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1-Methylcyclopentene can be converted into 2-methylcyclopentanol by (a) acid-catalyzed hydration (b) hydroboration oxidation (c) epoxide formation followed by reduction with LiAlH \(_{4}\) (d) oxymercuration-demercuration

Short Answer

Expert verified
The best method to convert 1-methylcyclopentene to 2-methylcyclopentanol is oxymercuration-demercuration.

Step by step solution

01

Evaluate Acid-Catalyzed Hydration

In acid-catalyzed hydration, alkenes react with water in the presence of an acid to form alcohols. For 1-methylcyclopentene, the double bond would open and the OH group from water would likely add to the more substituted carbon (Markovnikov addition), resulting in 2-methylcyclopentanol.
02

Evaluate Hydroboration-Oxidation

In hydroboration-oxidation, alkenes react with borane (BH₃) followed by oxidation with hydrogen peroxide in aqueous sodium hydroxide. This method typically adds an OH group to the less substituted carbon of the double bond (anti-Markovnikov addition). Since we need 2-methylcyclopentanol, this method wouldn't effectively yield the desired product from 1-methylcyclopentene.
03

Evaluate Epoxide Formation and Reduction with LiAlHâ‚„

Epoxide formation involves converting alkenes into epoxides, which can be subsequently reduced using lithium aluminum hydride (LiAlHâ‚„). However, reducing epoxides with LiAlHâ‚„ doesn't efficiently direct an OH group to the more substituted carbon of the original double bond, missing our target of forming 2-methylcyclopentanol.
04

Evaluate Oxymercuration-Demercuration

Oxymercuration-demercuration follows a Markovnikov addition mechanism without carbocation rearrangement. It involves using mercuric acetate in water followed by treatment with sodium borohydride. This approach adds OH to the more substituted carbon, yielding 2-methylcyclopentanol efficiently, which aligns well with our conversion goal from 1-methylcyclopentene.

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Key Concepts

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

Acid-Catalyzed Hydration
Acid-catalyzed hydration is a simple yet effective reaction for converting alkenes into alcohols. It involves the use of a strong acid, like sulfuric acid, to react with an alkene and water. The acid donates protons to the alkene, creating a carbocation intermediate. In this process, the carbon that forms the more stable carbocation—in other words, the more substituted carbon—attracts the hydroxyl (\( \text{OH} \)) group from water, resulting in a Markovnikov addition. For instance, when 1-methylcyclopentene is subjected to acid-catalyzed hydration, the \( \text{OH} \) group would prefer to attach to the more substituted carbon, forming 2-methylcyclopentanol. This method is popular for its straightforward approach and relatively mild reaction conditions.
Hydroboration-Oxidation
Hydroboration-oxidation is a two-step reaction that converts alkenes into alcohols in a unique way. In the first step, an alkene reacts with borane (\( \text{BH}_3 \)) or its derivatives. Borane acts as a catalyst by attaching to the less substituted carbon of the double bond, setting the reaction for an anti-Markovnikov addition. In the second step, the organic borane intermediate is treated with hydrogen peroxide (\( \text{H}_2\text{O}_2 \)) in a basic aqueous solution, like sodium hydroxide (\( \text{NaOH} \)). This step replaces the boron atom with a hydroxyl (\( \text{OH} \)) group. While hydroboration-oxidation is generally used to access less substituted alcohols, it wouldn't yield 2-methylcyclopentanol from 1-methylcyclopentene since the process inherently opposes the formation of a Markovnikov product.
Oxymercuration-Demercuration
Oxymercuration-demercuration is a valuable reaction for creating Markovnikov alcohols from alkenes without carbocation rearrangements, making it highly precise. In this reaction, mercuric acetate (\( \text{Hg(OAc)}_2 \)) in water adds to an alkene, forming a highly reactive mercurinium ion. The \( \text{OH} \) group then targets the more substituted carbon. Subsequently, sodium borohydride (\( \text{NaBH}_4 \)) reduces the organomercury compound, resulting in the formation of alcohol. This particular reaction fully aligns with the production of 2-methylcyclopentanol from 1-methylcyclopentene, as the \( \text{OH} \) group finds its way to the more substituted carbon, following Markovnikov's rule.
LiAlHâ‚„ Reduction
Lithium aluminum hydride (\( \text{LiAlH}_4 \)) is renowned for its use in reducing various functional groups. However, in the context of converting epoxides to alcohols, it has some limitations, especially concerning regioselectivity. When an epoxide generated from 1-methylcyclopentene is reduced by \( \text{LiAlH}_4 \), the aluminum hydride typically attacks the less hindered carbon atom. It fails to deliver the hydroxyl group (\( \text{OH} \)) to the more substituted carbon, which means that while \( \text{LiAlH}_4 \) showcases broad reducing capabilities, it does not effectively target the desired product of 2-methylcyclopentanol for this particular reaction. Nevertheless, \( \text{LiAlH}_4 \) remains vital for many other reductions in organic chemistry.
Markovnikov Addition
Markovnikov addition is a pivotal concept in understanding how substituents add across a double bond in alkenes. Named after the Russian chemist Vladimir Markovnikov, this rule states that the more electronegative atom or group among the reagents that adds to an alkene will attach itself to the more substituted carbon of the double bond. This trend arises because the more substituted carbon tends to form the more stable cation intermediate, facilitating further addition reactions. Processes such as acid-catalyzed hydration and oxymercuration-demercuration are classic examples where Markovnikov's rule is applied to form more substituted alcohols, like 2-methylcyclopentanol from 1-methylcyclopentene. Understanding Markovnikov's rule is essential for predicting the outcomes of many organic reactions.
Anti-Markovnikov Addition
Anti-Markovnikov addition provides a contrasting behavior to the traditional Markovnikov rule, offering a different regioselectivity during reactions. In this type of addition, the less substituted carbon of an alkene is favored for addition. The driving force behind this is contrary to forming a stable intermediate; rather, it often involves syn-addition processes, such as hydroboration. During hydroboration, for example, the boron atom is highly selective and adds to the less substituted carbon atom, leading to an anti-Markovnikov product. While anti-Markovnikov processes do not result in 2-methylcyclopentanol from 1-methylcyclopentene, they are important in synthesizing less substituted alcohols and integral in reactions where differing selectivity is desired. This nuanced understanding is crucial for planning synthesis routes in organic chemistry.

