The number of sp 3 hybridised carbons in an acyclic neutral compound with molecular formula C 4 H 5 N is

Step-by-step solution

Degree of unsaturation DU = 2 C + 2 + N - H - X 2 DU = 8 + 2 + 1 - 5 2 = 3 DU = Degree of unsaturation. It gives the information of number of pi bonds and rings in a compound. Given that acyclic compound, hence, the structure of the molecule is as follows, or CH 2 = C = CH = CH = NH Zero   sp 3   carbon