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Chapter 4: Q 46P (page 234)

In the presence of a small amount of bromine, the following light -promoted reaction has been observed.

(a) Write a mechanism for this reaction. Your mechanism should explain how both products are formed. (Hint: Notice which H atom has been lost in both products)

(b) Explain why only this one type of hydrogen atom has been replaced, in preference to any of the other hydrogen atoms in the starting material.

Short Answer

Expert verified

Mechanism showing how both products are formed

(b) The bond dissociation enthalpy for allylic hydrogen is lower due to the formation of more stable allylic free radical.

Step by step solution

01

Free radicals

An atom or group of atoms containing odd or unpaired electron is 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 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 decrease in the termination steps.

03

Mechanism and explanation

(a) The mechanism consists of three parts which are initiation step, propagation step I and propagation step II.

Mechanism showing how both products are formed

(b) The bond dissociation enthalpy for allylic hydrogen is lower due to the formation of more stable allylic free radical.

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Most popular questions from this chapter

What would be the product ratio in the chlorination of propane if all the hydrogens were abstracted at equal rates?

Iodination of alkanes using iodine (I2)is usually an unfavorable reaction. (See problem 4-17, for example). Tetraiodomethane (Cl4) can be used as the iodine source for iodination in the presence of a free-radical initiator such as hydrogen peroxide. Propose a mechanism (involving mildly exothermic propagation steps) for the following proposed reaction. Calculate the value of 螖贬for each of the steps in your proposed mechanism.

The following bond-dissociation energies may be helpful:

The chlorination of pentane gives a mixture of three monochlorinated products.

(a) Draw their structures.

(b) Predict the ratios in which these monochlorination products will be formed, remembering that a chlorine atom abstracts a secondary hydrogen about 4.5 times as fast as it abstracts a primary hydrogen.

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. Calculate螖贬0values 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).

(a) Draw the structure of the transition state for the second propagation step in the chlorination of methane.

Show whether the transition state is product-like or reactant-like and which of the two partial bonds is stronger.

(b) Repeat for the second propagation step in the bromination of methane.
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