What carbonyl compounds might you start with to prepare the Following compounds by Grignard reaction? List all possibilities. (a) 2-Methyl-2-propanol (b) 1-Ethylcyclohexanol (c) 3-Phenyl-3-pentanol (d) 2-Phenyl-2-pentanol

Question

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Answer 1

Textbook Question

Chapter 17.SE, Problem 44AP

What carbonyl compounds might you start with to prepare the Following compounds by Grignard reaction? List all possibilities.

(a) 2-Methyl-2-propanol

(b) 1-Ethylcyclohexanol

(c) 3-Phenyl-3-pentanol

(d) 2-Phenyl-2-pentanol

Chapter 17.SE, Problem 44AP, What carbonyl compounds might you start with to prepare the Following compounds by Grignard

Expert Solution

Interpretation Introduction

a) 2-Methyl-2-propanol

Interpretation:

All possible carbonyl compounds that will react in a Grignard reaction to yield 2-methyl-2-propanol are to be listed.

Concept introduction:

Grignard reagents react with ketones to give 3⁰ alcohols as the product. Esters also when treated with two molar equivalents of Grignard reagents yield 3⁰ alcohols.

To list:

All possible carbonyl compounds that will react in a Grignard reaction to yield 2-methyl-2-propanol.

Answer to Problem 44AP

2-methyl-2-propanol is a 3⁰ alcohol. It can be prepared by treating acetone with methylmagnesium bromide or an acetic ester with two molar equivalents of methylmagnesium bromide.

Explanation of Solution

A four carbon 3⁰ alcohol is required. Hence a three carbon ketone (acetone) is treated with methylmagnesium bromide. In the case of esters two carbons will be provided by methylmagnesium bromide since esters require two molar equivalents of the reagent. Hence an ester of the two carbon acid (acetic acid) is chosen.

Student Value Bundle: Organic Chemistry, + OWLv2 with Student Solutions Manual eBook, 4 terms (24 months) Printed Access Card (NEW!!), Chapter 17.SE, Problem 44AP , additional homework tip 1

Conclusion

2-methyl-2-propanol is a 3⁰ alcohol. It can be prepared by treating acetone with methylmagnesium bromide or an acetic ester with two molar equivalents of methylmagnesium bromide.

Expert Solution

Interpretation Introduction

b) 1-Ethylcyclohexanol

Interpretation:

All possible carbonyl compounds that will react in a Grignard reaction to yield 1-ethylcyclohexanol are to be listed.

Concept introduction:

Grignard reagents react with formaldehyde to produce 1⁰ alcohols, with other aldehydes to yield 2⁰ alcohols and with ketones to give 3⁰ alcohols as the product. Esters also when treated with two molar equivalents of Grignard reagents yield 3⁰ alcohols.

To list:

All possible carbonyl compounds that will react in a Grignard reaction to yield 1-ethylcyclohexanol.

Answer to Problem 44AP

1-Ethylcyclohexanol can be prepared by treating cyclohexanone with ethylmagnesium bromide.

Explanation of Solution

A six-membered cyclic 3⁰ alcohol with ethyl group on C₁ is required. Hence a six membered cyclic ketone (cyclohexanone) is treated with a two carbon Grignard reagent (ethylmagnesium bromide).

Conclusion

1-Ethylcyclohexanol can be prepared by treating cyclohexanone with ethylmagnesium bromide.

Expert Solution

Interpretation Introduction

c) 3-Phenyl-3-pentanol

Interpretation:

All possible carbonyl compounds that will react in a Grignard reaction to yield 3-phenyl-3-pentanol are to be listed.

Concept introduction:

Grignard reagents react with formaldehyde to produce 1⁰ alcohols, with other aldehydes to yield 2⁰ alcohols and with ketones to give 3⁰ alcohols as the product. Esters also when treated with two molar equivalents of Grignard reagents yield 3⁰ alcohols.

To list:

All possible carbonyl compounds that will react in a Grignard reaction to yield 3-phenyl-3-pentanol.

Answer to Problem 44AP

3-Phenyl-3-pentanol can be prepared by reacting i) ethylphenyl ketone with ethylmagnesium bromide ii) benzoic acid esters with two molar equivalents of ethylmagnesium bromide iii) diethyl ketone with phenylmagnesium bromide.

Explanation of Solution

3-Phenyl-3-pentanol is a 3⁰ alcohol with a five carbon straight chain with a –OH and phenyl groups on C₃. Hence an aromatic ketone (ethylphenyl ketone) is treated with ethylmagnesium bromide or the ester of benzoic acid is treated with two equivalents of ethylmagnesium bromide. The ring can come from the Grignard reagent also. Hence phenylmagnesium bromide is treated with diethyl ketone.

Conclusion

3-Phenyl-3-pentanol can be prepared by reacting i) ethylphenyl ketone with ethylmagnesium bromide ii) benzoic acid esters with two molar equivalents of ethylmagnesium bromide iii) diethyl ketone with phenylmagnesium bromide.

Expert Solution

Interpretation Introduction

d) 2-Phenyl-2-pentanol

Interpretation:

All possible carbonyl compounds that will react in a Grignard reaction to yield 2-phenyl-2-pentanol are to be listed.

Concept introduction:

Grignard reagents react with formaldehyde to produce 1⁰ alcohols, with other aldehydes to yield 2⁰ alcohols and with ketones to give 3⁰ alcohols as the product. Esters also when treated with two molar equivalents of Grignard reagents yield 3⁰ alcohols.

To list:

All possible carbonyl compounds that will react in a Grignard reaction to yield 2-phenyl-2-pentanol.

Answer to Problem 44AP

2-phenyl-2-pentanol can be prepared by reacting i) methylphenyl ketone with propylmagnesium bromide ii) phenylpropyl ketone with methylmagnesium bromide iii) methylpropyl ketone with phenylmagnesium bromide.

Explanation of Solution

2-Phenyl-2-pentanol is a 3⁰ alcohol with a five carbon straight chain with a –OH and phenyl groups on C₂. Hence an aromatic ketone like methylphenyl ketone is treated with propylmagnesium bromide or phenylpropyl ketone is treated methylmagnesium bromide. The ring can come from the Grignard reagent also. Hence phenylmagnesium bromide is treated with methylpropyl ketone.

Conclusion

2-phenyl-2-pentanol can be prepared by reacting i) methylphenyl ketone with propylmagnesium bromide ii) phenylpropyl ketone with methylmagnesium bromide iii) methylpropyl ketone with phenylmagnesium bromide.

Expert Solution

Interpretation Introduction

e)

Interpretation:

All possible carbonyl compounds that will react in a Grignard reaction to yield 2-p-tolylethanol are to be listed.

Concept introduction:

Grignard reagents react with formaldehyde to produce 1⁰ alcohols, with other aldehydes to yield 2⁰ alcohols and with ketones to give 3⁰ alcohols as the product. Esters also when treated with two molar equivalents of Grignard reagents yield 3⁰ alcohols.

To list:

All possible carbonyl compounds that will react in a Grignard reaction to yield 2-(p-tolyl) ethanol.

Answer to Problem 44AP

2-(p-tolyl) ethanol can be prepared by reacting formaldehyde with p-tolylmethylmagnesium bromide.

Explanation of Solution

2-(p-tolyl) ethanol is a 1⁰ alcohol having a p-tolyl group attached to C₂ of ethanol. Hence formaldehyde is required. The remaining part should come from the Grignard reagent. Hence formaldehyde is treated with p-tolylmethylmagnesium bromide.

Conclusion

2-(p-tolyl) ethanol can prepared by reacting formaldehyde with p-tolylmethylmagnesium bromide.

Expert Solution

Interpretation Introduction

f)

Interpretation:

All possible carbonyl compounds that will react in a Grignard reaction to yield 1-cyclopentyl-2-methyl-2-propanol are to be listed.

