3 Orgo Questions

[prsndwg]

Hi,

Can somebody help me out with these please..


Can you explain figure D to me? I am not understanding it! I know C is not planar and thats why it's not anti-planar!

How is choice A (R) and choice C (S)? I thought the exact opposite about both!!!

Why is this not A?

Thanks in advance

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[suhshi]

for the first question

i think C is anti aromatic because 4n+2=8
thus n is an integer, making it antiaromatic

im not sure about D


and the second question, the lowest number substituents are coming at you so you gotta switch the configurations


i dont kno bout third


correct me please if im wrong somebody

 

[Dental Jarry]

Hi,


How is choice A (R) and choice C (S)? I thought the exact opposite about both!!!

Thanks in advance

For choice A, shen you put the elements in order you should get 1)Br 2)COOH 3)CH3 4)H ..You want the H(least priority) to be facing away from you so you switch it with the CH3...When you do this your priority should be going counterclockwise(S), but since you interchanged one set you flip the orientation so it becomes R.

Same with choice C. The priority should be 1)F 2)NH2 3)COOH and 4)CH3
Again you want the one with least priority to face away from you so you interchange CH3 with COOH. You always want the one with least priority on the verticle line. After you do this the priority from 1 to 4 will be clockwise but since you interchanged one set you flip this and it becomes counterclockwise(S).

 

[loveoforganic]

The third question is B because it must proceed by antiperiplanar configuration (E2 reaction).

 

[Dental Jarry]

And for the first one, for something to be antiaromatic it must have 4n pi electrons, be cyclic, conjugated, and planar. Choice A is not conjugated, choice B is aromatic, Choice C should be correct and I don't really know whats going on with choice D

 

[loveoforganic]

Choice D is some sort of insanely unstable diradical. But it's antiaromatic I guess, 4 pi electrons, probably planar.

 

[Shinpe]

So cyclooctatetraene is not anti-aromatic because it's not planar?? It's just non-aromatic?

 

[r8 dds]

Hi,

Why is this not A?

Thanks in advance

Usually, you would get A because CH3OH/OCH3- is not sterically hindered, or bulky, meaning you would get a double bond in the most substituted position or the answer A - correct?

For this one, I thought because it's a B-elimination, you remove the Beta proton, giving the answer B, rather than A.

I don't know correct me if I'm wrong.

 

[loveoforganic]

E2 reactions proceed by antiperiplanar configuration, preferentially.

So cyclooctatetraene is not anti-aromatic because it's not planar?? It's just non-aromatic?

Correct. What I've gathered from the people here is that this is the only exception you need to know as far as planar/nonplanar.

 

[r8 dds]

E2 reactions proceed by antiperiplanar configuration, preferentially.



Correct. What I've gathered from the people here is that this is the only exception you need to know as far as planar/nonplanar.

Yeah that question kinda confused me too. How'd you know it's not planar?

 

[prsndwg]

Yeah that question kinda confused me too. How'd you know it's not planar?

i think when you draw the hydrogen bonds, they will push eachother away and casue the cyclo shape to get messed up!

 

[prsndwg]

I remember these from when I took that test... did you read the explanations? I seem to remember them explaining it fairly well.

yeah, but i didnt get it

 

[prsndwg]

E2 reactions proceed by antiperiplanar configuration, preferentially.



Correct. What I've gathered from the people here is that this is the only exception you need to know as far as planar/nonplanar.

can you please explain more the E2 proceeds by antiperiplaner please.. I still dont get why it does want the more subst double bond!

 

[Shinpe]

i think when you draw the hydrogen bonds, they will push eachother away and casue the cyclo shape to get messed up!

Hmm, do you mean C-H bonds? 'cause there's no hydrogen bonds here (no O,N,F).

 

[prsndwg]

Hmm, do you mean C-H bonds? 'cause there's no hydrogen bonds here (no O,N,F).

thats what i meant.. C-H bonds

 

[2PacClone23]

The second question kinda sucks because it didn't mention the fact that they were using Fischer Projections. Kinda had to assume that based on the look.

