QJEEM22Ch9d37: JEE Main 2022 Chemistry - Class 12 JEEM22Ch9d37 Solution: The number of sp 3 hybridised carbons in an acyclic neutral…

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$\mathrm{C}_{4} \mathrm{H}_{5} \mathrm{~N}$

$$\begin{array}{rl}\mathrm{DBE}& =(\mathrm{C}+1)-\left(\frac{\mathrm{H}+\mathrm{X}-\mathrm{N}}{2}\right)\\ \\ & =4+1-\left(\frac{5-1}{2}\right)=5-2=3\end{array}$$

3 double bond equivalent are present in compound


Only 1 $s p^{3}$ hybridised carbon is there

(Keeping compound as acyclic)