Concept introduction:

Grignard reagents react with formaldehyde to produce 1⁰ alcohols, with other aldehydes to yield 2⁰ alcohols and with ketones to give 3⁰ alcohols as the product. Esters also when treated with two molar equivalents of Grignard reagents yield 3⁰ alcohols.

To list:

All possible carbonyl compounds that will react in a Grignard reaction to yield 1-cyclopentyl-2-methyl-2-propanol are to be listed.

Answer to Problem 44AP

1-Cyclopentyl-2-methyl-2-propanol can be prepared by reacting i) cyclopentylmethyl methyl ketone with methylmagnesium bromide ii) an ester of cyclopentylacetic acid with two molar equivalents of methylmagnesium bromide iii) acetone with cyclopentylmethylmagnesium bromide.

Explanation of Solution

1-Cyclopentyl-2-methyl-2-propanol is a 3⁰ alcohol with a three carbon straight chain with a cyclopentyl group on C₁ and –OH on C₂. Hence cyclopentylmethyl methyl ketone is treated with methylmagnesium bromide or an ester of cyclopentylacetic acid is treated with two molar equivalents of methylmagnesium bromide. The ring can come from the Grignard reagent also. Hence cyclopentylmethylmagnesium bromide is treated with acetone.

Conclusion

1-Cyclopentyl-2-methyl-2-propanol can be prepared by reacting i) cyclopentylmethyl methyl ketone with methylmagnesium bromide ii) an ester of cyclopentylacetic acid with two molar equivalents of methylmagnesium bromide iii) acetone with cyclopentylmethylmagnesium bromide.

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Answer 2

Textbook Question

Chapter 17.SE, Problem 44AP

What carbonyl compounds might you start with to prepare the Following compounds by Grignard reaction? List all possibilities.

(a) 2-Methyl-2-propanol

(b) 1-Ethylcyclohexanol

(c) 3-Phenyl-3-pentanol

(d) 2-Phenyl-2-pentanol

Chapter 17.SE, Problem 44AP, What carbonyl compounds might you start with to prepare the Following compounds by Grignard

Expert Solution

Interpretation Introduction

a) 2-Methyl-2-propanol

Interpretation:

All possible carbonyl compounds that will react in a Grignard reaction to yield 2-methyl-2-propanol are to be listed.

Concept introduction:

Grignard reagents react with ketones to give 3⁰ alcohols as the product. Esters also when treated with two molar equivalents of Grignard reagents yield 3⁰ alcohols.

To list:

All possible carbonyl compounds that will react in a Grignard reaction to yield 2-methyl-2-propanol.

Answer to Problem 44AP

2-methyl-2-propanol is a 3⁰ alcohol. It can be prepared by treating acetone with methylmagnesium bromide or an acetic ester with two molar equivalents of methylmagnesium bromide.

Explanation of Solution

A four carbon 3⁰ alcohol is required. Hence a three carbon ketone (acetone) is treated with methylmagnesium bromide. In the case of esters two carbons will be provided by methylmagnesium bromide since esters require two molar equivalents of the reagent. Hence an ester of the two carbon acid (acetic acid) is chosen.

Conclusion

2-methyl-2-propanol is a 3⁰ alcohol. It can be prepared by treating acetone with methylmagnesium bromide or an acetic ester with two molar equivalents of methylmagnesium bromide.

Expert Solution

Interpretation Introduction

b) 1-Ethylcyclohexanol

Interpretation:

All possible carbonyl compounds that will react in a Grignard reaction to yield 1-ethylcyclohexanol are to be listed.

Concept introduction:

Grignard reagents react with formaldehyde to produce 1⁰ alcohols, with other aldehydes to yield 2⁰ alcohols and with ketones to give 3⁰ alcohols as the product. Esters also when treated with two molar equivalents of Grignard reagents yield 3⁰ alcohols.

To list:

All possible carbonyl compounds that will react in a Grignard reaction to yield 1-ethylcyclohexanol.

Answer to Problem 44AP

1-Ethylcyclohexanol can be prepared by treating cyclohexanone with ethylmagnesium bromide.

Explanation of Solution

A six-membered cyclic 3⁰ alcohol with ethyl group on C₁ is required. Hence a six membered cyclic ketone (cyclohexanone) is treated with a two carbon Grignard reagent (ethylmagnesium bromide).

Conclusion

1-Ethylcyclohexanol can be prepared by treating cyclohexanone with ethylmagnesium bromide.

Expert Solution

Interpretation Introduction

c) 3-Phenyl-3-pentanol

Interpretation:

All possible carbonyl compounds that will react in a Grignard reaction to yield 3-phenyl-3-pentanol are to be listed.

Concept introduction:

Grignard reagents react with formaldehyde to produce 1⁰ alcohols, with other aldehydes to yield 2⁰ alcohols and with ketones to give 3⁰ alcohols as the product. Esters also when treated with two molar equivalents of Grignard reagents yield 3⁰ alcohols.

To list:

All possible carbonyl compounds that will react in a Grignard reaction to yield 3-phenyl-3-pentanol.

Answer to Problem 44AP

3-Phenyl-3-pentanol can be prepared by reacting i) ethylphenyl ketone with ethylmagnesium bromide ii) benzoic acid esters with two molar equivalents of ethylmagnesium bromide iii) diethyl ketone with phenylmagnesium bromide.

Explanation of Solution

3-Phenyl-3-pentanol is a 3⁰ alcohol with a five carbon straight chain with a –OH and phenyl groups on C₃. Hence an aromatic ketone (ethylphenyl ketone) is treated with ethylmagnesium bromide or the ester of benzoic acid is treated with two equivalents of ethylmagnesium bromide. The ring can come from the Grignard reagent also. Hence phenylmagnesium bromide is treated with diethyl ketone.

Conclusion

3-Phenyl-3-pentanol can be prepared by reacting i) ethylphenyl ketone with ethylmagnesium bromide ii) benzoic acid esters with two molar equivalents of ethylmagnesium bromide iii) diethyl ketone with phenylmagnesium bromide.

Expert Solution

Interpretation Introduction

d) 2-Phenyl-2-pentanol

Interpretation:

All possible carbonyl compounds that will react in a Grignard reaction to yield 2-phenyl-2-pentanol are to be listed.

Concept introduction:

Grignard reagents react with formaldehyde to produce 1⁰ alcohols, with other aldehydes to yield 2⁰ alcohols and with ketones to give 3⁰ alcohols as the product. Esters also when treated with two molar equivalents of Grignard reagents yield 3⁰ alcohols.

To list:

All possible carbonyl compounds that will react in a Grignard reaction to yield 2-phenyl-2-pentanol.

Answer to Problem 44AP

2-phenyl-2-pentanol can be prepared by reacting i) methylphenyl ketone with propylmagnesium bromide ii) phenylpropyl ketone with methylmagnesium bromide iii) methylpropyl ketone with phenylmagnesium bromide.

Explanation of Solution

2-Phenyl-2-pentanol is a 3⁰ alcohol with a five carbon straight chain with a –OH and phenyl groups on C₂. Hence an aromatic ketone like methylphenyl ketone is treated with propylmagnesium bromide or phenylpropyl ketone is treated methylmagnesium bromide. The ring can come from the Grignard reagent also. Hence phenylmagnesium bromide is treated with methylpropyl ketone.

Conclusion

2-phenyl-2-pentanol can be prepared by reacting i) methylphenyl ketone with propylmagnesium bromide ii) phenylpropyl ketone with methylmagnesium bromide iii) methylpropyl ketone with phenylmagnesium bromide.

Expert Solution

Interpretation Introduction

e)

Interpretation:

All possible carbonyl compounds that will react in a Grignard reaction to yield 2-p-tolylethanol are to be listed.

