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Chapter 1: Problem 6

Write the enol forms and identify the more stable form in each of the following compounds. (i) 1-phenyl-2-butanone (ii) methyl isopropyl ketone

Short Answer

Expert verified
Answer: (i) Enol Form 2 is more stable for 1-phenyl-2-butanone, and (ii) Enol Form 4 is more stable for methyl isopropyl ketone.

Step by step solution

01

(i) 1-phenyl-2-butanone enol forms

To find the enol forms of 1-phenyl-2-butanone, first identify the carbonyl group. Next, find the α-hydrogen(s) attached to the carbon adjacent to the carbonyl carbon. Then, perform tautomerization by moving the α-hydrogen to the carbonyl oxygen and forming a double bond between the carbonyl carbon and the α-carbon. Here are the two possible enol forms of 1-phenyl-2-butanone: Enol Form 1: Ph-C(OH)=CH-CH3 (where Ph stands for Phenyl group) Enol Form 2: Ph-CH=CH-OH
02

(i) Stability of 1-phenyl-2-butanone enol forms

Now, identify the more stable enol form among the two options. Stability is usually determined by resonance and conjugation. Enol Form 1: There is no resonance in this form. However, there is conjugation between the phenyl ring and the alkene (C=C) part. Enol Form 2: There is resonance between the oxygen atom with the lone pair and the alkene (C=C) part. In this case, Enol Form 2 is more stable than Enol Form 1 due to resonance.
03

(ii) Methyl isopropyl ketone enol forms

To find the enol forms of methyl isopropyl ketone, follow the same process as above: Here are the two possible enol forms of methyl isopropyl ketone: Enol Form 3: (CH3)2C(OH)C=CH2 Enol Form 4: (CH3)2CHC(OH)=(CH2)
04

(ii) Stability of methyl isopropyl ketone enol forms

Now, identify the more stable enol form among these two options: Enol Form 3: There is no resonance or conjugation in this form. Enol Form 4: There is no resonance in this form. However, there is conjugation between the carbonyl group and the alkene (C=C) part. In this case, Enol Form 4 is more stable than Enol Form 3 due to conjugation. To summarize: - The more stable enol form for 1-phenyl-2-butanone is Enol Form 2. - The more stable enol form for methyl isopropyl ketone is Enol Form 4.

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Most popular questions from this chapter

Identify the correct statements among the following (I) Removal of a proton from \(\left(\mathrm{H}_{3} \mathrm{C}\right)_{2} \mathrm{CH}-\mathrm{CH}-\mathrm{CH}_{3}\) leads to a more stable carbocation. (II) The hydrogen bonding is strong in \(\mathrm{S}-\mathrm{H}-\mathrm{-} \mathrm{S}\). (III) The central carbon atom in a \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{C}^{\bullet}\) free radical possesses 5 electrons. (IV) The number of \(\sigma\) bonds in but-1-en-3-yne is 7 . (a) all are correct (b) (IV) only' (c) (I) and (IV) only (d) (I), (III) and (IV)

\(\mathrm{pK}_{\mathrm{a}}\) value is least for (a) \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{C}-\mathrm{COOH}\) (b) \(\mathrm{BrCH}_{2}-\mathrm{CH}_{2}-\mathrm{CH}_{2}-\mathrm{COOH}\) (c) \(\mathrm{CH}_{3}-\mathrm{CCl}_{2}-\mathrm{COOH}\) (d) \(\mathrm{CH}_{3}-\mathrm{CHCl}-\mathrm{COOH}\)

The relationship between and HO - will be CC(O)C(O)C(C)O (a) two representations of the same molecule (b) diastereomers (c) enantiomers (d) anomers

Identify the statement which is not correct among the following (a) \((+)\) and meso tartaric acids are diastereomers. (b) The enol percentage of acetyl acetone is more than that of acetone. (c) Meso butane- 2,3 -diol can easily be resolved to give \((+)\) and \((-)\) isomers. (d) \(\mathrm{CH}_{3} \mathrm{CH}(\mathrm{OH}) \mathrm{CH}\left(\mathrm{CH}_{3}\right)_{2}\) is known as 3 -methyl-2-butanol.

Account for the fact that when 3,3 -dimethyl-2-butanol is treated with HBr, the product formed is 2 -bromo-2,3dimethylbutane and not 3,3 -dimethyl-2-bromobutane as expected.
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