[billsauce]

so i think the major product will come from the aldehyde bonded to the secondary carbon, because the hydrogens on the seondary carbon are more acidic and will allow the carbonyl to be more positively charged? am i right?

[willug]

The answer is there is only one hit on scifinder for the preparation/reactant/reagent/etc of this compound;

(Cycloparaffins fused with heterocyclic rings. Part XXXII. A convenient synthesis of 2- or 3-alkyl- and 2,3-dialkylpyridines and cycloalkenopyridinesFull Text
    By Epsztajn, Jan et al
    From Polish Journal of Chemistry, 54(2), 341-7; 1980),

and the reaction in the question has more than likely never been done (i.e. it hasn't). So you can write whatever you please as the answer to this question (although I think there will be no difference between the two aldehydes). It is a stupid question, and the answer to it is clearly not obvious.

Basically be suspicious of everything you read because there is a lot of nonsense out there!

[billsauce]

well the answer is definitely not the one where it says neither right

[billsauce]

so what do you guys think

[OC pro]

You have the opinion from at least two "experts" with 1000+ reactions performed in the lab (Disco and me). Of course there is a slight favour to the attack of the least hindered aldehyde which is also electronically favoured. But we are talking about something like 60:40 which is really a bad selectivity. Therefore this will be the "major" product. For a chemist working on the bench this selectivity is absolutely useless. Therefore, the whole question is dumb and I don´t see any real learning benefit too.

[billsauce]

i agree. this is a bad question. I would say attack is more likely to happen at N, but they will both definitely be attacked

[OC pro]

In an examination you can answer Carbon N. Reasons are presented above (sterics, electronic effects).