Concept introduction:

Grignard reagents react with formaldehyde to produce 1⁰ alcohols, with other aldehydes to yield 2⁰ alcohols and with ketones to give 3⁰ alcohols as the product. Esters also when treated with two molar equivalents of Grignard reagents yield 3⁰ alcohols.

To list:

All possible carbonyl compounds that will react in a Grignard reaction to yield 2-(p-tolyl) ethanol.

Answer to Problem 44AP

2-(p-tolyl) ethanol can be prepared by reacting formaldehyde with p-tolylmethylmagnesium bromide.

Explanation of Solution

2-(p-tolyl) ethanol is a 1⁰ alcohol having a p-tolyl group attached to C₂ of ethanol. Hence formaldehyde is required. The remaining part should come from the Grignard reagent. Hence formaldehyde is treated with p-tolylmethylmagnesium bromide.

Conclusion

2-(p-tolyl) ethanol can prepared by reacting formaldehyde with p-tolylmethylmagnesium bromide.

Expert Solution

Interpretation Introduction

f)

Interpretation:

All possible carbonyl compounds that will react in a Grignard reaction to yield 1-cyclopentyl-2-methyl-2-propanol are to be listed.

Concept introduction:

Grignard reagents react with formaldehyde to produce 1⁰ alcohols, with other aldehydes to yield 2⁰ alcohols and with ketones to give 3⁰ alcohols as the product. Esters also when treated with two molar equivalents of Grignard reagents yield 3⁰ alcohols.

To list:

All possible carbonyl compounds that will react in a Grignard reaction to yield 1-cyclopentyl-2-methyl-2-propanol are to be listed.

Answer to Problem 44AP

1-Cyclopentyl-2-methyl-2-propanol can be prepared by reacting i) cyclopentylmethyl methyl ketone with methylmagnesium bromide ii) an ester of cyclopentylacetic acid with two molar equivalents of methylmagnesium bromide iii) acetone with cyclopentylmethylmagnesium bromide.

Explanation of Solution

1-Cyclopentyl-2-methyl-2-propanol is a 3⁰ alcohol with a three carbon straight chain with a cyclopentyl group on C₁ and –OH on C₂. Hence cyclopentylmethyl methyl ketone is treated with methylmagnesium bromide or an ester of cyclopentylacetic acid is treated with two molar equivalents of methylmagnesium bromide. The ring can come from the Grignard reagent also. Hence cyclopentylmethylmagnesium bromide is treated with acetone.

Conclusion

1-Cyclopentyl-2-methyl-2-propanol can be prepared by reacting i) cyclopentylmethyl methyl ketone with methylmagnesium bromide ii) an ester of cyclopentylacetic acid with two molar equivalents of methylmagnesium bromide iii) acetone with cyclopentylmethylmagnesium bromide.

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Answer 3

Textbook Question

Chapter 17.SE, Problem 44AP

What carbonyl compounds might you start with to prepare the Following compounds by Grignard reaction? List all possibilities.

(a) 2-Methyl-2-propanol

(b) 1-Ethylcyclohexanol

(c) 3-Phenyl-3-pentanol

(d) 2-Phenyl-2-pentanol

Chapter 17.SE, Problem 44AP, What carbonyl compounds might you start with to prepare the Following compounds by Grignard

Expert Solution

Interpretation Introduction

a) 2-Methyl-2-propanol

Interpretation:

All possible carbonyl compounds that will react in a Grignard reaction to yield 2-methyl-2-propanol are to be listed.

Concept introduction:

Grignard reagents react with ketones to give 3⁰ alcohols as the product. Esters also when treated with two molar equivalents of Grignard reagents yield 3⁰ alcohols.

To list:

All possible carbonyl compounds that will react in a Grignard reaction to yield 2-methyl-2-propanol.

Answer to Problem 44AP

2-methyl-2-propanol is a 3⁰ alcohol. It can be prepared by treating acetone with methylmagnesium bromide or an acetic ester with two molar equivalents of methylmagnesium bromide.

Explanation of Solution

A four carbon 3⁰ alcohol is required. Hence a three carbon ketone (acetone) is treated with methylmagnesium bromide. In the case of esters two carbons will be provided by methylmagnesium bromide since esters require two molar equivalents of the reagent. Hence an ester of the two carbon acid (acetic acid) is chosen.

Conclusion

2-methyl-2-propanol is a 3⁰ alcohol. It can be prepared by treating acetone with methylmagnesium bromide or an acetic ester with two molar equivalents of methylmagnesium bromide.

Expert Solution

Interpretation Introduction

b) 1-Ethylcyclohexanol

Interpretation:

All possible carbonyl compounds that will react in a Grignard reaction to yield 1-ethylcyclohexanol are to be listed.

Concept introduction:

Grignard reagents react with formaldehyde to produce 1⁰ alcohols, with other aldehydes to yield 2⁰ alcohols and with ketones to give 3⁰ alcohols as the product. Esters also when treated with two molar equivalents of Grignard reagents yield 3⁰ alcohols.

To list:

All possible carbonyl compounds that will react in a Grignard reaction to yield 1-ethylcyclohexanol.

Answer to Problem 44AP

1-Ethylcyclohexanol can be prepared by treating cyclohexanone with ethylmagnesium bromide.

Explanation of Solution

A six-membered cyclic 3⁰ alcohol with ethyl group on C₁ is required. Hence a six membered cyclic ketone (cyclohexanone) is treated with a two carbon Grignard reagent (ethylmagnesium bromide).

Conclusion

1-Ethylcyclohexanol can be prepared by treating cyclohexanone with ethylmagnesium bromide.

Expert Solution

Interpretation Introduction

c) 3-Phenyl-3-pentanol

Interpretation:

All possible carbonyl compounds that will react in a Grignard reaction to yield 3-phenyl-3-pentanol are to be listed.

Concept introduction:

Grignard reagents react with formaldehyde to produce 1⁰ alcohols, with other aldehydes to yield 2⁰ alcohols and with ketones to give 3⁰ alcohols as the product. Esters also when treated with two molar equivalents of Grignard reagents yield 3⁰ alcohols.

To list:

All possible carbonyl compounds that will react in a Grignard reaction to yield 3-phenyl-3-pentanol.

Answer to Problem 44AP

3-Phenyl-3-pentanol can be prepared by reacting i) ethylphenyl ketone with ethylmagnesium bromide ii) benzoic acid esters with two molar equivalents of ethylmagnesium bromide iii) diethyl ketone with phenylmagnesium bromide.

Explanation of Solution

3-Phenyl-3-pentanol is a 3⁰ alcohol with a five carbon straight chain with a –OH and phenyl groups on C₃. Hence an aromatic ketone (ethylphenyl ketone) is treated with ethylmagnesium bromide or the ester of benzoic acid is treated with two equivalents of ethylmagnesium bromide. The ring can come from the Grignard reagent also. Hence phenylmagnesium bromide is treated with diethyl ketone.

Conclusion

3-Phenyl-3-pentanol can be prepared by reacting i) ethylphenyl ketone with ethylmagnesium bromide ii) benzoic acid esters with two molar equivalents of ethylmagnesium bromide iii) diethyl ketone with phenylmagnesium bromide.

Expert Solution

Interpretation Introduction

d) 2-Phenyl-2-pentanol

Interpretation:

All possible carbonyl compounds that will react in a Grignard reaction to yield 2-phenyl-2-pentanol are to be listed.

Concept introduction:

Grignard reagents react with formaldehyde to produce 1⁰ alcohols, with other aldehydes to yield 2⁰ alcohols and with ketones to give 3⁰ alcohols as the product. Esters also when treated with two molar equivalents of Grignard reagents yield 3⁰ alcohols.

To list:

All possible carbonyl compounds that will react in a Grignard reaction to yield 2-phenyl-2-pentanol.

