Destroyer Orgo #94 - can someone explain it to me?

[FROGGBUSTER]

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I don't understand the principles behind this question.

I do understand that conjugation in a double-bond system requires light of lower frequency to excite electrons from one molecular orbital into the next. This was basically all I learned in undergrad about UV spectroscopy. So all I knew was that it definitely couldn't be B. I had no idea what to make of the alkanes (A & E) and the alkyne (C).

How would you apply UV spec toward alkanes and alkynes though? What are the guiding principles?

edit: sorry, meant #94 in 2011 version!

 

[LaughingGas]

I'll put my guess here:
UV/Vis you use to excite an electron from one molecular orbital to another M.O. of higher energy. By temporarily exciting the electron from HOMO to LUMO, during the moment of electron excitement, you are "temporarily trying" to break the bond. Which molecule would be hardest to break the bond?
A,E= Single Bond
B=Conjugated Double Bond
D= Double bond
C= Triple Bond.
You know why B is not.
The easier version of my explanation is that, the higher bond strength, you need the higher energy to excite the electron, higher energy means shorter wavelength, shorter wavelength means higher frequency.
Obviously you know the bond strength, triple bond highest, then double bond, conjugated double bond, then single bond.
Triple Bond needs highest energy, so has the higher frequency
Single Bond needs less energy, so has less frequency.
This is just my logic

 

[FROGGBUSTER]

I'll put my guess here:
UV/Vis you use to excite an electron from one molecular orbital to another M.O. of higher energy. By temporarily exciting the electron from HOMO to LUMO, during the moment of electron excitement, you are "temporarily trying" to break the bond. Which molecule would be hardest to break the bond?
A,E= Single Bond
B=Conjugated Double Bond
D= Double bond
C= Triple Bond.
You know why B is not.
The easier version of my explanation is that, the higher bond strength, you need the higher energy to excite the electron, higher energy means shorter wavelength, shorter wavelength means higher frequency.
Obviously you know the bond strength, triple bond highest, then double bond, conjugated double bond, then single bond.
Triple Bond needs highest energy, so has the higher frequency
Single Bond needs less energy, so has less frequency.
This is just my logic

That makes a lot of sense, thanks for that.

Is this something you learned though or are you speculating? Maybe my textbook is not great because it only talks about double bonds in there.

 

[LaughingGas]

I'm sorry. Maybe I should stop answering your Qs. I went through gen chem, ochem book and read through, websearch and came up to an explanation that made sense to me. If not had to go with MO orbital more detailed.

 

[loveoforganic]

I'll put my guess here:
UV/Vis you use to excite an electron from one molecular orbital to another M.O. of higher energy. By temporarily exciting the electron from HOMO to LUMO, during the moment of electron excitement, you are "temporarily trying" to break the bond. Which molecule would be hardest to break the bond?
A,E= Single Bond
B=Conjugated Double Bond
D= Double bond
C= Triple Bond.
You know why B is not.
The easier version of my explanation is that, the higher bond strength, you need the higher energy to excite the electron, higher energy means shorter wavelength, shorter wavelength means higher frequency.
Obviously you know the bond strength, triple bond highest, then double bond, conjugated double bond, then single bond.
Triple Bond needs highest energy, so has the higher frequency
Single Bond needs less energy, so has less frequency.
This is just my logic

I'm digging way back in the memory, but I'm pretty sure this isn't right. Your "trick" has the right idea, but you don't consider the bond strength between two atoms - you consider the actual bond. So for a triple bond, you're looking at the energy to excite the weakest pi bond. UV spec doesn't work for alkanes

 

[LaughingGas]

😕 haha yeah . it had to do something with exciting them to pi antibonding and blablabla. ok no more answering for me lol 😉

 

[loveoforganic]

no, please continue to answer q's if you were being serious, nothing the matter with (assuming I'm right here) being incorrect now and then. probably better ingrained now if nothing else!

 

[FROGGBUSTER]

Thanks guys. Weird question, I'll just ignore it lol if it's not in my textbook I guess.