# Does sodium borohydride reduce C=N group?

Q.15 the major product of the following reaction is:

I have read that $$\ce{NaBH4}$$ reduces only aldehyde, ketone and acid chlorides. In some cases it may reduce double bonds conjugated with carbonyl groups. So shouldn't the answer here be option 1?

• Sodium borohydride readily reduces isolated C=N bonds, read up on reductive amination. – Waylander Jan 16 at 8:28
• @Waylander Can you provide some reference in the literature? – tatan Jan 16 at 9:49
• – Waylander Jan 16 at 10:02

Waylander's comments answer the title question as well as I could. Sodium borohydride, and even modified versions where some of the hydride ligands are replaced by others like $$\ce{NaBH3CN}$$, can reduce imines.

But, if we have only a limited amount of borohydride we will likely see the carboxyl group primarily reduced, as it has a more polar pi bond and thus it is a more reactive electrophile. So you may in principle get mostly 1 with limited sodium borohydride, or 4 if you have enough sodium borohydride to attack both the carboxyl and less reactive imine functions.

The carbon-carbon bond here, not conjugated to either electronegatove function, is interactive to borohydride and thus products 2 and 3 are not reached.

• If the question is written exactly as I had uploaded in the question (nowhere excess or anything is mentioned) do you think I should be penalized for writing 1 and not 4 ? Note the question has specifically one correct answer and there are no other details than that I have given in my original question... please reply – tatan Jan 16 at 13:19
• Then either 1or 4 might be formed. I really can't decide because I do not have ESP and cannot figure whether they mean one or possibly two moles of borohydride reacting. Always ask precise questions! – Oscar Lanzi Jan 16 at 13:30
• Thanks for your reply. The question is exactly as written. No other details given. Then what do you think? Please share your opinions, it matters a lot – tatan Jan 16 at 13:33
• The examination commitee has decided that 4 is correct and decided to penalize option 1 . However they are ready to get their answers challenged if proper citations are provided. Please help me if you think that $1$ should not be penalized for the question written with no other specific details. Please sir. (This question is from an undergraduate entrance examination for high school students) – tatan Jan 16 at 13:35
• Then they apparently decided to assume excess borohulydride. I have no more to say. – Oscar Lanzi Jan 16 at 14:00