Answer to Problem 44AP

2-phenyl-2-pentanol can be prepared by reacting i) methylphenyl ketone with propylmagnesium bromide ii) phenylpropyl ketone with methylmagnesium bromide iii) methylpropyl ketone with phenylmagnesium bromide.

Explanation of Solution

2-Phenyl-2-pentanol is a 3⁰ alcohol with a five carbon straight chain with a –OH and phenyl groups on C₂. Hence an aromatic ketone like methylphenyl ketone is treated with propylmagnesium bromide or phenylpropyl ketone is treated methylmagnesium bromide. The ring can come from the Grignard reagent also. Hence phenylmagnesium bromide is treated with methylpropyl ketone.

Conclusion

2-phenyl-2-pentanol can be prepared by reacting i) methylphenyl ketone with propylmagnesium bromide ii) phenylpropyl ketone with methylmagnesium bromide iii) methylpropyl ketone with phenylmagnesium bromide.

Expert Solution

Interpretation Introduction

e)

Interpretation:

All possible carbonyl compounds that will react in a Grignard reaction to yield 2-p-tolylethanol are to be listed.

Concept introduction:

Grignard reagents react with formaldehyde to produce 1⁰ alcohols, with other aldehydes to yield 2⁰ alcohols and with ketones to give 3⁰ alcohols as the product. Esters also when treated with two molar equivalents of Grignard reagents yield 3⁰ alcohols.

To list:

All possible carbonyl compounds that will react in a Grignard reaction to yield 2-(p-tolyl) ethanol.

Answer to Problem 44AP

2-(p-tolyl) ethanol can be prepared by reacting formaldehyde with p-tolylmethylmagnesium bromide.

Explanation of Solution

2-(p-tolyl) ethanol is a 1⁰ alcohol having a p-tolyl group attached to C₂ of ethanol. Hence formaldehyde is required. The remaining part should come from the Grignard reagent. Hence formaldehyde is treated with p-tolylmethylmagnesium bromide.

Conclusion

2-(p-tolyl) ethanol can prepared by reacting formaldehyde with p-tolylmethylmagnesium bromide.

Expert Solution

Interpretation Introduction

f)

Interpretation:

All possible carbonyl compounds that will react in a Grignard reaction to yield 1-cyclopentyl-2-methyl-2-propanol are to be listed.

Concept introduction:

Grignard reagents react with formaldehyde to produce 1⁰ alcohols, with other aldehydes to yield 2⁰ alcohols and with ketones to give 3⁰ alcohols as the product. Esters also when treated with two molar equivalents of Grignard reagents yield 3⁰ alcohols.

To list:

All possible carbonyl compounds that will react in a Grignard reaction to yield 1-cyclopentyl-2-methyl-2-propanol are to be listed.

Answer to Problem 44AP

1-Cyclopentyl-2-methyl-2-propanol can be prepared by reacting i) cyclopentylmethyl methyl ketone with methylmagnesium bromide ii) an ester of cyclopentylacetic acid with two molar equivalents of methylmagnesium bromide iii) acetone with cyclopentylmethylmagnesium bromide.

Explanation of Solution

1-Cyclopentyl-2-methyl-2-propanol is a 3⁰ alcohol with a three carbon straight chain with a cyclopentyl group on C₁ and –OH on C₂. Hence cyclopentylmethyl methyl ketone is treated with methylmagnesium bromide or an ester of cyclopentylacetic acid is treated with two molar equivalents of methylmagnesium bromide. The ring can come from the Grignard reagent also. Hence cyclopentylmethylmagnesium bromide is treated with acetone.

Conclusion

1-Cyclopentyl-2-methyl-2-propanol can be prepared by reacting i) cyclopentylmethyl methyl ketone with methylmagnesium bromide ii) an ester of cyclopentylacetic acid with two molar equivalents of methylmagnesium bromide iii) acetone with cyclopentylmethylmagnesium bromide.

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Answer 4

Textbook Question

Chapter 17.SE, Problem 44AP

What carbonyl compounds might you start with to prepare the Following compounds by Grignard reaction? List all possibilities.

(a) 2-Methyl-2-propanol

(b) 1-Ethylcyclohexanol

(c) 3-Phenyl-3-pentanol

(d) 2-Phenyl-2-pentanol

Chapter 17.SE, Problem 44AP, What carbonyl compounds might you start with to prepare the Following compounds by Grignard

Expert Solution

Interpretation Introduction

a) 2-Methyl-2-propanol

Interpretation:

All possible carbonyl compounds that will react in a Grignard reaction to yield 2-methyl-2-propanol are to be listed.

Concept introduction:

Grignard reagents react with ketones to give 3⁰ alcohols as the product. Esters also when treated with two molar equivalents of Grignard reagents yield 3⁰ alcohols.

To list:

All possible carbonyl compounds that will react in a Grignard reaction to yield 2-methyl-2-propanol.

Answer to Problem 44AP

2-methyl-2-propanol is a 3⁰ alcohol. It can be prepared by treating acetone with methylmagnesium bromide or an acetic ester with two molar equivalents of methylmagnesium bromide.

Explanation of Solution

A four carbon 3⁰ alcohol is required. Hence a three carbon ketone (acetone) is treated with methylmagnesium bromide. In the case of esters two carbons will be provided by methylmagnesium bromide since esters require two molar equivalents of the reagent. Hence an ester of the two carbon acid (acetic acid) is chosen.

Conclusion

2-methyl-2-propanol is a 3⁰ alcohol. It can be prepared by treating acetone with methylmagnesium bromide or an acetic ester with two molar equivalents of methylmagnesium bromide.

Expert Solution

Interpretation Introduction

b) 1-Ethylcyclohexanol

Interpretation:

All possible carbonyl compounds that will react in a Grignard reaction to yield 1-ethylcyclohexanol are to be listed.

Concept introduction:

Grignard reagents react with formaldehyde to produce 1⁰ alcohols, with other aldehydes to yield 2⁰ alcohols and with ketones to give 3⁰ alcohols as the product. Esters also when treated with two molar equivalents of Grignard reagents yield 3⁰ alcohols.

To list:

All possible carbonyl compounds that will react in a Grignard reaction to yield 1-ethylcyclohexanol.

Answer to Problem 44AP

1-Ethylcyclohexanol can be prepared by treating cyclohexanone with ethylmagnesium bromide.

Explanation of Solution

A six-membered cyclic 3⁰ alcohol with ethyl group on C₁ is required. Hence a six membered cyclic ketone (cyclohexanone) is treated with a two carbon Grignard reagent (ethylmagnesium bromide).

Conclusion

1-Ethylcyclohexanol can be prepared by treating cyclohexanone with ethylmagnesium bromide.

Expert Solution

Interpretation Introduction

c) 3-Phenyl-3-pentanol

Interpretation:

All possible carbonyl compounds that will react in a Grignard reaction to yield 3-phenyl-3-pentanol are to be listed.

Concept introduction:

Grignard reagents react with formaldehyde to produce 1⁰ alcohols, with other aldehydes to yield 2⁰ alcohols and with ketones to give 3⁰ alcohols as the product. Esters also when treated with two molar equivalents of Grignard reagents yield 3⁰ alcohols.

To list:

All possible carbonyl compounds that will react in a Grignard reaction to yield 3-phenyl-3-pentanol.

Answer to Problem 44AP

3-Phenyl-3-pentanol can be prepared by reacting i) ethylphenyl ketone with ethylmagnesium bromide ii) benzoic acid esters with two molar equivalents of ethylmagnesium bromide iii) diethyl ketone with phenylmagnesium bromide.

Explanation of Solution

3-Phenyl-3-pentanol is a 3⁰ alcohol with a five carbon straight chain with a –OH and phenyl groups on C₃. Hence an aromatic ketone (ethylphenyl ketone) is treated with ethylmagnesium bromide or the ester of benzoic acid is treated with two equivalents of ethylmagnesium bromide. The ring can come from the Grignard reagent also. Hence phenylmagnesium bromide is treated with diethyl ketone.

