1. a)%i.%
O
OH
O
O
O
S
H2O
H
OH
N
%
SN2%reactions%are%most%hindered%by%polar%protic%solvents.%Polar%protic%
solvents%are%those%that%can%participate%in%hydrogen%bonding%and%stabilize,%and%
thus%weaken,%nucleophiles.%The%third,%fifth,%and%sixth%structures%cannot%
participate%in%this%hydrogen%bonding,%so%they%are%polar%aprotic%solvents,%and%
are%thus%better%solvents%for%SN2%reactions.%
%
ii.%
O
O Na
O
NaCN
O
OH
NaOEt
MeSH
H2O
SN2%reactions%occur%the%fastest%with%strong,%negatively%charged,%unhindered%
nucleophiles,%but%can%also%occur%with%moderate%nucleophiles.%They%will%
proceed%VERY%slowly,%if%at%all,%with%weak%nucleophiles.%%
%
NaCN%and%NaOEt%are%both%strong%nucleophiles%because%they%are%sterically%
unhindered%and%negatively%charged.%%
%
tBuONa,%acetic%acid,%and%water%are%all%poor%nucleophiles.%While%tBuONa%is%
negatively%charged,%it%is%too%sterically%hindered%to%act%as%a%good%nucleophile.%%
%
MeSH%and%acetate%are%both%moderate%nucleophiles,%so%they%were%neither%
circled%nor%crossed%out.%Remember%that%acetate%is%a%moderate%nucleophile,%
despite%the%negative%charge,%because%that%negative%charge%is%delocalized.%
%
For%a%summary%of%nucleophilicity%trends,%refer%to%Table%9.7%on%page%360%of%the%
textbook.%
%
b)%Remember%that%SN2%mechanisms%favor%polar%aprotic%solvents,%strong,%
unhindered%nucleophiles,%and%good%leaving%groups%on%an%unhindered%
electrophile.%SN1%mechanisms%favor%stable%carbocation%formation%after%a%
leaving%group%has%left%and%polar%protic%solvents.%
%%
NaCN
No Reaction
DMSO
OH
i.%
%
%
This%would%have%been%a%good%SN2%reaction,%as%it%has%a%strong%nucleophile%and%
a%polar%aprotic%solvent.%OHS%is%not%a%good%leaving%group,%however,%so%this%
would%not%proceed%via%an%SN2%mechanism.%
%
Br
NaCN
DMF
No Reaction
ii.%
%
This%reaction%would%not%proceed%as%SN1%or%SN2%because%it%has%a%primary%
leaving%group,%which%would%favor%SN2,%but%it%has%3˚%betaSbranching,%making%it%
too%hindered%for%an%SN2%reaction%to%take%place.%%
%
Br
NaCN
CN
CN
+
Et
Et
Et
Ph
Ph
Ph
iii.%
%
This%reaction%follows%an%SN1%mechanism.%It%has%a%good%leaving%group%that%
forms%a%stable%tertiary%carbocation%intermediate,%as%well%as%a%polar%protic%
solvent.%Remember%that%stereochemistry%is%scrambled%in%SN1%reactions.%
%
Br
OPh
NaOPh
DMSO
iv.%%
%
This%reaction%follows%an%SN2%mechanism,%as%it%has%a%good%leaving%group,%a%
strong%nucleophile,%and%a%polar%aprotic%solvent.%Remember%that%
stereochemistry%is%inverted%in%a%typical%SN2%mechanism!%
%
%
%
%
%
%
%
%
%
%
%
%
%
%
Trends:
a) Circle the electrophiles that will undergo an Sn2 reaction. Box the electrophiles that will
undergo an Sn1 reaction.
The fist structure has the leaving group on an sp2 carbon, but substitution reaction can
only occur on sp3 electrophilic carbons. Sn2 reactions require primary or secondary carbons as
electrophiles. This means that only the third and fourth structures could undergo Sn2 reactions.
The second structure cannot undergo an Sn2 reaction because there is beta-branching, which will
block the approach of the nucleophile.
Sn1 reactions proceed through a carbocation intermediate, and therefore only occur on
electrophilic carbons that can stabilize a carbocation (secondary, tertiary and benzylic).
Therefore, the second, third, and fifth structures could undergo Sn1 reactions.
i.
Which of the circled electrophiles will undergo the fastest Sn2 reaction?
The fourth structure will undergo the fastest Sn2 reaction since primary carbons have the most
backside access.
ii.
Which of the boxed electrophiles will undergo the fastest Sn1 reaction?
