Week5ProblemSet(Solutions)
(3/6,3/7,3/8)
ConceptsCovered
Acid/Base
ChairConformers
Stereochemistry
CyclicIsomers
NewmanProjections
CarbocationRearrangements
Reactions
HXAdditiontoAlkenes
BrominationofAlkenes
HalohydrinFormation
Acid-CatalyzedHydrationofAlkenes
Natalie Palaychuk
Natalie Palaychuk
Natalie Palaychuk
Natalie Palaychuk
Russell Shelp
3.#a)#These#two#compounds#are#diastereomers.#You#can#figure#this#out#by#assigning#
the#configuration#of#each#chiral#center.#When#you#do#that,#you#will#see#that#the#chiral#
center#with#the#alcohol#stayed#the#same,#while#the#chiral#center#with#the#Br#
switched.#
OH
Br
(S)
(S)
and
(S)
(R)
Br
#
b)#These#two#compounds#are#conformers.#In#chair#conformations,#you#will#notice#
that#everything#that#was#axial#becomes#equatorial,#and#vice#versa,#while#the#
substituents#stay#pointing#in#the#same#direction.#
OH
Br
Br
and
#
If#you#look#at#these#compounds,#and#use#the#arrow#to#indicate#the#same#carbon,#you#
can#see#that#the#substituent#on#the#first#carbon#is#down#and#equatorial#on#the#first#
compound,#and#down#and#axial#on#the#second.#The#other#substituent,#which#is#still#
two#carbons#away#in#each,#is#up#and#axial#in#the#first,#and#up#and#equatorial#in#the#
second.#Therefore,#these#are#conformers.#
#
c)#These#compounds#are-constitutional-isomers#because#they#have#the#same#
substituents,#but#different#connectivity.#
#
d)#These#two#compounds#are#identical.##When#you#assign#stereochemistry,#you#will#
see#that#they#both#have#the#same#configuration.#
OH
HO
(R)
(R)
and
#
Alternatively,#if#you#flip#over#the#yDaxis,#you#will#get#the#other#compound.#
OH
HO
#
e)#These#to#compounds#are#enantiomers.##
#
The#first#step#in#figuring#this#out#is#to#convert#the#Newman#projection#into#a#
sawhorse#configuration.##
#
In#this#configuration,#remember#that#the#substituents#on#the#left#side#point#away#
from#you,#and#the#substituents#on#the#right#point#towards#you.#
Ph
Ph
Et
Br
OH
Et
Br
#
#
Now#you#can#assign#stereochemistry#to#both#the#original#sawhorse,#and#the#new#one.#
In#doing#so,#you#will#see#that#both#chiral#centers#are#switched,#so#they#are#
enantiomers.#
#
OH
Ph
Et
Et
Ph
(S)
(R)
(S)
HO
#
#
#
#
#
#
#
#
#
#
#
#
#
#
#
#
#
#
#
#
(R)
Br
Br
OH
#
Ella Cohen
(4) There are a total of 11 isomers possible for this problem.
How do we go about finding these isomers? First let’s start with the constitutional isomers. From the
molecular formula and the condition that we must have a cyclohexane in the structure, we can deduce
that there are two substituents on the ring: Cl and OH. Once the substituents are known, we can place
the substituents increasingly apart from each other on the ring so find all the unique constitutional
isomers. Starting from both on the same carbon, we move to a 1,2 relationship, followed by a 1,3 and
then 1,4. 1,5 is not possible since it is actually just another form of 1,3. Also, it does not matter which
substituent you start at the 1 position as the relationship between the substituents will stay the same.
Chance Dunbar
Next, we have to determine if stereoisomers are possible for each constitutional isomer.
This molecule is achiral as there are no atoms that attached to 4 different substituents. The
sides of the ring on either side of the carbon at the 1 position are identical. This is the only
isomer possible for this constitutional isomer.
Both of these constitutional isomers can be chiral and each
has 2 chiral centers (indicated by the *). The number of
stereoisomers possible is 22 = 4 for each of them.
This constitutional isomer is achiral since there is an internal plane of symmetry down the
y-axis in the middle. However, since this is a ring, we can have cis/trans isomerism between
the substituents. So there are a total of two stereoisomers associated with this
constitutional isomer.
To find all the stereoisomers for the chiral constitutional isomers, start with the substituents either cis or
trans to one either. I’ll show the process for the 1,2 ring, but it’s the same for the 1,3 ring.
I randomly chose to have both substituents coming out of the page. This is one
of the possible stereoisomers. By definition, the enantiomer is the compound
that has the opposite absolute configuration for all stereocenters. The easiest
way to do this is to just invert both stereocenters. You can also draw a mirror
plane find the mirror image.
Chance Dunbar
Now we have one pair of enantiomers, we need to find the other pair of enantiomers. Each molecule in
this other pair of enantiomers will be a diastereomer to each of the enantiomers we already found. By
definition, a diastereomer is a stereoisomer in which one or more but not all stereocenters have their
configuration inverted. Since we only have 2 stereocenters, we need to change the configuration of only
one stereocenter to find its diastereomer. Another way to think of this is that we started with the
substituents cis to each other, but they can also be trans.
We now have one diastereomer, but this molecule has its own enantiomer. To find it, repeat the process
that was done previously by either inverting all the stereocenters or using the mirror image.
