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Chapter 27: Problem 68

In the chlorination of \(\mathrm{CH}_{4},\) some \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{Cl}\) is obtained as a product. Explain why this should be so.

Short Answer

Expert verified
The chlorination reaction of methane can continue past the initial formation of CH3Cl and result in the formation of CH3CH2Cl. This is due to the reactive nature of chlorine and its ability to replace hydrogen atoms in a chain reaction.

Step by step solution

01

Understand the Reactants and Products

In a methylation process, methane (CH4) reacts with chlorine (Cl). Chlorine atoms replace some of the hydrogen atoms in the methane molecule. CH3CH2Cl is formed when chlorine substitutes two hydrogen atoms in the CH4. The CH3CH2 represents adding an extra carbon and two hydrogens to the original methane molecule.
02

Explanation of the Formation of CH3CH2Cl

The chlorination of methane initially generates methyl chloride (CH3Cl). However, the process does not stop at this point. Methyl chloride can further undergo chlorination, leading to the formation of CH2Cl2. This reaction may occur multiple times, each time replacing a hydrogen atom with a chlorine atom.
03

Chain Reaction Mechanism

This process is essentially a chain reaction. Once a chlorine atom replaces a hydrogen atom, another chlorine atom replaces another hydrogen atom, forming CH3CH2Cl. This reaction mechanism can be summarized in three stages: initiation, propagation, and termination. In the initiation stage, a chlorine atom breaks forming two chlorine radicals. These radicals then react with methane in the propagation stage. In the termination stage, all radicals get consumed, forming stable molecules.

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Most popular questions from this chapter

To prepare methyl ethyl ketone, which of these compounds would you oxidize: 2 -propanol, 1 -butanol, 2-butanol, or tert-butyl alcohol? Explain.

Predict the major organic product obtained in each of the following reactions. Assume that [O] represents \(\mathrm{Na}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}\) in \(\mathrm{H}_{2} \mathrm{SO}_{4}\) (a) \(\left(\mathrm{CH}_{3}\right)_{2} \mathrm{CHCH}_{2} \mathrm{OH}+\mathrm{HBr} \longrightarrow\) (b) \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{COH}+\mathrm{K} \longrightarrow\) (c) \(\left(\mathrm{CH}_{3}\right)_{2} \mathrm{CHOH} \stackrel{[\mathrm{O}]}{\longrightarrow}\) (d) \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{OH}+\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{I} \longrightarrow\)

Write the name and structure of the benzene derivatives described below. (a) Formula: \(\mathrm{C}_{8} \mathrm{H}_{10} ;\) forms three monochlorination products when treated with \(\mathrm{Cl}_{2}\) and \(\mathrm{FeCl}_{3}\) (b) Formula: \(\mathrm{C}_{9} \mathrm{H}_{12}\); forms one monochlorination product when treated with \(\mathrm{Cl}_{2}\) and \(\mathrm{FeCl}_{3}\) (c) Formula: \(\mathrm{C}_{9} \mathrm{H}_{12}\); forms four monochlorination products when treated with \(\mathrm{Cl}_{2}\) and \(\mathrm{FeCl}_{3}\)

Answer the following questions for this E1 reaction: $$\begin{array}{c} \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{C}\left(\mathrm{CH}_{3}\right)_{2} \mathrm{Br}+\mathrm{CH}_{3} \mathrm{OH} \longrightarrow \mathrm{CH}_{3} \mathrm{CH}=\mathrm{C}\left(\mathrm{CH}_{3}\right)_{2}+\mathrm{Br}^{-}+\mathrm{CH}_{3} \mathrm{OH}_{2}^{+} \end{array}$$ (a) What is the rate expression for the reaction? (b) Draw the reaction profile for the reaction. Label all parts. Assume that the products are lower in energy than the reactants. (c) What is the effect on the rate of the reaction of doubling the concentration of \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{C}\left(\mathrm{CH}_{3}\right)_{2} \mathrm{Br} ?\) (d) What is the effect on the rate of the reaction of doubling the concentration of \(\mathrm{CH}_{3} \mathrm{OH} ?\)

Write the formulas of the products formed from the reaction of 2 -butene with each of the following substances: (a) \(\mathrm{H}_{2} ;\) (b) \(\mathrm{Cl}_{2} ;\) (c) \(\mathrm{HCl} ;\) (d) \(\mathrm{H}_{2} \mathrm{O}\) (in acid).
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