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Hydroboration of 2 -methyl-2-pentene at \(25^{\circ} \mathrm{C}\), followed by oxidation with alkaline \(\mathrm{H}_{2} \mathrm{O}_{2}\), yields 2 -methyl-3-pentanol, but hydroboration at \(160^{\circ} \mathrm{C}\) followed by oxidation yields 4 -methyl-1-pentanol. Suggest a mechanism.

Short Answer

Expert verified
At 25°C, it forms 2-methyl-3-pentanol; at 160°C, rearrangement forms 4-methyl-1-pentanol.

Step by step solution

01

Understand Hydroboration-Oxidation

Hydroboration-oxidation of alkenes is a two-step reaction that converts alkenes into alcohols. The hydroboration step adds a boron atom across the double bond, while the oxidation step replaces the boron with a hydroxyl group.
02

Identify the Structure of 2-Methyl-2-Pentene

2-Methyl-2-pentene is a five-carbon chain with a double bond between the second and third carbons, and a methyl group attached to the second carbon.
03

Hydroboration at 25°C

At 25°C, hydroboration of 2-methyl-2-pentene occurs via an anti-Markovnikov addition, where boron adds to the less substituted carbon. This results in the addition of boron to the third carbon, forming the intermediate 2-methyl-3-pentylborane.
04

Oxidation with Alkaline H2O2 at 25°C

Oxidation replaces the boron with a hydroxyl group, resulting in the production of 2-methyl-3-pentanol. This occurs because the boron is on the less hindered carbon.
05

Hydroboration at 160°C

At 160°C, the hydroboration is more likely to involve rearrangements due to the increased thermal energy. Here, boron may initially add to the less substituted carbon, but the intermediate can rearrange to form 4-methyl-1-pentylborane due to more stable isomer formation.
06

Oxidation with Alkaline H2O2 at 160°C

In this case, oxidation replaces the boron with a hydroxyl group, resulting in 4-methyl-1-pentanol, since the intermediate rearranged to a more thermodynamically stable form during hydroboration.

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Key Concepts

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

Anti-Markovnikov Addition
In the world of organic chemistry, predicting how molecules will react is crucial. One key concept involving alkenes is the 'Anti-Markovnikov Addition.' This principle describes how certain reactions lead to the addition of atoms or groups to a double bond in an unexpected manner. Specifically, during hydroboration, boron adds to the less substituted carbon atom of an alkene, contrary to what Markovnikov's rule would predict.

When we consider the example of 2-methyl-2-pentene, the hydroboration reaction at low temperatures (such as 25°C) follows this anti-Markovnikov pathway. This means that the boron atom prefers attaching itself to the third carbon in the chain, which is less hindered than the second carbon already burdened with a methyl group. Consequently, the initial product is 2-methyl-3-pentylborane.

This selectivity is influenced by a combination of steric and electronic factors which favor the less crowded carbon during the addition. It's a fascinating departure from conventional expectations in chemistry and serves as a basis for diverse synthetic applications.
Alkene to Alcohol Conversion
Transforming alkenes like 2-methyl-2-pentene to alcohols involves multiple steps, critical for creating various compounds in both laboratory and industrial settings. Through the process of hydroboration-oxidation, an alkene's double bond is efficiently turned into an alcohol group.

The course of this transformation begins with the hydroboration, where a boron-containing compound adds to the alkene. In the standard condition (25°C in our example), boron attaches anti-Markovnikov style to give an alkylborane, specifically 2-methyl-3-pentylborane.

Following this, the crucial oxidation step uses hydrogen peroxide in an alkaline medium to swap the boron atom out for a hydroxyl group. This feat allows chemists to generate 2-methyl-3-pentanol as a final product. The overall sequence avoids carbocation rearrangements and stereo complications, favoring a straightforward conversion from alkenes to their respective alcohols.
  • Boron attaches to less substituted carbon (anti-Markovnikov)
  • Oxidation exchanges boron with OH group
  • Resulting in alcohol formation
Temperature Effects in Reactions
Temperature can drastically influence the course of a chemical reaction. It dictates not only the rate but also the pathway and outcome of reactions due to the energy available to the reacting species.

In the mechanism of hydroboration-oxidation, the example with 2-methyl-2-pentene showcases how temperature plays a pivotal role. At a lower temperature of 25°C, the addition of boron to the alkene follows the anti-Markovnikov rule, leading to the expected product: 2-methyl-3-pentanol.

However, increasing the temperature to 160°C affords the reaction more energy. This additional thermal energy can facilitate unexpected rearrangements during hydroboration. At such higher temperatures, the intermediate may shift its structure to form a more stable isomeric form, resulting in 4-methyl-1-pentanol instead.
  • Lower temperatures favor predictable reactions
  • Higher temperatures can lead to rearrangements
  • Temperature influences both rate and selectivity
Understanding these temperature effects is crucial for organic synthesis, enabling chemists to tailor conditions for desired outcomes efficiently.

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Most popular questions from this chapter

Reaction of cyclohexene with mercury(II) acetate in \(\mathrm{CH}_{3} \mathrm{OH}\) rather than \(\mathrm{H}_{2} \mathrm{O},\) followed by treatment with \(\mathrm{NaBH}_{4},\) yields cyclohexyl methyl ether rather than cyclohexanol. Suggest a mechanism.

Compound \(\mathbf{A}, \mathrm{C}_{11} \mathrm{H}_{16} \mathrm{O},\) was found to be an optically active alcohol. Despite its apparent unsaturation, no hydrogen was absorbed on catalytic reduction over a palladium catalyst. On treatment of A with dilute sulfuric acid, dehydration occurred and an optically inactive alkene \(\mathbf{B}\), \(\mathrm{C}_{11} \mathrm{H}_{14},\) was the major product. Alkene \(\mathbf{B},\) on ozonolysis, gave two products. One product was identified as propanal, \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CHO}\). Compound \(\mathrm{C},\) the other product, was shown to be a ketone, \(\mathrm{C}_{8} \mathrm{H}_{8} \mathrm{O} .\) How many degrees of unsaturation does A have? Write the reactions, and identify \(\mathbf{A}, \mathbf{B},\) and \(\mathbf{C}\)

Compound A has the formula \(\mathrm{C}_{10} \mathrm{H}_{16}\). On catalytic hydrogenation over palladium, it reacts with only 1 molar equivalent of \(\mathrm{H}_{2}\). Compound \(\mathrm{A}\) also undergoes reaction with ozone, followed by zinc treatment, to yield a symmetrical diketone, \(\mathbf{B}\) ( \(\mathrm{C}_{10} \mathrm{H}_{16} \mathrm{O}_{2}\) ). (a) How many rings does A have? (b) What are the structures of \(\mathrm{A}\) and \(\mathrm{B}\) ? (c) Write the reactions.

What products would you expect from reaction of 1-methylcyclohexene with the following reagents? (a) Aqueous acidic \(\mathrm{KMnO}_{4}\) (b) \(\mathrm{O}_{3},\) followed by \(\mathrm{Zn}, \mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{H}\)

Simmons-Smith reaction of cyclohexene with diiodomethane gives a single cyclopropane product, but the analogous reaction of cyclohexene with 1,1 -diiodoethane gives (in low yield) a mixture of two isomeric methylcyclopropane products. What are the two products, and how do they differ?

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