 

[Shinpe]

can you please explain more the E2 proceeds by antiperiplaner please.. I still dont get why it does want the more subst double bond!

Double bonds on cyclohexenes can only undergo elimination when the leaving group (Br here) and the H being eliminated are both in axial position ('cause thats the only way they can be antiperiplanar) and wont happen if either of them is equitorial. As you can see in the figure, when Br is axial, the more substituted adjacent carbon has its H equitorial and CH3 axia, so that can't be eliminated because it's not axial. Only the other adjacent carbon has an H that is axial and can undergo elimination by anti-peri-planar.

 

[Shinpe]

The second question kinda sucks because it didn't mention the fact that they were using Fischer Projections. Kinda had to assume that based on the look.

Bind-line formulas don't have any information on 3D bond positions with respect to each other, so you would have to assume it's Fischer. There is no other option.

 

[loveoforganic]

Yeah that question kinda confused me too. How'd you know it's not planar?

That kind of stuff has to be determined as far as I know, through experiment. I knew it wasn't planar because I've seen that example a whole lot. The fact is that antiaromaticity rarely actually happens - even cyclobutadiene contorts itself in such a way as to avoid being antiaromatic.


The reason antiperiplanar conformation is preferred for E2 reactions is because that conformation taking place as the transition state most closely models the orbitals formed in the product.

 

[Dental Jarry]

Doesnt something have to have all sp2 hybridized carbons to be planar?

 

[r8 dds]

Double bonds on cyclohexenes can only undergo elimination when the leaving group (Br here) and the H being eliminated are both in axial position ('cause thats the only way they can be antiperiplanar) and wont happen if either of them is equitorial. As you can see in the figure, when Br is axial, the more substituted adjacent carbon has its H equitorial and CH3 axia, so that can't be eliminated because it's not axial. Only the other adjacent carbon has an H that is axial and can undergo elimination by anti-peri-planar.

Ahh it makes sense! Because the methyl group is in the axial position, it can't be eliminated! Thus, the other H, in the axial position, must be eliminated. The Br is positioned up, and the other H is pointing down. Bravo my friend, Bravo.

 

[Shinpe]

Bravo my friend, Bravo.

Lol Thanks and no problem at all .....

 

[Onix]

bump

 

[whawha]

How can D be cyclobutadiene? Can we draw it like that too??

 

[prsndwg]

Ahh it makes sense! Because the methyl group is in the axial position, it can't be eliminated! Thus, the other H, in the axial position, must be eliminated. The Br is positioned up, and the other H is pointing down. Bravo my friend, Bravo.

Oh s**t, now i get it.. thanks

 

[alanan84]

Correct. What I've gathered from the people here is that this is the only exception you need to know as far as planar/nonplanar.

My Kaplan book has cyclooctatetraene as antiaromatic. So just to be sure, cyclooctatetraene IS NOT antiaromatic right?

 

[prsndwg]

My Kaplan book has cyclooctatetraene as antiaromatic. So just to be sure, cyclooctatetraene IS NOT antiaromatic right?

It is not planar and so not aromatic

 

[enfuego]

Yeah that question kinda confused me too. How'd you know it's not planar?

Because if it were planar, it would be anti-aromatic. Molecules bend to avoid being anti-aromatic because it's unstable. Thus, it's folded and not planar. The answer is D, correct? I believe the double bond there forces that small molecule to be planar.

My Kaplan book has cyclooctatetraene as antiaromatic. So just to be sure, cyclooctatetraene IS NOT antiaromatic right?

It looks like this: http://en.wikipedia.org/wiki/File:COT-all-cis.png
It's not planar, so it's non-aromatic.

 

[The real deal]

#3 gives hoffman elimination because the Br and H are antiperiplanar- theyre both axial on opposite sides, whereas the CH3 is equatorial to the Br, so even though CH30- is not a bulky base the H is still less sterically hindered than the CH3 so it goes for the H