Download Week 10 Problem Set (Answers) (4/17, 4/18, 4/19) Reactions and

Week 10 Problem Set (Answers)
(4/17, 4/18, 4/19)
Reactions and concepts covered
Substitution and elimination reactions
Generating alkoxides from alcohols
Oxymercuration and hydroboration of alkynes
Thinking through syntheses
1) Box Problems
a)
1) HBr
2) tBuO- / tBuOH
3) Br2
4) xs NaNH2
Hg+2
O
H2SO 4, H2O
1) NaH
2)
I
1) HBR2
2) H2O2, H2O
Catalytic H2SO 4
H2O
1) H2/Lindlar
2) O3 / MesS
O
O
O
O
+
b) Please fill in the boxes so as to form the following esters. Please provide reaction
conditions in the boxes provided over the arrows and identify the type of reaction. For the
first two, provide two different methods for forming the desired molecules.
ThefirstboxescanbecompletedthrougheitheranSn1orSn2mechanism.Inthe
Sn2mechanism,theoxygenanionmustbeattackingaprimarycarbonsoastoavoid
aneliminationreaction.
O
Br
+
(Sn1)
OH
EtOH
O
(Sn2)
+
Br
O
DMSO
ThesecondreactioncanbecompletedthrougheitheranSn1orSn2reaction.Phenol
hasapkaof10,soitcandoanSn2reactiononasecondaryleavinggroupwithout
significanteliminationproduct.Thebenzeneringcannotcontaintheleavinggroup
sinceitissp2hybridized.
Br
OH
O
(Sn1)
+
iPrOH
Br
O
+
O
(Sn2)
DMSO
ThethirdboxcanonlybecompletedthroughanSn1reaction,sinceanSn2would
giveelimination.
(Sn1)
OH
Br
O
+
tBuOH
2)Provideamechanismforthefollowingtransformation.
O
H2SO 4, H2O
HgSO4
Thisisatextbookacid-catalyzedhydrationofanalkyne.Themercuryionactsasanelectrophile
andcreatesabridgedmercuriniumionintermediate.
Hg2+
Hg +
Amoleculeofwaterattacksthepositivelychargedbridgestructure.Deprotonationyieldsa
morestableintermediate.
H
H
Hg +
H
O
O+
H
Hg +
H
O
H
H
O
Hg +
Theenolimmediatelytautomerizesintoaketone.
H
O
O
Hg +
Hg +
Aprotonisaddedtotheketoneandthemercuryleavestoformaneutralspecies.
H
O
H
H
H
O
O+
O
Hg +
Hg +
Onelastketo-enoltautomerizationyieldsthedesiredproduct.
H
O
O
3.#The#first#thing#to#do#here#is#number#our#carbons!#
#
OH
O
3
4
2
1
2
1
5
7
4
6
7
5
3
O
6
#
Now#let’s#look#at#the#bonds#formed#and#broken:#
Bonds#formed#
#
Bonds#broken#
C2?O#double#bond# #
C2?C7#single#bond#
C7?O#double#bond#
#
These#analyses#show#us#that#we#have#the#same#number#of#carbons,#and#that#we#
broke#a#ring,#and#put#carbonyls#on#the#carbons#that#had#been#bonded#together.#
This#suggests#that#we#probably#cleaved#the#C2?C7#double#bond#via#ozonlysis.#
#
O
O
#
#
We’re#already#really#close#to#our#starting#material!#Our#last#step#is#to#create#a#double#
bond!#Our#starting#material#had#an#alcohol,#and#we#know#two#ways#of#converting#an#
alcohol#to#a#good#leaving#group#and#then#eliminating:#turning#the#alcohol#into#a#
tosylate,#or#using#a#strong#acid#and#heat#to#eliminate#the#alcohol.#The$important$
thing$here$is$to$look$at$the$step$requirements.$The#requirements#said#the#
synthesis#must#take#fewer$than$3$steps.#So,#although#both#of#these#synthetic#
pathways#are#valid,#only#the#strong#acid#and#heat#elimination#route#fits#the#
requirements.#
#
OH
#
#
Full#synthesis:#
OH
O
H3PO4
#
1. O3
2. Me2S
O