Conclusion

3-Phenyl-3-pentanol can be prepared by reacting i) ethylphenyl ketone with ethylmagnesium bromide ii) benzoic acid esters with two molar equivalents of ethylmagnesium bromide iii) diethyl ketone with phenylmagnesium bromide.

Expert Solution

Interpretation Introduction

d) 2-Phenyl-2-pentanol

Interpretation:

All possible carbonyl compounds that will react in a Grignard reaction to yield 2-phenyl-2-pentanol are to be listed.

Concept introduction:

Grignard reagents react with formaldehyde to produce 1⁰ alcohols, with other aldehydes to yield 2⁰ alcohols and with ketones to give 3⁰ alcohols as the product. Esters also when treated with two molar equivalents of Grignard reagents yield 3⁰ alcohols.

To list:

All possible carbonyl compounds that will react in a Grignard reaction to yield 2-phenyl-2-pentanol.

Answer to Problem 44AP

2-phenyl-2-pentanol can be prepared by reacting i) methylphenyl ketone with propylmagnesium bromide ii) phenylpropyl ketone with methylmagnesium bromide iii) methylpropyl ketone with phenylmagnesium bromide.

Explanation of Solution

2-Phenyl-2-pentanol is a 3⁰ alcohol with a five carbon straight chain with a –OH and phenyl groups on C₂. Hence an aromatic ketone like methylphenyl ketone is treated with propylmagnesium bromide or phenylpropyl ketone is treated methylmagnesium bromide. The ring can come from the Grignard reagent also. Hence phenylmagnesium bromide is treated with methylpropyl ketone.

Conclusion

2-phenyl-2-pentanol can be prepared by reacting i) methylphenyl ketone with propylmagnesium bromide ii) phenylpropyl ketone with methylmagnesium bromide iii) methylpropyl ketone with phenylmagnesium bromide.

Expert Solution

Interpretation Introduction

e)

Interpretation:

All possible carbonyl compounds that will react in a Grignard reaction to yield 2-p-tolylethanol are to be listed.

Concept introduction:

Grignard reagents react with formaldehyde to produce 1⁰ alcohols, with other aldehydes to yield 2⁰ alcohols and with ketones to give 3⁰ alcohols as the product. Esters also when treated with two molar equivalents of Grignard reagents yield 3⁰ alcohols.

To list:

All possible carbonyl compounds that will react in a Grignard reaction to yield 2-(p-tolyl) ethanol.

Answer to Problem 44AP

2-(p-tolyl) ethanol can be prepared by reacting formaldehyde with p-tolylmethylmagnesium bromide.

Explanation of Solution

2-(p-tolyl) ethanol is a 1⁰ alcohol having a p-tolyl group attached to C₂ of ethanol. Hence formaldehyde is required. The remaining part should come from the Grignard reagent. Hence formaldehyde is treated with p-tolylmethylmagnesium bromide.

Conclusion

2-(p-tolyl) ethanol can prepared by reacting formaldehyde with p-tolylmethylmagnesium bromide.

Expert Solution

Interpretation Introduction

f)

Interpretation:

All possible carbonyl compounds that will react in a Grignard reaction to yield 1-cyclopentyl-2-methyl-2-propanol are to be listed.

Concept introduction:

Grignard reagents react with formaldehyde to produce 1⁰ alcohols, with other aldehydes to yield 2⁰ alcohols and with ketones to give 3⁰ alcohols as the product. Esters also when treated with two molar equivalents of Grignard reagents yield 3⁰ alcohols.

To list:

All possible carbonyl compounds that will react in a Grignard reaction to yield 1-cyclopentyl-2-methyl-2-propanol are to be listed.

Answer to Problem 44AP

1-Cyclopentyl-2-methyl-2-propanol can be prepared by reacting i) cyclopentylmethyl methyl ketone with methylmagnesium bromide ii) an ester of cyclopentylacetic acid with two molar equivalents of methylmagnesium bromide iii) acetone with cyclopentylmethylmagnesium bromide.

Explanation of Solution

1-Cyclopentyl-2-methyl-2-propanol is a 3⁰ alcohol with a three carbon straight chain with a cyclopentyl group on C₁ and –OH on C₂. Hence cyclopentylmethyl methyl ketone is treated with methylmagnesium bromide or an ester of cyclopentylacetic acid is treated with two molar equivalents of methylmagnesium bromide. The ring can come from the Grignard reagent also. Hence cyclopentylmethylmagnesium bromide is treated with acetone.

Conclusion

1-Cyclopentyl-2-methyl-2-propanol can be prepared by reacting i) cyclopentylmethyl methyl ketone with methylmagnesium bromide ii) an ester of cyclopentylacetic acid with two molar equivalents of methylmagnesium bromide iii) acetone with cyclopentylmethylmagnesium bromide.

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Answer 5

Textbook Question

Answer 6

Textbook Question

Answer 7

Expert Solution

Interpretation Introduction

a) 2-Methyl-2-propanol

Interpretation:

All possible carbonyl compounds that will react in a Grignard reaction to yield 2-methyl-2-propanol are to be listed.

Concept introduction:

Grignard reagents react with ketones to give 3⁰ alcohols as the product. Esters also when treated with two molar equivalents of Grignard reagents yield 3⁰ alcohols.

To list:

All possible carbonyl compounds that will react in a Grignard reaction to yield 2-methyl-2-propanol.

Answer to Problem 44AP

2-methyl-2-propanol is a 3⁰ alcohol. It can be prepared by treating acetone with methylmagnesium bromide or an acetic ester with two molar equivalents of methylmagnesium bromide.

Explanation of Solution

A four carbon 3⁰ alcohol is required. Hence a three carbon ketone (acetone) is treated with methylmagnesium bromide. In the case of esters two carbons will be provided by methylmagnesium bromide since esters require two molar equivalents of the reagent. Hence an ester of the two carbon acid (acetic acid) is chosen.

Conclusion

2-methyl-2-propanol is a 3⁰ alcohol. It can be prepared by treating acetone with methylmagnesium bromide or an acetic ester with two molar equivalents of methylmagnesium bromide.

Expert Solution

Interpretation Introduction

b) 1-Ethylcyclohexanol

Interpretation:

All possible carbonyl compounds that will react in a Grignard reaction to yield 1-ethylcyclohexanol are to be listed.

Concept introduction:

Grignard reagents react with formaldehyde to produce 1⁰ alcohols, with other aldehydes to yield 2⁰ alcohols and with ketones to give 3⁰ alcohols as the product. Esters also when treated with two molar equivalents of Grignard reagents yield 3⁰ alcohols.

To list:

All possible carbonyl compounds that will react in a Grignard reaction to yield 1-ethylcyclohexanol.

Answer to Problem 44AP

1-Ethylcyclohexanol can be prepared by treating cyclohexanone with ethylmagnesium bromide.

Explanation of Solution

A six-membered cyclic 3⁰ alcohol with ethyl group on C₁ is required. Hence a six membered cyclic ketone (cyclohexanone) is treated with a two carbon Grignard reagent (ethylmagnesium bromide).

Conclusion

1-Ethylcyclohexanol can be prepared by treating cyclohexanone with ethylmagnesium bromide.

Expert Solution

Interpretation Introduction

c) 3-Phenyl-3-pentanol

Interpretation:

All possible carbonyl compounds that will react in a Grignard reaction to yield 3-phenyl-3-pentanol are to be listed.

Concept introduction:

Grignard reagents react with formaldehyde to produce 1⁰ alcohols, with other aldehydes to yield 2⁰ alcohols and with ketones to give 3⁰ alcohols as the product. Esters also when treated with two molar equivalents of Grignard reagents yield 3⁰ alcohols.

