[Dolphinsiu]

Describe the change in stretching frequency for CO upon the reaction with BF₃. Given v (free CO) = 2143 cm^-1

I truly know it is adduct formation, CO contains one sigma bond, one pi bond and one dative covalent bond from O to C, leaving both a lone pair in C and O. BF3 has low lying vacant 2p orbital. They truly reacts via adduct formation

However, it seems the probability of CO: --> BF₃ and OC: --> BF₃ are the same. The problem is the energy required to break B - F p(pi) - p(pi) interaction. Is the above adduct reaction still feasible? If yes, does the stretching frequency increase due to steric crowding (sp² to sp³)? Thank you!

My Attempt Answer:
OC: + BF₃ --> OCBF₃
CO: + BF₃ --> COBF₃

B in BF₃, +3 oxidation state, sp² hybrid.
B – F is covalent and multiple bond in nature due to p(pi) –p (pi) interaction. B in BF₃ is 6 valence electrons which is incomplete octet.
C – O involves one sigma bond, one pi bond and one dative covalent bond from O to C, leaving a lone pair electrons on both C and O atom.
As B in BF₃ is electron deficient and has low lying vacant pz orbital. The reaction proceeds via adduct formation.
CO can form stable octet configuration with BF₃ with the breakage of p(pi) –p (pi) interaction due to strong sigma bonding is formed in B – C bond and B – O bond which makes the product more stable. Steric hindrance of 3F’s gives frontal strain on CO. Weakening of CO bond means decrease in CO stretching frequency.

[nj_bartel]

OC + BF₃ seems much more probably to me on a precursory glance simply because the carbon in carbon monoxide is strongly nucleophilic and I would assume the B in BF₃ is pretty strongly electrophilic, so OCBF₃ seems much more probable.

My knowledge level of the chemistry you're doing is far lower than yours, so take that with a grain of salt, but it just makes sense to me