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Roman55 [17]
3 years ago
6

4-methyl-3-hexanol was prepared by reacting an alkene with either hydroboration-oxidation or oxymercuration-reduction. Draw the

structure of the alkene that was used to prepare the alcohol in highest yield.

Chemistry
1 answer:
Ivenika [448]3 years ago
3 0

Answer:

Structure in attachment.

Explanation:

The oxymercuration-demercuration of an asymmetric alkene usually produces the  Markovnikov orientation of an addition. The electrophile ⁺Hg(OAc), formed by the electrophile attack of the mercury ion, remains attached to least substituted group at the end of the double bond. This electrophile has a considerable amount of positive charge on its two  carbon atoms, but there is more positive charge on the more substituted carbon atom,  where it is more stable. The water attack occurs on this more electrophilic carbon, and the Markovnikov orientation occurs.

In hydroboration, borane adds to the double bond in one step. Boron is added to the less  hindered and less substituted carbon, and hydrogen is added to the more substituted carbon. The electrophilic boron atom adds to the less substituted end of the double bond, positioning the positive charge (and the hydrogen atom) at the more substituted end. The result is a product with the anti-Markovnikov orientation.

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A more electronegative atom A) will have more attraction to the electrons in a chemical bond. B) is more likely to lose an elect
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Answer: A more electronegative atom will have more attraction to the electrons in a chemical bond.

Explanation:

An atom that is able to attract electrons or shared pair of electrons more towards itself is called an electronegative atom.

For example, fluorine is the most electronegative atom.

Due to its high electronegativity it is able to attract an electropositive atom like H towards itself. As a result, both fluorine and hydrogen will acquire stability by sharing of electrons.

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gavmur [86]

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For the reaction PC15 (8) PC13 (g) + Cl2 (g) K = 0.0454 at 261 °C. If a vessel is filled with these gases such that the initial
Fiesta28 [93]

Answer:

The correct answer is B.

The K_{eq} is samller than Q of the reaction . So,the reaction will shift towards the left i.e. towards the reactant side.

Explanation:

Equilibrium constant is defined as the ratio of concentration of products to the concentration of reactants each raised to the power their stoichiometric ratios. It is expressed as K_{eq}

K is the constant of a certain reaction when it is in equilibrium, while Q is the quotient of activities of products and reactants at any stage other than equilibrium of a reaction.

For the given chemical reaction:

PCl_5(g)\rightleftharpoons PCl_3(g)+Cl_2(g)

The expression for Q is written as:

Q=\frac{[PCl_3][Cl_2]}{[[PCl_5]^1}

Q=\frac{0.20 M\times 2.5 M}{0.20 M}

Q=2.5

Given : K_{eq} = 0.0454

Thus as K_{eq}, the reaction will shift towards the left i.e. towards the reactant side.

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