Download Week 6 Solutions - Brown University Wiki

(1)
Russell Shelp
Russell Shelp
3.
a) Provide the chair structure of the product of the following reaction IMMEDIATELY after the
reaction occurs. What is the lowest energy chair conformer?
Br2
OH
H2O
OH
+
Br
Br
Br2/R-OH addition occurs through an anti-periplanar stereochemistry, so we know the bromine
and OH group must initially be anti-periplanar right after the reaction occurs. In order to be antiperiplanar, they must be axial.
Initial conformation:
OH
Br
or
OH
Br
After a while, equilibrium is reached between the other chair conformers and the lowest
energy chair conformers prevail:
Br
Br
OH
OH
OH
or
OH
Br
Br
b) Label the following structures as conformers, enantiomers, or diastereomers.
I find the best way to approach this problem is to first convert both Newmans and chair
diagrams to planar cyclohexane rings. This makes it easier to identify stereochemical
relationships.
i.
Cl
H
H
Me
Me
vs.
H
H
H
H
Converting to cyclohexane rings gives us:
Cl
Cl
Cl
vs.
Me
Me
The left structure shows the methyl equatorial and the right structure shows the chlorine
equatorial, but the structures are otherwise identical. These structures are conformers.
ii.
H
Et
H
H
H
H
Me
H
Br
Me
Et
vs.
Br
The problems showed the Ethyl of the right hand molecule in an incorrect axial
position, sorry about that!
Converting to planar cyclohexanes gives us:
Et
Et
vs.
Me
Me
Br
Br
These structures are mirror images of one another and are therefore enantiomers.
iii.
Br
Br
H
H
H
H
H
H
vs.
Me
Converting to a planar cyclohexane gives us:
Me
Me
vs.
Br
Br
Only one of the three stereocenters have been switched, making these structures
diastereomers.
Mary O'Connor
4.
5. Mechanism:
As always, let’s start out with numbering the starting material and the product and making a bonds
broken/bonds made list.
Bonds Broken
C4-C5 pi
O-H sigma
Bonds Made
O-C4 sigma
C5-H sigma
We know that the oxygen has to make a bond to form a quaternary center. The likely candidate for this
is carbon number 4 as it already has three bonds to carbon. If the alcohol oxygen made a bond with
carbon 4, it would form the ring system in the product. But how do we make a bond with carbon 4? So
far, we have seem bonds formed between electrophiles and nucleophiles. Since oxygen has lone pairs, it
will be the nucleophilic atom. The question then becomes, how do we make carbon 4 electrophilic? One
way to make carbons electrophilic is to turn them into carbocations. That means we will need a way to
break the C4-C5 pi bond.
Another consideration is to look at the reagent. Our sole reagent is H2SO4, which is a strong acid. We
know that alkenes in strong acids can be protonated. Given this, we can start pushing arrows:
Remember that the tertiary carbocation forms preferentially over the secondary carbocation. Since the
alcohol is the only source of oxygen, it attacks the carbocation. Once deprotonated by the conjugate
base, the product is formed.
Chance Dunbar
6.
9)
H 2SO 4
H 2O
HO
This mechanism may at first seem daunting – but if you approach it in a systematic
manner, you will be able to figure out whats going on.
Like in all mechanism problems the first thing we want to do is to number carbons
and make a bonds broken/bonds formed table.
In this case numbering carbons is actually pretty difficult but lets try it out. We know that
we’ve had a ring expansion – thus one of the carbons in the 5 membered ring that was
formed has to come from one of the carbons outside the initial 4 membered ring.
Furthermore, we know one of the branched carbons in the product has to come from the
branched carbon in the starting materal (shown with C in both molecules).
a
1
a
C
1
C
C
1
HO
Also note that the carbon that is labeled 1 in the first compound (note that they are the
same as the molecule is symmetric) must have been the one to attack the carbon
currently labeled a to make a 5 membered ring. If it is attacking carbon a – we know that
the other branched carbon has to correspond to the carbon labeled in red in the starting
material.
This may all seem confusing currently. It takes some practice – especially in a pretty
challenging example like this one. The full numbering is shown below:
7
5
6
6
1
1
5
H 2SO 4
2
4
H 2O
3
4
7
3
2
HO
If the numbering didn’t make sense before – try to see if you can reason out why the
numbering scheme above works. Notice that C1,2, and 3 is part of both rings and is
unbranched in both. Furthermore note that C4 still has the connection to C7 and C5 is
still connected to C6.
