Week 2 Problem Set (Solutions)
(2/14, 2/15, 2/16)
Concepts Covered
Nomenclature
Constitutional Isomers
Acid and Bases
Resonance Structures
Additions to Alkenes
Conformational Analysis of Linear Alkanes
(1)
(2)
The equilibrium lies to the left. Remember, in acid-base equilibrium problems, the goal is to find the
weaker acid/base. The equilibrium lies to the side with the weaker acid (higher pKa). One way to find
the side with the weakest acid is to determine which conjugate base is stronger. The stronger conjugate
base with have the weaker conjugate acid. In this case, NH- is a stronger base than O-, which means NH2
is a weaker acid than OH. How do we determine the strength of acid? One way is to evaluate the
stability of the base. The more stable the base, the weaker it is. Factors that increase stability include
the electronegativity of the atom holding the negative charge and charge delocalization. More
electronegative atoms can hold negative better than less electronegative atoms, increasing stability of
the base. Additionally, delocalization of the negative charge onto neighboring atoms (demonstrated via
resonance) also increases stability of the base. Since O is more electronegative than N, it can stabilize
negative charge better. This means that O- is the weaker base while NH- is the stronger base. Hence, the
equilibrium lies to the left.
The equilibrium lies to the right. The goal is to look for the weaker acid and base. The base on the left,
ethoxide, and the base on the right, phenoxide, both are negatively charged oxygens. How do we
distinguish between them? To determine which one is stronger/weaker, we have to consider the factors
for stability of bases. The electronegativity of the negatively charged atom doesn’t change between
bases as both are oxygen. The other factor is delocalization of the negative charge. Another way of
saying this is which base has resonance structures that shows the spreading out of negative charge?
Phenoxide, the negatively charged oxygen species on the right, has delocalization of the negative charge
as shown through resonance:
Because the charge is delocalized, the base is said to be more stable. Since it is more stable, the base is
weaker. Conversely, because it is stable, ethoxide, the other base on the left of the equilibrium, is
stronger. Its conjugate acid is weaker, and thus the equilibrium lies to the right.
The equilibrium lies to the right. This might confusing at first since one would think that the neutral
species would be more stable. After all, doesn’t the right side have more charge and don’t molecule
want to neutral, not charged? While this statement is logical, it does not consider the relative pKa’s and
the naturally tendency for reactions to occur. Remember, chemistry happens! If everything stayed
neutral, there wouldn’t be any fun orgo to talk about. This problem is very similar to the carboxylic acid
and ammonia equilibrium. The first step is to identify which molecules are the acid and base.
Amines are weak bases, so the protonated amine on the right must be an acid. Notice how the sulfuric
acid derivative is giving up a proton to the amine, meaning it must be an acid (proton donor). The amine
is the proton acceptor, so it is the base. Now the question is which acid is weaker, the sulfuric acid or the
protonated amine? Let’s look at the stability of their respective conjugate bases. The conjugate base of
the sulfuric acid is a negatively charged oxygen species. Oxygen is fairly electronegative and can hold
negative charge. What about delocalization? The sulfuric acid has two other oxygens that can resonate
with the existing negative charge, just like the carboxylate anion. Because the negative charge is
delocalized over three oxygens, the base is very stable, and thus weak. The cyclic amine (pyridine) is also
a weak base. It has no formal charge, so it stable in that regard. However, it does have a free lone pair,
which allows it to act as a base by picking up protons. Additionally, nitrogen is less electronegative than
oxygen and is more lenient of holding positive charges than oxygen. The negative charge on oxygen in
the sulfuric conjugate base is involved in resonance and is not as available to pick up protons as the
amine lone pair. All these factors mean that cyclic amine is more of a reactive base than the sulfuric
conjugate base. Therefore, the protonated amine is the weaker acid and the equilibrium lies to the right.
It’s a good idea to know common pKas. This sulfur acid derivate has a pKa of -0.6, while the protonated
amine has a pKa of about 5. I would recommend becoming familiar with other pKas.
