Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

Match the following: List I (Reaction) 1\. \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CHO}+\mathrm{NaOH} \longrightarrow \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COOH}\) 2\. \(\mathrm{C}_{6} \mathrm{H}_{6}+\mathrm{CH}_{3} \mathrm{Cl} \stackrel{\text { Anhyd. } \mathrm{AICl}_{3}}{+\mathrm{Cd} / \mathrm{BaSO}} \stackrel{\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CH}_{2} \mathrm{OH}}{\longrightarrow} \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CH}_{3}\). 3\. \(\mathrm{RCOCl}+\mathrm{H}_{2} \stackrel{\mathrm{Pd} / \mathrm{BaSO}_{4}}{\longrightarrow} \mathrm{RCHO}\) 4\. \(\mathrm{C}_{6} \mathrm{H}_{3} \mathrm{OH}+\mathrm{NaOH}+\mathrm{CO}_{2} \stackrel{\text { 1. heat, 2. } \mathrm{H}^{\prime}}{\longrightarrow}\) List II (Name of reactions) (i) Rosenmund reduction (ii) Kolbe's reaction (iii) Claisen reaction (iv) Cannizzaro reaction (v) Friedal-Craft's reaction The correct matching is: The correct matching is: 1 2 3 4 (a) (iv) (v) (i) (ii) (b) (v) (i) (ii) (iii) (c) (v) (iii) (ii) (i) (d) (iv) (ii) (iii) (i)

Short Answer

Expert verified
1 (iv), 2 (v), 3 (i), 4 (ii) is the correct match, option (a).

Step by step solution

01

Identify each type of reaction from List I

1. The reaction \( \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CHO}+\mathrm{NaOH} \rightarrow \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COOH} \) is known as Cannizzaro's reaction, which involves the disproportionation of a non-enolizable aldehyde in the presence of a base.2. The reaction \( \mathrm{C}_{6} \mathrm{H}_{6}+\mathrm{CH}_{3} \mathrm{Cl} \stackrel{\text{Anhyd. } \mathrm{AlCl}_{3}}{\longrightarrow} \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CH}_{3} \) is known as Friedel-Crafts alkylation, where an alkyl halide reacts with an aromatic compound in the presence of a Lewis acid.3. The reaction \( \mathrm{RCOCl}+\mathrm{H}_{2} \stackrel{\mathrm{Pd} / \mathrm{BaSO}_{4}}{\longrightarrow} \mathrm{RCHO} \) is the Rosenmund reduction, which reduces an acid chloride to an aldehyde using a palladium-barium sulfate catalyst.4. The reaction \( \mathrm{C}_{6} \mathrm{H}_{3} \mathrm{OH}+\mathrm{NaOH}+\mathrm{CO}_{2} \stackrel{\text{1. heat, 2. } \mathrm{H}^{+}}{\longrightarrow} \) is Kolbe's reaction, involving the carboxylation of phenolate ions to form salicylic acid.
02

Match each reaction with its name from List II

1. \( \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CHO}+\mathrm{NaOH} \rightarrow \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COOH} \) is Cannizzaro's reaction - (iv).2. \( \mathrm{C}_{6} \mathrm{H}_{6}+\mathrm{CH}_{3} \mathrm{Cl} \) with \( \mathrm{AlCl}_{3} \) is Friedel-Crafts reaction - (v).3. \( \mathrm{RCOCl}+\mathrm{H}_{2} \rightarrow \mathrm{RCHO} \) is Rosenmund reduction - (i).4. \( \mathrm{C}_{6} \mathrm{H}_{3} \mathrm{OH}+\mathrm{NaOH}+\mathrm{CO}_{2} \rightarrow \) is Kolbe's reaction - (ii).
03

Write down the correct matching sequence

Using the identified reaction types and their corresponding matches from List II: - Reaction 1 corresponds to (iv). - Reaction 2 corresponds to (v). - Reaction 3 corresponds to (i). - Reaction 4 corresponds to (ii). Therefore, the correct matching sequence is: 1 (iv), 2 (v), 3 (i), 4 (ii).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

