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

Nucleophilic substitution of halogenoalkane by hydroxide ion (hydrolysis with OH^-)

mechanism 1 - nucleophilic substitution of a halogenoalkane by hydroxide ion (S_N1 'unimolecular' via carbocation)

• S_N1 unimolecular, a two step ionic mechanism via carbocation formation [mechanism 1 above]

  • In step (1) the C^δ+-X^δ- polar bond of the halogenoalkane splits heterolytically to form a carbocation and a free halide ion (X^-, e.g. chloride or bromide) and this is a reversible reaction. The C-Cl bond is most likely to break because it is a weaker bond than the C-C or C-H bond AND breaks heterolytically because of the big difference in electronegativity between carbon (2.1) and chlorine (3.0) giving a polar C^δ+-Cl^δ- bond. So the C-Cl bonding pair of electrons leaves with the Cl atom as the Cl^- ion.

  • In step (2) the hydroxide ion is a negative electron pair donor and rapidly combines with the carbocation, forming the C-O bond in the alcohol product.

  •  Step (1) is the rate determining step with the much larger activation energy (see reaction profile diagram 45)

  • This mechanism is most likely with tertiary halogenoalkanes. Primary halogenoalkanes tend to react by the S_N2 mechanism NOT involving a carbocation. Secondary halogenoalkanes react via both mechanisms at the same time! The relative rate order for the S_N1 mechanism is tert > sec > prim i.e. following the pattern of decreasing stability of the carbocation (see extra discussion points).

style (a)

mechanism 2 - nucleophilic substitution of a halogenoalkane by hydroxide ion (S_N2 'bimolecular')

style (b)

mechanism 34 - nucleophilic substitution of a halogenoalkane by hydroxide ion (S_N2 'bimolecular')

style (c)

mechanism 33 - nucleophilic substitution of a halogenoalkane by hydroxide ion (S_N2 'bimolecular')

• S_N2 'bimolecular', a one step bimolecular collision mechanism [mechanisms 2, 34 and 33 above]

  • The C^δ+-X^δ- bond is polar because of the difference in electronegativity between carbon (2.1) and chlorine (3.0), so the electron rich nucleophile, the hydroxide ion, attacks the slightly positive carbon.

  • The nucleophile acts as an electron pair donor (Lewis base) to bond with the C^δ+ carbon to make the C-O bond in the newly formed C-OH alcohol group.

  • Simultaneously the chlorine atom is ejected, taking with it the C-X bond pair, so forming the chloride ion on expulsion.

  • This mechanism is most likely with primary halogenoalkanes. Tertiary halogenoalkanes tend to react by the S_N1 mechanism involving a carbocation, secondary halogenoalkanes react via both mechanisms (see extra discussion points).

• Diagram styles and explanation:

  • Style (a) Does not show the 'activated complex' or 'transition state' at the point where the hydroxide ion ('incoming') and the chloride ion ('outgoing') are 'half-bonded' to the central carbon atom of the functional group.

  • Style (b) Shows a simplified version of the 'activated complex' or 'transition state'. The blue ^.... represent 'half-bonds' (not weak inter-molecular forces) in the sense that the hydroxy/hydroxide group is 'coming in' and the chloro/chloride group is 'going out'!

    • The change can also be represented as a simple reaction progress-profile diagram 41.

    • Note that 'activated complex' or 'transition state' is not the same as an intermediate like a carbocation which is a definite entity in its own right, however short its lifetime.

  • Style (c)  Shows the reaction in terms of the stereochemistry. This is not a particularly important 'style' unless the original halogenoalkane is chiral (see note on chirality below).

Also read the FURTHER COMMENTS for this reaction * mechanism index

mechanism 10 - nucleophilic substitution of a halogenoalkane by water (S_N1 unimolecular via carbocation)

• S_N1 unimolecular, a two step ionic mechanism via carbocation formation. [mechanism 10 above]

  • In step (1) the halogenoalkane splits heterolytically to form a carbocation and a free halide ion (e.g. chloride or bromide) and this is a reversible reaction. The C-X bond is most likely to break because it is a weaker bond than C-C or C-H AND breaks heterolytically because of the big difference in negativity between carbon (2.1) and chlorine (3.0) giving a polar C^δ+-X^δ- bond. So the C-Cl bonding pair of electrons leaves with the Cl atom as the Cl^- ion.  Step (1) is the rate determining step (rds) and the rate effectively only depends on the halogenoalkane concentration.

