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Topic: Identifying IMFs in Oxycodone  (Read 1930 times)

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Offline GrendelWendel

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Identifying IMFs in Oxycodone
« on: May 08, 2019, 08:29:45 AM »
https://imgur.com/a/GCpCbEk

Hello, I'm new to this forum. Can someone please help me identify the IMFs that are found in oxycodone and how these IMFs can be identified through its structure. I know that hydrogen bonds and ldfs exist in it, but I can't figure out how to relate them to its structure, and if there are more IMFs I'm missing.

Thank you!

Offline Babcock_Hall

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Re: Identifying IMFs in Oxycodone
« Reply #1 on: May 08, 2019, 10:36:06 AM »
IMO it is important to distinguish between hydrogen bond donors and hydrogen bond acceptors.  No doubt you were taught general rules for identifying them, so feel free to restate those rules now.  Which atoms in this structure could donate a hydrogen in a hydrogen bond?  Which atoms could accept a hydrogen bond?

Offline GrendelWendel

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Re: Identifying IMFs in Oxycodone
« Reply #2 on: May 08, 2019, 10:55:38 AM »
Wouldn't every H atom be a donor and over O and N atom be the acceptors?

Offline Babcock_Hall

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Re: Identifying IMFs in Oxycodone
« Reply #3 on: May 08, 2019, 11:15:10 AM »
Carbon-hydrogen bonds do not generally participate in hydrogen bonds.  Typical hydrogen bond donor atoms are O, N, and F.  There are some exceptions, but those are usually not discussed in introductory courses.  Not every O or N can accept a hydrogen bond; only those that have a lone pair of electrons.

Offline GrendelWendel

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Re: Identifying IMFs in Oxycodone
« Reply #4 on: May 08, 2019, 11:27:36 AM »
How can you determine if they have lone pairs of electrons based on the structure?

Offline Babcock_Hall

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Re: Identifying IMFs in Oxycodone
« Reply #5 on: May 08, 2019, 11:49:16 AM »
The number of lone pairs can be calculated using the formal charge on the atom, the number of valence electrons and the number of bonds.  All of the oxygen and nitrogen atoms have formal charges of zero in this structure.  FC = (number of valence electrons) - (number of bonds + number of electrons in lone pairs).

In practice some situations are so commonly encountered that one does not need to do the calculation.  For example an oxygen atom that has two bonds and a formal charge of zero has two lone pairs.  Water is an example.

EDT
There are other important intermolecular forces.  Can you define or explain what dipole-dipole forces and London dispersion forces are?
« Last Edit: May 08, 2019, 12:38:10 PM by Babcock_Hall »

Offline GrendelWendel

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Re: Identifying IMFs in Oxycodone
« Reply #6 on: May 08, 2019, 01:10:15 PM »
I know london dispersion forces are the weakest forces and are always present and that they would occur in the nonpolar areas of the molecule
dipole-dipole forces are when two polar molecules interact

Offline Babcock_Hall

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Re: Identifying IMFs in Oxycodone
« Reply #7 on: May 08, 2019, 01:37:33 PM »
London dispersion forces occur in all parts of a molecule.  Which bonds, if any, are polar in oxycodone, and how does one decide?

Offline GrendelWendel

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Re: Identifying IMFs in Oxycodone
« Reply #8 on: May 08, 2019, 02:09:41 PM »
A bond is polar when the electronegativity difference between the two atoms in a covalent bond is between .4 and 1.8. I know O-H is one polar bond in oxycodone.

Offline Babcock_Hall

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Re: Identifying IMFs in Oxycodone
« Reply #9 on: May 08, 2019, 04:50:59 PM »
What about C-O and C-N bonds?

Offline GrendelWendel

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Re: Identifying IMFs in Oxycodone
« Reply #10 on: May 08, 2019, 05:12:18 PM »
Yes those would also be polar bonds in the structure

Offline Babcock_Hall

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Re: Identifying IMFs in Oxycodone
« Reply #11 on: May 08, 2019, 05:41:04 PM »
So dipole-dipole forces come into play.

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