MCQ
The number of $sp ^{3}$ hybridised carbons in an acyclic neutral compound with molecular formula $C _{4} H _{5} N$ is.
  • $0$ or $1$
  • B
    $2$
  • C
    $3$
  • D
    $4$

Answer

Correct option: A.
$0$ or $1$
a
$DU =4+1-\left(\frac{5-1}{2}\right)=3$

$CH _{2}= C = CH = CH = NH$

Zero $sp ^{3}$ carbon

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