The fifth structure will undergo the fastest Sn1 reaction because it offers the most carbocation
stability, as it is tertiary benzylic.
b) Circle the electrophiles with good leaving groups.
A good leaving group is something that will be stable once it has left. This means that a good
leaving group is generally the conjugate base of a strong acid. Definitions vary, but good leaving
groups generally have an acid form with a pka less than or equal to 0.
HI has a pka of about -10
H2O has a pka of 16
Carboxylic acids have a pka of about 5
NH3 has a pka of about 35
Me2OH+ has a pka of about -3
Therefore, only the first and fifth structures have good leaving groups.
c) Circle the solvents that would be used in an Sn2 reaction. Box the solvents that would be
used in an Sn1 reaction.
Sn2 reactions occur most readily in polar aprotic (not proton-donating) solvents. Sn1 reactions
occur most readily in polar protic (proton-donating) solvents. Generally, the solvent for the Sn1
reaction also serves as the nucleophile. Polar protic solvent are bad for Sn2 reactions since they
hydrogen bond to the nucleophile, reducing its reactivity. The first (THF), third (DMSO), and
fifth (DCM) are polar aprotic solvents. The second (methanol) and the fourth (acetate) are polar
protic solvents.
d) Are the following reactions Sn2 or Sn1?
i.
Which reaction will be faster and why?
These two reactions are Sn2 reactions, since they employ a strong nucleophile against a good
leaving group on a secondary carbon in a polar protic solvent. In Sn2 reactions, it is imperative
that the nucleophile approaches the leaving
group from 180 degrees behind it (known as
‘back side attack’). Converting both starting
materials into chair structures shows that, in
the first, the bromine is equatorial and, in the
second, the bromine is axial. When the
bromine is equatorial, the axial hydrogens on
the underside of the ring block the attack on
the nucleophile and slow down the reaction
significantly.
In the second structure, the bromine is
axial. This allows the CN- the approach the
backside of the electrophilic carbon easily.
The second reaction is faster.
3)Let’sgothroughthesereactionsonebyone,startingatthetopandgoing
clockwise.
1) O3
2) Me2S
O
H
O
Thisisaclassicozonlysisreaction.Topredicttheproduct,allweneedtodoisbreak
thecarbon-carbondoublebondandreplacethemwithcarbon-oxygendouble
bonds.
H2
Pt
Hydrogenonplatinum(andsomeothertransitionmetals)willreducealkenes,
resultinginsynadditionoftwohydrogenatoms(thestereochemistryisn’t
importantforthisreactionsincetheproductisachiral).
1) BH3
2) H2O2, NaOH
OH
+ enantiomer Weknowseveraldifferentwaystoaddanalcoholgrouptoanalkene,butbecareful-
notallofthemhavetherightregiochemistryorstereochemistry.Forthisreaction,
weseethatthealcoholisattachedtothelesssubstitutedcarbononthealkene.This
meansthatweneedanti-Markovnikovaddition-hydroboration!Let’sdoublecheck
thestereochemistry.Weseethatforthedrawnproduct,thehydroxideand
hydrogenattachedfromthetopofthemolecule(samesideaddition=>syn)andin
theenantiomer,thehydroxideandhydrogenattachedfromthebottomofthe
molecule(stillsamesideaddition=>syn).Thisleaveshydroborationasthebest
choicetosynthesizethegivenproduct.
1) OsO4
OH
2) H2O2
OH
+enantiomer
Osmiumtetroxidegivessynadditionoftwoalcoholstoanalkene.Thedrawn
producthasbothalcoholsabovethecyclohexanewhiletheenantiomerhasthe
alcoholsbelowthecyclohexane.
1) Hg(OAc)2, H2O
2) NaBH4
OH
OR
H2SO 4, H2O
Thistimewhendeterminingreactants,weseethatwehaveMarkovnikovaddition
ofthealcoholandnostereochemicalrestrictions.Bothoxymercuration/reduction
andacid-catalyzedhydrationwillproducethedesiredproduct.
Thisnextoneisalittlebitmoretricky-let’slookatwhathappensaftereachstepof
thereaction.
1) HBr
Br
2) H2O
OH
First,westartwithclassicHXaddition.Wehaveahydrogenatomaddedby
Markovnikov’sruleattachedtoonesideofthealkeneandagoodleavinggroup(our
bromine)attachedtotheother.We’realsoaddinginwater,whichisadecent
nucleophile.Goodleavinggroup(attachedtoatertiarycarbon)+decentnucleophile
(that’salsoaweakbase)=SN1!It’sgoodtorememberthatSN1reactionsscramble
stereochemistry,eventhoughitdoesn’tmatterinthisparticularproblem(our
productisachiral).