Now lets make a bonds broken/bonds formed table:
Bonds Broken
C5-C6 pi
C1-C4 sigma
Bonds Formed
C1-C5 sigma
C4-O sigma
C6-H sigma
Now lets think about the mechanism – we have an alkene in out starting material and
we are subjecting it to a very strong acid. We know that it will get protonated and form
the more stable carbocation – in this case secondary.
7
5
7
5
6
H-HSO4
1
4
6
1
4
3
2
3
2
In that step we broke the C5-C6 pi and formed the C6-H sigma. Now we know that to
form the 5 membered ring we broke the C1-C4 sigma and made the C1-C5 sigma. This
is a case of a rearrangement.
6
7
5
1
5
6
1
2
4
4
3
2
3
7
Now there is a positive charge on C4, which means the remaining water in solution will
attack here, forming the C4-O sigma.
6
1
5
2
4
3
7
HO
H 2O
H
And finally, we have to deprotonate the alcohol group. And now you can also check
your BB/BF table to see if you did everything you needed to in the mechanism.
HO
OH
H
B
Full Mechanism
7
5
6
1
4
3
H-HSO4
6
7
5
6
1
2
4
2
1
5
3
4
3
7
2
H 2O
HO
OH
H
B
Natalie Palaychuk
7.
Please provide a mechanism for the following reactions:
a) Start this problem by numbering the starting material and the product and creating a
“bonds made” and “bonds broken” list.
1
4
1
8
OH
6
3
5
Hg(OAc)2
AcOHg
4
H 2O
7
3
5
2
Bonds Made
C3-O
C4-HgOAc
2
O
8
7
6
Bonds Broken
C3-C4 pi
O-H
(HgOAc)-OAc
This is the mechanism for oxymercuration of a double
bond. Like Br2 addition, this mechanism proceeds
through a three-membered ring intermediate. The
tertiary carbon of the double bond (C3) is more
electrophilic since it has more carbocation character
than the secondary carbon of the double bond (C4). Once the three-membered ring is
formed, it is opened by the OH within the molecule, not water.
But, Why doesn’t water attack C3?! Because: intramolecular reactions (electrophile and
nucleophile in the same molecule) are much faster than intermolecular reactions
(electrophile and nucleophile are on separate molecules) due to the increased proximity.
AcO
OAc
Hg
AcO
OAc
Hg
Hg
OH
OH
OAc
Hg
OH
AcOHg
AcOHg
H
O
O
BH4
NaBH4 is the second of two reagents used during oxymercuration. It removes the Hg(OAc) from
C4 and replaces it with a hydrogen, in a process know as “reduction”.
OH
Hg(OAc)2
H 2O
AcOHg
O
NaBH4
O
b) CHALLENGE
*Note: D means deuterium, an isotope of hydrogen
H 2O
HO
D
H2SO 4
D
As with any mechanism, it is best to start by numbering the carbons and constructing a “bonds
made” and “bonds broken” list.
3
1
2
4
5
2
H 2O
7
4
8
1
10
6
HO
D
H2SO 4
9
3
5
6
8
7
9
D
10
If you find the numbering difficult, consider these tips:
• You know that carbon 7 had the deuterium, so you can generally assume that it retains
the deuterium in the product. While this is not always the case (maybe there was a
deuterium shift), it’s a good place to start.
• The terminal alkene seems to have been retained from starting material to product, so we
can assume that it is still 1. Again, this is not always the case but it’s a good place to start.
• The methyl groups in the starting material are 4 and 10. When numbering the product,
assign the methyls to 4 or 10, depending on the surrounding numbers.
Bonds
Made
C3-C9
C5-C6 pi
C9-OH
C7-H
C8-H
Bonds
Broken
C3-C5 pi
C6-C7 pi
C8-C9 pi
There are a couple of starting moves we could take. The best way to get used to doing
mechanisms is to go down unsuccessful paths and learn to recognize patterns. For example:
• Protonating the 3,5 double bond can give us a cyclic product, but the cyclic double bond
in the product will end up in the wrong location
•
•
Protonating the 8,9 double bond can also give us a cyclic product, but the cyclic double
bond in the product will end up in the wrong location. Also, the resulting carbocation
(and OH addition) will be in wrong location.
