Chemical Forums
Chemistry Forums for Students => Organic Chemistry Forum => Topic started by: newkiddonblock on December 08, 2012, 08:22:56 AM
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:'(
Would using Br and H2O work for the first one?
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Please post your attempts in order to receive any help. Forum rules (http://www.chemicalforums.com/index.php?topic=33740.0).
And what do you mean Br and H2O? what brominating agent are you using here?
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Please post your attempts in order to receive any help. Forum rules (http://www.chemicalforums.com/index.php?topic=33740.0).
And what do you mean Br and H2O? what brominating agent are you using here?
Sorry about that...
This is what I did for the first reaction:
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What about the second reaction though?
I don't get it
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You have "epoxidised" the double bond. But in your subsequent products it is still there. Remove it and see how you get on.
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You have "epoxidised" the double bond. But in your subsequent products it is still there. Remove it and see how you get on.
Oh oops...ok I removed it...
Wait but for the second one.
I know it's an epoxide ring opening with a base. But how do I determine the missing components of that reaction??
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This is what i did...
How did they get the CH3O on the top of the product though??
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Oxygen has to be protonated from H+(compound-)
The other missing component (after Br2) would have to be H+CH3-?
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Two things, watch your stereochemistry, and if you have a methyl ether formed then there must be methanol present.
Why are you using Br2? Do you have to? Can you use HBr?
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Two things, watch your stereochemistry, and if you have a methyl ether formed then there must be methanol present.
Why are you using Br2? Do you have to? Can you use HBr?
Well I just wouldn't be using H-Br because the oxygen won't get protonated right away... Also, this isn't an acid epoxide reaction...
The nucleophile from Br-Br will readily attack the less sterically hindered carbon.
And about the whole methyl ether produced from methanol thing...
So, whenever I have a problem like this, where I have something at the top right beside oxygen (O), do I always ask myself what compound that can be derived from?
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Two things, watch your stereochemistry, and if you have a methyl ether formed then there must be methanol present.
Why are you using Br2? Do you have to? Can you use HBr?
Also, why would the answer be CH3OH instead of CH3-H??
I don't get it.
Because if the Oxygen in the diagram I drew wants to be protonated, then you'd just need CH3 (the molecule in the product) attached to H+, no?
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Why should bromine attack an epoxide? To get ring opening you will have to protonate the epoxide oxygen atom.
And again if you form a methyl ether methanol must be present.
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Two things, watch your stereochemistry, and if you have a methyl ether formed then there must be methanol present.
Why are you using Br2? Do you have to? Can you use HBr?
Also, why would the answer be CH3OH instead of CH3-H??
I don't get it.
Because if the Oxygen in the diagram I drew wants to be protonated, then you'd just need CH3 (the molecule in the product) attached to H+, no?
CH3 attached to H is methane!!
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Why should bromine attack an epoxide? To get ring opening you will have to protonate the epoxide oxygen atom.
And again if you form a methyl ether methanol must be present.
Idk :(
That's what my teacher said in his solution key: the answer on top of the arrow says
Br2/CH3OH
So that's what I assumed...
Is that assumption wrong?
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Why should bromine attack an epoxide? To get ring opening you will have to protonate the epoxide oxygen atom.
And again if you form a methyl ether methanol must be present.
Ok so that's like a general rule...
If you form methyl ether (CH3O) :rarrow: methanol (CH3OH) must be present
Is it the same as...
If you form OH :rarrow: an acid (H-Nucleophile) must be present
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What would methanolic HBr solution give you?
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What would methanolic HBr solution give you?
You mean an epoxide reaction with HBr and methanol?
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Yes.
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Yes.
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Wait but you didn't answer my question though :(
Was using Br2 to attack the less sterically hindered carbon correct/incorrect?
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Hint, methylcyclohexene plus Br2/H2O gives 2-bromo-1-methylcyclohexanol with the bromo and hydroxyl trans.
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Hint, methylcyclohexene plus Br2/H2O gives 2-bromo-1-methylcyclohexanol with the bromo and hydroxyl trans.
For which one?
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Hint, methylcyclohexene plus Br2/H2O gives 2-bromo-1-methylcyclohexanol with the bromo and hydroxyl trans.
For which one?
That illustrates a principle that needs to be applied. What reagents answers #1? That should help to answer #2.