#
4.#Let’s#number#our#carbons#and#make#a#list#of#the#bonds#formed#and#the#bonds#
broken!#
OH
3
1
2
4
7
2
6
5
1
4
Br
3
OH
+ enantiomer
#
#
Bonds#formed# #
#
Bonds#broken#
C4?C5#single#bond#
#
C4?Br#single#bond#
*C5?O#single#bond#
*C6?O#single#bond#
#
Now#let’s#try#to#piece#things#together#a#little#bit#before#we#begin#with#
retrosynthesis.#We#added#3#carbons#to#our#starting#material#to#get#to#our#
product.#It’s#important#to#remember#that#we#only#know#1#way#to#add#carbons:#an#
acetylide#anion#with#a#primary#haloalkane.#Given#that#our#starting#material#is#a#
primary#haloalkane,#it#is#reasonable#to#assume#that#this#must#have#been#one#of#
the#steps#we#performed.#
#
Because#we#know#we#had#to#add#an#alkyne#anion,#we#know#that#the#alcohols#on#
C5#and#C6#couldn’t#have#been#present#initially,#because#we#wouldn’t#be#able#to#
deprotonate#the#alkyne#to#make#it#a#good#nucleophile???instead#we#would#
probably#deprotonate#an#alcohol.#
#
With#this#information,#let’s#try#retrosynthesis.#We#only#know#1#way#to#create#a#
1,2#diol#system:#Osmium#tetroxide#on#an#alkene.#So#it’s#reasonable#to#assume#
that#this#was#probably#our#last#step.#But$the$stereochemistry$here$is$
important,$and$determines$the$type$of$alkene$that$must$have$preceeded$
this$1,2$diol.#
#
We#have#two#options:#a#cis#or#a#trans#alkene.#Let’s#try#adding#the#diols#to#each.#
Remember#that#the#osmium#tetroxide#adds#the#alcohols#syn.#
#
OH
1. OsO4
2. NaHSO3
H2O
+enantiomer
or#
OH
#
OH
1. OsO4
2. NaHSO3
H2O
OH
+enantiomer
#
#
At#first#glance,#neither#of#these#looks#right.#Upon#further#inspection,#you#can#see#
that#the#final#product#shows#the#alcohols#in#opposite#planes,#which#we#know#isn’t#
possible,#and#the#carbons#in#the#chain#look#“trans”#to#one#another#(i.e.#if#you#put#a#
double#bond#between#those#carbons,#it#would#make#a#trans#double#bond).#This#
suggests#that#we#probably#added#the#alcohols#to#a#cis#alkene#and#rotated#around#
the#single#bond.#
#
OH
OH
OH
+enantiomer
+ enantiomer
OH
#
#
We#now#know#that#we#added#the#alcohols#to#a#cis#alkene.#We#also#know#that#the#
only#way#for#us#to#add#carbons#is#using#an#acetylide#anion.#This#means#that#we#
probably#had#an#alkyne#that#we#reduced#to#a#cis#alkene.#H2/Lindlar#are#the#
reagents#we#use#to#reduce#an#alkyne#to#a#cis#alkene.#
#
#
#
And#we#know#we#add#acetylide#anions#to#primary#haloalkanes,#so#we#have#gotten#
back#to#the#starting#material!#
#
Br
#
#
#
#
We#also#need#to#remember#to#show#the#formation#of#the#acetylide#anion!#
#
#
Full#synthesis:#
#
NaH
Na
H2/Lindlar
Na
Br
OH
OH
1. OsO4
2. NaHSO3
H2O
OH
+enantiomer
+ enantiomer
OH
#
6) Synthesis
Please propose a synthesis for the following transformation:
Lets start this problem with retrosynthesis. If we count our carbons, we see that we
are not changing the number of carbons in our molecule. Therefore, we are probably
adding to an alkyne to get the carbonyl. Since the carbonyl is on the less substituted
of the two alkyne carbons, we must have used hydroboration. We can get to the
alkyne through an Br2 addition to an alkene followed by a double elimination.