To list:

All possible carbonyl compounds that will react in a Grignard reaction to yield 3-phenyl-3-pentanol.

Answer to Problem 44AP

3-Phenyl-3-pentanol can be prepared by reacting i) ethylphenyl ketone with ethylmagnesium bromide ii) benzoic acid esters with two molar equivalents of ethylmagnesium bromide iii) diethyl ketone with phenylmagnesium bromide.

Explanation of Solution

3-Phenyl-3-pentanol is a 3⁰ alcohol with a five carbon straight chain with a –OH and phenyl groups on C₃. Hence an aromatic ketone (ethylphenyl ketone) is treated with ethylmagnesium bromide or the ester of benzoic acid is treated with two equivalents of ethylmagnesium bromide. The ring can come from the Grignard reagent also. Hence phenylmagnesium bromide is treated with diethyl ketone.

Conclusion

3-Phenyl-3-pentanol can be prepared by reacting i) ethylphenyl ketone with ethylmagnesium bromide ii) benzoic acid esters with two molar equivalents of ethylmagnesium bromide iii) diethyl ketone with phenylmagnesium bromide.

Expert Solution

Interpretation Introduction

d) 2-Phenyl-2-pentanol

Interpretation:

All possible carbonyl compounds that will react in a Grignard reaction to yield 2-phenyl-2-pentanol are to be listed.

Concept introduction:

Grignard reagents react with formaldehyde to produce 1⁰ alcohols, with other aldehydes to yield 2⁰ alcohols and with ketones to give 3⁰ alcohols as the product. Esters also when treated with two molar equivalents of Grignard reagents yield 3⁰ alcohols.

To list:

All possible carbonyl compounds that will react in a Grignard reaction to yield 2-phenyl-2-pentanol.

Answer to Problem 44AP

2-phenyl-2-pentanol can be prepared by reacting i) methylphenyl ketone with propylmagnesium bromide ii) phenylpropyl ketone with methylmagnesium bromide iii) methylpropyl ketone with phenylmagnesium bromide.

Explanation of Solution

2-Phenyl-2-pentanol is a 3⁰ alcohol with a five carbon straight chain with a –OH and phenyl groups on C₂. Hence an aromatic ketone like methylphenyl ketone is treated with propylmagnesium bromide or phenylpropyl ketone is treated methylmagnesium bromide. The ring can come from the Grignard reagent also. Hence phenylmagnesium bromide is treated with methylpropyl ketone.

Conclusion

2-phenyl-2-pentanol can be prepared by reacting i) methylphenyl ketone with propylmagnesium bromide ii) phenylpropyl ketone with methylmagnesium bromide iii) methylpropyl ketone with phenylmagnesium bromide.

Expert Solution

Interpretation Introduction

e)

Interpretation:

All possible carbonyl compounds that will react in a Grignard reaction to yield 2-p-tolylethanol are to be listed.

Concept introduction:

Grignard reagents react with formaldehyde to produce 1⁰ alcohols, with other aldehydes to yield 2⁰ alcohols and with ketones to give 3⁰ alcohols as the product. Esters also when treated with two molar equivalents of Grignard reagents yield 3⁰ alcohols.

To list:

All possible carbonyl compounds that will react in a Grignard reaction to yield 2-(p-tolyl) ethanol.

Answer to Problem 44AP

2-(p-tolyl) ethanol can be prepared by reacting formaldehyde with p-tolylmethylmagnesium bromide.

Explanation of Solution

2-(p-tolyl) ethanol is a 1⁰ alcohol having a p-tolyl group attached to C₂ of ethanol. Hence formaldehyde is required. The remaining part should come from the Grignard reagent. Hence formaldehyde is treated with p-tolylmethylmagnesium bromide.

Conclusion

2-(p-tolyl) ethanol can prepared by reacting formaldehyde with p-tolylmethylmagnesium bromide.

Expert Solution

Interpretation Introduction

f)

Interpretation:

All possible carbonyl compounds that will react in a Grignard reaction to yield 1-cyclopentyl-2-methyl-2-propanol are to be listed.

Concept introduction:

Grignard reagents react with formaldehyde to produce 1⁰ alcohols, with other aldehydes to yield 2⁰ alcohols and with ketones to give 3⁰ alcohols as the product. Esters also when treated with two molar equivalents of Grignard reagents yield 3⁰ alcohols.

To list:

All possible carbonyl compounds that will react in a Grignard reaction to yield 1-cyclopentyl-2-methyl-2-propanol are to be listed.

Answer to Problem 44AP

1-Cyclopentyl-2-methyl-2-propanol can be prepared by reacting i) cyclopentylmethyl methyl ketone with methylmagnesium bromide ii) an ester of cyclopentylacetic acid with two molar equivalents of methylmagnesium bromide iii) acetone with cyclopentylmethylmagnesium bromide.

Explanation of Solution

1-Cyclopentyl-2-methyl-2-propanol is a 3⁰ alcohol with a three carbon straight chain with a cyclopentyl group on C₁ and –OH on C₂. Hence cyclopentylmethyl methyl ketone is treated with methylmagnesium bromide or an ester of cyclopentylacetic acid is treated with two molar equivalents of methylmagnesium bromide. The ring can come from the Grignard reagent also. Hence cyclopentylmethylmagnesium bromide is treated with acetone.

Conclusion

1-Cyclopentyl-2-methyl-2-propanol can be prepared by reacting i) cyclopentylmethyl methyl ketone with methylmagnesium bromide ii) an ester of cyclopentylacetic acid with two molar equivalents of methylmagnesium bromide iii) acetone with cyclopentylmethylmagnesium bromide.

Answer 8

Expert Solution

Interpretation Introduction

a) 2-Methyl-2-propanol

Interpretation:

All possible carbonyl compounds that will react in a Grignard reaction to yield 2-methyl-2-propanol are to be listed.

Concept introduction:

Grignard reagents react with ketones to give 3⁰ alcohols as the product. Esters also when treated with two molar equivalents of Grignard reagents yield 3⁰ alcohols.

To list:

All possible carbonyl compounds that will react in a Grignard reaction to yield 2-methyl-2-propanol.

Answer to Problem 44AP

2-methyl-2-propanol is a 3⁰ alcohol. It can be prepared by treating acetone with methylmagnesium bromide or an acetic ester with two molar equivalents of methylmagnesium bromide.

Explanation of Solution

A four carbon 3⁰ alcohol is required. Hence a three carbon ketone (acetone) is treated with methylmagnesium bromide. In the case of esters two carbons will be provided by methylmagnesium bromide since esters require two molar equivalents of the reagent. Hence an ester of the two carbon acid (acetic acid) is chosen.

Conclusion

2-methyl-2-propanol is a 3⁰ alcohol. It can be prepared by treating acetone with methylmagnesium bromide or an acetic ester with two molar equivalents of methylmagnesium bromide.

Answer 9

Expert Solution

Answer 10

Expert Solution

Answer 11

Expert Solution

Answer 12

Interpretation Introduction

a) 2-Methyl-2-propanol

Interpretation:

All possible carbonyl compounds that will react in a Grignard reaction to yield 2-methyl-2-propanol are to be listed.

Concept introduction:

Grignard reagents react with ketones to give 3⁰ alcohols as the product. Esters also when treated with two molar equivalents of Grignard reagents yield 3⁰ alcohols.

To list:

All possible carbonyl compounds that will react in a Grignard reaction to yield 2-methyl-2-propanol.

Answer 13

Answer to Problem 44AP

2-methyl-2-propanol is a 3⁰ alcohol. It can be prepared by treating acetone with methylmagnesium bromide or an acetic ester with two molar equivalents of methylmagnesium bromide.

