The reason corresponding to the fact that the reaction between 1 , 3 − diphenyl − 1 , 3 − propanedione with I 2 and NaOH forms a precipitate of iodoform even though it is not a methyl ketone is to be stated. Concept introduction: In organic chemistry , the carbonyl compound is a class of functional group which contains a C = O group. They are of two types aldehyde − CHO and ketone RC = OR . They are the oxidized product of alcohol. Iodoform test used to confirm the presence of the carbonyl group. The compounds which contain methyl ketone CH 3 C = O or secondary alcohol CH 3 CH ( OH ) group gives positive iodoform test.

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Answer 1

Question

Chapter 22, Problem 22.69AP

Interpretation Introduction

Interpretation:

The reason corresponding to the fact that the reaction between 1,3−diphenyl−1,3−propanedione with I2 and NaOH forms a precipitate of iodoform even though it is not a methyl ketone is to be stated.

Concept introduction:

In organic chemistry, the carbonyl compound is a class of functional group which contains a C=O group. They are of two types aldehyde −CHO and ketone RC=OR. They are the oxidized product of alcohol. Iodoform test used to confirm the presence of the carbonyl group. The compounds which contain methyl ketone CH3C=O or secondary alcohol CH3CH(OH) group gives positive iodoform test.

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Answer 2

Question

Chapter 22, Problem 22.69AP

Interpretation Introduction

Interpretation:

The reason corresponding to the fact that the reaction between 1,3−diphenyl−1,3−propanedione with I2 and NaOH forms a precipitate of iodoform even though it is not a methyl ketone is to be stated.

Concept introduction:

In organic chemistry, the carbonyl compound is a class of functional group which contains a C=O group. They are of two types aldehyde −CHO and ketone RC=OR. They are the oxidized product of alcohol. Iodoform test used to confirm the presence of the carbonyl group. The compounds which contain methyl ketone CH3C=O or secondary alcohol CH3CH(OH) group gives positive iodoform test.

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Answer 3

Question

Chapter 22, Problem 22.69AP

Interpretation Introduction

Interpretation:

The reason corresponding to the fact that the reaction between 1,3−diphenyl−1,3−propanedione with I2 and NaOH forms a precipitate of iodoform even though it is not a methyl ketone is to be stated.

Concept introduction:

In organic chemistry, the carbonyl compound is a class of functional group which contains a C=O group. They are of two types aldehyde −CHO and ketone RC=OR. They are the oxidized product of alcohol. Iodoform test used to confirm the presence of the carbonyl group. The compounds which contain methyl ketone CH3C=O or secondary alcohol CH3CH(OH) group gives positive iodoform test.

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Answer 4

Question

Chapter 22, Problem 22.69AP

Interpretation Introduction

Interpretation:

The reason corresponding to the fact that the reaction between 1,3−diphenyl−1,3−propanedione with I2 and NaOH forms a precipitate of iodoform even though it is not a methyl ketone is to be stated.

Concept introduction:

In organic chemistry, the carbonyl compound is a class of functional group which contains a C=O group. They are of two types aldehyde −CHO and ketone RC=OR. They are the oxidized product of alcohol. Iodoform test used to confirm the presence of the carbonyl group. The compounds which contain methyl ketone CH3C=O or secondary alcohol CH3CH(OH) group gives positive iodoform test.

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Answer 5

Chapter 22, Problem 22.69AP

Interpretation Introduction

Interpretation:

The reason corresponding to the fact that the reaction between 1,3−diphenyl−1,3−propanedione with I2 and NaOH forms a precipitate of iodoform even though it is not a methyl ketone is to be stated.

Concept introduction:

In organic chemistry, the carbonyl compound is a class of functional group which contains a C=O group. They are of two types aldehyde −CHO and ketone RC=OR. They are the oxidized product of alcohol. Iodoform test used to confirm the presence of the carbonyl group. The compounds which contain methyl ketone CH3C=O or secondary alcohol CH3CH(OH) group gives positive iodoform test.

Answer 6

Chapter 22, Problem 22.69AP

Interpretation Introduction

Interpretation:

The reason corresponding to the fact that the reaction between 1,3−diphenyl−1,3−propanedione with I2 and NaOH forms a precipitate of iodoform even though it is not a methyl ketone is to be stated.

Concept introduction:

In organic chemistry, the carbonyl compound is a class of functional group which contains a C=O group. They are of two types aldehyde −CHO and ketone RC=OR. They are the oxidized product of alcohol. Iodoform test used to confirm the presence of the carbonyl group. The compounds which contain methyl ketone CH3C=O or secondary alcohol CH3CH(OH) group gives positive iodoform test.

Answer 7

Chapter 22, Problem 22.69AP

Interpretation Introduction

Interpretation:

The reason corresponding to the fact that the reaction between 1,3−diphenyl−1,3−propanedione with I2 and NaOH forms a precipitate of iodoform even though it is not a methyl ketone is to be stated.

Concept introduction:

In organic chemistry, the carbonyl compound is a class of functional group which contains a C=O group. They are of two types aldehyde −CHO and ketone RC=OR. They are the oxidized product of alcohol. Iodoform test used to confirm the presence of the carbonyl group. The compounds which contain methyl ketone CH3C=O or secondary alcohol CH3CH(OH) group gives positive iodoform test.

Answer 8

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Answer 11

Ch. 22 - Prob. 22.1P Ch. 22 - Prob. 22.2P Ch. 22 - Prob. 22.3P Ch. 22 - Prob. 22.4P Ch. 22 - Prob. 22.5P Ch. 22 - Prob. 22.6P Ch. 22 - Prob. 22.7P Ch. 22 - Prob. 22.8P Ch. 22 - Prob. 22.9P Ch. 22 - Prob. 22.10P

Solution Summary: The author explains that a reaction between 1,3-diphenyl-1,3–propanedione with I_–_2'NaOH

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Solution Summary: The author explains that a reaction between 1,3-diphenyl-1,3–propanedione with I_–_2&#39;NaOH

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Solution Summary: The author explains that a reaction between 1,3-diphenyl-1,3–propanedione with I_–_2'NaOH