4)Fillintheemptyboxeswiththeappropriateproductorreagent.
1) BH3
TsCl
2) H2O2, NaOH
CH2Cl2
D
NaOMe
NaH
MeOH
Let’stakethisboxproblemonestepatatime.Mostofthereactantsarefilledinfor
us,soallwehavetodoispredicttheproduct.Let’slookatthefirststep:
1) BH3
2) H2O2, NaOH
D
Thereactantsshouldtellusthatwe’reusinghydroboration-oxidation,soweneedto
addanalcohol.Hydroborationhasanti-Markovnikovregioselectivity,soweshould
addtheOHatthelesssubstitutedcarbon.Wemustalsoconsiderstereochemistry-
hydroborationcausessynaddition,sowecanexpectthealcoholandprotonto
attachtothesamesideofthealkene.Wegettheleftproductiftheboronattacks
fromthetopofthealkene(comingoutofthepage)andtherightproductifthe
boronattacksfromthebottomofthealkene(goingintothepage).Thesetwo
moleculesareenantiomers,soit’sokaytodrawoneofthemandwrite
“+enantiomer”.
OH
OH
D
H
H
D
Let’slookatthenextreaction:
OH
D
H
+enantiomer
TsCl
CH2Cl2
Tosylchlorideisausefulreagentthatletsusreplacehydroxides(normallyavery
poorleavinggroup)withatosylate(whichisaverygoodleavinggroup).It’snot
reactingwithastereocenterinourstartingmaterialandit’snotproducinganynew
stereocentersintheproduct,sowedon’tneedtoworryaboutthestereochemistry
ofthisreaction.
Theproductofthisreaction:
OTs
D
H
+enantiomer Let’scontinuewiththisproductfornow(we’llgetbacktothesodiumhydride
reactioninaminute).
OTs
D
H
NaOMe
MeOH
+enantiomer
Rememberhowtosylatesaregoodleavinggroups?Methoxideisastrong
nucleophile.Astrongnucleophile+terminalleavinggroupmeansthatwe’relooking
foranSN2reaction.Methoxidereplacestosylateandinvertsthestereochemistry
(thoughthatdoesn’taffecttheproductsofthisreaction).Afterdeprotonatingthe
oxygen,we’releftwithanether:
O
D
H
+enantiomer Let’sreturntothatsodiumhydridereaction:
OH
D
H
NaH
+enantiomer
Thinkbacktoacid-basechemistry.H2isaveryweakacid,soitsconjugatebase(H-)
mustbeverystrong.Whenanalcoholisinthepresenceofaverystrongbase,itcan
actlikeanacidandgiveupitsproton-recallthattheprotonontheoxygenisthe
mostacidic,sothat’stheprotonthat’sremoved.Thehydrideandprotonuniteto
formhydrogengas,whichbubblesoutofthesolutionanddrivesthereactionto
completion.
O
D
H
+enantiomer Nowwecanlookatthelastreaction-theonewhereyouaren’tgiventhestarting
material,youaren’tgiventheproduct,andyouaren’tgiventhereactionconditions,
andyetyou’restillsomehowexpectedtogettherightanswer.
Weknowthestartingmaterialandproductfrompreviousreactionswe’vedonein
thisproblem,soallthat’slefttosolvearethereactionconditions.Let’sdrawout
whatweknowfirst.
O
D
H
+enantiomer
O
D
H
+enantiomer Goodnews:wedon’thavetoworryaboutstereochemistry,sincetheonly
stereocenterinthismoleculeisn’tparticipatinginthereaction.
Badnews:wehavetofigureouthowtoturnthisalkoxideintoanether.Sincewe
don’talreadyknowanyreactionsthatcandothis,weshouldconsideranSN1or
SN2reaction.Ifhydroxideisabadleavinggroup,though,thenoxideisevenworse,
sowecan’tusethatoxygenasourleavinggroup.However,alkoxidesarevery
powerfulnucleophiles,soallweneedisareagentwithagoodleavinggroupand
we’reset.Methylchloride,methylbromide,andmethyliodideareallgoodchoices.
CH3I
O
D
O
H
+enantiomer
D
H
+enantiomer ThisisanSN2reaction,sincemethylcarbocationsareneverformed.Thismeansthat
ifwewerereactingatastereocenter,wewouldneedtoinvertthestereochemistry.
