Reaction mechanism of rearrangement

Question

I am currently studying for my organic chemistry exam, but there is one problem I do not understand. Unfortunately, I do not have any solutions.

See image for the problem.

My first thought was the Dienone-phenol-rearrangement. But in order to do so, I need to add the alkyl substituent and I need a dienone. So can I deprotonate the phenol using potassium carbonate? (Or is it too weak as a base? The pKa of bicarbonate is around 10, so maybe it is too weak) But if it is possible, then the free electron pair of the oxygen can form a ketone... I could then use the alkene with a chloride substituent, so that the aromatic ring can attack it. Then I would have the dienone. But with the Dienone-phenol-rearrangement, the alkyl substituent can only move by one carbon-atom in the ring...

Is my thought process wrong? Can you please help me?

I did not have any lectures on rearrangement reactions, I had to learn them by myself, but by now, I have not found any similar reactions.

Thank you!

Answer 1

The first reaction is O-alkylation of p-cresol to give a 4-methylphenyl allyl ether derivative 3. The reagent in the first box should be 1-bromo-3-methylbut-2-ene (1; see the top box in the picture), which would undergo $\mathrm{S_N2}$ reaction with phenolic anion (2) in refluxing acetone. Note that potassium carbonate is a strong enough base to complete this reaction (this is surely a strong base than potassium bicarbonate).

The product 3 from alkylation would undergo Claisen rearrangement upon heating to give final alkylated cresol 4 in solvent-free condition. Remember, Claisen found his famous rearrangement first in solvent-free conditions, while he was trying to find the melting point of newly synthesized naphthyl allyl ether! (Picture is from https://www.ncbi.nlm.nih.gov/pmc/articles/PMC3873100/ and modified accordingly)

Answer 2

Potassium carbonate is a perfectly good base for the alkylation of phenol ($\mathrm{p}K_\mathrm{a} = 10$) with a good electrophile, in this case 3,3-dimethylallylbromide. The reaction you are looking for is Claisen rearrangement which proceeds by a 3,3-sigmatropic rearrangement mechanism.

Image from ref 1

Answer 3

The first step is nucleophilic substitution.

It is possible for nucleophile to attack directly at the allylic position, displacing the leaving group in a single step, in a process referred to as $\mathrm{S_N2'}$ substitution. This is likely in cases when the allyl compound is unhindered, and a strong nucleophile is used. The products will be similar to those seen with $\mathrm{S_N1'}$ substitution. Thus reaction of 1-chloro-2-butene with sodium hydroxide gives a mixture of 2-buten-1-ol and 3-buten-2-ol:

Step 1: Williamson Ether synthesis

• para-Cresol reacts with 3,3-dimethylallylbromide to give 1 and 2 using potassium carbonate in acetone as shown below.

Step 2: 3,3-Sigmatropic rearrangement (Aromatic Claisen rearrangement)

• Compounds 1 and 2 undergo 3,3-sigmatropic rearrangement (Aromatic Claisen rearrangement)${^1}$. Compound 1 gives 3 via sigmatropic shift followed by tautomerism to give 4 (as shown below).

Similarly, 2 gives 5 via sigmatropic shift followed by tautomerism to give 6 (as shown below).

Net reaction

As per the question given, the following is the scheme.

References

${^1}$:https://www.ncbi.nlm.nih.gov/pmc/articles/PMC3873100/ ${^2}$:https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Map%3A_Organic_Chemistry_(McMurry)/Chapter_18%3A_Ethers_and_Epoxides%3B_Thiols_and_Sulfides/18.04_Reactions_of_Ethers_-_Claisen_Rearrangement

${^3}$:J. Am. Chem. Soc.19891112511-519 Publication Date:January 1, 1989 https://doi.org/10.1021/ja00184a018