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Chapter 6: Problem 96

Column I (reaction) (a) Wurtz reaction (b) Dehydration of alcohol with Conc. \(\mathrm{H}_{3} \mathrm{PO}_{4}\) (c) Kolbe's electrolysis (d) Diazotisation Column II (intermediate) (p) Carbocation (q) Electrophile (excluding free radicals) (r) Free radical (s) Carbanion (t) \(\mathrm{N}_{2}\) gas liberated in intermediate step

Short Answer

Expert verified
(a) - (r), (b) - (p), (c) - (r), (d) - (t)

Step by step solution

01

Understand the Wurtz Reaction

The Wurtz reaction typically involves two alkyl halides reacting in the presence of sodium metal to form a coupled hydrocarbon product. The key intermediate in this process is a free radical, formed when the alkyl halide reacts with sodium to create an alkyl radical.
02

Dehydration of Alcohol with Conc. \(\mathrm{H}_{3}\mathrm{PO}_{4}\)

In the dehydration of alcohol using concentrated \(\mathrm{H}_{3}\mathrm{PO}_{4}\), the alcohol undergoes protonation and water is eliminated to form a carbocation intermediate. This carbocation can then lead to the formation of an alkene.
03

Analyze Kolbe's Electrolysis

Kolbe's electrolysis involves the electrochemical decarboxylation of carboxylic acids to produce hydrocarbons. It's a process where a free radical is formed during the reaction when the carboxylate ion loses carbon dioxide at the anode to form a radical.
04

Identify Diazotisation

Diazotisation is the process of forming a diazonium salt from an amine and nitrous acid. A notable intermediate step in this reaction is the liberation of \(\mathrm{N}_{2}\) gas, as it exits the reaction as a stable gas.
05

Match Reactions to Intermediates

- (a) Wurtz reaction corresponds to (r) Free radical. - (b) Dehydration of alcohol corresponds to (p) Carbocation. - (c) Kolbe's electrolysis corresponds to (r) Free radical. - (d) Diazotisation corresponds to (t) \(\mathrm{N}_{2}\) gas liberated.

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Key Concepts

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

Wurtz Reaction
The Wurtz reaction is a fascinating method employed in organic chemistry to couple two alkyl halide molecules into a larger alkane. This transformation occurs in the presence of sodium metal. Here's how it works:
  • When the alkyl halide contacts sodium metal, it transforms into a free radical. This is due to the homolytic cleavage of the carbon-halogen bond.
  • The alkyl radical generated is highly reactive. These radicals quickly couple, forming the new carbon-carbon bond in the resultant alkane.
  • This reaction is particularly useful when forming symmetrical alkanes, as two identical alkyl halides are employed.
The formation of free radicals is an energetically demanding step but crucial for the success of this reaction. Understanding the creation and behavior of these intermediaries is key to mastering the Wurtz reaction.
Dehydration of Alcohol
In dehydration reactions, alcohols are transformed into alkenes when water is eliminated. Concentrated \(H_3PO_4\) serves as the dehydrating agent. Let's explore this:
  • The alcohol's hydroxyl group is protonated by \(H_3PO_4\) which enhances its leaving ability as water.
  • This results in the formation of a carbocation. This positively charged intermediate is pivotal in dictating the structure and stability of the resulting alkene.
  • Ultimately, the formation of the alkene occurs as a hydrogen atom is removed nearby the carbocation, creating the double bond characteristic of alkenes.
The dehydration process is quite useful environmentally and economically, being a simple pathway to produce key alkene building blocks.
Kolbe's Electrolysis
Kolbe's electrolysis is an electrochemical method used to convert carboxylic acids into hydrocarbons via decarboxylation. Here鈥檚 how it works:
  • Carboxylate ions are oxidized at the anode, leading to carbon dioxide gas evolution and forming a free radical intermediate.
  • These radicals subsequently couple, forming the desired hydrocarbon chain.
  • The nature of the newly formed hydrocarbon depends on the original carboxylic acid used.
Kolbe's electrolysis is especially unique because it not only leverages free radical chemistry but does so in an electrochemical cell. This makes it different from the classical chemical radical reactions found elsewhere.
Diazotisation
Diazotisation is the process through which diazonium salts are formed from primary amines using nitrous acid. Here's a breakdown of this interesting reaction:
  • Initially, a primary amine reacts with nitrous acid forming a diazonium ion and water.
  • A gas, nitrogen gas \(N_2\), is then liberally released, a hallmark of this reaction and an indication that it proceeds smoothly.
  • The resultant diazonium salt can participate in further reactions, making this process an essential building block in making a wide array of complex chemical structures.
The liberation of \(N_2\) gas is not only key for the progression of this reaction, but it is also a stable and useful trait for synthetic strategies.
Reaction Mechanisms
Understanding reaction mechanisms is crucial for understanding how and why reactions occur. This involves:
  • Identifying reactants, intermediates, and products.
  • Understanding the step-by-step pathway of a reaction.
  • Visualizing how bonds break and form during the reaction.
  • Using arrow pushing to represent the movement of electrons.
For students, comprehending these mechanisms helps demystify complex reactions and make them more predictable.
Intermediate Species
In organic reactions, intermediates are transient species formed during the conversion of reactants to products. These can include:
  • Carbocations: Positively charged ions often seen in reactions like dehydration.
  • Radicals: Neutral species with an unpaired electron, prevalent in reactions like Wurtz and Kolbe's electrolysis.
  • Understanding these can help predict reaction pathways and product formation.
  • Key to simplifying complex multi-step reactions by breaking them into manageable parts.
Intermediates are crucial for understanding the flow and outcome of organic reactions.

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Most popular questions from this chapter

The factor which is expected to promote ipso substitution is (a) The group \(\mathrm{G}\) should be a strong electron withdrawing group (b) A group which is highly electron withdrawing should be attached at ortho or para position with respect to the group \(\mathrm{G}\) (c) The group G should leave as \(G^{+}\) easily, i.e., \(G^{+}\) should be highly stable (d) The group G should leave easily, i.e., \(\mathrm{G}^{+}\) should be highly unstable

Which of the following is/are more reactive towards nitration than benzene? (a) \(\mathrm{CH}_{2}=\mathrm{CH}-\mathrm{CH}_{3} / \mathrm{HF}\) (b) \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{Cl} /\) Anh. \(\mathrm{AlCl}_{3}\) (c) \(\mathrm{CH}_{3}-\mathrm{CHCH}_{3} / \mathrm{Anh} . \mathrm{AlCl}_{3}\) (d) All of these

Match the Column I with Column II. Column I (a) Toluene (b) Cumene (c) Benzene (d) Mesitylene Column II (p) On oxidation by \(\mathrm{KMnO}_{4}\) in acid medium gives benzoic acid (q) Used in the manufacture of phenol (r) Symmetrical trimethyl benzene (s) On oxidation \(\left(\mathrm{V}_{2} \mathrm{O}_{5} / \mathrm{O}_{2}\right)\) at \(250-450^{\circ} \mathrm{C}\) gives maleic anhydride

What one of the following is the best description for the overall shape of benzene molecule? (a) Linear (b) Triangular (c) Tetrahedron (d) Planar (e) Goofy

Regarding the nitration of phenol with Dil. \(\mathrm{HNO}_{3}\) as described above, the incorrect statement is (a) \(\mathrm{HNO}_{3}\) acts as an acid and also as an oxidant. (b) Nitrous acid acts as a base. (c) The reaction intermediate is an arenium ion. (d) Amount of \(\mathrm{HNO}_{2}\) goes on decreasing with the progress of nitration.
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