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Hint, methylcyclohexene plus Br2/H2O gives 2-bromo-1-methylcyclohexanol with the bromo and hydroxyl trans.
For which one?
That illustrates a principle that needs to be applied. What reagents answer #1? That should help to answer #2.
1st one: HBr and H3CCO3H
That's what confuses me for the second one because the answers says:
Br2/H3COH
???
What about H3CCO3H?? Is that not in there as well? I thought this was also an epoxidation reaction.
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The first one is to make the epoxide first. Open the epoxide with HBr because Br2/H2O gives the opposite isomer. Opening a bromonium results in attack at the more stable carbocation. If methanol replaces water, then methanol can attacked the bromonium ion at the more stable carbocation position.
Nucleophilic opening are sensitive to steric effects.
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The first one is to make the epoxide first. Open the epoxide with HBr because Br2/H2O gives the opposite isomer. Opening a bromonium results in attack at the more stable carbocation. If methanol replaces water, then methanol can attacked the bromonium ion at the more stable carbocation position.
Nucleophilic opening are sensitive to steric effects.
Wait wait...i have to visualize this...The part I bolded...you're talking about the second one right?
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The first one is to make the epoxide first. Open the epoxide with HBr because Br2/H2O gives the opposite isomer. Opening a bromonium results in attack at the more stable carbocation. If methanol replaces water, then methanol can attacked the bromonium ion at the more stable carbocation position.
Nucleophilic opening are sensitive to steric effects.
Ok I'm lost :(
Can you tell me step by step where things go in the second reaction (the diagram I drew)
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Oh my God this is getting more and more confusing...
Ok ok...
I'm a very logical thinker and I need to put all this stuff into steps rather than a paragraph (if you don't mind)
First thing first, WHICH one attacks the less sterically hindered carbon at the bottom? methanol or Br-?
Because these are my notes for epoxidation reactions with BASES (from many youtube videos)
1) the nucleophile readily attacks the less substituted carbon atom
2) the C-O bond at THAT site is broken and a ring opened species is formed
3) the oxygen is then protonated by the conjugate acid of the nucleophile (?)
4) You end up with a final ring opened product with the nucleophile at the bottom right corner, and the OH on the top
If that's right, then how do I translate that into what you just said and into pictures.
Focusing on the second reaction
1) Br2 is Br+ + Br-
2) Br-, the nucleophile, attacks the less sterically hindered carbon (the most stable), but the ring can only be opened by an H atom (according to the user "discoder"???)
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Oh my God this is getting more and more confusing...
Ok ok...
I'm a very logical thinker and I need to put all this stuff into steps rather than a paragraph (if you don't mind)
First thing first, WHICH one attacks the less sterically hindered carbon at the bottom? methanol or Br-?
Because these are my notes for epoxidation reactions with BASES (from many youtube videos)
1) the nucleophile readily attacks the less substituted carbon atom
2) the C-O bond at THAT site is broken and a ring opened species is formed
3) the oxygen is then protonated by the conjugate acid of the nucleophile (?)
4) You end up with a final ring opened product with the nucleophile at the bottom right corner, and the OH on the top
If that's right, then how do I translate that into what you just said and into pictures.
Focusing on the second reaction
1) Br2 is Br+ + Br-
2) Br-, the nucleophile, attacks the less sterically hindered carbon (the most stable), but the ring can only be opened by an H atom (according to the user "discoder")
I too am becoming confused by the above. Reaction 1 and 2 are electrophile controlled and 3 & 4 are nucleophile controlled. If electrophile controlled (SN1-like), then nucleophile attacks carbon best able to stabilize carbocation (even if a bromonium, protonated epoxide, or other equivalents). If nucleophile controlled (SN2-like), the steric hindrance controls regio and stereochemistry. Attack at least substituted carbon.
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I too am becoming confused by the above. Reaction 1 and 2 are electrophile controlled and 3 & 4 are nucleophile controlled. If electrophile controlled (SN1-like), then nucleophile attacks carbon best able to stabilize carbocation (even if a bromonium, protonated epoxide, or other equivalents). If nucleophile controlled (SN2-like), the steric hindrance controls regio and stereochemistry. Attack at least substituted carbon.
I'm asking if you could clarify the pushing mechanism for me...
Hold on let me upload a picture
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Ok...these are my questions...
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