[OrganicDan96]

Hello everyone,
I recently undertook a lithium-halogen exchange with 8-bromo-4-methyl quinoline, the reaction gave a mixture of products but i couldn't see anything recognisable in the ¹NMR, I know the methyl group is rather acidic, I have been told that I am not correct becausee rate of lithium-halogen exchange is faster that deprotonation, however; the lithiate its self is a strong base so i don't see any reason why it cannot deprotonate the methyl group to give an un-desired product with R on the methyl group. I would like to know weather my reasoning is correct here or if i am missing something?

 

[pgk]

You don’t see anything recognisable in the 1H-NMR because:
1). Deprotonation of the methyl group leads to the less basic methylene 2H-quinolinium imide (dihydroquinolone nitranion) via aromatic conjugation, which will react to the alkyl chloride and form the unstable methylene dihydroquinoline N-alkyl; and which by its turn, will be transformed in situ (via aromatic conjugation again and a further methyl sn2 substitution with alkylhalide or bromoquinoline) to the stable quaternary alkyl(alkylmethyl-bromoquinolinium) halide that has quite different 1H-NMR shifts of the Hα-N and Hβ-N aromatic protons and is accompanied with disappearance of the aromatic methyl protons shift, as well as with different alkyl protons integration than the mother compound:
NMR Spectra of Pyridine, Picolines and Hydrochlorides and of Their Hydrochlorides and Methiodides, Bulletin de l' Académie Polonaise des Sciences - Serie des Sciences Chimiques, 16,(7), 347-350, (1968)
http://bcpw.bg.pw.edu.pl/Content/3747/bulletin_de_lacademie_polonaise_des_sciences_1968_nr7_s347.pdf
(Attention because proton shifts are measured in τ old units and not in δ units, in this article. The conversion is δ = 10 – τ .)
2). But this it not finished yet. Unreacted bromoquinoline and the formed alkyl quinoline main product, form π-charge complexes with the above quaternary alkyl(alkylmehyl bromoquinolinium) halide that also have different 1H-NMR aromatic proton shifts than the uncomplexed starting compounds: 
Pi-Pi complexation of bupivacaine and analogues with aromatic receptors: Implications for overdose remediation, International Journal of Nanomedicine, 2(3), 449–459, (2007) 
https://www.ncbi.nlm.nih.gov/pmc/articles/PMC2676660/

As a consequence, you have a mixture of five different products (plus the starting alkyl chloride, dimerization products and a little alkylmehyl-bromoquinoline formed) that the most of them, are not recognisable in the 1H-NMR.

PS: Indeed, lithium-halogen exchange is faster that deprotonation but this is translated to a mixture of obtained products and not that the product formed via deprotonation, will not be formed at all.

[OrganicDan96]

thats a great explanation, thanks for the detailed reply

[rolnor]

I wonder, even if you get the wanted lithiated material, will this be a good nucleophile? Dont you need to make a cuprate to substitute a alkyl halide?
There is a big problem with the acidic methyl-group, could you make a less basic metalated material, zinc-cuprate? Or make reductive elimination with Pd(0) to get the aldehyde, then react with a Wittig-reagent to introduce the R-group? If this could work depends on what R-group you want to use, offcourse.

[clarkstill]

What is the R-group you need to introduce? The simpler route may be to form R-[M] (e.g. R₂Zn) and do a transition metal-catalysed cross-coupling with your aryl bromide - e.g. negishi

[BobfromNC]

The other problem is that the nBuLi can also react with the aryl bromide to form the butyl aromatic as well, I see that in a similar reaction with a bromobenzene as a major side product, even at -78 C.    These are not always clean reactions, unfortunately.   tBuLi (2 eqcan be better, in that it reacts and