For the carbon in part D with the +1 formal charge, what would it\'s hybridizati
ID: 850816 • Letter: F
Question
For the carbon in part D with the +1 formal charge, what would it's hybridization be?
I was thinking that it would be sp because the lone pair from the nitrogen could create a triple bond with the carbon in a resonance structure of this compound. Or is that wrong since it's in a ring structure?
And if it is sp then would its geometry be linear? (unless sp is wrong)
Beckman rearrangement from oxime to amide. This reaction unveils four discrete steps to transform the oxime starting material to the amide product in which the ring has expended. See the detailed mechanism below with the three key intermediates B, C, and D. The first and final steps are extremely fast so we will not consider them in the overall mechanism. Therefore two steps remain, step 1: proton transfer from N O and then step 2: ring expansion through the carbon shift (jump).Explanation / Answer
No. There wouldnt be any triple bond. Hybridisation of that carbon is sp2. goemetry is planar but not linear.