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Oxymercuration-demercuration of 1 -methylcyclopentene gives which of the following products?

Short Answer

Expert verified
1-methylcyclopentanol is the final product of oxymercuration-demercuration of 1-methylcyclopentene.

Step by step solution

01

Understand the Reaction

Oxymercuration-demercuration is a method for converting alkenes into alcohols. The process involves the addition of mercury(II) acetate to the alkene in the presence of water, followed by reduction with sodium borohydride.
02

Identify the Alkene and Substituents

The compound in question is 1-methylcyclopentene. This is a cyclopentene ring with a methyl group attached to one of the carbons in the double bond.
03

Predict the Oxymercuration Addition Site

During oxymercuration, the more substituted carbon in the alkene will bond with the hydroxyl group (OH). For 1-methylcyclopentene, the more substituted carbon is the one with the methyl group.
04

Complete the Oxymercuration

Attach the OH group to the more substituted carbon on the cyclopentene ring, which holds the methyl substituent. The result is a Markovnikov addition, meaning the OH goes to the more substituted position.
05

Perform Demercuration

The final step involves removing the mercury group, generally using sodium borohydride (NaBH4), to result in the alcohol. After this step, the mercury moiety is replaced with a hydrogen.
06

Identify the Final Product

The final product of the oxymercuration-demercuration of 1-methylcyclopentene is 1-methylcyclopentanol.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Markovnikov Addition
Markovnikov addition is named after the Russian chemist Vladimir Markovnikov who discovered this rule. It is a fundamental concept in organic chemistry that dictates how certain molecules add to alkenes. During a chemical reaction, particularly when hydrogen halides or other polar reagents are added to an unsaturated carbon compound like an alkene, the rule predicts that the substituent of the reagent '(e.g., OH in oxymercuration)' will attach to the more substituted carbon atom. This is because the more substituted carbon tends to be more stable, allowing for a more favorable transition state.

For instance, in the case of 1-methylcyclopentene, when undergoing oxymercuration, the hydroxyl (OH) group will attach to the carbon that has the methyl group, because it is more substituted than the other carbon of the double bond. This ensures that the reaction proceeds in a way that maximizes stability and efficiency. Understanding Markovnikov's rule can substantially clarify why certain reaction products are formed over others.
Alkene to Alcohol Conversion
The conversion of alkenes to alcohols is a crucial reaction in organic chemistry. One efficient method for achieving this is through oxymercuration-demercuration. Starting with an alkene, the process involves two main steps. First, during oxymercuration, the alkene reacts with mercury(II) acetate in water. This step allows the attachment of a mercuric acetate group and a hydroxyl group (from water) to what was the double bond. The double bond is temporarily replaced by these groups, a critical part of turning the alkene into an alcohol.

The demercuration step follows, using sodium borohydride (NaBHâ‚„). In this step, the mercury group is replaced by a hydrogen atom. This completes the transformation into an alcohol, leaving behind the newly formed hydroxyl group bonded to the more substituted carbon, according to Markovnikov's rule. This whole process is advantageous because it provides the alcohol without rearrangements or multiple side products, making it a cleaner reaction pathway.
1-Methylcyclopentene
1-Methylcyclopentene is a specific type of alkene. It consists of a five-membered carbon ring, known as cyclopentene, with a double bond. Importantly, it has a methyl group (CH₃) attached directly to one of the carbons of the double bond, making it a substituted alkene.

This structural aspect is crucial when considering reactions like oxymercuration-demercuration. The presence of the methyl group makes one carbon in the double bond more substituted than the other, which directly influences the site of the reaction according to Markovnikov's rule. During the conversion to alcohol, the hydroxyl group will attach to this more substituted carbon.

Understanding the structure of 1-methylcyclopentene helps in predicting the outcome of chemical reactions, ensuring that students understand why certain products are formed based on the molecular composition and arrangement.

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Most popular questions from this chapter

Write the structure of the major organic product formed in the reaction of 1 -pentene with each of the following: (a) Hydrogen chloride (b) Dilute sulfuric acid (c) Diborane in diglyme, followed by basic hydrogen peroxide (d) Bromine in carbon tetrachloride (e) Bromine in water (f) Peroxyacetic acid (g) Ozone (h) Product of part (g) treated with zinc and water (i) Product of part (g) treated with dimethyl sulfide \(\left(\mathrm{CH}_{3}\right)_{2} \mathrm{~S}\)

Does the presence or absence of a catalyst such as finely divided platinum, palladium, or nickel affect the equilibrium constant for the ethylene-ethane conversion?

Alcohol A \(\left(\mathrm{C}_{10} \mathrm{H}_{18} \mathrm{O}\right)\) is converted to a mixture of alkenes \(\mathrm{B}\) and \(\mathrm{C}\) on being heated with potassium hydrogen sulfate \(\left(\mathrm{KHSO}_{4}\right)\). Catalytic hydrogenation of \(\mathrm{B}\) and \(\mathrm{C}\) yields the same product. Assuming that dehydration of alcohol A proceeds without rearrangement, deduce the structures of alcohol A and alkene \(\mathrm{C}\).

(a) Which primary alcohol of molecular formula \(\mathrm{C}_{5} \mathrm{H}_{12} \mathrm{O}\) cannot be prepared from an alkene by hydroboration-oxidation? Why? (b) Write equations describing the preparation of three isomeric primary alcohols of molecular formula \(\mathrm{C}_{5} \mathrm{H}_{12} \mathrm{O}\) from alkenes. (c) Write equations describing the preparation of the tertiary alcohol of molecular formula \(\mathrm{C}_{5} \mathrm{H}_{12} \mathrm{O}\) by acid- catalyzed hydration of two different alkenes.

Compounds \(\mathrm{A}\) and \(\mathrm{B}\) are isomers of molecular formula \(\mathrm{C}_{9} \mathrm{H}_{19} \mathrm{Br}\). Both yield the same alkene \(\mathrm{C}\) as the exclusive product of elimination on being treated with potassium tert-butoxide in dimethyl sulfoxide. Hydrogenation of alkene C gives \(2,3,3,4\) -tetramethylpentane. What are the structures of compounds \(\mathrm{A}\) and \(\mathrm{B}\) and alkene \(\mathrm{C}\) ?

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