Textbook Question
Hydration of alkynes (via oxymercuration) gives good yields of single compounds only with symmetrical or terminal alkynes. Show what the products would be from hydration of each compound.
a. hex-3-yne
1593
views
Ch. 18 - Ketones and Aldehydes
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 18, Problem 64b
Hydration of alkynes (via oxymercuration) gives good yields of single compounds only with symmetrical or terminal alkynes. Show what the products would be from hydration of each compound.
b. hex-2-yne
Verified step by step guidance1
Identify the reaction type: The problem involves the hydration of an alkyne via oxymercuration. This reaction adds water (H₂O) across the triple bond of the alkyne in the presence of a catalyst, typically mercuric sulfate (HgSO₄) and sulfuric acid (H₂SO₄). The product is a ketone due to tautomerization of the initially formed enol.
Analyze the structure of the alkyne: Hex-2-yne is an internal alkyne with the triple bond located between the second and third carbons in a six-carbon chain. Its structure can be represented as CH₃-C≡C-CH₂-CH₂-CH₃.
Determine the regioselectivity: In the hydration of internal alkynes, the addition of water is not regioselective because both carbons of the triple bond are equally substituted. This means the enol intermediate can form on either carbon of the triple bond.
Describe the enol formation: The addition of water across the triple bond forms an enol intermediate. The enol has a hydroxyl group (-OH) attached to one of the carbons of the former triple bond and a double bond between the two carbons. For hex-2-yne, the enol structures would be CH₃-C(OH)=CH-CH₂-CH₂-CH₃ and CH₃-CH=CH(OH)-CH₂-CH₂-CH₃.
Explain tautomerization: The enol intermediates are unstable and undergo keto-enol tautomerization to form ketones. For hex-2-yne, the final products would be two ketones: CH₃-CO-CH₂-CH₂-CH₃ (pentan-2-one) and CH₃-CH₂-CO-CH₂-CH₃ (pentan-3-one). These are structural isomers, and both are formed in equal amounts due to the lack of regioselectivity.
Verified video answer for a similar problem:
This video solution was recommended by our tutors as helpful for the problem above.
Video duration:
7mWas this helpful?
Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Hydration of Alkynes
Hydration of alkynes involves the addition of water (H2O) across the triple bond of an alkyne, resulting in the formation of an alcohol. This reaction can be catalyzed by acids or through oxymercuration, which uses mercuric acetate to facilitate the addition. The regioselectivity of the reaction is influenced by the structure of the alkyne, particularly whether it is symmetrical or terminal.
Recommended video:
Guided course
09:11
Alkyne Hydration
Oxymercuration-Demercuration
Oxymercuration-demercuration is a two-step reaction used to convert alkenes and alkynes into alcohols. In the first step, mercuric acetate reacts with the alkyne to form a mercurial intermediate, which allows for Markovnikov addition of water. The second step involves the reduction of the mercurial compound to yield the corresponding alcohol, typically using sodium borohydride (NaBH4).
Recommended video:
Guided course
05:General properties of oxymercuration-reduction.
Markovnikov's Rule
Markovnikov's Rule states that in the addition of HX (where X is a halogen or OH) to an alkene or alkyne, the hydrogen atom will attach to the carbon with the greater number of hydrogen atoms already attached. This rule helps predict the major product of hydration reactions, particularly in unsymmetrical alkynes, where the position of the hydroxyl group in the final alcohol product is determined by the stability of the carbocation formed during the reaction.
Recommended video:
Guided course
03:54The 18 and 16 Electron Rule
Related Practice
Textbook Question
Hydration of alkynes (via oxymercuration) gives good yields of single compounds only with symmetrical or terminal alkynes. Show what the products would be from hydration of each compound.
c. hex-1-yne
640
views
Textbook Question
Hydration of alkynes (via oxymercuration) gives good yields of single compounds only with symmetrical or terminal alkynes. Show what the products would be from hydration of each compound.
d. cyclodecyne
703
views
Textbook Question
Two structures for the sugar glucose are shown on page 914. Interconversion of the open-chain and cyclic hemiacetal forms is catalyzed by either acid or base.
(b) The cyclic hemiacetal is more stable than the open-chain form, so very little of the open-chain form is present at equilibrium. Will an aqueous solution of glucose reduce Tollens reagent and give a positive Tollens test? Explain.
719
views
Textbook Question
Hydration of alkynes (via oxymercuration) gives good yields of single compounds only with symmetrical or terminal alkynes. Show what the products would be from hydration of each compound.
e. 3-methylcyclodecyne
717
views
Textbook Question
Two structures of the sugar fructose are shown next. The cyclic structure predominates in aqueous solution.
(a) Number the carbon atoms in the cyclic structure. What is the functional group at C2 in the cyclic form?
(b) Propose a mechanism for the cyclization, assuming a trace of acid is present.
1712
views