Question

The enolic form of ethyl acetoacetate as below has

• A. 9 sigma bonds and 2 pi-bonds
• B. 9 sigma bonds and 1 pi-bond
• C. 18 sigma bonds and 2 pi-bonds
• D. 16 sigma bonds and 1 pi-bond

Answer 1

The Correct Option is C

 To determine the number of sigma and pi bonds in the enolic form of ethyl acetoacetate, let's first understand its molecular structure. The enolic form of ethyl acetoacetate is typically represented by the structure where one hydrogen from the keto form migrates to the oxygen, forming an -OH group, with the accompanying formation of a C=C (carbon-carbon double bond).

Here's a step-by-step analysis of the bond types in the enolic form of ethyl acetoacetate:

1. The molecule primarily consists of:
   • An -CH₃ group (3 sigma bonds)
   • A -CH₂- group (2 sigma bonds)
   • A C=C bond (1 sigma and 1 pi bond)
   • An -OH group (1 sigma bond)
   • One C=O group (1 sigma and 1 pi bond, since it's in enol form, it doesn’t transform entirely to a keto form)
   • An ethoxide component (containing an O-C bond, counted as 1 sigma bond)
2. Let us count all the sigma bonds in the structure:
   • 3 C-H bonds from the -CH₃ group
   • 2 C-H bonds from the -CH₂ group
   • 1 C-C bond between the -CH₃ and -CH₂ group
   • 1 C=C bond (counting only the sigma portion)
   • 1 O-H bond
   • 2 C-C bonds from the remaining chain
   • 1 C-O bond in the ethoxide
3. Adding these up gives us: 3 + 2 + 1 + 1 + 1 + 2 + 1 = 18 sigma bonds.
4. Next, count the pi bonds:
   • 1 bond from the C=C enolic group
   • 1 bond from the C=O group
5. This results in a sum of 2 pi bonds.

Based on this structural analysis, the enolic form of ethyl acetoacetate contains 18 sigma bonds and 2 pi bonds. Therefore, the correct option is:

18 sigma bonds and 2 pi-bonds

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