The final synthesis is as follows:
(7) Mechanism:
As with many mechanisms, it’s helpful to label the carbons on the starting material and the product. This
makes it easier to recognize and visualize the transformations taking place in the mechanism. Based on
the numbering scheme, we can make a list of bonds broken and formed:
Bonds Broken
C2-N sigma
C1-H sigma
C6-Br sigma
C4-O pi
Bonds Formed
C1-C2 pi
C4-N pi
C6-O sigma
The key transformation occurring is the breaking of the C2-N bond and the formation of the C6-O bond.
There is an inversion of stereochem at the Br position, indicating an SN2 type reaction. A new pi bond is
also made between C1-C2. This is a big clue. We know that we have to break the bond between C2-N and
also form a new pi bond between C1-C2. What other reactions do we know of that form a pi bond as
another bond is broken? Well, E2 reactions occur via this same mechanism. A new pi bond is formed at
the same time a leaving group leaves. Something similar may be occurring here.
Additionally, we have to examine the reagent. tBuOK is a strong bulky base that favors E2 elimination
towards the less-substituted alkene. Isn’t Br a good leaving group? We would expect that an E2 reaction
would occur directly with the secondary Br. However, upon closer inspection we discover that the
starting material actually doesn’t have any hydrogens that are anti to the Br. This means that E2 can’t
occur with the secondary Br. So, E2 is out with the Br as our first step. But tBuOK is a base, which means
it can still deprotonate elsewhere. Where else do we have acidic protons? Given our previous
discussion that we have to make a pi bond between C1-C2 while also breaking the C2-N bond, it starts
to make sense to deprotonate at C1. This does not seem unreasonable since the leaving group is an Nspecies that is attached to a tosyl group, which can be expected to be analogous to –OTs (but not
exactly). The only other site that is possible to deprotonate at is C3 since this can resonate with the
carbonyl oxygen. However, C3 is more difficult to access than C1 and since tBuOK is bulky, it makes more
sense to deprotonate at C1. Also, this is a mechanism, and deprotonating at C3 will not generate the
correct product since C2-N is still intact and the C1-C2 hasn’t been formed.
Therefore, let’s begin by deprotonating at C1 and having nitrogen leave as –N:
This does not seem unreasonable since the nitrogen anion that is formed is resonance stabilized by both
the tosyl group and the adjacent carbonyl oxygen. It also satisfies the requirement that the bond
between C2-N is broken and a new pi bond is formed between C1-C2. What’s left? We have to make a
new bond between the carbonyl oxygen and C6, which is the carbon attached the bromine. The product
shows that C6 experiences an inversion of configuration, which is strong evidence of an SN2 reaction.
We also know that a new pi bond was formed between C4-N. Based on these observations, it seems that
negative charge on the nitrogen resonates to the oxygen, which then does an SN2 reaction on the C6
secondary bromine.
That is what that would look like:
All transformations have been accounted for. The O- attacks at the C6, inverting the configuration and
forming a C6-O sigma bond. Additionally, a new pi bond is formed between C4-N.
Full Mechanism:
(8) Synthesis
The first step is to identify where the starting materials have ended up in the product. Since all carbons
must be derived from the starting materials, look for two-carbon synthetic “units”.
There are 3 two-carbon “units” in the product, which suggests that the starting materials have been
used three times. The combination of which starting materials is unknown at this point.
Let’s begin the synthesis with a retrosynthetic analysis. There are multiple places to start making
disconnections, and sometimes it’s not obvious which disconnections are more important than others,
or which ones will be helpful. The intuition of where to start a retrosynthesis will become more natural
the longer you practice. For this synthesis, I will begin the retrosynthesis at the ketone instead of the
ether. The decision for this is two-fold: 1) we only know very specific ways of generating ketones, so this
can narrow down the synthesis more quickly; (2) it makes sense that a ketone would be generated in the
last step as they can react with other reagents, so reserving it for last minimizes unwanted reactivity.
The only ways we know how to generate ketone are either through ozonolysis of an alkene or through
oxymercuration of an alkyne (generates the enol which tautomerizes to the ketone). If pursue the
ozonolysis route, the previous compound must have been a 7-carbon unit, which is an unlikely
intermediate product given that the starting materials are both sources of even-numbered carbons.
Additionally, we do not know how to back a carbon-carbon bond between an alkene and a primary
carbon, and the alkene can’t have been formed via a reduction of an alkyne since it is disubstituted at
one end. The oxymercuration route makes the most sense because it requires the use of alkyne, which is
an available starting material for use.
