Side chain hydroxylation of pyrrole

Question

What reagents are required to form 1-[1,5-dimethyl-3-phenyl-4-(propan-2-yl)-1⁠H-pyrrol-2-yl]propan-1-ol with 1,2-dimethyl-4-phenyl-3-(propan-2-yl)-5-propyl-1⁠H-pyrrole as starting material?

Is it possible for the starting material to react with a halogen gas e.g. bromine to form a bromine on the carbon and then react with e.g. NaOH to form the OH with Br as the leaving group?

Answer 1

I'd like to propose a dark, non-radical pathway :)

Pyrroles, supposed that they are not fully substituted, undergo electrophilic aromatic substitutions with halogens.

For the given starting compound, a complete bromination isn't possible due to the lack of a proton that would have to be eliminated in the final step.

However, the addition of an electrophile is conceivable. Addition of "$\ce{Br+}$" at C-2 of the starting material, usually the prefered position in the reaction of unsubstituted pyrroles, is a dead end here. Addition at C-3 yields an $\alpha$-enamine cation intermediate from which a proton might be eliminated. The resulting species is both

• a double enamine and
• an allylic halide

This intermediate might rearrange upon elimination and addition of a nucleophile $\ce{Br-}$ in a way that the aromaticity of the pyrrole is reestablished and bromine is introduced in the side chain.

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Edit 1

Supposed that you are mainly interested in the hydroxyalkyl-substituted pyrrole as the target molecule, you might want to consider an alternative strategy. Let's have a second look:

1. The target compound (1) has an $\ce{OH}$ group in a side chain. Alcohols may be prepared by different routes, such as

   • reduction of the corresponding ketone
   • reaction an aldehyde with carbon nucleophile

   
   Let's focus on the second idea. This would leave us with the 2-formyl pyrrole (2), that could be transformed into (1) by a Grignard reaction with $\ce{EtMgBr}$.

2. Again, several methods are conceivable for the synthesis of the aldehyde (2), such as

   • selective reduction of a $\ce{-CO2R}$ substituent (or reduction to $\ce{-CH2OH}$ and subsequent oxidation to $\ce{-CHO}$)
   • direct introduction of a $\ce{-CHO}$ substituent at carbon 2 of pyrrole (3)

   
   The latter seems promising and you will probably realize that this an application of the Vilsmeier-Haack reaction, where the reactive intermediate, the $\ce{CHO}$ synthon, is prepared in situ from N,N-dimethylformamide and $\ce{POCl3}$.

This would leave you with a strategy for the synthesis of (3), but that's another question.

Answer 2

Though what you suggested may work to some extent (bromine is a liquid at room temperature, though, mind you), I feel it would be good to discuss the best conditions to maximize the yield of this synthesis and the reasons for them.

Let us first consider free-radical halogenation as a means to introducing a leaving group on the n-propyl side chain. Though we might first be worried that the isopropyl radical would propagate faster than the n-propyl radical, which would be unanimously true for benzylic radicals, the position of the substituents is much more important in pyrrole:

$\hspace{3.4cm}$

Hydrogen is abstracted preferentially from substituents at the 2 and 5 positions because the $\pi$ system the resulting unpaired electron occupies is more conjugated and thereby lower in energy. Given this, we need now only maximize abstraction from $n$-propyl over methyl.

Since more substituted radicals are more stable (due to hyperconjugative effects) and thereby react faster, by lowering the temperature of the reaction, the number of molecules with sufficient energy to form the methyl radical is decreased. Though this does the same for the n-propyl radical, the effect is less drastic due to the nature of the Maxwell-Boltzmann distribution, and a greater proportion of n-propyl halide is formed.

Lastly, the halide needs to be replaced with a hydroxyl group. Though you could try an $\mathrm{S_N2}$ using $\ce{NaOH}$ or $\ce{KOH}$, $\mathrm{E2}$ would be a major competitor, especially considering that its product only further extends the conjugation of the ring. Instead, just try microwaving the molecule in water for 3-5 minutes and you'll more than likely have converted it nearly quanitiatvely to the alcohol.

Using only water, we almost completely remove the possibility of elimination as it is such a weak base. Using the microwave allows us to thoroughly and quickly heat the solution to temperatures >$100º\mathrm{C}$ even, increasing the number of water molecules with sufficient energy to successfully collide with the reactant and the rate at which these collisions occur. Thus, one possible pathway would be: