I am asked to identify the "best catalyst" for the reaction:
$$\ce{CH3(CH2)8CH2Br->[\ce{KCN}][\text{benzene}]CH3(CH2)8CH2CN}$$
The options are (A) benzyl chloride (B) aniline (C) N,N,N-trimethylbenzylamine chloride, and (D) N-phenylethanamide.
Now, at first, I immediately thought that the catalyst chosen may have to prevent a possible ring formation, given there are already ten carbon atoms. However, I am not sure how any of those catalysts help prevent that (I doubt if a ring would even form though, because there isn't any nucleophile, but those ten carbons look suspicious!)
Anyway, I realized that this catalyst may have to be polar aprotic for the given SN2 to occur. But, all of the given catalysts are pretty protic (resonance delocalisation of the benzyl carbanion with the ring) I would rank them on proticity as (D) > (C) > (B) > (A).
Hence, although all are protic, I may make a vague guess that (A) is the least protic and hence the answer. However, the answer given is (C) and the solution instead states that:
Solvation of nucleophile is very less in presence of quaternary ammonium salt.
So, how? Where did the part about acidity go? And why worry about solvation? There doesn't appear to be any factor related to it at all (I read from here). Even so, how exactly would quaternary ammonium salts be poorer "solvators" than the others, say option (A)? Can I even predict this theoretically?
Source: MS Chouhan; Advanced Problems In Organic Chemistry; 11th ed; Q11 in Amines