Textbook Question
Show how you would add Grignard reagent to acid chloride or ester to synthesize the following alcohols.
a. Ph3C–OH
1711
views
Ch.10 - Structure and Synthesis of Alcohols
Wade9th EditionOrganic ChemistryISBN: 9780135213728
Not the one you use?Change textbookSelect a different textbook
Table of contents
- Ch.1 - Structure and Bonding107
- Ch. 2 - Acids and Bases; Functional Groups115
- Ch.3 - Structure and Stereochemistry of Alkanes100
- Ch.4 - The Study of Chemical Reactions99
- Ch.5 - Stereochemistry93
- Ch.6 - Alkyl Halides; Nucleophilic Substitution118
- Ch. 7 - Structure and Synthesis of Alkenes; Elimination158
- Ch.8 - Reactions of Alkenes171
- Ch.9 - Alkynes91
- Ch.10 - Structure and Synthesis of Alcohols143
- Ch.11 - Reactions of Alcohols104
- Ch. 12 - Infrared Spectroscopy and Mass Spectrometry22
- Ch. 13 - Nuclear Magnetic Resonance Spectroscopy45
- Ch. 14 - Ethers, Epoxides, and Thioethers72
- Ch. 15 - Conjugated Systems, Orbital Symmetry, and Ultraviolet Spectroscopy89
- Ch. 16 - Aromatic Compounds74
- Ch. 17 - Reactions of Aromatic Compounds126
- Ch. 18 - Ketones and Aldehydes193
- Ch. 19 - Amines129
- Ch. 20 - Carboxylic Acids77
- Ch. 21 - Carboxylic Acid Derivatives141
- Ch. 22 - Condensations and Alpha Substitutions of Carbonyl Compounds175
- Ch. 23 - Carbohydrates and Nucleic Acids93
- Ch. 24 - Amino Acids, Peptides, and Proteins54
Chapter 10, Problem 18a
A formate ester, such as ethyl formate, reacts with an excess of a Grignard reagent to give (after protonation) secondary alcohols with two identical alkyl groups.
![]()
(a) Propose a mechanism to show how the reaction of ethyl formate with an excess of allylmagnesium bromide gives, after protonation, hepta-1,6-dien-4-ol.
Verified step by step guidance1
Step 1: Begin by recognizing that ethyl formate is an ester with the structure HCOOCH₂CH₃. The Grignard reagent, allylmagnesium bromide (CH₂=CH-CH₂MgBr), will act as a nucleophile and attack the carbonyl carbon of the ester.
Step 2: In the first nucleophilic addition, the allyl group (CH₂=CH-CH₂) from the Grignard reagent attacks the carbonyl carbon of ethyl formate. This breaks the C=O double bond and forms a tetrahedral intermediate with a negatively charged oxygen.
Step 3: The intermediate formed in Step 2 undergoes elimination of the ethoxy group (-OCH₂CH₃), regenerating a carbonyl group and forming an aldehyde intermediate (CH₂=CH-CH₂CHO).
Step 4: Since an excess of allylmagnesium bromide is present, the aldehyde intermediate undergoes a second nucleophilic attack by another allyl group from the Grignard reagent. This forms another tetrahedral intermediate with a negatively charged oxygen.
Step 5: Finally, protonation of the negatively charged oxygen occurs during the acidic workup (H₃O⁺), resulting in the formation of the secondary alcohol, hepta-1,6-dien-4-ol (CH₂=CH-CH₂-CH(OH)-CH₂-CH=CH₂).
Verified video answer for a similar problem:
This video solution was recommended by our tutors as helpful for the problem above.
Video duration:
6mWas this helpful?
Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Grignard Reagents
Grignard reagents are organomagnesium compounds that act as nucleophiles in organic reactions. They are formed by reacting alkyl or aryl halides with magnesium in dry ether. In the context of the question, the Grignard reagent (allylmagnesium bromide) reacts with ethyl formate to form a new carbon-carbon bond, leading to the formation of a secondary alcohol after protonation.
Recommended video:
Guided course
02:12
Carbonation of Grignard Reagents
Ester Reactivity
Esters, such as ethyl formate, are susceptible to nucleophilic attack due to the presence of a carbonyl group. In this reaction, the nucleophilic Grignard reagent attacks the carbonyl carbon of the ester, leading to the formation of a tetrahedral intermediate. This intermediate can then undergo further reactions, ultimately resulting in the formation of a secondary alcohol after protonation.
Recommended video:
Guided course
02:39Ester Nomenclature
Protonation and Alcohol Formation
After the nucleophilic addition of the Grignard reagent to the ester, the reaction mixture is typically treated with an acid (like H3O+) to protonate the alkoxide intermediate formed. This step is crucial as it converts the alkoxide into a stable alcohol. In this case, the final product is hepta-1,6-dien-4-ol, a secondary alcohol with two identical alkyl groups, demonstrating the importance of protonation in alcohol synthesis.
Recommended video:
Guided course
01:01How to name alcohols
Related Practice
Textbook Question
Show how you would synthesize the following alcohol by adding Grignard reagents to ethylene oxide.
(c)
609
views
Textbook Question
Show how you would add Grignard reagent to acid chloride or ester to synthesize the following alcohols.
c. dicyclohexylphenylmethanol
612
views
Textbook Question
Propose a mechanism for the reaction of acetyl chloride with phenylmagnesium bromide to give 1,1-diphenylethanol.
1143
views
Textbook Question
A formate ester, such as ethyl formate, reacts with an excess of a Grignard reagent to give (after protonation) secondary alcohols with two identical alkyl groups.
(b) Show how you would use reactions of Grignard reagents with ethyl formate to synthesize the following secondary alcohols.
(i) pentan-3-ol
(ii) diphenylmethanol
(iii) trans,trans-nona-2,7-dien-5-ol
2291
views
Textbook Question
Show how you would synthesize the following alcohol by adding Grignard reagents to ethylene oxide.
(a) 2-phenylethanol
1420
views