[newkiddonblock]

What would methanolic HBr solution give you?

You mean an epoxide reaction with HBr and methanol?

[discodermolide]

Yes.

[newkiddonblock]

Yes.

[newkiddonblock]

Wait but you didn't answer my question though

Was using Br₂ to attack the less sterically hindered carbon correct/incorrect?

[orgopete]

Hint, methylcyclohexene plus Br2/H2O gives 2-bromo-1-methylcyclohexanol with the bromo and hydroxyl trans.

[newkiddonblock]

Hint, methylcyclohexene plus Br2/H2O gives 2-bromo-1-methylcyclohexanol with the bromo and hydroxyl trans.

For which one?

[orgopete]

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.

[newkiddonblock]

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.

[orgopete]

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.

[newkiddonblock]

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?

[newkiddonblock]

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)

[newkiddonblock]

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")

[orgopete]

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.

[newkiddonblock]

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

[newkiddonblock]

Ok...these are my questions...