Transcript Slide 1
Part 3ii Substitution Reactions: Solvent Me CN HH Cl Polar Protic Solvent DMSO MeOH SN1 Me Cl HH N C Polar Non-Protic Solvent MeOH DMSO SN2 Me CN HH Cl Content of Part 3ii Solvent polarity and its effect on the SN1 reaction mechanism Solvent polarity and its effect on the SN2 reaction mechanism Protic and non-protic solvents and their effect on the SN2 mechanism Protic and polar non-protic solvents and change of mechanism from SN1 to SN2 Increasing solvent polarity and change of mechanism from SN2 to SN1 CHM1C3 – Introduction to Chemical Reactivity of Organic Compounds– – Learning Objectives Part 3ii – Substitution Reactions: Solvent After completing PART 5ii of this course you should have an understanding of, and be able to demonstrate, the following terms, ideas and methods. (i) Understand how changes in the polarity of the solvent can change the rate of reaction in both SN2 and SN1 reaction mechanisms, (ii) Understand what is meant by the dielectric constant of a solvent, (iii) Understand what is meant by protic and non-protic solvents, (iv) Understand what a hydrogen bond is, (v) Understand how protic solvents reduce the rate of reaction of nucleophiles with substrates, and (vi) Understand how increasing the polarity of the solvent can change the mechanism of substitution from SN2 to SN1. Effect of Solvent Changing the solvent in which a reaction is carried out in can (i) Effect the reaction rate (ii) Result in a change in reaction mechanism Solvent Polarity and SN1 Reactions Relative Rate Br Solvent: OH Rate = k[R-Br] 1 H2O/EtOH SN1 Br Solvent: OH Rate = k[R-Br] H2O SN1 Why is reaction much faster in water alone? 30 000 Both reactions must involve the sp2 hybridised carbocation. The carbocation is obviously a very polar species. Polar species are most stable in polar solvents. The dielectric constant (e) of a solvent is a measure of the polarity of a solvent. The larger the e the more polar the solvent. e (H2O) = 78 e (EtOH) = 30 Thus, higher e solvent solvate and stabilise the carbocation more easily than lower e solvents. Thus, cations are generated more rapidly in higher e solvents. Solvent Polarity and SN2 Reactions Increasing solvent polarity on substrates that undergo SN2 substitution reactions has a much less dramatic effect. As there is no cation formed there is nothing to stabilise. However, it is found that there is usually a slight decrease in rate with increasing polarity, especially when the nucleophile is charged, i.e. an anion. In much the same way as a carbocation is stabilised by higher polarity solvents, so are anions. Thus, in higher polarity solvents the nucleophile is more solvated and more stable, and therefore less reactive and the rate drops in higher polarity solvents. Protic Solvents and Nucleophile Interaction A protic solvent is one in which there is hydrogen capable of forming a hydrogen bond to the nucleophile, i.e. a hydrogen attached to an electronegative heteroatom (O and N). e.g. H2O, MeOH, EtOH Me O H + :Nu This results in a highly solvated and stable nucleophile, and therefore relative unreactive. Non-Protic Solvents and the Nucleophile Interaction A non-protic solvent is one in which there is no hydrogen capable of forming a hydrogen bond to the nucleophile. This then results in the nucleophile being poorly solvated and relatively more reactive. O N O O O S Me Me DMSO O H Me2N DMF H MeCN Acetone Ethylacetate H O Cl Cl Cl Cl Cl H Chloroform Dichloromethane Et Et Ether Alkanes Rel. Rate H N3 I HH I HH H N3 Non-Protic Solvent DMF I Rate = k[R-I][N3-] 1 HH MeOH H N3 Protic Solvent SN2 H N3 I HH Rate = k[R-I][N3-] 45 000 SN2 It should be noted MeOH and DMF have similar e values, 33 and 37 respectively. Thus, any rate differences are not a result of solvating ability alone. - Me Polar Protic Solvent O H N + N N H O Hydrogen Bond DMF H Me Lone pairs are rendered less effective in attacking the electrophilic centre Polar Non-Protic Solvent O N CH3 CH3 None of the Hs are attached to electronegative atoms, therefore no H-bond donors, as above. Therefore, lone pairs on N3- are available for attaching the electrophilic centre Protic and Polar Non-Protic Solvents and Change of Mechanism from SN1 to SN2 Me MeOH Me Rate = k[R-Cl] Cl N C CN e = 33 HH DMSO Me HH HH e = 46 SN1 Rate = k[R-I][N3-] Me Cl N C Cl CN HH Cl SN2 MeOH:Protic, e = 33 Me MeOH hydrogen bonds to CNvery efficiently, thus making the CN- a relatively poor nucleophile, under these conditions. O H C N H Unreactive O The C-Cl bond then has time