Side chain hydroxylation of pyrrole
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.



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:




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 $\ce{EtMgBr}$.
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.