CR₂ + H₂O + NaBr

    • as well as/versus the

    • substitution reaction: R₂CH-CBrR₂ + NaOH ==> R₂CH-C(OH)R₂ + NaBr

      • In both cases R = H, alkyl or aryl and 'HBr' is eliminated or 'Br' substituted.

  • Elimination versus substitution

    • Elimination E2 is favoured by ethanol solvent and substitution is favoured by water/aqueous media.

    • Although ethanol is polar, it is less polar than water and a more polar solvent favours substitution.

    • However, although ethanolic KOH gives a larger yield of the elimination product than aqueous NaOH, the yield compared to that of the substitution product, may still be low because the structure of the halogenoalkane (haloalkane) greatly affects the relative amounts of substitution and elimination products (see next).

  • Effect of structure on elimination

    • Elimination E1 is more favoured by increase in alkyl groups attached to the carbon of the C-halogen bond, therefore yield of alkene increases in the order tert RX >sec RX > prim RX

    • This is due to the greater stability of the intermediate carbocation which is more likely to be formed in the order tert RX > sec RX > prim RX (as is the preference of the S_N1 mechanism!),

      • so carbocation stability is tert R₃C⁺ > sec R₂CH⁺ > prim RCH₂⁺ (R = alkyl),

      • therefore increasing stability of the carbocation, increases the chance of proton loss from the carbocation to give the alkene (see carbocation stability discussion).

  • Elimination E2 is favoured by an even stronger base, particularly in the less polar ethanol, than sodium hydroxide e.g. like potassium hydroxide since in the E2 mechanism the hydroxide ion abstracts a proton from the halogenoalkane.

    • Its worth noting that with weak bases like ammonia, very little elimination occurs.

  • Examples of data on con-current nucleophilic substitution versus elimination reactions

    • (i) Refluxing bromoethane (prim) with ethanolic sodium hydroxide (CH₃CH₂OH/NaOH) solution, gives 1% ethene (elimination product) and 99% ethanol (nucleophilic substitution product.

      • Refluxing with aqueous sodium hydroxide will give ~100% ethanol.

    • (ii) Refluxing 2-bromopropane (sec) with ethanolic sodium hydroxide gives 80% propene (elimination product) and 20% propan-2-ol (nucleophilic substitution product).

      • Comparing (i) and (ii), the secondary haloalkane gives a much higher yield of alkenes than a primary haloalkane.

      • With ethanolic potassium hydroxide, 2-bromobutane gives a mixture of but-2-ene and but-1-ene in the ratio 4:1, as well some butan-2-ol. 1-bromobutane would only give one alkene, but-1-ene, as well as butan-1-ol.

    • (iii) Refluxing 2-bromopropane (sec) with aqueous sodium hydroxide gives 5% propene (elimination product) and 95% propan-2-ol (nucleophilic substitution product).

      • Comparing (ii) and (iii), the aqueous medium gives a much lower yield of alkene.

    • (iv) Refluxing 2-bromo-2-methylpropane (tert) with aqueous NaOH gives 19% methylpropene (elimination product) and 81% 2-methyl-propan-2-ol (nucleophilic substitution product).

      • Even with an aqueous solvent, the tertiary haloalkane gives an appreciable yield of alkene.

      • Ethanolic KOH will give a much higher % of alkene (over 80% I would expect, comparing it to 2-bromopropane, but I couldn't find any data).

    • Note that in all example, using aqueous NaOH or KOH will reduce the alkene yields, i.e. more nucleophilic substitution, compare (ii) and (iii).

    • Further note that ethanol is often added to the aqueous NaOH or KOH to increase the solubility of the halogenoalkane.

  • Summary of elimination versus substitution:

    • The 'classic' conditions to maximise the yield of an alkene by elimination are refluxing the halogenoalkane with ethanolic potassium hydroxide.

    • Compared to water, it involves the less polar ethanol even though its refluxing at lower reaction temperature.

    • The method also uses the strongest 'common' base and further more, the yield of alkene will increase in the order tert RX > sec RX > prim RX.

keywords phrases: reaction conditions formula intermediates organic chemistry reaction mechanisms elimination R2CH-CBrR2 + KOH ==> R2C=CR2 + H2O + KBr R2CH-CBrR2 + OH- ==> R2C=CR2 + H2O + Br- R2CH-CBrR2 + NaOH ==> R2C=CR2 + H2O + NaBr R2CH-CBrR2 + NaOH ==> R2CH-C(OH)R2 + NaBr CH3CH2OH + OH- <=> CH3CH2O- + H2O tert RX >sec RX > prim RX tert R3C+ > sec R2CH+ > prim RCH2+

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APPENDIX

 COMPLETE MECHANISM and Organic Synthesis INDEX (so far!)

• PART 10 An Introduction to organic chemical reaction mechanisms and technical terms explained

• Organic Synthesis - MECHANISMS INDEX

• Part 10.2 ALKANES - an introduction to their chemistry

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

  • Free radical mechanism for cracking hydrocarbons to give shorter alkanes and alkenes

  • Ionic mechanism for cracking hydrocarbons

• Part 10.3 ALKENES - introduction to their chemistry

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

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

  • Free radical polymerisation to give poly(alkene) polymers e.g. ethene ==> poly(ethene)

  • Hydrogenation to give saturated alkanes

• Part 10.4 HALOGENOALKANES - introduction to the chemistry of 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, rate expression, 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 to their chemistry

  • Conversion of an alcohol to a halogenoalkane

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

• Part 10.6 Carbonyl compounds - ALDEHYDES and KETONES - introduction to their chemistry

  • 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

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

  • Hydrolysis of acid chlorides with water to give a carboxylic acid

  • Esterification of acid chlorides with alcohols to give an ester

  • Amide formation from reaction of acid chlorides 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)

• Other important associated links

• Isomerism - chain isomerism, positional isomerism, functional group isomerism & Tautomerism
• Stereoisomerism introduction, E/Z isomerism ('ex' Geometric/Geometrical cis/trans Isomerism)
• R/S Optical Isomerism and chiral auxiliary synthesis
• All Advanced Organic Chemistry Notes
• Summary of Organic Functional Groups
• Quiz on Organic Structure Recognition
• Summary of organic chemistry functional group tests
• The shapes and bond angles of simple organic molecules
• Aliphatic structure & nomenclature m/c BUMPER QUIZ
• Type in name aliphatic structure and nomenclature BUMPER QUIZ

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