Chemical Forums
Chemistry Forums for Students => Organic Chemistry Forum => Topic started by: theanonymous on October 16, 2012, 07:03:16 PM
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5. Please provide a mechanistic explanation for the following observation:
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This is what I wrote down (I took it from my notes, but I'm not 100% sure if it's entirely accurate...also, I'm guessing my teacher also wants me to briefly explain what's happening in words)
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How would I go about explaining in words what's going on?
Second structure I've drawn: Hydrogen atom (H+) moves at the bottom right corner to make way for the incoming nucleophile Br-; a carbocation (C+) is formed in the process
Third structure I've drawn: You now have 2 hydrogen atoms at the bottom (one pointing towards you and one pointing away from you); nucleophile attacks and a C-Br bond is made
Fourth structure of drawn: The result
^
Is that fine (and accurate)?
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Your reaction mechanism isn't correct. I'll give you the first step, protonate the alcohol which will ultimately allow dehydration and the formation of a pi bond between carbons 1 and 2. See what you can do from there.
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Your reaction mechanism isn't correct. You need to show the alcohol on carbon 1
Wait what?
I don't understand :(
Can you show me?
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Sorry, I accidentally hit enter, look at my post above!!
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Your reaction mechanism isn't correct. I'll give you the first step, protonate the alcohol which will ultimately allow dehydration and the formation of a pi bond between carbons 1 and 2. See what you can do from there.
Uhhh... :-[
I'll try
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ughghghgh
I'm not doing this right...
What do you mean by protonate and making the C1 and C2 bonds form a pi bond??
Is there a way for you to draw it on here?
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So your alcohol doesn't have negative charge, it is neutral originally. The hydrogen in HBr leaves Br (leaving a Br anion and a H cation or proton) and is accepted by the alcohol.
So using your drawing, show an arrow from the oxygen on the alcohol to the H in HBr, while showing that the H in HBr is leaving Br. This will give you OH2 as you have shown, but the oxygen will be protonated.
What step would occur next to cause the water to leave (dehydration) and a pi bond to form between carbons 1 and 2 at the same time?
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So your alcohol doesn't have negative charge, it is neutral originally. The hydrogen in HBr leaves Br (leaving a Br anion and a H cation or proton) and is accepted by the alcohol.
So using your drawing, show an arrow from the oxygen on the alcohol to the H in HBr, while showing that the H in HBr is leaving Br. This will give you OH2 as you have shown, but the oxygen will be protonated.
What step would occur next to cause the water to leave (dehydration) and a pi bond to form between carbons 1 and 2 at the same time?
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^
The methy groups will move/shift to the bottom (?)
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Nope! What is on carbon 2 besides methyl? Hydrogen, right? Could the bonding electrons of that hydrogen form the pi bond and then that hydrogen become a proton in solution? Show how that would work.
Note: Make sure to show the positive charge on the oxygen of the -OH2 group while it is still attached to the parent structure!
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Nope! What is on carbon 2 besides methyl? Hydrogen, right? Could the bonding electrons of that hydrogen form the pi bond and then that hydrogen become a proton in solution? Show how that would work.
Note: Make sure to show the positive charge on the oxygen of the -OH2 group while it is still attached to the parent structure!
Ok, I have a quick question... "Carbon 2"... is that the carbon on the bottom of the vertical line.. you know where ––––– OH2 is.... Carbon 2 would be located here right?: C2 ––––––– OH2
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If you were naming this compound, the alcohol would be considered more electronegative or polar than the methyl group, thus the carbon where the alcohol is would be considered #1. Then you rank other R-groups as necessary, here the methyl group is attached to carbon #2.
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If you were naming this compound, the alcohol would be considered more electronegative or polar than the methyl group, thus the carbon where the alcohol is would be considered #1. Then you rank other R-groups as necessary, here the methyl group is attached to carbon #2.
Ok...
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Ughh...
:(
This is hopeless.
I'm not good at this yet and this is probably taking longer than it should because my teacher never elaborated on /how/ to do these problems
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You still there?
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You are closer!
In your drawing, you have the bonding electrons moving to make that pi bond. But, as that pi bond is forming there would be 5 bonds on carbon 1, which means that the water would have to be leaving as the pi bond is forming.