Alternatively, you can also find the above cis isomer’s other diastereomer by inverting the other
stereocenter (the carbon with the chloro group), which will give you the missing 4 th isomer.
Repeat this process to for the 1,3 cyclohexane to find all of its isomers. For the 1,4 ring, there are only
two ways to place the substituents: cis or trans.
Since these are achiral, they have no enantiomers. You can try to find
the enantiomers, but you will realize that they are the same compound
as the first one.
Chance Dunbar
(5) This problem can seem overwhelming at first, so let’s break things down.
a) It’s difficult to compare structures that are in different perspectives. My recommendation is to
convert either all the chairs into Newman conformers or the Newmans into chairs. I will do the latter in
this instance, but try the other method as well.
To interpret cyclohexane Newman projections, pretend you are looking at the ring straight on from a
level field. Due to the puckering of the ring, one end will be pointing down and the other will be pointing
up. The Newman projection in this case shows the end pointing down in front (carbon 1) and the end
pointing up in back (carbon 4). The “eye” in the above diagram shows the perspective being used. Using
this method, we can convert the Newman projections into rings:
Now let’s compare these to the compounds A-D. What exactly are we looking for to make sure they are
an appropriate match? The first thing to check for is that they have the same connectivity, which means
that they are the same constitutional isomer. Next, we must look for to see if the two structures share
the same stereochemistry, in that the same substituents are either up or down when looking at the
same view of the ring. Finally, we look to see that a ring flip has taken place, which means that all groups
that were equatorial become axial, and all groups that are axial become equatorial (but retain their
stereochemistry).
Chance Dunbar
Below each structure are listed the relationship of the substituents, whether they are up or down, and
whether they are equatorial or axial.
Remember, for chair flips we are looking for all axial groups to become equatorial and vice versa. Note:
the chair does not have to be drawn in the opposite pucker as proof of a chair flip since it can be viewed
from a different angle. Looking at compound A, we see that I and the tBu are both equatorial up. Its
chair flip should then have I and tBu both be axial up. Therefore, compound A matches with compound
II. Using this logic, we can match the rest of the compounds:
Chance Dunbar
(b) Now that we have the compounds matched, we can find the more stable conformer. The stability of
the conformer will be dictated by the position of the tBu group as it is by far the bulkiest group. Bulky
groups on chairs are most stable in the equatorial position, so the more stable conformer will be the
conformer that has the tBu as equatorial.
For each pair the more stable conformer has been circled (look for equatorial tBu):
c) To determine the stereochemistry of compounds A-D, I find it easiest to change all of the chair
structures into flat cyclohexane depictions. To do this, pretend you are looking at the chair from above.
You should get something like this:
Chance Dunbar
Now we can assign absolute configuration to the stereocenters more easily.
We will only be comparing A with D and B with C since those pairs share the same connectivity.
Remember, if they are constitutional isomers, then they cannot be enantiomers or diastereomers. A and
D have a 1,3 substituent relationship while B and C have a 1,2 relationship.
A and D are enantiomers since both stereocenters have been inverted between them. B and C are
diastereomers since only one stereocenter has changed between them.
Chance Dunbar
Natalie Palaychuk
Natalie Palaychuk
7.#a)#
HBr
Br
#
There#are#two#key#things#to#remember#about#this#transformation:#Br#adds#to#the#
more#substituted#side,#and#rearrangements#will#occur#if#they#can#provide#a#more#
stable#carbocation#intermediate.#
#
The#hydrogen#will#add#to#the#right#side#of#the#alkene,#leaving#a#2˚#carbocation#right#
next#to#a#quaternary#carbon.#Therefore,#you#can#shift#a#methyl#from#the#quaternary#
carbon,#to#create#a#3˚#carbocation.#
1
2
3
1
#
3
2
#
#
From#there,#the#Br#will#attack#the#3˚#carbocation.#There#are#no#stereocenters#in#this#
molecule,#so#only#one#product#is#formed.#
#
#
b)#
Br2
Br
Br
Br
Br
+
(enantiomers)
#
For#this#transformation,#it#is#important#to#remember#that#Br2#always#adds#anti,#so#
the#product#will#be#a#mix#of#enantiomers.#
#
When#determining#the#starting#material,#you#must#look#at#the#product,#and#decide#
what#existed#before#the#addition,#and#what#was#added.#In#this#case,#the#Bromines#
were#added,#meaning#that#the#two#methyls#already#existed.#In#the#product,#the#
methyls#are#in#the#same#plane,#and#also#on#the#same#side#of#the#molecule,#indicating#
that#the#product#originated#from#a#cis#alkene.#
#
#
Ella Cohen
c)#
Cl
Cl2
+
Cl
OH
H2O
OH
(enantiomers)
#
In#the#halohydrin#addition,#the#halogen#and#the#water#add#anti,#and#the#water#adds#
to#the#more#substituted#side.#There#is#a#mix#of#enantiomers#because#the#water#can#
attack#from#above#or#below#the#carbocation#intermediate.#
#
d)#
H2SO 4
#
OH
H2O
#
#
In#acid#catalyzed#hydration,#water#adds#to#the#more#substituted#side#of#the#alkene.#In#
this#case,#the#product#formed#does#not#have#any#stereocenters,#so#there#is#only#1#
product.##
Ella Cohen
Russell Shelp
Russell Shelp
Russell Shelp