  • In step (2) the water is an electron pair donor and rapidly combines with the carbocation, forming the C-O bond in the protonated alcohol.

  • In step (3) a water molecules rapidly removes a proton to leave the free alcohol product.

  • The overall change can also be represented as a reaction progress-profile diagram 42.

  • This mechanism above is most likely with tertiary halogenoalkanes. Primary halogenoalkanes tend to react by the S_N2 mechanism NOT involving a carbocation, though very slow with water if it reacts at all. Secondary halogenoalkanes tend to react via both mechanisms at the same time (see extra discussion points).

mechanism 35 nucleophilic substitution of a halogenoalkane by water (S_N2 bimolecular)

• S_N2 'bimolecular', a two step bimolecular collision mechanism but NOT involving a carbocation. [mechanism 35 above]

  • Step (1): The C^δ+-X^δ- bond is polar because of the difference in electronegativity between carbon (2.1) and chlorine (3.0), so the electron rich nucleophile, the water molecule, attacks the slightly positive carbon. The nucleophilic water acts as an electron pair donor (Lewis base) to bond with the 'delta positive' carbon to give the C-O bond of the protonated alcohol. Simultaneously the chlorine atom is ejected, taking with it the C-X bond pair, so forming the chloride ion on expulsion. Step (1) is the rate determining step (rds) and the rate effectively only depends on the halogenoalkane concentration.

  • In step (2) another water molecule rapidly accepts the proton from the protonated alcohol to leave the free alcohol product.

  • The reaction profile would be similar to diagram 45 

    • BUT the intermediate is [R₃COH₂]⁺ NOT an R₃C⁺ carbocation.

  • This mechanism is most likely with primary halogenoalkanes, but very slow if at all with water, much faster with the hydroxide ion, a more powerful nucleophile (negative ion as well as an electron pair donor). Tertiary halogenoalkanes tend to react by the S_N1 mechanism involving a carbocation, secondary halogenoalkanes react via both mechanisms (see extra discussion points).

• FURTHER COMMENTS on these halogenoalkane (haloalkane) nucleophilic substitution reactions

  • Comparison of aliphatic halogenoalkane and aromatic halide hydrolysis.

    • Halogenoalkanes hydrolyse much more readily than aryl halides which are aromatic compounds with a halogen atom directly bonded to the benzene ring. The carbon-halogen bond is stronger and less polar in aromatic compounds compared to halogenoalkanes. This is because in the aromatic halogen compounds there is some overlap/delocalisation of the lone pairs of electrons of the chlorine with the ∏ electrons of the benzene ring. This makes the aromatic ring C-Hal bond stronger and less easy to break, hence less polar and less susceptible to nucleophilic attack.

      • However, if the halogen is in an alkyl side chain off the benzene ring, hydrolysis will readily take place.

        • e.g. refluxing chloromethylbenzene (phenylchloromethane, above left) with aqueous/ethanolic sodium hydroxide gives phenylmethanol.

        • C₆H₅CH₂Cl + NaOH ==> C₆H₅CH₂OH + NaCl

      • or or Whereas hydrolysis of any of the three other isomeric aromatic halogen compounds, namely chloro-2/3/4-methylbenzene (above left) where the Cl atom is attached to the ring, is very difficult to achieve to produce a phenol (where the OH hydroxy group is directly attached to a benzene ring) by hydrolysis.

      • ClC₆H₄CH₃ + NaOH ==> HOC₆H₄CH₃ + NaCl

      • is NOT EASY! though it can be brought about at higher temperatures and high pressures than normal reflux conditions in the school laboratory, and in an 'industrial' sealed reaction vessel.

  • Rate of reaction, OH^- versus H₂O:

    • Only applies to the S_N2 mechanism: Irrespective of the structure of the halogenoalkane, the hydrolysis will be faster with aqueous sodium hydroxide than just water because the hydroxide ion is a more powerful nucleophile. Although they are both electron pair donors, the hydroxide ion carries a full negative charge compared to the electrically neutral water molecule.

    • The rate determining formation of the carbocation in the S_N1 mechanisms means the rate is not affected by using alkali or just water (read about the kinetics below).

  • Reaction kinetics: The possibility of two reaction mechanisms has consequences for the rate expressions when the rates of halogenoalkane (RX) nucleophilic substitution reactions are studied.

