The enolic form of ethyl acetoacetate as shown below has [figure]
- 9 sigma bonds and 2 pi-bonds
- 9 sigma bonds and 1 pi-bond
- 18 sigma bonds and 2 pi-bonds
- 16 sigma bonds and 1 pi-bond.
Answer
(3) 18 sigma bonds and 2 pi-bonds
Solution
Enol form: CH3-C(OH)=CH-COOC2H5 ( C6H10O3)
Every bond has one σ bond; each double bond has one extra π bond.
σ bonds: C–H = 9, O–H = 1, C–C = 4, C–O = 4 ⇒ total = 18
(check: open-chain molecule with 19 atoms ⇒ 19 − 1 = 18 σ bonds)
π bonds: one C=C and one C=O ⇒ 2
Answer: (3)