Video Transcript
It is possible for a molecule
containing an alcohol and a carboxylic acid group to react with itself to form
cyclic esters known as lactones. Which structure would result from
the reaction of 5-hydroxypentanoic acid with itself?
We have been given five possible
solutions to this question labeled from (A) to (E). The question tells us that cyclic
esters are formed. All options from (A) to (E) are
cyclic because they contain a ring in their structure. But do they all contain the ester
functional group? An ester has the following general
displayed formula, where this R group, which represents the rest of the molecule,
can be an alkyl group, an aryl group, or a hydrogen atom. And this R group can be an alkyl
group or an aryl group.
Of the five possible answers, only
two of them, (D) and (E), contain the ester functional group. The structures in (A), (B), and (C)
contain the ether functional group. (A) also contains the carboxylic
acid functional group, and the structures in (B) and (C) also contain the alcohol
functional group. Therefore, structures (A), (B), and
(C) cannot be the answer. Let’s remove these answer choices
from the screen so that we have a bit more space.
Both (D) and (E) are cyclic
esters. So, we need to figure out which one
of these two structures would result from the reaction of 5-hydroxypentanoic acid
with itself. The question tells us that to form
these esters, the molecule must contain an alcohol and carboxylic acid functional
group. Before we look at how to make
cyclic esters, let’s look at how to make linear esters.
In this reaction, the carboxylic
acid and alcohol functional groups would be on separate molecules. They react reversibly in the
presence of a strong acid catalyst to produce an ester and the side product
water. In this reaction, the hydroxy part
of the carboxylic acid forms part of the water molecule, and the remaining part of
the carboxylic acid forms part of the ester. The other part of the ester comes
from the alkoxide part of the alcohol.
Now that we know how to form a
linear ester, let’s look at how to form a cyclic ester. This is the structure of
5-hydroxypentanoic acid. Let’s start by identifying the key
parts of the molecule. This is the hydroxy part that forms
part of the water molecule. This is the other part of the
carboxylic acid functional group and the start of the R group that will form part of
the ester. And this is the alkoxide group,
which forms the other part of the ester molecule. The part of the molecule that’s
drawn in black is where the two R groups meet.
As shown in the formation of a
linear ester, a bond will form between the oxygen atom in the alkoxide group of the
alcohol and the carbonyl carbon in the carboxylic acid. In 5-hydroxypentanoic acid, these
parts of the molecule are quite far away from each other. So, we should redraw the molecule
to move these parts of the molecule closer together. To ensure that we don’t miss any
parts of the molecule when it’s redrawn, let’s number the carbon atoms. There are five carbon atoms in this
molecule. So when we redraw it, we need to
make sure there are still five carbon atoms. Since we know that the end product
is a cyclic ester, it makes sense for us to redraw the ester in a cyclical form.
When 5-hydroxypentanoic acid is
reacted with itself, a bond forms between the oxygen atom in the alkoxide group of
the alcohol and the carbonyl carbon in the carboxylic acid functional group. In this diagram, the carbon is
labeled with the number five. These bonds break and the side
product of water is formed. We now have the structure of the
final product. And if we rotate that structure, we
see that it matches the structure found in (D), where the ring is six-membered as it
contains five carbon atoms and one oxygen atom, as opposed to the structure found in
(E), which contains a five-membered ring as one of the carbon atom forms a side
group.
Therefore, the answer to the
question “which structure would result from the reaction of 5-hydroxypentanoic acid
with itself?” is (D).
