Molecular Orbital for Sn2 reaction

  • Chemistry
  • Thread starter Yokoko
  • Start date
  • #1
15
1

Homework Statement:

Question as attached to this thread. I'd like confirmation whether or not my solution to a) is correct. But I am mainly confused with b)

Relevant Equations:

/
IMG_20200802_151912__01.jpg


The equation can be seen in the picture. I was assuming it was an Sn2 reaction given the polar aprotic solvent and strong nucleophile.

For b), I assumed that the question asked me to draw the orbitals as shown above (as opposed to orbital diagram) because it asked me to identify the HOMO and LUMO.

I'm confused as to where SEt (-) attacks/how to draw the attack i.e. the reaction itself as both steps occur at the same time. Also I'm assuming that the orbital around S is the HOMO and the orbital around C the LUMO. Is that correct?

Thanks a lot!
 

Attachments

Answers and Replies

  • #2
TeethWhitener
Science Advisor
Gold Member
1,887
1,271
I was assuming it was an Sn2 reaction given the polar aprotic solvent and strong nucleophile.
Yes.
For b), I assumed that the question asked me to draw the orbitals as shown above (as opposed to orbital diagram) because it asked me to identify the HOMO and LUMO.
Honestly, it’s probably best to do both.
I'm confused as to where SEt (-) attacks/how to draw the attack i.e. the reaction itself as both steps occur at the same time.
Well, you got the stereochemistry inversion correct, so where do you think the thiolate attacks?
Also I'm assuming that the orbital around S is the HOMO and the orbital around C the LUMO. Is that correct?
It’s probably more useful to only draw the portions of the orbitals that are important for the changes that occur during the reaction.

It’s kind of tough to give you more help without seeing you put in some more effort first.
 
  • #3
15
1
15964345860285810145332600895344.jpg

15964349434356802911227774981391.jpg

Thanks for your input!
I know it's not a lot but I started working upon your prompts. In the first one is the orbital attacking the C centre with the positive charge after Br has left. Maybe it would be good to but Br- on the side too?

The second one is my attempt at the orbital diagram. Again, I'm not sure how this question wants me to answer this. This one shows the c centre bonding to a hydrogen and another sp3 hybridised C (again unsure how to draw in the second one as both methyl and ethyl are bonded by a sp3 hybridised C, right? Are they all of equal energy?), The one orbital bonding with SEt is missing too as of now...

Thanks again
 
  • #4
TeethWhitener
Science Advisor
Gold Member
1,887
1,271
after Br has left.
It’s SN2. Should Br have left?


The second one is my attempt at the orbital diagram. Again, I'm not sure how this question wants me to answer this.
I agree, the question isn’t particularly clear. I’ll try to focus on the most important conceptual takeaways. Instead of focusing on the hybridization of the orbitals, it’s probably better to focus on whether the interacting orbitals are bonding, non-bonding, or antibonding. Bring those ideas into your answer and see what you can come up with.
 
  • #5
15
1
15965384040621560329406741974335.jpg

15965384787463571689068384547697.jpg

I think I've come a bit farther this time. The bonding/anti bonding hint really helped! :)

Two questions: is there only one or two e- in the orbital of SEt- ? It is a lone pair, so there would be 2 e-, correct?

And would the orbital diagram drawn enough to answer the question combined with the sketch?

Thanks again!
 
  • #6
TeethWhitener
Science Advisor
Gold Member
1,887
1,271
Getting closer. Look at the MO diagram in your second picture. Is the sigma antibonding orbital higher or lower in energy than the sigma bonding orbital?
Also, a Lewis structure should give you an idea of whether the thiolate anion has a lone pair or an unpaired electron.
 
  • #7
15
1
Oh you are right, the C would be lower in energy, therefore under the SEt orbital.

And since the bond with sodium breaks, it seems like it only has an unpaired electron left
 
  • #8
TeethWhitener
Science Advisor
Gold Member
1,887
1,271
Oh you are right, the C would be lower in energy, therefore under the SEt orbital.
I meant the MOs that are formed by the interaction of the sulfur with the carbon. You have the antibonding orbital lower in energy than the bonding orbital in your picture.
And since the bond with sodium breaks, it seems like it only has an unpaired electron left
Look closer. The S has a negative charge. It really might help to draw out the Lewis dot structure and actually count the electrons around the sulfur.
 
  • #9
15
1
IMG_20200804_140228__01.jpg

So, this should be right now. I forgot about the negative charge on the sulfur, I think that was my mistake.
 
  • Like
Likes TeethWhitener
  • #10
TeethWhitener
Science Advisor
Gold Member
1,887
1,271
Better. I might have confused you before: the occupied sulfur HOMO is definitely lower in energy than the C LUMO. Otherwise it would be energetically favorable simply to transfer electrons to the lower energy empty orbital than to form a new set of MOs.

The biggest conceptual takeaway is this: the orbital that’s being created by the interaction of the thiolate with the carbon centered nucleofuge is a combination of the lone pair sulfur orbital and the C-Br antibonding orbital. The “bonding” orbital that ends up being formed by this combination has strong bonding character with the sulfur and strong antibonding character with the bromine. This is ultimately the reason that the C-Br bond breaks and the C-S bond forms.
 
  • #11
15
1
That makes sense! Thank you so much for help, I know it took a while!
 

Related Threads on Molecular Orbital for Sn2 reaction

  • Last Post
Replies
5
Views
12K
  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
7
Views
4K
Replies
2
Views
1K
Replies
1
Views
2K
  • Last Post
Replies
1
Views
6K
Replies
1
Views
781
  • Last Post
Replies
2
Views
2K
Replies
0
Views
4K
  • Last Post
Replies
6
Views
4K
Top