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Arrange the following pyridine derivatives in the order of increasing stability of adduct formed with $\ce{BBr3}.$ Explain briefly.

I: 3‐methylpyridine; II: 2‐methylpyridine; III: 3,5‐dimethylpyridine

I'm pretty confused about which structure is the most stable. I'd say structure III is the most stable as it has two methyl groups to donate electrons to nitrogen and increase its Lewis basicity.

However, I think III may not be the most stable because of steric hindrance. In fact, structure II could be the most stable as the methyl group is closer to nitrogen and there's no steric hindrance. Therefore, the answer should be III < I < II.

Is my reasoning correct? Are there any factors to consider for stability?

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I believe that the order of stability is II > III > I. The adduct should be formed between the lone pair on nitrogen and the Lewis acid. From there, I would apply the M+, I+ effects of the methyl groups as are known from benzene derivatives (although the substitution chemistry of pyridine is different from benzene).

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    $\begingroup$ I see. Just clarifying, does the distance between N and the methyl group play a part in stability? If the methyl group is closer to N, it'd be easier for N to receive electrons from the methyl group, hence higher lewis basicity. I'm not sure if it's a misconception. $\endgroup$
    – poorchemstudent
    Commented Nov 20, 2020 at 5:18
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    $\begingroup$ I think it is a misconception, take a look at the mechanism of the aromatic electrophilic substitution on benzene, toluene/phenol, and nitro-benzene. I think the same concepts apply here. $\endgroup$
    – TAR86
    Commented Nov 20, 2020 at 5:21

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