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2,3-Dimethylbutanereacts with bromine in the presence of light to give a mono brominate product. Further reaction gives a good yield of a dibrominated product. Predict the structures of these products, and propose a mechanism for the formation of the monobrominated product.

Short Answer

Expert verified

The Themonobrominated product formed is 2-bromo-2, 3-dimethylbutane and the dibrominated product formed is 2, 3-dibromo-2, 3-dimethylbutane.

The mechanism for the formation of mono brominated product.


The mechanism for the formation of dibrominated product

Step by step solution

01

Free radicals

An atom or group of atoms containing odd or unpaired electrons is known as free radical. The unpaired electron is represented by a single unpaired dot in the formula. Free radicals are electrically neutral. They are highly reactive species formed by homolytic fission of a covalent bond.

02

Steps involved in free radical chain reaction

In a free-radical chain reaction, free radicals are generally created in the initiation step. A free radical and a reactant are combined to yield a product in the propagation step. Lastly, the number of free radicals generally decreases in the termination step

03

Reaction of 2,3-Dimethylbutane  with bromine in presence of light

Reaction of 2,3-Dimethylbutanewith bromine in presence of light

Themonobrominated product formed is 2-bromo-2,3-dimethylbutaneand the dibrominated product formed is2,3-dibromo-2,3-dimethylbutane .

04

Mechanism for the formation of monobrominated product

The mechanism consists of three parts—initiation step, propagation step I, and propagation step II.

Mechanism for the formation of monobrominated product

05

Mechanism for the formation of dibrominated product

The mechanism consists of three parts—initiation step, propagation step I, and propagation step II.

Mechanism for the formation of dibrominated product

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Most popular questions from this chapter

Question: The bromination of methane proceeds through the following steps:

(a) Draw a complete reaction-energy diagram for this reaction.

(b) Label the rate-limiting step.

(c) Draw the structure of each transition state

(d) Compute the overall value of∆H0for the bromination.

When a small piece of platinum is added to a mixture of ethene and hydrogen, the following reaction occurs:

Doubling the concentration of hydrogen has no effect on the reaction rate. Doubling the concentration of ethene also has no effect.

(a) What is the kinetic order of this reaction with respect to ethene? With respect to hydrogen? What is the overall order?

(b) Write the unusual rate equation for this reaction.

(c)Explain this strange rate equation, and suggest what one might do to accelerate the reaction.

Question: Each of the following proposed mechanisms for the free-radical chlorination of methane is wrong. Explain how the experimental evidence disproves each mechanism.

(a)

CI2+³ó±¹â†’C±õ2(anactivatedformofCI2)CI2+CH4→H°ä±ô+°ä±á3Cl(b)CH4+hv→CH3+HCH3+Cl2→CH3Cl+ClCl+H→HCl

When a small amount of iodine is added to a mixture of chlorine and methane, it prevents chlorination from occurring. Therefore, iodine is a free-radical inhibitor for this reaction. CalculateΔ±á0values for the possible reactions of iodine with species present in the chlorination of methane and use these values to explain why iodine inhibits the reaction. (The I-Clbond-dissociation enthalpy is 211 kJ/molor 50 kcal/mol).

The following reaction is a common synthesis used in the organic chemistry laboratory course.

When we double the concentration of methoxide ion (CH3O-) , we find that the reaction rate doubles. When we triple the concentration of 1-bromopropane , we find the reaction rate triples.

(a) What is the order of this reaction with respect to 1-bromopropane? What is the order with respect to methoxide ion? Write the rate equation for this reaction. What is the overall order?

(b) One lab textbook recommends forming the sodium methoxide in methanol solvent, but before adding 1-bromopropane ,it first distills off enough methanol to reduce the mixture to half of its original volume. What difference in rate will we see when we run the reaction (using the same amounts of reagents) in half the volume of solvent?

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