JEE MAIN - Chemistry (2022 - 25th July Morning Shift - No. 19)
The number of sp3 hybridised carbons in an acyclic neutral compound with molecular formula C4H5N is ___________.
Answer
1
Explanation
$\mathrm{C}_{4} \mathrm{H}_{5} \mathrm{~N}$
$$ \begin{aligned} \mathrm{DBE} & =(\mathrm{C}+1)-\left(\frac{\mathrm{H}+\mathrm{X}-\mathrm{N}}{2}\right) \\\\ & =4+1-\left(\frac{5-1}{2}\right)=5-2=3 \end{aligned} $$
3 double bond equivalent are present in compound
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Only 1 $s p^{3}$ hybridised carbon is there
(Keeping compound as acyclic)
$$ \begin{aligned} \mathrm{DBE} & =(\mathrm{C}+1)-\left(\frac{\mathrm{H}+\mathrm{X}-\mathrm{N}}{2}\right) \\\\ & =4+1-\left(\frac{5-1}{2}\right)=5-2=3 \end{aligned} $$
3 double bond equivalent are present in compound
Only 1 $s p^{3}$ hybridised carbon is there
(Keeping compound as acyclic)
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