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

 

Degree of unsaturation$\left(\mathrm{DU}\right)=\frac{2\mathrm{C}+2+\mathrm{N}-\mathrm{H}-\mathrm{X}}{2}$

$\mathrm{DU}=\frac{8+2+1-5}{2}=3$

$\mathrm{DU}$ = 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  $\underset{\mathrm{Zero} {\mathrm{sp}}^3 \mathrm{carbon}}{{\mathrm{CH}}_2=\mathrm{C}=\mathrm{CH}=\mathrm{CH}=\mathrm{NH}}$