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Lab 7 SN1+ SN2 reactions

by: Puck Reeders

Lab 7 SN1+ SN2 reactions CHM 206

Puck Reeders
GPA 4.0
Organic Chemistry lab 2
T. Eve

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Complete organic chemistry 2 lab report for lab number 7 (SN1 and SN2 reactions) + the post lab questions
Organic Chemistry lab 2
T. Eve
Class Notes
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This 8 page Class Notes was uploaded by Puck Reeders on Thursday February 5, 2015. The Class Notes belongs to CHM 206 at Geneeskunde Universiteit Utrecht taught by T. Eve in Fall2014. Since its upload, it has received 796 views. For similar materials see Organic Chemistry lab 2 in Chemistry at Geneeskunde Universiteit Utrecht.

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Date Created: 02/05/15
SN 1 and SNZ reactions Puck Reeders Lab 7 Date September 16 2014 Objective To study the relationship of structure and the reactivity in nucleophilic substitution reactions by using qualitative tests of reactivity Introduction In SNlSN2 reactions a nucleophile is used to displace a leaving group If the substrate is an alkyl halide the nucleophilic substitution depends of the reaction conditions and the substrate structure Reactivity can be shown as 1 Leaving group a I gt Br gt C1 gt F 2 Steric Effect a Tertiary gt Secondary gt Primary 3 Solvent a Protic or Aprotic SN2 Mechanism This mechanism is a one step direct displacement process A reagent composed of N31 in Acetone favors the SN2 reaction Acetone is nonprotic and relatively non polar The Iodine anion is an excellent nucleophile NaBr or NaCl are both not soluble in acetone but NaI is Therefore When a substitution Will occur Where I39 is exchanged for C139 or Br39 there Will be a precipitation formed during the reaction Removing the precipitation makes the reaction move to the right RCl Na I39 I R1 NaCl s RBr Na I39 I R1 NaBr s SNl Mechanism A reagent composed of AgN03 favors the SNl reaction Ethanol is a protic and moderately ionizing solvent and the nitrate anion is a poor nucleophile The silver anion can coordinate With a leaving haline ion Which weakend the carbonhalogen bond so the alkyl halinde ionizes to form the alkyl carbocation Therefore the silver ion helps in the ionization the alkyl halide and also the silver halide formed is insoluble and precipitates Removing the precipitation makes the reaction move to the right R EtOHgt ROet RCl or Br I C139 or Br39 Aggt Ag Cl s Bulky molecules usually do not undergo SN2 reactions because steric hindrance prevents the backside attack from occurring Molecules that form primary carbocations generally do not undergo SNl reactions because the intermediate is too unstable The leaVing group on the alkyl halide also affects substitution reactions Weak bases are more stable bases and therefore make better leaVing groups The larger halides Br and I are better able to bear a negative charge weakermore stable bases and therefore make better leaVing groups than Cl and F In fact uorine usually does not undergo SNl reactions because it is too unstable Please see Table l for an explanation of the results of the experiment P roc e d u re 1 Make two sets of test tubes With the 9 different reagents 99 testtubes a l 39F PHPPPP 1 chlorobutane 1 bromobutane 2 chlorobutane 2 bromobutane bromocyclopentane bromocyclohexane tbutylchloride benzyl chloride bromobenzene 2 In the first set a Add NaIAcetone i Note in Which testtubes precipitation has formed ii Heat the ones Where no precipitation has formed to 50 C iii Note Whether or not precipitation has formed 3 In the second set a Add AgN03EtOH i Note in Which testtubes precipitation has formed ii Heat the ones Where no precipitation has formed to 90 C iii Note Whether or not precipitation has formed Results Table 1 Substrate Predictio Actual Why Reactivity 11 Result lchlorobutane SN2 Both slow Slowly Similar to 1 because Cl is bromobutane but Cl not the best not as good of leaving group leaving group more reactive 1bromobutane SN2 Both Quickly Similar to 1 SN2 faster because Br is chlorobutane but Br than SNl a good a better leaving leaving group group more reactive 2 chlorobutane SNl or SN2 SNl only SNl because Similar to 2 slow secondary bromobutane but Cl carbocation not as good of a leaving group less reactive 2 bromobutane SNl or SN2 Both Both because Similar to 2 SNl much secondary chlorobutane but Br faster carbocation a better leaving and Br is group more good leaving reactive group Bromocyclopentan SNl Both SNl faster Similar to e SNl because bromocyclohexane immediately secondary but less steric SN2 after carbocation hindrance heating SN2 slow because steric hindrance Bromocyclohexane SNl SNl quickly SNl only Similar to because too bromocyclopentane much steric but more steric hindrance for hindrance SN2 tbutylchloride SNl SNl quickly SNl only Not similar to others because too only tertiary halide much steric hindrance for SN2 and tertiary Stable carbocation Benzyl chloride SN2 Both Both because Similar reactivity to 8N2 more stable crotyl chloride quickly than carbocation because benzylic SNl intermediate and allylic