The equilibrium lies to the right. The question here is the acetylide anion (-C triple bond CH) a stronger
or weaker base than cyanide (-C triple bond N). Both have the negative charge directly on a carbon
atom, so there is no difference between them there. They both have triple bonds, so that isn’t different.
The only difference is the other atom in the triple bond. The nitrogen in cyanide is more electronegative
than the carbon in acetylide, meaning that cyanide is better at stabilizing negative charge overall.
Nitrogen can “pull” the negative charge slightly away from the carbon so that the carbon doesn’t have
to bear the full negative charge. This makes cyanide more stable and thus a weaker base. The
equilibrium must lie to the right.
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(6)
(7)
a) There are seven isomers that fit the parameters.
The key to isomer problems is to stay organized! It is tempting to draw structures as soon as they come
into your head and then try to decide if it matches the limitations for the problem. However, this
method is random and will not ensure that you have accounted for all the possible isomers. The first
step is to draw all the possible carbon “backbones” with the given molecular formula. Since we have
only 4 carbons to work with, there are only two possible backbones we can use.
Now that we have our carbon backbones, we can add functional groups systematically at different
positions to make different isomers. We first need to identify the unique positions for each backbone.
Unique in this context means the position is not identical when the molecule is flipped or can’t be
accounted for another position.
The unique positions have been indicated by the arrows. There are only two per structure as no third
position exists that isn’t already covered. Now we can add functional groups, in this case oxygen in the
form of an ether or alcohol, to these positions. Let’s start with the alcohols.
Notice how the –OH was attached to each of the indicated unique positions. This should cover all the
possible alcohol isomers. Now, what about ethers? If we remember, an ether consists of an oxygen
bonded to two carbons through single bonds. This must mean that the oxygen lies in between carbons
and cannot exist on the end of the molecule, or else it would be an alcohol. We need to change our
unique positions slightly when thinking about ethers. Instead of identifying unique carbons, we need to
identify unique positions between carbons (unique bonds).
Now that we have identified our unique positions, we can insert oxygens in each of them to create the
ether isomers.
It might look confusing since adding the oxygen adds an extra “kink” in the carbon chain, but if you
count you can identify that the number of carbons hasn’t changed. Remember, isomer problems require
organization to account for all the possible different connections and configurations. Try to keep this
approach in mind when it comes to the exam, otherwise panic sets in and you might have a difficult time
coming up with all of the correct structures.
b) This problem is actually similar to the previous one. The key difference is the chemical nature of
nitrogen. Nitrogen is able to form up to three bonds without a formal charge, unlike oxygen which can
only form two. This changes the number of available structures available to us.
Again, just like a), we have to be systematic in our approach or else we risk missing out on some isomers
and losing time that could be spent on other problems. You don’t want to spend a large amount of time
on this kind of problem when it comes to exam time, trust me. Since we have 4 carbons, we will have
the same two backbones available to us.
This is the same from the last problem. There are two unique sites on each backbone where the
nitrogen can form a single bond uniquely to those carbons. Now we can add a primary amine (Primary =
only bond to one carbon) to each of these sites.
We’ve covered primary amines. What about secondary amines (bound to two carbons)? This will be very
similar to the process used to find the ethers in the last problems. Instead of finding unique carbons, we
need to identify the unique bonds where we can insert a nitrogen between two carbons.
Same diagram from a). These are the unique bonds where a nitrogen can be inserted. Now let’s find the
available secondary amines.
Great! Now we have secondary amines covered. Lastly, we have to find any tertiary (bound to three
carbons) amines.
Instead of finding a site in the backbone to place our nitrogen, let’s think about this logically. We know
that a tertiary amine has to be bound to at least three carbons. The simplest tertiary amine looks like
this:
Three carbons are already “used” just to make this simple tertiary amine base structure. We know we
have to have 4 carbons, so let’s just add one carbon to one of the existing carbons. Each existing carbon
is identical since this base structure is symmetrical, so it does not matter where we tack on the last
carbon.
And now we have all the isomers. Nitrogen isomers can be tricky because the can form up to three
bonds with carbon. Try organizing by backbone and primary, secondary, and tertiary amines.
(8)