Answer 14

Explanation of Solution

A four carbon 3⁰ alcohol is required. Hence a three carbon ketone (acetone) is treated with methylmagnesium bromide. In the case of esters two carbons will be provided by methylmagnesium bromide since esters require two molar equivalents of the reagent. Hence an ester of the two carbon acid (acetic acid) is chosen.

Answer 15

Conclusion

2-methyl-2-propanol is a 3⁰ alcohol. It can be prepared by treating acetone with methylmagnesium bromide or an acetic ester with two molar equivalents of methylmagnesium bromide.

Answer 16

Expert Solution

Interpretation Introduction

b) 1-Ethylcyclohexanol

Interpretation:

All possible carbonyl compounds that will react in a Grignard reaction to yield 1-ethylcyclohexanol are to be listed.

Concept introduction:

Grignard reagents react with formaldehyde to produce 1⁰ alcohols, with other aldehydes to yield 2⁰ alcohols and with ketones to give 3⁰ alcohols as the product. Esters also when treated with two molar equivalents of Grignard reagents yield 3⁰ alcohols.

To list:

All possible carbonyl compounds that will react in a Grignard reaction to yield 1-ethylcyclohexanol.

Answer to Problem 44AP

1-Ethylcyclohexanol can be prepared by treating cyclohexanone with ethylmagnesium bromide.

Explanation of Solution

A six-membered cyclic 3⁰ alcohol with ethyl group on C₁ is required. Hence a six membered cyclic ketone (cyclohexanone) is treated with a two carbon Grignard reagent (ethylmagnesium bromide).

Conclusion

1-Ethylcyclohexanol can be prepared by treating cyclohexanone with ethylmagnesium bromide.

Answer 17

Expert Solution

Answer 18

Expert Solution

Answer 19

Expert Solution

Answer 20

Interpretation Introduction

b) 1-Ethylcyclohexanol

Interpretation:

All possible carbonyl compounds that will react in a Grignard reaction to yield 1-ethylcyclohexanol are to be listed.

Concept introduction:

Grignard reagents react with formaldehyde to produce 1⁰ alcohols, with other aldehydes to yield 2⁰ alcohols and with ketones to give 3⁰ alcohols as the product. Esters also when treated with two molar equivalents of Grignard reagents yield 3⁰ alcohols.

To list:

All possible carbonyl compounds that will react in a Grignard reaction to yield 1-ethylcyclohexanol.

Answer 21

Answer to Problem 44AP

1-Ethylcyclohexanol can be prepared by treating cyclohexanone with ethylmagnesium bromide.

Answer 22

Explanation of Solution

A six-membered cyclic 3⁰ alcohol with ethyl group on C₁ is required. Hence a six membered cyclic ketone (cyclohexanone) is treated with a two carbon Grignard reagent (ethylmagnesium bromide).

Answer 23

Conclusion

1-Ethylcyclohexanol can be prepared by treating cyclohexanone with ethylmagnesium bromide.

Answer 24

Expert Solution

Interpretation Introduction

c) 3-Phenyl-3-pentanol

Interpretation:

All possible carbonyl compounds that will react in a Grignard reaction to yield 3-phenyl-3-pentanol are to be listed.

Concept introduction:

Grignard reagents react with formaldehyde to produce 1⁰ alcohols, with other aldehydes to yield 2⁰ alcohols and with ketones to give 3⁰ alcohols as the product. Esters also when treated with two molar equivalents of Grignard reagents yield 3⁰ alcohols.

To list:

All possible carbonyl compounds that will react in a Grignard reaction to yield 3-phenyl-3-pentanol.

Answer to Problem 44AP

3-Phenyl-3-pentanol can be prepared by reacting i) ethylphenyl ketone with ethylmagnesium bromide ii) benzoic acid esters with two molar equivalents of ethylmagnesium bromide iii) diethyl ketone with phenylmagnesium bromide.

Explanation of Solution

3-Phenyl-3-pentanol is a 3⁰ alcohol with a five carbon straight chain with a –OH and phenyl groups on C₃. Hence an aromatic ketone (ethylphenyl ketone) is treated with ethylmagnesium bromide or the ester of benzoic acid is treated with two equivalents of ethylmagnesium bromide. The ring can come from the Grignard reagent also. Hence phenylmagnesium bromide is treated with diethyl ketone.

Conclusion

3-Phenyl-3-pentanol can be prepared by reacting i) ethylphenyl ketone with ethylmagnesium bromide ii) benzoic acid esters with two molar equivalents of ethylmagnesium bromide iii) diethyl ketone with phenylmagnesium bromide.

Answer 25

Expert Solution

Answer 26

Expert Solution

Answer 27

Expert Solution

Answer 28

Interpretation Introduction

c) 3-Phenyl-3-pentanol

Interpretation:

All possible carbonyl compounds that will react in a Grignard reaction to yield 3-phenyl-3-pentanol are to be listed.

Concept introduction:

Grignard reagents react with formaldehyde to produce 1⁰ alcohols, with other aldehydes to yield 2⁰ alcohols and with ketones to give 3⁰ alcohols as the product. Esters also when treated with two molar equivalents of Grignard reagents yield 3⁰ alcohols.

To list:

All possible carbonyl compounds that will react in a Grignard reaction to yield 3-phenyl-3-pentanol.

Answer 29

Answer to Problem 44AP

3-Phenyl-3-pentanol can be prepared by reacting i) ethylphenyl ketone with ethylmagnesium bromide ii) benzoic acid esters with two molar equivalents of ethylmagnesium bromide iii) diethyl ketone with phenylmagnesium bromide.

Answer 30

Explanation of Solution

3-Phenyl-3-pentanol is a 3⁰ alcohol with a five carbon straight chain with a –OH and phenyl groups on C₃. Hence an aromatic ketone (ethylphenyl ketone) is treated with ethylmagnesium bromide or the ester of benzoic acid is treated with two equivalents of ethylmagnesium bromide. The ring can come from the Grignard reagent also. Hence phenylmagnesium bromide is treated with diethyl ketone.

Answer 31

Conclusion

3-Phenyl-3-pentanol can be prepared by reacting i) ethylphenyl ketone with ethylmagnesium bromide ii) benzoic acid esters with two molar equivalents of ethylmagnesium bromide iii) diethyl ketone with phenylmagnesium bromide.

Answer 32

Expert Solution

Interpretation Introduction

d) 2-Phenyl-2-pentanol

Interpretation:

All possible carbonyl compounds that will react in a Grignard reaction to yield 2-phenyl-2-pentanol are to be listed.

Concept introduction:

Grignard reagents react with formaldehyde to produce 1⁰ alcohols, with other aldehydes to yield 2⁰ alcohols and with ketones to give 3⁰ alcohols as the product. Esters also when treated with two molar equivalents of Grignard reagents yield 3⁰ alcohols.

To list:

All possible carbonyl compounds that will react in a Grignard reaction to yield 2-phenyl-2-pentanol.

Answer to Problem 44AP

2-phenyl-2-pentanol can be prepared by reacting i) methylphenyl ketone with propylmagnesium bromide ii) phenylpropyl ketone with methylmagnesium bromide iii) methylpropyl ketone with phenylmagnesium bromide.

Explanation of Solution

2-Phenyl-2-pentanol is a 3⁰ alcohol with a five carbon straight chain with a –OH and phenyl groups on C₂. Hence an aromatic ketone like methylphenyl ketone is treated with propylmagnesium bromide or phenylpropyl ketone is treated methylmagnesium bromide. The ring can come from the Grignard reagent also. Hence phenylmagnesium bromide is treated with methylpropyl ketone.

Conclusion

2-phenyl-2-pentanol can be prepared by reacting i) methylphenyl ketone with propylmagnesium bromide ii) phenylpropyl ketone with methylmagnesium bromide iii) methylpropyl ketone with phenylmagnesium bromide.

