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The enolic form of ethyl acetoacetate as below has:
A
18 sigma bonds and 2 pi-bonds
B
16 sigma bonds and 1 pi-bond
C
9 sigma bonds and 2 pi-bonds
D
9 sigma bonds and 1 pi-bond
Detailed Solution
Enol form: $CH_3-C(OH)=CH-COOC_2H_5$ ($C_6H_{10}O_3$, 19 atoms, open chain).
Number of $\sigma$ bonds in an acyclic molecule = atoms − 1 = 18.
$\pi$ bonds: one C=C and one C=O, so 2.
So the enol form has 18 $\sigma$ and 2 $\pi$ bonds.
Number of $\sigma$ bonds in an acyclic molecule = atoms − 1 = 18.
$\pi$ bonds: one C=C and one C=O, so 2.
So the enol form has 18 $\sigma$ and 2 $\pi$ bonds.