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Explain why the product distribution in the chlorination of propane by sulfuryl chloride is expected to differ according to whether the hydrogen- abstraction step is accomplished by \(\mathrm{Cl} \cdot\) or \(\cdot \mathrm{SO}_{2} \mathrm{Cl}\)

Short Answer

Expert verified
The product distribution changes due to the different reactivities of radicals, with Cl· forming a broader range and ·SO₂Cl forming more selective products.

Step by step solution

01

Understanding Chlorination of Propane

In this reaction, propane (C₃H₈) can be chlorinated by sulfuryl chloride (SO₂Cl₂) through the assistance of radical initiators. The chlorination typically involves hydrogen abstraction to form chlorinated products.
02

Role of Cl· in Hydrogen Abstraction

The chlorine radical (\(\mathrm{Cl} \cdot\) ) is a very reactive species. It abstracts hydrogen atoms from propane leading to the formation of alkyl radicals efficiently across different available hydrogen sites. This can lead to the formation of both primary, secondary, and even tertiary chlorinated products.
03

Role of ·SO₂Cl in Hydrogen Abstraction

The radical \(\cdot \mathrm{SO}_{2} \mathrm{Cl}\) is less reactive compared to the chlorine radical. Therefore, it has a preference for abstracting hydrogen atoms from more substituted carbon atoms, like secondary or tertiary ones, which have weaker C-H bonds.
04

Comparing Product Distribution

When the more reactive \(\mathrm{Cl} \cdot\) abstracts hydrogen atoms, it leads to a broader range of alkyl radicals and thus a wider product distribution. In contrast, \(\cdot \mathrm{SO}_{2} \mathrm{Cl}\) tends to selectively form fewer products by abstracting hydrogen atoms mainly from secondary or tertiary carbons in propane.
05

Expected Outcome

Based on the above reactivity differences, the chlorination with \(\mathrm{Cl} \cdot\) is likely to produce a mixture of primary, secondary, and tertiary chlorinated products, while \(\cdot \mathrm{SO}_{2} \mathrm{Cl}\) will favor secondary over primary chlorinated products due to its selectivity.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Sulfuryl Chloride
Sulfuryl chloride, with the chemical formula \(\mathrm{SO}_2\mathrm{Cl}_2\), is an interesting compound used in various chlorination reactions. It's a colorless liquid that releases chlorine radicals upon thermal decomposition or when exposed to radical initiators. This property makes it quite valuable in organic synthesis and specifically for the chlorination of hydrocarbons like propane. In the reaction, sulfuryl chloride can generate two different radicals: the reactive chlorine radical (\(\mathrm{Cl} \cdot\)) and the less reactive sulfuryl chloride radical (\(\cdot \mathrm{SO}_2 \mathrm{Cl}\)). Each of these radicals affects the reaction pathway and the final product distribution differently because they differ in their reactivity and selectivity when abstracting hydrogen atoms from propane.
Hydrogen Abstraction
Hydrogen abstraction is a fundamental step in radical reactions. It involves the removal of a hydrogen atom from a molecule, which in this case is propane \(\mathrm{C}_3\mathrm{H}_8\). This step is crucial as it forms different types of alkyl radicals depending on the site from which hydrogen is abstracted.
With chlorination, the nature of the radical performing the abstraction plays a critical role:
  • \(\mathrm{Cl} \cdot\) radical is very reactive and can efficiently abstract hydrogen from primary, secondary, and tertiary C-H bonds.
  • \(\cdot \mathrm{SO}_2 \mathrm{Cl}\) radical is selective and prefers weaker, more substituted C-H bonds, typically targeting secondary or tertiary carbons.
These differences in hydrogen abstraction lead to variations in the chlorinated product's types and distribution.
Radical Initiators
In radical reactions such as the chlorination of propane with sulfuryl chloride, radical initiators play a crucial role. These are substances that can create radicals readily, often by thermally decomposing themselves or by reacting with other molecules. They expedite the reaction by providing the initial radicals needed to kickstart the chain mechanism.
Some common radical initiators include:
  • Peroxides, which decompose to form oxygen-centered radicals.
  • Azo compounds, breaking down into nitrogen-centered radicals.
In the presence of these initiators, sulfuryl chloride can effectively generate \(\mathrm{Cl} \cdot\) and \(\cdot \mathrm{SO}_2 \mathrm{Cl}\) radicals necessary for the hydrogen abstraction process, steering the chlorination of propane.
Chlorine Radicals
Chlorine radicals, or \(\mathrm{Cl} \cdot\), are highly reactive and play a pivotal role in the chlorination process. Their high reactivity arises from the presence of an unpaired electron, making them eager to pair up by attacking hydrogen atoms in other molecules. In the chlorination of propane, chlorine radicals are responsible for creating a broader range of products as they are not selective about which hydrogen atoms they abstract.
This reactivity trend leads to:
  • Formation of multiple types of alkyl radicals.
  • Chlorinated derivatives ranging from primary to tertiary products.
It's the lack of selectivity that results in diverse product distribution in reactions involving chlorine radicals, contrasting with the more selective sulfuryl chloride radicals.

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Most popular questions from this chapter

The heat of combustion of cyclopropane, \(\left(\mathrm{CH}_{2}\right)_{3}\), to give carbon dioxide and liquid water is \(499.8 \mathrm{kcal} \mathrm{mol}^{-1}\). Show how this value, assuming normal \(\mathrm{C}-\mathrm{H}\) bond strengths, can be used to calculate the average \(\mathrm{C}-\mathrm{C}\) bond energy of cyclopropane.

a. N-Bromosuccinimide (NBS) is an excellent brominating reagent and is used widely to prepare bromoalkenes from alkenes (Wohl-Ziegler reaction): The reaction is initiated with chemical initiators (peroxides) and is as selective as bromination with molecular bromine. Write plausible propagation steps (three of them) for this reaction, given the fact that the actual brominating agent appears to be molecular bromine that is generated from NBS by HBr. b. What products would you expect to be formed on bromination of 2 -methylbutane with N-bromosuccinimide?

A \(C-F\) bond energy can be computed from thermochemical studies of the vapor- phase reaction $$ \mathrm{CH}_{4}+4 \mathrm{~F}_{2} \rightarrow \mathrm{CF}_{4}+4 \mathrm{HF} \quad \Delta H^{0}=-460 \mathrm{kcal} $$ Show how the \(\Delta H^{0}\) value for this reaction may be used to calculate the energy of the \(\mathrm{C}-\mathrm{F}\) bond if all the other bond energies are known.

tert-Butyl hypobromite is a radical brominating agent that is similar to tert- butyl hypochlorite (Exercise 4 \(18^{*}\) ), but is less easily prepared than the hypochlorite. A good substitute, provided radical bromination is possible, is a mixture of \(\mathrm{BrCCl}_{3}\) and \(\mathrm{ce}\left(\mathrm{CH}_{3}\right)_{3} \mathrm{COCl}\) Thus, bromination of cyclohexene results if a high ratio of bromotrichloromethane to hypochlorite is used. O=C(O)CCC1CCCCC1Br Suggest how this reaction is initiated and propagated, and explain why it is necessary to have an excess of bromotrichloromethane.

The heat of combustion of 1 mole of liquid decane to give carbon dioxide and liquid water is \(1620.1\) kcal The heat of vaporization of decane at \(25^{\circ}\) is \(11.7\) kcal \(\mathrm{mol}^{-1}\). Calculate the heat of combustion that would be observed for all the participants in the vapor phase.

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