Answer 33

Expert Solution

Answer 34

Expert Solution

Answer 35

Expert Solution

Answer 36

Interpretation Introduction

d) 2-Phenyl-2-pentanol

Interpretation:

All possible carbonyl compounds that will react in a Grignard reaction to yield 2-phenyl-2-pentanol are to be listed.

Concept introduction:

Grignard reagents react with formaldehyde to produce 1⁰ alcohols, with other aldehydes to yield 2⁰ alcohols and with ketones to give 3⁰ alcohols as the product. Esters also when treated with two molar equivalents of Grignard reagents yield 3⁰ alcohols.

To list:

All possible carbonyl compounds that will react in a Grignard reaction to yield 2-phenyl-2-pentanol.

Answer 37

Answer to Problem 44AP

2-phenyl-2-pentanol can be prepared by reacting i) methylphenyl ketone with propylmagnesium bromide ii) phenylpropyl ketone with methylmagnesium bromide iii) methylpropyl ketone with phenylmagnesium bromide.

Answer 38

Explanation of Solution

2-Phenyl-2-pentanol is a 3⁰ alcohol with a five carbon straight chain with a –OH and phenyl groups on C₂. Hence an aromatic ketone like methylphenyl ketone is treated with propylmagnesium bromide or phenylpropyl ketone is treated methylmagnesium bromide. The ring can come from the Grignard reagent also. Hence phenylmagnesium bromide is treated with methylpropyl ketone.

Answer 39

Conclusion

2-phenyl-2-pentanol can be prepared by reacting i) methylphenyl ketone with propylmagnesium bromide ii) phenylpropyl ketone with methylmagnesium bromide iii) methylpropyl ketone with phenylmagnesium bromide.

Answer 40

Expert Solution

Interpretation Introduction

e)

Interpretation:

All possible carbonyl compounds that will react in a Grignard reaction to yield 2-p-tolylethanol are to be listed.

Concept introduction:

Grignard reagents react with formaldehyde to produce 1⁰ alcohols, with other aldehydes to yield 2⁰ alcohols and with ketones to give 3⁰ alcohols as the product. Esters also when treated with two molar equivalents of Grignard reagents yield 3⁰ alcohols.

To list:

All possible carbonyl compounds that will react in a Grignard reaction to yield 2-(p-tolyl) ethanol.

Answer to Problem 44AP

2-(p-tolyl) ethanol can be prepared by reacting formaldehyde with p-tolylmethylmagnesium bromide.

Explanation of Solution

2-(p-tolyl) ethanol is a 1⁰ alcohol having a p-tolyl group attached to C₂ of ethanol. Hence formaldehyde is required. The remaining part should come from the Grignard reagent. Hence formaldehyde is treated with p-tolylmethylmagnesium bromide.

Conclusion

2-(p-tolyl) ethanol can prepared by reacting formaldehyde with p-tolylmethylmagnesium bromide.

Answer 41

Expert Solution

Answer 42

Expert Solution

Answer 43

Expert Solution

Answer 44

Interpretation Introduction

e)

Interpretation:

All possible carbonyl compounds that will react in a Grignard reaction to yield 2-p-tolylethanol are to be listed.

Concept introduction:

Grignard reagents react with formaldehyde to produce 1⁰ alcohols, with other aldehydes to yield 2⁰ alcohols and with ketones to give 3⁰ alcohols as the product. Esters also when treated with two molar equivalents of Grignard reagents yield 3⁰ alcohols.

To list:

All possible carbonyl compounds that will react in a Grignard reaction to yield 2-(p-tolyl) ethanol.

Answer 45

Answer to Problem 44AP

2-(p-tolyl) ethanol can be prepared by reacting formaldehyde with p-tolylmethylmagnesium bromide.

Answer 46

Explanation of Solution

2-(p-tolyl) ethanol is a 1⁰ alcohol having a p-tolyl group attached to C₂ of ethanol. Hence formaldehyde is required. The remaining part should come from the Grignard reagent. Hence formaldehyde is treated with p-tolylmethylmagnesium bromide.

Answer 47

Conclusion

2-(p-tolyl) ethanol can prepared by reacting formaldehyde with p-tolylmethylmagnesium bromide.

Answer 48

Expert Solution

Interpretation Introduction

f)

Interpretation:

All possible carbonyl compounds that will react in a Grignard reaction to yield 1-cyclopentyl-2-methyl-2-propanol are to be listed.

Concept introduction:

Grignard reagents react with formaldehyde to produce 1⁰ alcohols, with other aldehydes to yield 2⁰ alcohols and with ketones to give 3⁰ alcohols as the product. Esters also when treated with two molar equivalents of Grignard reagents yield 3⁰ alcohols.

To list:

All possible carbonyl compounds that will react in a Grignard reaction to yield 1-cyclopentyl-2-methyl-2-propanol are to be listed.

Answer to Problem 44AP

1-Cyclopentyl-2-methyl-2-propanol can be prepared by reacting i) cyclopentylmethyl methyl ketone with methylmagnesium bromide ii) an ester of cyclopentylacetic acid with two molar equivalents of methylmagnesium bromide iii) acetone with cyclopentylmethylmagnesium bromide.

Explanation of Solution

1-Cyclopentyl-2-methyl-2-propanol is a 3⁰ alcohol with a three carbon straight chain with a cyclopentyl group on C₁ and –OH on C₂. Hence cyclopentylmethyl methyl ketone is treated with methylmagnesium bromide or an ester of cyclopentylacetic acid is treated with two molar equivalents of methylmagnesium bromide. The ring can come from the Grignard reagent also. Hence cyclopentylmethylmagnesium bromide is treated with acetone.

Conclusion

1-Cyclopentyl-2-methyl-2-propanol can be prepared by reacting i) cyclopentylmethyl methyl ketone with methylmagnesium bromide ii) an ester of cyclopentylacetic acid with two molar equivalents of methylmagnesium bromide iii) acetone with cyclopentylmethylmagnesium bromide.

Answer 49

Expert Solution

Answer 50

Expert Solution

Answer 51

Expert Solution

Answer 52

Interpretation Introduction

f)

Interpretation:

All possible carbonyl compounds that will react in a Grignard reaction to yield 1-cyclopentyl-2-methyl-2-propanol are to be listed.

Concept introduction:

Grignard reagents react with formaldehyde to produce 1⁰ alcohols, with other aldehydes to yield 2⁰ alcohols and with ketones to give 3⁰ alcohols as the product. Esters also when treated with two molar equivalents of Grignard reagents yield 3⁰ alcohols.

To list:

All possible carbonyl compounds that will react in a Grignard reaction to yield 1-cyclopentyl-2-methyl-2-propanol are to be listed.

Answer 53

Answer to Problem 44AP

1-Cyclopentyl-2-methyl-2-propanol can be prepared by reacting i) cyclopentylmethyl methyl ketone with methylmagnesium bromide ii) an ester of cyclopentylacetic acid with two molar equivalents of methylmagnesium bromide iii) acetone with cyclopentylmethylmagnesium bromide.

Answer 54

Explanation of Solution

1-Cyclopentyl-2-methyl-2-propanol is a 3⁰ alcohol with a three carbon straight chain with a cyclopentyl group on C₁ and –OH on C₂. Hence cyclopentylmethyl methyl ketone is treated with methylmagnesium bromide or an ester of cyclopentylacetic acid is treated with two molar equivalents of methylmagnesium bromide. The ring can come from the Grignard reagent also. Hence cyclopentylmethylmagnesium bromide is treated with acetone.

Answer 55

Conclusion

1-Cyclopentyl-2-methyl-2-propanol can be prepared by reacting i) cyclopentylmethyl methyl ketone with methylmagnesium bromide ii) an ester of cyclopentylacetic acid with two molar equivalents of methylmagnesium bromide iii) acetone with cyclopentylmethylmagnesium bromide.