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

Cannizzaro Reaction
The Cannizzaro Reaction is an intriguing chemical process involving the simultaneous oxidation and reduction of a molecule. Specifically, it deals with the reaction of a non-enolizable aldehyde in the presence of a strong base like sodium hydroxide (NaOH). During this reaction, one molecule of the aldehyde is oxidized to form a carboxylic acid, while another is reduced to an alcohol.
This type of reaction is particularly useful when dealing with aldehydes that lack an alpha-hydrogen, making them non-enolizable. Phenylmethanal, commonly known as benzaldehyde, is a classic example, where it undergoes self-disproportionation, resulting in the conversion to benzoic acid and benzyl alcohol.
A general representation of the Cannizzaro reaction is:
\[ 2 ext{ RCHO} + ext{ NaOH} ightarrow ext{ RCOONa} + ext{ RCH}_2 ext{OH} \]
This reaction has practical significance because it provides a method for synthesizing alcohols and carboxylic acids from aldehydes. It's noteworthy that not all bases can facilitate the Cannizzaro reaction effectively; typically, the reaction requires strong bases.
Friedel-Crafts Alkylation
Friedel-Crafts Alkylation is a pivotal reaction in organic chemistry crucial for adding alkyl groups to aromatic rings. This allows the transformation of a simple aromatic compound into a more complex one through the introduction of carbon chains.
The process utilizes an alkyl halide and a Lewis acid catalyst, such as aluminum chloride (AlCl₃), to create a highly reactive carbocation. This carbocation attacks the aromatic ring, forming an alkylated aromatic compound. For instance, in the aromatic compound benzene, you might add a methyl group using methyl chloride in the presence of AlCl₃ to produce toluene.
One needs to be cautious, though. Friedel-Crafts reactions can lead to polyalkylation, where more than one alkyl group attaches, altering the reaction's expected outcomes. Additionally, Friedel-Crafts alkylation is not suitable for deactivating groups, as they hinder the mechanism.
This reaction is highly versatile, allowing chemists to synthesize complex aromatic structures that are essential in developing drugs and dyes.
Rosenmund Reduction
The Rosenmund Reduction is a specialized reaction used to convert acid chlorides to aldehydes, involving the use of a special catalyst, palladium on barium sulfate (Pd/BaSO₄). This transformation is particularly noteworthy because aldehydes are reactive intermediates that are valuable in various syntheses.
A critical aspect of this reaction is achieving hydrogenation selectively to aldehydes, without further reducing them to alcohols. The addition of barium sulfate helps to moderate the reaction, ensuring a smooth reduction process.
Generally, the reaction is expressed as:\[ \text{RCOCl} + \text{H}_2 \xrightarrow[{BaSO_4}]{Pd} \text{RCHO} \]
This specificity is vital in synthetic pathways where delicate control over reaction products is necessary. However, one needs to maintain careful control over the reaction conditions to prevent further reduction. This makes the Rosenmund Reduction an indispensable tool for synthesizing vital aldehyde functionalities in complex organic molecules.
Kolbe's Reaction
Kolbe's Reaction is a fascinating electrochemical process used primarily for synthesizing salicylic acid, an important precursor to aspirin, from phenolates. This reaction exemplifies the use of electrolysis in organic synthesis, where phenoxide ions are carboxylated through the introduction of carbon dioxide (\(CO_2\)).
During the reaction, phenol is first converted into its sodium salt, phenoxide, using sodium hydroxide. The key step involves heating this mixture with CO₂ under controlled conditions, facilitated by the generation of electricity through electrolysis. This process yields salicylate ions, which are then acidified to obtain salicylic acid.
This reaction can be denoted as:\[ \text{C}_6\text{H}_5\text{ONa} + \text{CO}_2 \xrightarrow[]{heat} \text{C}_6\text{H}_4\text{OHCOOH} \]
Kolbe's Reaction stands out by bridging organic chemistry with electrochemical techniques. Understanding the intricacies of this reaction helps in appreciating how classical organic reactions can be adapted for industrial-scale production of significant medicinal and aromatic compounds.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

In the reaction \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{C} \equiv \mathrm{CH}$$\stackrel{\mathrm{H}_{2} \mathrm{SO}_{4} \mathrm{HgSO}_{4}}{\longrightarrow}[\mathrm{A}]\) The compound [A] is (a) \(\mathrm{CH}_{3}-\mathrm{CH}_{2}-\mathrm{CH}_{2}-\mathrm{COOH}\) (b) \(\mathrm{CH}_{3}-\mathrm{CH}_{2}-\mathrm{CH}_{2}-\mathrm{CHO}\) (c) \(\mathrm{CH}_{3}-\mathrm{CH}_{2}-\mathrm{CO}-\mathrm{CH}_{3}\) (d) none of these

Treatment of propionaldehyde with dil. \(\mathrm{NaOH}\) solution gives (a) \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{COCH}_{2} \mathrm{CH}_{2} \mathrm{CHO}\) (b) \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CHOHCH}_{2} \mathrm{CH}_{2} \mathrm{CHO}\) (c) \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CHOHCH}\left(\mathrm{CH}_{3}\right) \mathrm{CHO}\) (d) \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{COOCH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{3}\)

Compound 'A' (molecular formula \(\mathrm{C}_{3} \mathrm{H}_{8} \mathrm{O}\) ) is treated with acidified potassium dichromate to form a product 'B' (molecular formula \(\mathrm{C}_{3} \mathrm{H}_{6} \mathrm{O}\) ). 'B' forms a shining silver mirror on warming with ammonical silver nitrate. 'B' when treated with an aqueous solution of \(\mathrm{H}_{2} \mathrm{NCONHNH}_{2}\) HCl and sodium acetate gives a product 'C'. Identify the structure of ' \(\mathrm{C}^{\prime}\). (a) \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}=\mathrm{NNHCONH}_{2}\) (b) \(\mathrm{CH}_{3}-\mathrm{C}=\mathrm{NNHCONH}_{2}\)

Which one of the following pairs is not correctly matched? (a) \(>\mathrm{C}=\mathrm{O} \longrightarrow \mathrm{CH}_{2}\) Clemmensen reduction (b) \(-\mathrm{COCl} \longrightarrow \mathrm{CHO}\) Rosenmund reduction (c) \(>\mathrm{C}=\mathrm{O} \longrightarrow>\mathrm{CHOH}\) Wolff - Kishner reduction (d) \(-\mathrm{C} \equiv \mathrm{N} \longrightarrow-\mathrm{CHO}\) Stephen reduction

Which reaction product is formed from benzaldehyde and methyl amine? (a) \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{2}\) (b) \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CH}=\mathrm{NCH}_{3}\) (c) \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CH}_{2} \mathrm{NH}_{2}\) (d) \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CONH}_{2}\)

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free