Skip to main content
Ch. 8 - Alkenes 1: Properties and Electrophilic Additions
Mullins - Organic Chemistry: A Learner Centered Approach 1st Edition
Mullins1st EditionOrganic Chemistry: A Learner Centered ApproachISBN: 9780137566471Non è quello che usi tu?Cambia libro di testo
Capitolo 7, Problema 54a

Predict the products you would get when the following alkenes undergo (i) hydroboration–oxidation (1. BH3 2. NaOH, H2O2 or (ii) oxymercuration–reduction [1. Hg(OAc)2, H2O 2. NaBH4].
(a)

Guida verificata passo dopo passo
1
Step 1: Analyze the given alkene structure. The molecule contains a phenyl group (Ph) attached to a carbon chain with a double bond located between the second and third carbons in the chain.
Step 2: For hydroboration–oxidation (1. BH3, 2. NaOH, H2O2), the reaction proceeds via anti-Markovnikov addition of water across the double bond. The boron atom initially adds to the less substituted carbon of the double bond, followed by oxidation to replace the boron with a hydroxyl group (-OH). This results in the alcohol being formed at the less substituted carbon.
Step 3: For oxymercuration–reduction (1. Hg(OAc)2, H2O, 2. NaBH4), the reaction proceeds via Markovnikov addition of water across the double bond. The mercury intermediate forms at the more substituted carbon, and during reduction, the mercury is replaced with a hydroxyl group (-OH). This results in the alcohol being formed at the more substituted carbon.
Step 4: Consider stereochemistry. Hydroboration–oxidation typically results in syn addition (both the hydrogen and hydroxyl group add to the same face of the double bond), while oxymercuration–reduction does not involve stereospecificity.
Step 5: Predict the products. For hydroboration–oxidation, the alcohol will form at the less substituted carbon of the double bond. For oxymercuration–reduction, the alcohol will form at the more substituted carbon of the double bond. Ensure the phenyl group remains unaffected in both reactions.

Risposta video verificata per un problema simile:

Questa soluzione video è stata consigliata dai nostri tutor come utile per risolvere questo problema.
Durata del video:
7m

Concetti chiave

Ecco i concetti essenziali che devi comprendere per rispondere correttamente alla domanda.

Hydroboration-Oxidation

Hydroboration-oxidation is a two-step reaction that converts alkenes into alcohols. In the first step, BH₃ adds to the alkene, forming a trialkylborane intermediate. This is followed by oxidation with hydrogen peroxide (H₂O₂) and a base (NaOH), which converts the boron to an alcohol. This reaction is stereospecific and results in anti-Markovnikov addition, meaning the hydroxyl group attaches to the less substituted carbon.
Video consigliato:
06:38
General properties of hydroboration-oxidation.

Oxymercuration-Reduction

Oxymercuration-reduction is another method for converting alkenes to alcohols, involving two steps. Initially, the alkene reacts with mercuric acetate (Hg(OAc)₂) in the presence of water, leading to the formation of a mercurial intermediate. The subsequent reduction with sodium borohydride (NaBH₄) replaces the mercury with a hydrogen atom, yielding an alcohol. This reaction follows Markovnikov's rule, where the hydroxyl group attaches to the more substituted carbon.
Video consigliato:
05:
General properties of oxymercuration-reduction.

Markovnikov's Rule

Markovnikov's rule is a principle in organic chemistry that predicts the outcome of electrophilic addition reactions to alkenes. It states that when HX (where X is a halogen or hydroxyl group) adds to an asymmetric alkene, the more stable carbocation will form, leading to the more substituted carbon receiving the electrophile. This rule is crucial for understanding the regioselectivity of reactions like oxymercuration-reduction, where the product distribution is influenced by the stability of the intermediates formed.
Video consigliato:
03:54
The 18 and 16 Electron Rule