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.