[xyjiryo]

Hey guys,

Just trying to figure out this problem. I've tried a Michael addition, which creates a tricyclic molecule with a 4 membered ring (shown below), but I can't seem to figure out how it gets from that to the product or if that's even the right pathway. Also, is it correct to assume the Br substitutes with an OH via SN2?

[pgk]

1). Are you sure for the structure and the presence of bromine in the final product?
If yes, then take a look to the mechanism of the Darzens reaction, which has a similar beginning with the mechanism that explains the formation of your product.
2). HBr is a stronger acid than H2O. As a consequence, the conjugate base Br- is a weaker nucleophile than OH- and therefore, Br- cannot replace OH- via SN2 mechanism.

[clarkstill]

Hey guys,

Just trying to figure out this problem. I've tried a Michael addition, which creates a tricyclic molecule with a 4 membered ring (shown below), but I can't seem to figure out how it gets from that to the product or if that's even the right pathway. Also, is it correct to assume the Br substitutes with an OH via SN2?

I don't think the intramolecular Michael reaction will get you there. Think about doing a 1,4-addition of hydroxide, remembering that this reaction will be reversible.

If it helps, I've put a star to show where a 13C label in the SM would end up in the product; see if you can work it out from that.

[rolnor]

pgk,
The reason why Br- cannot substitute OH is more because OH is a very poor leavingroup?

[pgk]

1). Yes! Indeed, OH- is a poor nucleophyge group (= leaving group) because it is a stronger nucleophile than Br-, as being the conjugate base of a weaker acid. 
2). Getting back to the point:
The top product is formed from the bottom one by a Favorskii type rearrangement. Nevertheless, the so formed multibridged ring favors the formation of an aldehyde that is accompanied with ring expansion, instead of the typical formation of a carboxylic group.
As a consequence, the top product can be obtained in good yields by adding the double amount (plus a slight excess) of the base.

[rolnor]

pgk,
I feel something is not completely right, would you not say that for example I- is a stronger nucleophile than OH-? Wich acid is weakest is not the only factor in play, also how strong bond OH or Br (or I) form with carbon is important.

[pgk]

Indeed, nucleophilic strength does not always coincide with the strength of the corresponding base. Various other factors also play an important role, such as the corresponding C bond strength, the number of electrons of the attacking/leaving group, etc. However, the strength of the corresponding base is the most important parameter, regarding SN2 substitution (OH-/I- comparison, included).

[rolnor]

Another example would be to compare HS- and OH- again the stronger acid H2S gives a better nucleophile compared to H2O, are you really sure about your reasoning?

[pgk]

As mentioned above, nucleophilic strength does not always coincide with the strength of the corresponding base and various other factors are involved therein. But by the chance, let’s clarify the issue:
Basicity refers to the offering of the electrons pair to a proton and/or an electron cavity that leads to the formation a stable salt and/or adduct; while nucleophilicity refers to the displacement of the electrons pair in another atom (often, carbon atom) that also leads to the formation a stable salt and/or adduct. As a consequence, basicity plays an important role in nucleophilicity.
On the other hand, basicity is not affected by the steric hindrance, while nucleophilicity is highly affected by the relative size of the attacking molecule. As a consequence, the base hardness/softness also plays an important role in nucleophilicity, too.
By application of all above in the SN2 substitution, we can assume that:
1). For bases of similar hardness, nucleophilic strength follows the order of basicity strength.
OH- > Br-
2). For bases of different hardness, the softer the base the stronger the nucleophilicity is.
HS- > OH-
3). Usually, SN2 departure follows the inverse order of nucleophilicity. However, the soft base character sometimes leads to paradox behavior. As an indicative example, I- is both a soft base and a large anion and therefore, I- is both a good attacking and a good leaving group and consequently, its nucleophilicity also depends on the concentration of reactants.
For further reading, please take a look to the following or a similar textbook:
Peter Sykes: A guidebook to mechanism in organic chemistry, 6th Longman LTD, London, (1985), Chapter 4: Nucleophilic substitution at a saturated carbon atom - Subchapter 4.5: Effect of entering and leaving group, pp. 96-99

[clarkstill]

2). Getting back to the point:
The top product is formed from the bottom one by a Favorskii type rearrangement. Nevertheless, the so formed multibridged ring favors the formation of an aldehyde that is accompanied with ring expansion, instead of the typical formation of a carboxylic group.
As a consequence, the top product can be obtained in good yields by adding the double amount (plus a slight excess) of the base.

Can you give a detailed mechanism for what you are saying here? Do you mean 1,4-addition of hydroxide followed by cyclopropanone formation? Then some rearrangement?

I've attached what I think is a much more plausible mechanism - beta elimination of the phenol should be facile due to its low pKa, so I think it makes sense.

[rolnor]

That was better pgk, but your reference is very old, -85?

[pgk]

The principles and the essentials of mechanisms in organic reactions, have not significantly changed since that time.
Anyway, for further reading, please take a look to a similar and newer textbook.

[pgk]

Please, see the attached file and let's discuss it.

[pgk]

Please, also see the more complete version.