[pfnm]

My lecture notes say, in acid, that an epoxide is first protonated, then the Nucleophine attacks at the most substituted carbon of the unsymmetrical epoxide by SN2.

However, the notes say later, 'However, Br- attacks the LEAST substituted carbon."

So why is this? Are the notes wrong? Why does Br- behave differently, from say, I- which would attack the most substituted carbon?

Thanks

[Dan]

The dominant factor is whether the conditions are acidic or basic. Where the notes appear to contradict themselves are they still describing the reaction with HBr?

In acidic conditions, the epoxide is protonated resulting in C-O bond weakening at the C which can best stabilise the build up of positive charge (normally the most substituted). The nucleophile attacks this partial positive charge. This mechanism can be viewed as somewhere between S_N1 and S_N2 - Clayden/Greeves/Warren/Wothers call this a "loose S_N2".

In base (or the absence of an acid) this bond weakening does not occur and the nucleophile will displace at the less hindered position at a higher rate by a straight S_N2 reaction.

You might find this video helpful.

[pfnm]

Thanks Dan,

Yes the notes appear to contradict themselves. Here's the slide that says Br attacks at the least substituted position.


Thanks for the video link too. I've made some flash cards using that info, to help me remember.

[pfnm]

I asked the lecturer today about this:

He said:

In epoxide ring opening, even with acid, a strong nucleophile attacks the least-hindered position.

So the epoxide is protonated, then, the Br- ion attacks the least hindered carbon of the epoxide (for steric reasons). The reaction occurs by SN2. He said it's the same for all good nucleophiles in acid.

He said weak nucleophiles, eg MeOH would first protonate the oxygen. Then the C-O bond breaks to form the most substituted carbocation, and the MeO- nucleophile reacts there.

He said nucleophiles in the absence of acid, react at the least hindered carbon.

---

In our tutes there's a question:
1,2-epoxypentane reacts with HBR, what's the product?

I said 1-bromo-2-pentanol (eg the Br having attacked at the less substituted, less hindered carbon).

He said that was correct....

But every other place I search, says good Nucleophiles in acid attack the least substituted carbon in epoxide-ring opening reactions.

Do they attack at the least, or the most substituted carbon? Or do they attack at the least hindered position, regardless of how substituted it is?

[azmanam]

Do what your prof says, cuz he's giving you the grade... but I agree with Dan that you're prof's wrong, and epoxide + HBr = Br- attack at most substituted carbon atom.

I also disagree when your prof says that under acidic conditions the epoxide opens, then the nucleophile attacks the resulting carbocation.  If that were true, we'd expect a mixture of stereocenters at that position when the "weak" nucleophile attacks the carbocation from either the top face or the bottom face.  We typically do not see this, rather we see the sterospecifically anti arrangement of products - suggesting no free carbocation.

But, again, do whatever your prof says to do on the exam.  Know in your heart that something else is probably going on, then move on with your life   And check back here frequently if your notes contradict the book or other websites some more

ps: GAA!  she's writing on sheet music!  OH THE HUMANITY!

[orgopete]

I'm with azmanam. Your professor is wrong. If bromide is used in the absence of acid, it won't react with an epoxide. The reverse reaction is preferred. A bromo alcohol when treated with hydroxide will generate an epoxide. The bromide leaving group will not reverse the reaction (in the absence of acid).

[pfnm]

As Dan mentioned, in Clayden, Greeves, et al, pg 513, they say in acid-catalysed ring opening, "protonation by acid produces a positively charged intermediate" which they call a loose SN2 transition state.

(The reaction is of 2,3-epoxy-2-methylpropane with MeOH, HCl)

Says the book, "with the acid catalysed reaction...opening occurs at the most substituted end...the two alkyl groups make possible a build up of charge on the carbon at the tertiary end of the protonated epoxide, and methanol attacks there, just as it does in the bromonium ion."

The diagram says the positive charge in the transition state (on the tertiary carbon) is stabilised by the alkyl groups.

On page 514,

"But with epoxides, regioselectivity is not as simple as this because, even with acid catalysts, SN2 substitution at a primary centre is very fast.

For example, Br- in acid attacks this epoxide (now they talk of 2,3-epoxypropane) mainly at the less substituted end, and only 24% of the product is produced by the 'cation-stabilised' pathway. It is very difficult to override the preference of epoxides unsubstituted at one end to react at that end."

The major product here is 3-bromo-2-propanol and the minor product shown is 2-bromo-1-propanol

---

So could I say:

- Nucleophiles in acid attack the MOST substituted end of an epoxide....if it's the same substitution, then Nucleophiles attack the least hindered position....BUT

- in an epoxide that's unsubstituted at one end, eg 2,3-epoxypropane, a Nucleophile in acid will attack the unsubstituted end? Or just Br- in acid?

Or is this an error perhaps in the book? (Its a large book, there are probably a few errors in it)

The text adds,
"For most substitution reactions of epoxides, then, regioselectivity is much higher if you give in to the epoxide's desire to open at the less substituted end, and enhance it with a strong nucleophile under basic conditions."


Thanks very much - and yep I'll do that azmanam if there are more contradictions

[pfnm]

McMurry's 'Organic Chemistry' (pg 651 of the 'International Student Edition, 6e') says:

"When both epoxide carbon atoms are either primary or secondary, attack of the nucleophile occurs primarily at the LESS substituted site - an SN2 like result.

When one of the epoxide carbon atoms is tertiary, however, a nucleophilic attack occurs primarily at the MORE highly substituted site - an SN1 like result."

[orgopete]

Well, well, well, in light of data, I stand corrected. Your professor was correct and I was wrong. So if the potential carbocation is secondary, it does not offer enough stability to dominate the product ratio. I have made up two schemes to show the reactions in question. (While I think the opening of the tertiary carbocation is SN2-like, I don't have actual data to support it.)

     

The difference between these two systems reflects the stability of the carbocations. If the potential carbocation is tertiary (I° v III°), then attack occurs at the tertiary carbon. If the potential carbocation is secondary, then the selectivity of I° v II° is lower and most of the product is from the opening at the primary carbon.