Thewholeboxproblemdrawnout:
1) BH3
2) H2O2, NaOH
OH
D
H
D
+enantiomer
TsCl
CH2Cl2
OTs
D
H
+enantiomer
NaOMe
NaH
MeOH
O
D
H
CH3I
O
D
+enantiomer
H
+enantiomer
5)
Draw a mechanism for the following reaction:
As with any mechanism, let’s start by numbering our carbons and making a ‘bonds made’ and
‘bonds broken’ list. Additionally, lets assign the absolute stereochem of the chiral center in our
reactant and product.
We know that this reaction likely proceeds through an Sn2
mechanism since 1) there is inversion at C2, 2) the reaction is
run in DMSO, and 3) our nucleophile (HS-) is a strong
nucleophile.
Bonds Made
S-C5
S-C2
Bonds Broken
C2-Br
C5-Br
S-H
There are two electrophilic centers in this molecule, both with the same leaving group (Br). The
one on the right is primary, while the one on the left is secondary. Since primary electrophiles
react most readily in an Sn2 reaction, we will attack this center first.
In order to regenerate a strong nucleophile, -SH will be deprotonated by excess HS-.
Now we need to invert the stereochemistry of C2 and close the ring. This is done by and Sn2
reaction between C2 and the terminal thiol ion.
The full mechanism is shown below:
(9) Reaction Schemes.
Let’s look at these schemes one at a time.
Scheme A: The first reaction involves Br2 and MeOH. From what we know about the halohydrin reaction,
we should remember that this is an anti-addition reaction. This means that the –Br and –OMe
substituents should end up on opposite sides of the ring. A pair of enantiomers should be produced. So
far we have:
Now that we have intermediate A, we can find the product. The first question is to identify the reactivity
of the reagent NaSPh. Na+ is a common positive counter ion, which means that reagent is really –SPh,
with the negative charge on sulfur. Sulfurs with negative charges are good nucleophiles, which means
that the reaction should proceed through an Sn2-type mechanism. What about the substrate?
Intermediate A has a Br, which is a good leaving group, on a secondary carbon. Since SN2 reactions can
occur on 2o carbons, this should be an SN2 reaction. SN2 reactions invert the stereocenter, so the product
should show the –SPh substuent with a dashed bond.
The correct Product A is shown above. This does not match the proposed product which shows the –SPH
coming out the page (wedged bond), so the proposed product is incorrect.
Scheme B: The first step in this scheme is oxymercuration, which we know adds an alcohol to the more
substituted carbon. The intermediate B should look like:
The second reaction uses HCl and heat as reagents. The next question is, what type of reaction is this?
We have a tertiary alcohol for our substrate with a strong acid for our reagent. The tertiary carbon
center rules out the possibility of SN2 due to steric crowding. We also know that in strong acid, alcohols
can be protonated to become a good leaving group. After HCl dissociates into H+ and Cl-, the Cl- can act
as a weak nucleophile. All of these factors point to an SN1 reaction. Therefore, product B should be:
Scheme C: The first reaction is with HBr, which we know adds a Br to the more substituted side of a
double bond. In this case, both carbons are secondary, but one has a phenyl substituent which gives it a
more stable carbocation intermediate. Therefore, the Br will add to the carbon attached to the phenyl:
The second reaction is with NaCN. Similar to scheme A, Na+ is a common positive counter ion, which
means the reagent is really Na+ -CN (cyanide ion) in solution, which the negative charge on carbon. Is
this a good or poor nucleophile? Due to the concentrated negative charge, we can assume cyanide acts
as a good nucleophile. What about the substrate? The intermediate product C shows a bromine, which
is a good leaving group, on a secondary carbon. This observation coupled with the good nucleophile
cyanide suggests that SN2 will occur in this reaction. Therefore, product C should be:
This matches the proposed product, so the proposed product is correct.
Scheme D: The first reaction is hydroboration, which we know to add an alcohol and hydrogen syn
across a double bond, with the alcohol ending up on the least substituted carbon. Given this, we can
draw the intermediate D:
The alcohol should end up trans to the methyl group due to the syn mechanism of the reaction. Looking
at the next set of reactions after intermediate D, there are two transformations occurring. The first uses
tosyl chloride (TsCl) to convert the alcohol into a good leaving group. The second uses I- as a
nucleophile. I- is a good nucleophile and can react with good electrophiles (atoms with good leaving
groups) in an SN2 reaction.
Remember that TsCl retains the stereochemistry at the alcohol. Since the alcohol was pointed up on the
ring, the OTs group is still up. I- attacks the carbon at the –OTs position in an SN2 fashion, which inverts
the stereocenter at that carbon.