Your first drawing should have an arrow showing water leaving. Your second drawing should not have water, but should have a pi bond and no charge on carbon 2.
Now, if you draw what I said correctly, you can show the pi bond electrons attacking a proton in solution (which formed from the first drawing you have here), which will put the hydrogen on carbon 1. This will leave a positive charge on carbon 2, and will show that the last segment of your drawing here is correct giving the final product.
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You are closer!
In your drawing, you have the bonding electrons moving to make that pi bond. But, as that pi bond is forming there would be 5 bonds on carbon 1, which means that the water would have to be leaving as the pi bond is forming.
Your first drawing should have an arrow showing water leaving. Your second drawing should not have water, but should have a pi bond and no charge on carbon 2.
Now, if you draw what I said correctly, you can show the pi bond electrons attacking a proton in solution (which formed from the first drawing you have here), which will put the hydrogen on carbon 1. This will leave a positive charge on carbon 2, and will show that the last segment of your drawing here is correct giving the final product.
Ohhh :D
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Awesome!!
Thanks!!
I have one more question like this - Imma try to do it on my own first and get back to you if youlll still be here!
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95% Correct.
Make sure that arrow in the first drawing is from the BOND (electrons) to the oxygen, showing that those electrons went to the oxygen, not ambiguously that the -OH2 just left.
The second drawing should show those pi electrons reaching out and grabbing a H+, not that the pi electrons just jumped to what was carbon 1 and a hydrogen just all the sudden appeared (show where from).
Also, after that first drawing there is no need to show stereochemistry (no dashed line on the methyl) because that methyl is now in a trigonal planar electronic geometry. The bromine anion could attack from the front or the back, which is why the product given in the question is just solid lines, since you cannot be completely confident as to whether the bromine or methyl is in the front or the back (there will be a racemic mixture).
I will check back periodically to help you out.
And don't give up, it isn't that you won't ever understand. O-chem takes time, never feel dumb or that you shouldn't ask questions (especially ask a lot to your professor), because I can bet you that there are many many others that feel the same.
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95% Correct.
Make sure that arrow in the first drawing is from the BOND (electrons) to the oxygen, showing that those electrons went to the oxygen, not ambiguously that the -OH2 just left.
The second drawing should show those pi electrons reaching out and grabbing a H+, not that the pi electrons just jumped to what was carbon 1 and a hydrogen just all the sudden appeared (show where from).
Also, after that first drawing there is no need to show stereochemistry (no dashed line on the methyl) because that methyl is now in a trigonal planar electronic geometry. The bromine anion could attack from the front or the back, which is why the product given in the question is just solid lines, since you cannot be completely confident as to whether the bromine or methyl is in the front or the back (there will be a racemic mixture).
I will check back periodically to help you out.
And don't give up, it isn't that you won't ever understand. O-chem takes time, never feel dumb or that you shouldn't ask questions (especially ask a lot to your professor), because I can bet you that there are many many others that feel the same.
Ohh ok!
Like this?
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That's it!
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Thanks!!!!
Last question:
6. 6. The reaction of 2,2-dimethyl-1-propanol with HBr is very slow and gives 2-bromo-2-methylbutane as the major product.
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Give a mechanistic explanation for these observations.
EDIT:
OH WAIT!!!!
I think I got the answer!
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Does that look right?? ;D
I'm so happy right now!
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Always remember that you have the arrowheads pointing towards the positive charges, so the oxygen should be accepting that proton (not the proton accepting the oxygen, since which is bigger and more electronegative? - oxygen is).
That oxygen becomes protonated. The bond (electrons) between the oxygen and carbon go to the oxygen to give water and a positive charge on that carbon.
A carbocation rearrangement occurs where the "bottom" methyl shifts to the positive carbon and ends up giving a positive charge on the "middle" carbon.
This allows for the bromine anion to attack that "middle" carbon and form the final product.
Show what I just said in words with pictures.
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Always remember that you have the arrowheads pointing towards the positive charges, so the oxygen should be accepting that proton (not the proton accepting the oxygen, since which is bigger and more electronegative? - oxygen is).
That oxygen becomes protonated. The bond (electrons) between the oxygen and carbon go to the oxygen to give water and a positive charge on that carbon.
