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.

enter image description here

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!


3 Answers 3


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).

Claisen Rearrangement

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)


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.

Claisen rearrangement Image from ref 1


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:

Allyl halide reaction with NaOH

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.

Allyl phenyl ether preparation

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).

3,3-sigmatropic rearrangement-1

3,3-sigmatropic rearrangement-2

Net reaction

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

Reaction Scheme


${^1}$: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


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.