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Question: Free-radical chlorination of hexane gives very poor yields of 1 clorohexyane, while cyclohexane can be converted to chlorocyclohexane in good yield.

(a)How do you account for this difference?

(b) What ratio of reactants (cyclohexane and chlorine)

Short Answer

Expert verified

Answer

(a) There are twelveequivalent protons attached to the secondary carbon atom, and only sixprotonsareattached to the primary carbons in hexene. The desired product with cyclohexane is more likely to occur.

(b) Cyclohexane should be used in excess to the synthesis of chlorocyclohexane.

Step by step solution

01

Free radicals

An atom or group of atoms containing odd or unpaired electronsis known as the 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 a free radical chain reaction

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

03

Explanation

(hexane) (poor yield)

(cyclohexane) chlorocyclohexane (good yield)

(a) There are twelveequivalent protons attached to the secondary carbon atom, and only sixprotonsareattached to the primary carbons in hexene. The desired product with cyclohexane is more likely to occur.

(b)Cyclohexane should be used in excess for the synthesis of chlorocyclohexane.

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Most popular questions from this chapter

Question: When ethene is mixed with hydrogen in the presence of a platinum catalyst, hydrogen adds across the double bond to form ethane. At room temperature, the reaction goes to completion. Predict the signs โˆ†H0andโˆ†S0for this reaction. Explain these signs in terms of bonding and freedom of motion.

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?

For each compound, predict the major product of free-radical bromination. Remember that bromination is highly selective, and only the most stable radical will be formed.

(a) cyclohexane

(b) methylcyclopentane

(c) decalin

(d) hexane

(e)

(f)

Draw the important resonance forms of the following free radicals.

(a)

(b)

(c)

(d)

(e)

(f)

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 ofโˆ†H0for the bromination.

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