A carbocation rearrangement occurs where the "bottom" methyl shifts to the positive carbon and ends up giving a positive charge on the "middle" carbon.
This allows for the bromine anion to attack that "middle" carbon and form the final product.
Show what I just said in words with pictures.
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^ Like that?
Should the CH2 still be there though?
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1) No -- in your first picture the arrow should go from the oxygen to the proton because the oxygen is accepting the proton because of it's electronegativity.
There are five carbons in a "wheel" like formation, just for understanding, lets call them top, left, bottom, right, and center for the names of the carbon (you should be able to see which is which).
2) Now that the oxygen (alcohol) is protonated, the bond (electrons) between the right carbon and the alcohol go to the oxygen to form water and a positive charge on the right carbon.
3) The bond (electrons) between the center and bottom carbons now move in a carbocation rearrangement to the right carbon (the one with the positive charge).
4) There should now be a positive charge on the middle carbon that the bromine anion attacks (like you have shown in your last mechanistic step).
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1) No -- in your first picture the arrow should go from the oxygen to the proton because the oxygen is accepting the proton because of it's electronegativity.
There are five carbons in a "wheel" like formation, just for understanding, lets call them top, left, bottom, right, and center for the names of the carbon (you should be able to see which is which).
2) Now that the oxygen (alcohol) is protonated, the bond (electrons) between the right carbon and the alcohol go to the oxygen to form water and a positive charge on the right carbon.
3) The bond (electrons) between the center and bottom carbons now move in a carbocation rearrangement to the right carbon (the one with the positive charge).
4) There should now be a positive charge on the middle carbon that the bromine anion attacks (like you have shown in your last mechanistic step).
Like this?? :o
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Almost, In your second picture there should be no positive charge on the carbon and that bottom methyl should not be doing anything yet -- you should only show the water leaving.
One extra step between your second picture and third picture would be the formation of a positive charge on that right carbon due to the water leaving. Then, you should show the electrons of that bottom methyl moving to the positive charge on the right carbon. Do not show it going from carbon to carbon, it is the ELECTRONS (the bond!!!) going to carbon.
See if you can do that.
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Almost, In your second picture there should be no positive charge on the carbon and that bottom methyl should not be doing anything yet -- you should only show the water leaving.
One extra step between your second picture and third picture would be the formation of a positive charge on that right carbon due to the water leaving. Then, you should show the electrons of that bottom methyl moving to the positive charge on the right carbon. Do not show it going from carbon to carbon, it is the ELECTRONS (the bond!!!) going to carbon.
See if you can do that.
Like this?
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You got it!
*Note: It would be more accurate if you showed the bond (the electrons) in HBr going to bromine leaving just the proton.
So H-Br show an arrow from the dash to bromine giving a bromine anion and a proton which is then used in the mechanism.*
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You got it!
*Note: It would be more accurate if you showed the bond (the electrons) in HBr going to bromine leaving just the proton.
So H-Br show an arrow from the dash to bromine giving a bromine anion and a proton which is then used in the mechanism.*
Awesome!!!! ;D
Thanks a bunch!!!
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Anytime!
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Your reaction mechanism isn't correct. I'll give you the first step, protonate the alcohol which will ultimately allow dehydration and the formation of a pi bond between carbons 1 and 2. See what you can do from there.
So, you are saying that if 2-deutero-2-methylcyclohexanol were used, the product would lose all deuterium atoms and no migration would occur?
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So, you are saying that if 2-deutero-2-methylcyclohexanol were used, the product would lose all deuterium atoms and no migration would occur?
No, in the step where the pi electrons undergo electrophilic attack to a proton, this could have been a deuterium if the original reactant had deuterium. Deuterium wasn't present though. Isn't that right?
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He is probably referring to the fact that there will never be a double bond in that mechanism, but a hydride shift similar to the methyl shift you explained in the second example.
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He is probably referring to the fact that there will never be a double bond in that mechanism, but a hydride shift similar to the methyl shift you explained in the second example.
Correct, just because BetaAmyloid said it goes by the alkene does not make that true. It could have been true, but if a methyl group and hydrogen migrate similarly, do you think an alkene is the intermediate?