Protonating the 6,7 bond is just right!
Notice that there is a hydride shift moving the carbocation downhill from a secondary carbon to a
tertiary carbon. This gives up OH addition at the correct location.
H+
D
D
D
H
H
O
H
D
H 2O
D
B
D
HO
D
Mary O’Connor
8. Box$problems!$
$
a)$
OH
1. BH3
2. H2O2, NaOH
1. Hg(OAc)2, H2O
2. NaBH4
H2SO 4, H2O
OH
HO
*Scrambled stereochemistry
$
As$I$indicated,$remember$that$the$reduction$step$in$oxymercuration:reduction$
scrambles$stereochemistry!$Also,$remember$that$water$will$always$add$to$the$more$
substituted$side$of$the$alkene.$
$
With$acid$catalyzed$hydration,$water$still$always$adds$to$the$more$substituted$side$
of$the$alkene,$but$thermodynamically$favored$carbocation$rearrangements$will$
occur$whenever$possible!$
$
With$hydroboration:oxidation,$water$adds$to$the$less$substituted$side$of$the$alkene.$
$
It’s$really$useful$to$start$recognizing$the$patterns$of$the$products$for$each$of$these$
reactions.$I$would$also$recommend$starting$a$reaction$chart,$where$you$summarize$
each$reaction$and$any$of$its$important$features$(when$doing$this,$it’s$useful$to$go$to$
the$reaction$summary$at$the$end$of$chapter$6—there’s$one$of$these$summaries$at$
the$end$of$each$chapter$going$forward!).$$
$
$
$
$
$
b)$
O O
1. O3
2. Me2S
H2, Pd/C
1. OsO4
2. NaHSO3, H2O
These products are diastereomers.
OH OH
+
OH OH
+
(diastereomers)
$
Remember$that$ozonolysis$is$like$molecular$scissors;$you$use$it$to$cut$an$alkene$in$
half,$and$create$a$carbonyl$on$each$side.$It$creates$a$ketone$or$an$aldehyde$
depending$on$the$substitution$of$the$alkene.$
$
$
To$go$from$an$alkene$to$an$alkane,$you$can$use$a$transition$metal$catalyst$(Pd/C$or$
Pt/C)$with$H2!$The$hydrogens$will$add$syn.$The$product$is$a$mix$of$diastereomers$
because$the$chiral$centers$in$the$starting$material$do$not$change,$but$the$hydrogens$
can$both$add$to$either$the$top$or$bottom$face$of$the$ring.$
$
Osmylation$adds$two$alcohols$syn$to$one$another!$In$this$case,$the$product$is$a$mix$
of$diastereomers$because$the$starting$material$has$two$other$chiral$centers$that$
remain$the$same$in$the$product,$while$the$alcohols$can$add$to$either$face$of$the$ring.$
$
Ella Cohen
9. Box Problems. Indicate stereochem when necessary.
a)
Things to remember:
Osmylation and Hydrogenation are both syn additions
Halohydrin is an anti addition
For the osmylation reaction, it looks like the wrong product. Why are the alcohols on opposite sides of
the molecule if osmylation is a syn addition? It’s actually just a conformer of the product. Here’s a
different perspective on this:
As we can see, the methyl is just rotated so that it is anti to the ethyl group.
For the halohydrin reaction, the water will preferentially attack the carbon with the phenyl group as this
carbon as better able to stable the partial positive charge in the transition state. A benzylic tertiary
carbocation is more stable than just a normal tertiary carbocation due to resonance. This reaction will
form a pair of enantiomers.
Hydrogenation is a simple syn addition and forms a pair of enantiomers in this reaction.
b)
Sometimes it’s hard to think about going backwards. However, we can use the 2 equivalents of product
to our advantage. Since ozonolysis cleaves double bonds, then the both sides of the double bond must
be identical. There must also be two double bonds since there are two carbonyls in the product.
c)
Hydroboration is a syn addition, which means the alcohol and the hydrogen will add to the same face of
the double bond. Since the stereochemistry of the alkene is different between starting materials, each
reaction will produce a different pair of enantiomers. The pairs of enantiomers are diastereomers to
each other, as shown by the indicated absolute configuration.
Chance Dunbar