  • The S_N1 mechanism is referred to as a 'unimolecular', despite it being a two/three step mechanism of bimolecular collisions, because the rate is only dependent on one reactant, the R₃C-X is shown in step (1), but it still has to collide with the solvent!

    • [see mechanism 1 or mechanism 10 and reaction profile 42]

    • Experimental results produce the overall 1st order rate expression: rate = k₁[RX] 

    • This is because the activation energy of the 1st step, forming the carbocation by heterolytic bond fission, is so high, that the speed is relatively low, so step (1) alone determines the speed of the reaction. This is referred to as the rate determining step (or rds in shorthand!). Step (2) has a much lower activation energy and is much faster (see reaction profile diagram 42). You would register zero order for the order of reaction with respect to sodium hydroxide (or more specifically, the hydroxide ion concentration).

  • The S_N2 mechanism is referred to as a 'bimolecular'.

    • [see mechanisms 2/33/34 or mechanism 35 and reaction profile 41]

    • Experimental results produce the overall 2nd order rate expression: rate = k₂[RX][OH] 

    • This is because it is a one step mechanism involving the bimolecular collision of the two reactant molecules/ions (see also reaction profile diagram 41). The rate depends on both the halogenoalkane and hydroxide ion concentrations (individually, 1st order with respect to both). 

  • The relative reactivity of the C-X bond, where X = F, Cl, Br or I.

    • In general the reactivity order is C-I > C-Br > C-Cl > C-F.

    • This is due to the decrease in bond enthalpy as the halogen atom radius increases the bond gets longer and weaker, i.e. easier to form the carbocation in the S_N1 mechanism or easier to release a X^- ion from the 'activated complex' in the S_N2 mechanism.

      • Average bond enthalpies/kJmol^-1: C-F 484, C-Cl 338, C-Br 276, C-I 238

    • This bond strength factor overrides any increase in bond polarity, which on face value, since the order of becoming less polar is: C-F > C-Cl > C-Br > C-I, which would suggest an increase in susceptibility to nucleophilic attack at the delta positive carbon, but that's not what is observed!

      • So bond enthalpies override the bond polarity trend.

  • Carbon chain structure and relative reactivity.

    • Generally speaking the reactivity order is tert > sec > prim halogenoalkane, but unfortunately it isn't quite that simple!

      • For the S_N1 mechanism the reactivity trend is tert > sec > prim halogenoalkane.

        • The reason for this is that alkyl groups have a small electron charge donating inductive effect (+I effect). Its actually the positive charge on the carbon (from the C-X bond) attracting the electron charge of the alkyl group(s) which helps stabilise the carbocation by 'spreading' the charge and lowering the potential energy of the intermediate carbocation.

        • So a typical reactivity series is (CH₃)₃CBr > (CH₃)₂CHBr > CH₃CH₂Br > CH₃Br

          • because the order of carbocation stability would be ...

          • (CH₃)₃C⁺ > (CH₃)₂CH⁺ > CH₃CH₂⁺ > CH₃⁺

          • where 3, 2, 1 and 0 methyl groups are available to stabilise the carbocation.

        • S_N1 is also favoured by polar solvents like water or ethanol, which help stabilise the carbocation by solvation.

      • For the S_N2 mechanism the reactivity trend is prim > sec > tert halogenoalkane because of steric hindrance, that is alkyl groups get in the way of the incoming nucleophile such as the hydroxide ion or water molecule. It is usually very slow with water, but no problem if the R₃C-X is warmed/refluxed with aqueous sodium hydroxide, though some alcohol can be added to increase the solubility of R₃C-X to increase the rate of hydrolysis.

      • The S_N1 mechanism  is favoured by tertiary halogenoalkanes because of the greater stability of tertiary carbocations formed.

      • Secondary halogenoalkanes often react by either the S_N1 or S_N2 mechanisms simultaneously.

      • Primary halogenoalkanes tend to react by the S_N2 mechanism, because on heterolytic C-X bond fission, they would form the least stable primary carbocations.

      • For a given halogen, overall tertiary halogenoalkanes tend to be the most reactive and tend to undergo nucleophilic substitution via the S_N1 carbocation mechanism and primary halogenoalkanes tend to be the least reactive and react via the S_N2 mechanism BUT it is a very complicated situation!

  • Con-current nucleophilic substitution AND elimination reactions are discussed at the end of the section on the elimination mechanism where things get even more complicated.