halides delocalized are similar electrons and less steric Bromobenzene SN2 Neither Unreactive Not similar to others unreactive because aryl only aryl halide halide Crotyl chloride SNl Both Both because Similar reactiVity to Quickly stable benzyl chloride carbocation because benzylic delocalized and allylic halides electrons and are similar primary not much steric hindrance Conclusion The reactions were identified as either SN2 or SNl by carrying out the reaction and measuring the speeds at which they took place and whether they required hot water baths or ice baths to complete Further experiments would focus on proper heating and cooling of the reactions to ensure that temperature does not have a different effect on the reactions than intended Discussion A possible error in the experiment includes the transfer of reagent to the test tube incomplete transfer of the reagent could lower the concentration of the nucleophile and therefore slow down the rate of the SN2 reactions Another source of error was the heating of the test tubes It was difficult to keep the temperature of the water bath constant at 50 C or 90 C This would affect the rate of reactions lower temperature means slower reaction or stop it from occurring at all if the temperature was not high enough lchlorobutane lbromobutane 2bromobutane Benzyl chloride and Crotyl chloride reacted in both the SNl reaction with the NaI in Acetone and in the SN2 reaction with the AgN03 in EtOH Some reacted quicker in SNl and some reacted quicker in SN2 Bromobenzene did not react because it is an aryl halide 2chlorobutane Bromocyclohexane tbutylchloride took place quicker in SNl reaction because there is too much steric hinderance to undergo a 8N2 reaction Reference 1 Department of Chemistry University of Miami Chemistry 205206 Laboratory Manual 2nd Ed Hayden McNeil 20102012 PostLab Questions 1 For both the SNl and SN2 reactions 2bromobutane reacts faster than 2 chlorobutane because Br is a weaker base than Cl is This is because Br is a larger atom than Cl and so it is better able to bear a negative charge This makes it a more stable base and therefore a weaker base The weaker the basicity of a group the better its leaving ability so 2bromobutane reacts faster than 2 chlorobutane because it has a better leaving group 2 The Reactivity a Benzyl chloride is more reactive for both SNl and SN2 reactions because it is a benzylic halide This means that it will form a relatively stable carbocation intermediate for SNl reactions delocalized electrons and since it is a primary carbocation the backside of attack of an SN2 reaction will not be blocked by steric hindrance b Bromobenzene is nonreactive for under both conditions because it is an aryl halide Under SN2 conditions the nucleophile is repelled by the pi electron cloud of the aromatic ring Under SNl conditions it does not react because the carbocation intermediate would be more unstable than a primary carbocation the positive charge would be on an sp carbon which is more electronegative and therefore more resistant to a positive charge and because sp2 carbons form stronger bonds than sp3 carbons so it would be harder for the carbonbromine bond to break 3 Benzyl chloride reacts much faster than lchlorobutane because the benzyl chloride would form a more stable carbocation 4 1methyl lbromocyclohexane would react by SNl because it will form a tertiary carbocation more stable and because steric hindrance would block the backside attack of an SN2 reaction lbromopropane would react by SN2 because it would form an unstable primary carbocation too unstable for SNl 2bromohexane would react by both mechanisms It is stable enough to form a carbocation intermediate for the SNl mechanism secondary carbocation and it does not have enough steric hindrance to prevent an SN2 reaction 5 R2 bromo2methylhexane will react the fastest because it will form a tertiary carbocation which is the most stable Bromocyclopentane will react the slowest because although it is not a secondary carbocation it does not have delocalized electrons from double bonds like 4bromo13pentadiene 6 Bromocyclohexane reacts faster than chlorocylohexane because Br is a weaker base than Cl This is because Br is a larger atom than Cl and so it is better able to bear a negative charge This makes it a more stable base ie a weaker base The weaker the basicity of a group the better its leaving ability so bromocyclohexane reacts faster than chlorocyclohexane because it has a better leaving group 7 1chlorobutane NaI I 1iodobutane NaCl 90mL d Acetone 90mL 07925 gmL 71325 g Acetone 15 g NaI100 g acetone X g NaI71325 g acetone I X 107 g NaI 107 g NaI 1mol NaI1499 g NaI 0719 mol NaI 25mL 886gmL 1mol9245 g 024 g 1chlorobutane 1chlorobutane is limiting reagent because there are only 024 moles compared to 0714 moles NaI yield 13g44g 295 8 This is true because a more polar solvent will help stabilize the transition state and will increase the rate of the reaction 9 True because the rate kalkyl halide nucleophile because both reactants are involved in transition state of ratedetermining step bimolecular step


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