The terminal alkyne could been generated via an SN2 reaction with a primary good leaving group (GLG)
and the acetylide anion starting material.
This is where it gets a little tricky. There are a couple of routes we could take to generate this product.
We know that we can form a terminal leaving group with hydroboration followed by TsCl to form OTs.
However, this route would require a terminal alkene bonded to an oxygen in an enol-like fashion. Up to
this point, we have not seen a reaction that will generate this type of molecule, so the route is a deadend. Another route is the halohydrin reaction using ethanol (instead of water) and Br2 on an alkene.
Visualizing this type of disconnection is difficult and I would not expect it to be the first thing you think
of when approaching this problem. If, however, you are able to use a process of elimination on reaction
pathways that give dead-ends, you should eventually make your way to this reaction route. It’s a
convenient reaction in that you functionalize both ends of a molecule in one synthetic step.
Ethanol can be generated from a starting alkene using either oxymercuration or hydroboration.
Putting it all together, a full synthesis can be proposed:
9) Solutions
Compound A and Compound E both have a pKa of 25.
Must be a terminal alkyne.
Compound B and Compound D give the same major product upon reaction with HCl, even
though they are constitutional isomers of one another. They both have only one pi bond.
This is hard, but the only possible answers will have a ring and a pi bond; given that these
are distinct compounds, you’ll realize that the only possibilities are cyclopropanes that have an
alkene on them.
Compound C yields one, symmetrical compound upon reaction with ozone and Me2S.
This must be a ring, since ozonolysis of a linear compound would yield two compounds. For
it to be symmetrical, the ring itself should be symmetrical.
Upon being subjected to oxymercuration-reduction conditions, Compound B yields at least one
chiral compound. This is untrue for Compound D.
Now you can separate the two possible cyclopropanes you came up with earlier for B and D.
Upon reaction with H2 and platinum metal, Compound E yields a product with 4 unique
carbons, whereas Compound A yields a product with 2 unique carbons.
Now you can separate the two different alkynes you have for B and D.
Subjecting Compound F to hydroboration-oxidation conditions yields two different ketones
which are formed in relatively equivalent amounts.
The implication here is that you have an alkyne, but trying to perform hydroboration-oxidation
with this alkyne doesn’t give you much selectivity. The only possibility is the one shown below.
COMPOUND A
COMPOUND B
COMPOUND C
COMPOUND D
COMPOUND E
COMPOUND F
10)Foreachpair,circlethereactionmixturethatwillgivethedesiredproduct.
a)
NaNH 2
+
H
OR
+
OH
Br
Normally,eliminationreactionsrequireaprotonantitotheleavinggroup.However,during
alkeneeliminations,astrongbasewillstillproducethealkyneviasynelimination(albeitslower).
Contrastthiswiththeleftreaction.Inorderforthisreactiontoproducethedesiredproduct,
we’dneedanSN2reaction.However,hydroxideisaterribleleavinggroup,sothatreaction
won’tproceed.Whatwillhappenisanacid-basereaction.Rememberthatalkyneanionsare
strongbasesandthatalcoholsareweakacids:theycannotbeputintosolutiontogetherunless
youwanttheprotontransfertooccur.
b)
O
O
+
Br
OR
Br
+
O
Bothoptionshaveastrongnucleophile(analkoxide)andagoodleavinggroup(bromide).For
eachofthem,iftheSN2occurs,thenthey’llproducethedesiredproduct.However,recallthat
strongbasesfavoreliminationoversubstitutionreactionsatsecondaryhalides.Forthisreason,
thesecondreactionmixturewouldprimarilyproduceeliminationproductinsteadofthedesired
ether.
c)
O
OH
+
Na
OR
OH
+
PhO
Recallthatphenol(pKaofabout10)isamuchstrongeracidthanisopropanol(pKaofabout16)
becauseitsconjugatebasecandelocalizethenegativechargearoundthebenzenering.Since
phenoxideismorestablewiththenegativecharge,thennoreactionoccursandtheproton
remainsontheisopropanol.However,sodiummetalcaneasilydeprotonateanalcoholby
releasinghydrogengas.