to fragment to generate the carbocation and Cl-. Therefore, reaction proceeds via SN1 mechanism DMSO:Non-Protic, e = 47 There are no strong hydrogen bonding interactions between the cyanide and the DMSO. Thus, the cyanide anion lone pairs are ‘naked’ and, therefore is a relatively good nuclephile. i.e. the cyanide can react with R-Cl before the C-Cl bond fragments. Thus, reaction proceeds via an SN2 mechanism DMSO O N C Reactive S H 3C CH3 Me Increasing Solvent Polarity and Change of Mechanism from SN2 to SN1 I- Solvent: Br I Br- Rate = k[I-][R-Br] SN2 Br- SN1 EtOEt e=3 Solvent: I- Br I Rate = k[R-Br] H2O e = 78 This is fairly simple to conceptualise: If your start with a alkyl halide in a low dielectric solvent then the carbocation will not be formed as the dielectric of the solvent will not support the polar carbocation. Thus, SN2 reaction mechanism will predominate at low e. As the polarity of the solvent increases it becomes increasingly easier to support a carbocation, and the I- is hydrogen bonded as is a poor nuclephile. Thus, SN1 reaction mechanism will predominate at high e. Polar Protic Solvent Polar Non-Protic Dielectric Constant H2O Solvent 80 Non-Polar Non-Protic Dielectric Constant HCONMe2 33 MeCOMe 21 acetone EtOH 24 1-PrOH MeCOEt 20 1-BuOH MeCO2Et 18 MeS(O)Me 18 6 47 DMSO 1 PeOH 15 MeNO2 39 nitromethane HCO2H 58 CH3CO2H 6 MeCN acetonitrile Polar? O O Me H NMe2 Me DMF dimethyformaldehyde O Me 9 CHCl3 5 Polar? CCl4 2 THF 9 tetrahydrofuran Et2O 3 CCl4 2 n-Octane 2 n-Hexane 2 n-Pentane 2 Benzene 2 37 acetonitrile O N CH2Cl2 N Me acetone Dielectric Constant 38 DMF MeOH Solvent All values rounded to nearest integer S O nitromethane Me Me DMSO dimethylsulfoxide O THF tetrahydrofuran http://www.asiinstruments.com/technical.asp High e solvents: Cation formation promoted (if Nu is poor, i.e. attacks slowly Low e solvents: Cation formation prevented SN1 Promoted Nu R X SN2 Promoted Nu R Polar Protic solvents: Hydrogen bond to Nu, Ineffective Nu, R-X is able to cleave, before Nu attacks SN1 Promoted R Nu X Nu R X Polar Non-protic solvents Nu lone pair not hydrogen bonded Effective Nu, and attacks before R-X cleaves SN2 Promoted X CHM1C3 – Introduction to Chemical Reactivity of Organic Compounds– – Summary Sheet Part 3ii – Substitution Reactions: Solvent If one considers an SN1 reaction in which a carbocation is formed, then increasing the polarity of the solvent will lead to better solvation of the carbocation, and therefore greater stability. Thus, increasing the solvent polarity from, for example EtOH (e = 30) to H2O (e = 78), the rate of reaction increases by several orders of magnitude as the stability of the carbocation is so much greater in H2O than EtOH. Indeed, a reaction in a low dielectric solvent may go via an SN2 reaction (i.e. no requirement for stabilisation of a carbocation), but when transferred to a higher dielectric solvent may change mechanism to SN1. Not only is the stabilisation of the carbocation important by the solvent, but also the interaction of the nucleophile with the solvent. Nucleophiles have by definition a lone pair of electrons for donation to electrophilic centres, i.e. the carbon attached to the leaving group. However, if the solvent has a hydrogen atom attached to a electronegative heteroatom (O, S, N, such as an alcohol, amine or thiol) then this hydrogen atom will carry a partial positive charge and interact as hydrogen bond acceptor with the lone pair of electrons on the nucleophile. Thus, the lone pair of electrons will now be less effective in a nucleophilic sense. Thus, SN2 mechanisms might be slowed down on transferring to protic solvents, or even change to SN1 if the media is even just slight more polar due to the increased time available for scission of the C-leaving group bond. Exercise 1: Substitution Reactions Identify the ethers B and C, and rationalise the difference in reaction outcomes. NaOEt NaOEt DMSO EtOH B Br rate = k[R-Br][NaOEt] A rate = k[R-Br] B+C Answer 1: Substitution Reactions Identify the ethers D and E, and rationalise the difference in reaction outcomes. NaOEt NaOEt DMSO EtOH B SN2 Br rate = k[R-Br][NaOEt] B+C SN1 rate = k[R-Br] A DMSO does not hydrogen bond to the EtO- lone pairs, making EtOa good nucleophile. Thus, EtO- attacks electrophilic centre before C-Br bond cleave. Me Me Br Br EtO Br Me carbocationic reactive intermediate Br OEt EtOH hydrogen bonds to the EtO-, making EtO- a poor nucleophile. Thus, C-Br bond cleaves before attack of EtO-. EtO Me Me Br EtO Me Me OEt EtO E D EtO