  • Some points concerning the STEREOCHEMISTRY of the reaction :-

  • What happens if the original haloalkane has chirality?

  • AND what is the optical activity of the product?

    • If the haloalkane has three different R groups on the carbon of the C-Hal bond, i.e. RR'R''CX, then there are four different groups bonded to the carbon of the C-Hal bond. Therefore the molecule is chiral and can exhibit optical isomerism (non-super imposable mirror image forms). If the initial halogenoalkane is an optical isomer, the stereochemical consequences depend on which mechanism by which the halogenoalkane reacts. The results can be very complex and their full explanation goes beyond this level. However there are two product formation trends.

      • 

      • (Apologies for repeating diagrams but it helps to appreciate these stereochemical points)

      • In the S_N1 carbocation mechanism (e.g. mechanism 1 above), the three bonds of the R groups of the carbocation formed in step (1), are in a trigonal planar arrangement >C-. This means the nucleophile (e.g. OH^- or H₂O) can attack the carbocation with equal probability on each side. This results in a tendency for a racemic mixture to form, that is an optically inactive mixture of equal amounts of the two optical isomers.

        • (What you actually get in practice is a significant reduction in optical activity in the product)

      • For an optically active halogenoalkane reactant, the considerable reduction in optical activity in a nucleophilic substitution reaction is evidence of the S_N1 unimolecular mechanism i.e. the formation of a trigonal planar carbocation.

      • 

      • However, in the case of the S_N2 mechanism (e.g. mechanism 33 above), racemisation does NOT take place and chirality and optical activity is completely preserved in the molecule, BUT inversion takes place i.e. the absolute 3D configuration of the product is completely opposite to that of the reactant. Stereochemically the most successful line of attack for S_N2 substitution, is if the nucleophile hits the carbon of the C-Hal bond on the opposite side to the halogen atom. The result has been likened to an umbrella being blown inside out in a gale! The three single bonds for the -CRR'R'' are pushed through and so the configuration inverted! [see mechanism 33 style (c)]

      • For an optically active halogenoalkane reactant, the retention of complete optical activity in a nucleophilic substitution reaction is evidence of the S_N2 bimolecular mechanism.

  • Reaction progress profiles to show the progress of S_N1 and S_N2 nucleophilic substitutions.

  S_N1 with OH^-

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

• E2 mechanism via single step bimolecular collision process. [mechanism 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, 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. RX = halogenoalkane/haloalkane/alkyl halide/halogenated alkane.

  • i.e. nucleophilic substitution to give an alcohol versus elimination 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.

    • e.g. R₂CH-CBrR₂ + NaOH ==> R₂C=CR₂ + H₂O + NaBr

    • as well as/versus R₂CH-CBrR₂ + NaOH ==> R₂CH-C(OH)R₂ + NaBr

      • In both cases R = H, alkyl or aryl.

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

    • One explanation is that in ethanol, the ethoxide ion is formed, and is a stronger base than water, hence better at abstracting a proton in the E2 mechanism and competes with the nucleophilic attack of the OH^- ion at the C^δ+-X^δ- bond in the S_N2 mechanism.

      • CH₃CH₂OH + OH^- CH₃CH₂O^- + H₂O

    • Also, although ethanol is polar, it is less polar than water and a more polar solvent favours substitution. A lower temperature also favours elimination and ethanol boils at 79^oC, 21^oC lower than water (bpt. 100^oC).

  • 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 with 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).

    • e.g. (i) after refluxing ethanolic sodium hydroxide (CH₃CH₂OH/NaOH), bromoethane (prim) gives 1% ethene, whereas 2-bromopropane (sec) gives 80% propene. (ii) At 80^oC in aqueous-ethanol/NaOH 2-bromo-2-methylpropane (tert) gives 19% methylpropene and 2-bromopropane (sec) gives 5% propene. Note in the 2nd example the presence of water significantly reduces the alkene yields, but the tert yield is greater than the sec.

  • 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.

    • See comments under using ethanol solvent, and probably more significant for prim > sec > tert halogenoalkanes.

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

  • Summing up: The 'classic' conditions to maximise the yield of an alkene are refluxing the halogenoalkane with ethanolic potassium hydroxide. Compared to water, it involves the less polar ethanol, a lower reaction temperature. The method also uses the strongest 'common' base and further more, the yield of alkene will increase for tert RX > sec RX > prim RX.

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