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SAL BARRIGA: Hi, I'm Sal.
Today, we're going to solve Problem 8 of the Final Exam of
2009 for 3.091.
The problem reads as follows.
The phospholipid--
the chemical formula is right here--
is added to hexane--
which is C6H14, which is a hydrocarbon--
in sufficient quantities to enable interaction between
phospholipid molecules.
Draw a cartoon depicting one possible molecular arrangement
that results.
You may stylize your parts of the phospholipid molecule with
the squiggly line-- which is shown in the exam--
for a hydrocarbon chain.
So if we look at our molecule here, the first thing that we
want to understand is what the structure is.
And as you can see, the molecule has a nonpolar
component and it has a polar component.
So for the problem, it will be good to isolate the two with
different colors, and then we'll stylize them such that
the nonpolar part is the zigzag line and the polar
component is the red circle.
So we want to go ahead and draw the arrangement together
when it's added to hexane.
So the easiest thing to do is to draw a picture.
So we come over here to the right.
And this is part a.
I'm going to draw a beaker with hexane.
And I know that as soon as I look at my chemical formula
for hexane--
which is C6H14 and is a hydrocarbon--
I know that it's nonpolar.
So I know from my knowledge of 3.091, when chemical bonding,
that like dissolves like.
So the molecules that are polar are going to want to
stick together, and molecules that are nonpolar are going to
want to stick together.
So by looking at my phospholipid, I
already know that--
the problem tells me that--
I'm adding a sufficient amount such that the molecules can
interact with each other.
So the only thing that comes to mind for me that will be a
possible arrangement, would be that--
in the hexane, which is nonpolar, and if we have a
distribution of these phospholipid molecules--
then one possible arrangement will be that you have a bunch
of your polar heads interacting together with the
hydrocarbon components out here
interacting with the hexane.
Because the hydrocarbon component again is nonpolar.
The hexane is nonpolar.
So those want to interact with each other.
But the polar component, which are the
heads for the molecule--
the phospholipid as a molecule--
are going to go ahead and interact with each other.
And pretty much if you were to write this down on the exam,
you would get full points for part a.
Now we'll move on to part b.
Part b says, to the beaker described in part a, an equal
volume of liquid ammonia--
NH3--
is added.
The mixture of hexane-NH3--
or ammonia containing the phospholipid--
is now subjected to vigorous agitation for several minutes
and then allowed to equilibrate.
Draw a schematic representation depicting one
possible molecular arrangement that results.
The values of density are 0.65 grams per centimeter cubed for
hexane and 0.6 grams per centimeter
cubed for liquid ammonia.
Assume that hexane and liquid ammonia are mutually
immiscible.
So there's a lot of
information within the paragraph.
And the first thing that it gives us is the densities
between the two components.
It tells us they're immiscible, so immiscibility
means that they're not going to mix.
The question that you want to answer now is, which layer is
going to be at the top?
Which layer is going to be at the bottom?
Well, the more dense layer should be at the bottom, and
the more dense happens to be hexane.
So if I go ahead and draw my beaker again for part b, and
then I mix my components.
And now I have two components that are separated by an
immiscible boundary.
And I know that the density of hexane is given by 0.65 grams
per centimeter cubed.
And the density of liquid ammonia--
which is NH3--
is given as 0.60 grams per centimeter cubed.
So I know that the hexane is going to be down here, so I'll
go ahead and label this C6H14.
And I'll label this as NH3.
So one important thing to know is, what is the
structure of ammonia?
Well, we know that ammonia happens
to be a polar molecule.
And its structure is simply given by-- you have a central
nitrogen atom, and you have three hydrogens and a lone
pair of electrons-- and the lone pair of electrons is what
gives the ammonia the structure.
Hence, you have a polar dipole moment on your molecule.
The fact that this is polar means that it's going to
interact with the polar head of your phospholipid molecule.
And we already know we have hexane plus your phospholipid.
Well, if in our arrangement with the absence of ammonia
the heads group together because that was the only
polar component in the mixture, then over here I
should expect the heads to rearrange themselves such that
they are in contact with the ammonia.
So one possible arrangement for the system would be that I
would have my polar part of my molecule interacting with the
ammonia, and the hydrocarbon chain interacting--
which is the nonpolar component-- interacting with
the nonpolar mixture.
So if you were to draw this schematic, then you would get
full points on the exam as well.
So that's again, like dissolves like.
And if you fully understand the nature of secondary
bonding, then this problem shouldn't be that difficult to
go through.
Now there's a final part to this problem, which is part c.
And part c says, to the beaker described in part b, a small
amount of iodine--
I2--
is added to the mixture, and then subjected to vigorous
agitation for several minutes, and then allowed to
equilibrate.
Explain what you expect to observe.
Again, we're going on this whole issue of like dissolves
like, or that polar molecules want to be with polar
molecules because they should be miscible together.
And we are given that now we're going to add I2--
iodine.
Well, iodine is a linear molecule that's nonpolar.
There are no other atoms that are bonded to it that can give
any polarity.
So if I was to go ahead and draw this diagram, the first
thing that comes to my mind is that the iodine should mix
with the hexane, because essentially it's a nonpolar
component mixing with another nonpolar component.
So if I draw my diagram, I'm going to go ahead and draw the
two immiscible regions.
And one question that a lot of students have is, how do you
know that the iodine is not going to be
on top of the ammonia?
Well, iodine is pretty dense.
It's actually more dense than these two molecules over here.
So the fact that it's more dense, I know it's going to
want to sit at the bottom where the other nonpolar
molecule is at.
So I would argue that down here we would have hexane--
C6H14--
mixed with iodine.
Over here, we will still have our ammonia.
And then our polar heads would still be--
the polar part of the phospholipid molecule will be
in contact or bonding with the ammonia.
And the nonpolar tail would be with the nonpolar components
which are hexane and iodine.
So if you did this on the exam, you get full points.
And being able to fully understand secondary bonding
will help you solve this problem in no
less than 10 minutes.
So that's it.