# Effect of substituents on arene oxide-phenol rearrangement

When I was learning about the reactions of epoxides from my textbook, I came across the following mechanism for arene oxide-phenol rearrangement in the presence of acid:

NIH Shift - National Institute of Health Shift (c.f. Wikipedia)

In the above example, there is no substituents on the ring and hence the final product remains irrespective of which $$\ce{C-O}$$ bond in the arene oxide cleaves to open the ring. Later, the author presented the following reaction where the ring has an electron donating group - methyl group:

Here the final product depends on which of the two $$\ce{C-O}$$ bonds in the arene oxide is broken. Based on the general reaction mechanism provided by the textbook, I derived the following mechanism to form the product prescribed by the author:

Before writing the above mechanism, I first arrived at a different product by following the mechanism given below:

The reason I initially chose to cleave the left $$\ce{C-O}$$ bond in "Reaction Mechanism - 2" was, it gives rise to the carbocation 2(a) where the methyl group is at the para position. I've learnt that methyl group stabilizes a positive charge by hyperconjugation. Since, resonance effect is felt only at the ortho and para positions and not at the meta position, the carbocation 2(a) is more stable than 1(a) from "Reaction Mechanism - 1". Moreover, in the first mechanism, NIH shift involves a 1,2 Deuterium shift whereas in the second it's a 1,2 Hydrogen shift. I find a deuterium shift to be slightly energy expensive process due to its higher mass and stronger bond with carbon compared to a hydrogen shift.

The only reason why I think the first mechanism is better than the second is the product 1(b) is more stable than 2(b). But I don't understand why is the second reaction not preferred over the first even though it has two factors supporting (stability of 2(a) over 1(a) and hydrogen shift over deuterium shift) it.

I faced similar contradictory results while writing the mechanisms for the following reactions (from the same textbook):

My questions are:

• Are the reactions provided by my textbook correct?

• If yes, why is the cleavage of one $$\ce{C-O}$$ bond of the epoxide preferred over the other even though it produces a comparatively less stable carbocation? Are there any other factors which determine which bond cleaves?

• Is arene-oxide phenol rearrangement "specific" or highly "selective"? Or in simpler terms, does the rearrangement exclusively form only one product or does it form some small amount of the other product too?

Image source: My own work :) | Made using Marvin Sketch.
Textbook: Organic Chemistry by Solomons, Fryhle, and Snyder | Chapter 10: Alcohols and Ethers | Section 10.14: Reactions of Epoxides

What your book says is correct, and the issue arose due to your premature judgement of the intermediates' stability without actually working it out.

First of all, [1,2] Wagner-Meerwin shifts are intramolecular rearrangements which are primarily driven by thermodynamic concerns. Therefore, if their is a possibility of a reasonable cyclic transition state being formed, then the presence of the deuterium isotope hardly makes much of a change for the anionotropic shift, especially since the $$\ce{C-D}$$ is only marginally stronger than the $$\ce{C-H}$$ bond as stated here. So, if their is enough motivation for the rearrangement to occur, it can happen via either of the pathways.

Now, if you actually try drawing out the resonance structures of 1(a) and 2(a), you will actually find that the cation on the former will receive stabilisation by the +H effect, and not the second. The positive charge will only be able to align suitably for hyperconjugation in 1(a) as shown below:

Hence, mechanism 1 should be the preferred pathway

I think the rule you are stating refers to the increment in electron density at the ortho and para positions of an aromatic ring by the virtue of a methyl substituent as shown below:

However, this thing is only valid when the methylation has been done at a completely aromatic ring and not at some broken fragment of the same.

Only when their is complete conjugation and the ring itself is aromatic does it even make sense to use ortho and para descriptors as per definition. When you are considering a part of the same with incomplete conjugation, then it's pretty much up to the bonding schematic of the current molecule to see where the positive charge is getting delocalised.

o-/p-/m- descriptors lose their value in such cases, and along with them, their delocalisation rules as well

## Clarification

I guess the OP had some confusion regarding the mechanism after the formation of these intermediates as well. The competition of each pathway occurs stepwise:

1. First, the competition will be between the cleavage of either one of the two $$\ce{C-O}$$ bonds, one adjacent to the hydrogen and the other adjacent to the deuterium. As the intermediate 1(a) is more stable than 2(a) due to reasons stated above, the cleavage will occur as shown in the second step of Reaction Mechanism-1

1. Now, the intermediate 1(a) will try go for rearrangements that will generate species which are more energetically favorable. Due to the electron donating(+M effect) of the hydroxyl present adjacent to the cationic part in 1(a), a 1,2 deuterium shift will occur leading to the formation of 1(b) as shown in the third step of Reaction Mechanism-1

Note that as soon as the cleavage step occurred as described above, all considerations of Reaction Mechanism -2 and it's intermediates after 2(a) were rendered moot. 2(a) lost the stability battle against 1(a), and after that, none of the intermediates or products of Mechanism-2 would matter at all

• Thanks for the answer! Now, I understand why is 1(a) more stable than 2(a). Similarly, we can say that, 1(b) is less stable than 2(b). If the presence of deuterium doesn't matter, how to decide which of the two pathways is feasible? For the labelled intermediates, relative potential energies follows this order: 2(b)<1(a)<1(b)<2(a). Product (b)s are more stable than (a)s due to the electron donating nature of hydroxyl group. I'm unable to figure out which reaction will be preferred from this data. Did you consider any other things to arrive at the conclusion - reaction 1 is the preferred one? – Guru Vishnu Apr 15 '20 at 3:59
• Useful extended discussion in chat: Transcript – Guru Vishnu Apr 20 '20 at 4:38