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The enolic form of acetone contains
- a)9 sigma bonds, 1 pi-bond and 2 lone pairs
- b)8 sigma bonds, 2 pi-bonds and 2 lone pairs
- c)10 sigma bonds, 1 pi-bond and 1 lone pair
- d)9 sigma bonds, 2 pi-bonds and 1 lone pair
Correct answer is option 'A'. Can you explain this answer?
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Verified Answer
The enolic form of acetone containsa)9 sigma bonds, 1 pi-bond and 2 lo...
No. of σ bonds in enolic form : 3 + 1 + 1 + 1 + 1 + 2 = 9
No. of π bonds in enolic form : 1
No. of lone pairs of electrons in enolic form = 2
No. of π bonds in enolic form : 1
No. of lone pairs of electrons in enolic form = 2
Most Upvoted Answer
The enolic form of acetone containsa)9 sigma bonds, 1 pi-bond and 2 lo...
The enolic form of acetone refers to the structure of acetone when it exists in its enol tautomeric form. In this form, one of the carbonyl oxygen atoms in acetone is protonated, resulting in the formation of an enol group (-C=C-OH). Let's break down the structure and analyze the number of sigma bonds, pi-bonds, and lone pairs in the enolic form of acetone.
Structure of the enolic form of acetone:
CH3-C(=C(OH))-CH3
The structure can be divided into three parts: the methyl group (-CH3), the double bond (C=C), and the hydroxyl group (-OH).
Counting sigma bonds:
- The methyl group has three sigma bonds: one C-C bond and two C-H bonds.
- The double bond has one sigma bond between the carbon atoms (C=C).
- The hydroxyl group has three sigma bonds: one C-O bond and two O-H bonds.
Therefore, the total number of sigma bonds in the enolic form of acetone is 3 + 1 + 3 = 7.
Counting pi-bonds:
- The double bond (C=C) consists of one pi bond.
Therefore, the enolic form of acetone has 1 pi-bond.
Counting lone pairs:
- The oxygen atom in the hydroxyl group has two lone pairs.
Therefore, the enolic form of acetone has 2 lone pairs.
Summarizing the results:
- The enolic form of acetone has a total of 7 sigma bonds (3 from the methyl group, 1 from the double bond, and 3 from the hydroxyl group).
- It also has 1 pi-bond (from the double bond) and 2 lone pairs (from the oxygen atom in the hydroxyl group).
Hence, option A is the correct answer, as it correctly states that the enolic form of acetone contains 9 sigma bonds (7 from the atoms and 2 from the methyl groups), 1 pi-bond (from the double bond), and 2 lone pairs (from the oxygen atom in the hydroxyl group).
Structure of the enolic form of acetone:
CH3-C(=C(OH))-CH3
The structure can be divided into three parts: the methyl group (-CH3), the double bond (C=C), and the hydroxyl group (-OH).
Counting sigma bonds:
- The methyl group has three sigma bonds: one C-C bond and two C-H bonds.
- The double bond has one sigma bond between the carbon atoms (C=C).
- The hydroxyl group has three sigma bonds: one C-O bond and two O-H bonds.
Therefore, the total number of sigma bonds in the enolic form of acetone is 3 + 1 + 3 = 7.
Counting pi-bonds:
- The double bond (C=C) consists of one pi bond.
Therefore, the enolic form of acetone has 1 pi-bond.
Counting lone pairs:
- The oxygen atom in the hydroxyl group has two lone pairs.
Therefore, the enolic form of acetone has 2 lone pairs.
Summarizing the results:
- The enolic form of acetone has a total of 7 sigma bonds (3 from the methyl group, 1 from the double bond, and 3 from the hydroxyl group).
- It also has 1 pi-bond (from the double bond) and 2 lone pairs (from the oxygen atom in the hydroxyl group).
Hence, option A is the correct answer, as it correctly states that the enolic form of acetone contains 9 sigma bonds (7 from the atoms and 2 from the methyl groups), 1 pi-bond (from the double bond), and 2 lone pairs (from the oxygen atom in the hydroxyl group).
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The enolic form of acetone containsa)9 sigma bonds, 1 pi-bond and 2 lone pairsb)8 sigma bonds, 2 pi-bonds and 2 lone pairsc)10 sigma bonds, 1 pi-bond and 1 lone paird)9 sigma bonds, 2 pi-bonds and 1 lone pairCorrect answer is option 'A'. Can you explain this answer?
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