(A) The degree of unsaturation $(DU)$ for $C_{4}H_{5}N$ is calculated as:$DU = C + 1 - \frac{H - N}{2} = 4 + 1 - \frac{5 - 1}{2} = 5 - 2 = 3$.$A$ poss

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(A) The degree of unsaturation $(DU)$ for $C_{4}H_{5}N$ is calculated as:$DU = C + 1 - \frac{H - N}{2} = 4 + 1 - \frac{5 - 1}{2} = 5 - 2 = 3$.$A$ possible acyclic structure with $DU = 3$ is $CH_{2}=C=CH-CH=NH$.In this structure,all carbon atoms are either $sp$ or $sp^{2}$ hybridised.Specifically,the carbons are $sp^{2}$,$sp$,$sp^{2}$,and $sp^{2}$ hybridised respectively.Therefore,the number of $sp^{3}$ hybridised carbons is $0$.

Class 11 Chemistry 8-1.Organic Chemistry : Classification and Nomenclature of Organic compounds Bonding and hybridisation in organic compounds

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The number of $sp^{3}$ hybridised carbons in an acyclic neutral compound with molecular formula $C_{4}H_{5}N$ is.

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A
$0$ or $1$
B
$2$
C
$3$
D
$4$

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8-1.Organic Chemistry : Classification and Nomenclature of Organic compounds

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Bonding and hybridisation in organic compounds

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The number of $sp^{3}$ hybridised carbons in an acyclic neutral compound with molecular formula $C_{4}H_{5}N$ is.

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