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Most popular questions from this chapter

The aldol self-condensation of acetone is in equilibrium that favours acetone over its corresponding product. Which of the following conditions is most likely to shift the position of equilibrium toward product A? (a) By using a catalytic amount of \(\mathrm{NaOH}\). (b) By using only a catalytic amount of acetone. (c) By removing product \(A\) as soon as it is formed. (d) By increasing reaction temperature.

Alkyl halides and alcohols easily undergo nucleophilic substitution either through \(\mathrm{S}_{\mathrm{N}} 1\) or \(\mathrm{S}_{\mathrm{N}} 2\) mechanism. The relative case of these two processes depends upon the nature of the substrate (alkyl group as well as leaving group), nature of nucleophile and also upon the nature of solvent. \(\mathrm{S}_{\mathrm{N}} 1\) mechanism involves the formation of carbocation as intermediate while \(\mathrm{S}_{\mathrm{N}} 2\) mechanism involves the formation of a transition pentavalent state. \(\mathrm{S}_{\mathrm{N}} 1\) is the main mechanism in \(3^{\circ}\) alkyl halides and alcohols, while \(\mathrm{S}_{\mathrm{N}} 2\) mechanism is the path adopted by most of the \(1^{\circ}\) alkyl halides and \(2^{\circ}\) alkyl halides may follow \(\mathrm{S}_{\mathrm{N}} 1\) as well as \(\mathrm{S}_{\mathrm{N}} 2\). Neopentyl alcohol, \(\mathrm{Me}_{3} \mathrm{CCH}_{2} \mathrm{OH}\), reacts with HX according to (a) \(\mathrm{S}_{\mathrm{N}} 1\) mechanism (b) \(\mathrm{S}_{\mathrm{N}} 2\) mechanism (c) Both (a) and (b) (d) None

Arrange the following in increasing acidic character (i) Phenol (ii) m-nitrophenol (iii) \(\mathrm{m}\) -chlorophenol (iv) \(\mathrm{m}\) -cresol (a) iv \(<\mathrm{i}<\mathrm{iii}<\mathrm{ii}\) (b) iv \(<\mathrm{i}<\mathrm{ii}<\mathrm{iii}\) (c) \(i<\) iv \(<\) iii \(<\) ii (d) iii \(<\mathrm{ii}<\mathrm{iv}<\mathrm{i}\)

Choose the correct statement regarding acidic character of acetic acid, \(\mathrm{CH}_{3} \mathrm{COOH}\) and peroxyacetic acid, \(\mathrm{CH}_{3} \mathrm{COOOH}\). (a) Peroxyacetic acid is stronger acid than acetic acid since the former has one extra oxygen, an electronegative element. (b) Peroxyacetic acid is stronger than acetic acid because its conjugate base is a weaker base than acetate. (c) Peroxyacetic acid is weaker than acetic acid because its conjugate base is less stable than that of acetate ion. (d) Both are equally strong.

The correct order of decreasing basic character is (i) \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{2}\) (ii) \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CH}_{2} \mathrm{NH}_{2}\) (iii) \(\left(\mathrm{C}_{6} \mathrm{H}_{5}\right)_{2} \mathrm{NH}\) (iv) \(\mathrm{C}_{6} \mathrm{H}_{11} \mathrm{NH}_{2}\) (a) \(\mathrm{ii}>\mathrm{i}>\mathrm{iii}>\mathrm{iv}\) (b) \(\mathrm{iv}>\mathrm{ii}>\mathrm{i}>\mathrm{iii}\) (c) \(\mathrm{iv}>\mathrm{iii}>\mathrm{ii}>\mathrm{i}\) (d) \(\mathrm{iv}>\mathrm{ii}>\mathrm{iii}>\mathrm{i}\)

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