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Chapter 7: Problem 5

Esterification of salicylic acid cannot be carried out by reaction with (A) \(\mathrm{CH}_{3}-\mathrm{C}-\mathrm{Cl}\) (B) \(\mathrm{CH}_{3}-\mathrm{CH}_{2}-\mathrm{OH}\) (C) \(\left(\mathrm{CH}_{3}-\mathrm{CO}\right)_{2} \mathrm{O}\) (D) \(\mathrm{CH}_{3}-\mathrm{CH}_{2}-\mathrm{NH}_{2}\)

Short Answer

Expert verified
The reactant that cannot be used for esterification of salicylic acid is (A) \(\mathrm{CH}_{3}-\mathrm{C}-\mathrm{Cl}\) as it is an alkyl halide and less reactive than amines in nucleophilic acyl substitution reactions.

Step by step solution

01

Identify the functional groups in the given reactants

In every given reactant, we need to identify if they have either an alcohol functional group (-OH) or another group compatible with esterification. (A) \(\mathrm{CH}_{3}-\mathrm{C}-\mathrm{Cl}\): This molecule has a chlorine atom attached to the carbon, so it is an alkyl halide. (B) \(\mathrm{CH}_{3}-\mathrm{CH}_{2}-\mathrm{OH}\): This molecule has a hydroxyl group (-OH) attached to the carbon, so it is an alcohol. (C) \(\left(\mathrm{CH}_3-\mathrm{CO}\right)_2\mathrm{O}\): This molecule has two acyl groups joined by an oxygen atom, so it is an anhydride. (D) \(\mathrm{CH}_{3}-\mathrm{CH}_{2}-\mathrm{NH}_{2}\): This molecule has an amine group (-NH鈧�) attached to the carbon, so it is an amine.
02

Determine which reactants can participate in esterification

Carboxylic acids can react with alcohols and anhydrides to form esters, so we can eliminate those options that don't have these functional groups. (A) \(\mathrm{CH}_{3}-\mathrm{C}-\mathrm{Cl}\): Cannot participate in the esterification as it is an alkyl halide. (B) \(\mathrm{CH}_{3}-\mathrm{CH}_{2}-\mathrm{OH}\): Can participate in the esterification as it is an alcohol. (C) \(\left(\mathrm{CH}_3-\mathrm{CO}\right)_2\mathrm{O}\): Can participate in the esterification as it is an anhydride. (D) \(\mathrm{CH}_{3}-\mathrm{CH}_{2}-\mathrm{NH}_{2}\): Cannot participate in the esterification as it is an amine.
03

Select the reactant that cannot be used for esterification of salicylic acid

As we determined in Step 2, the reactants that cannot participate in esterification are A and D, but since only one of the options is correct, we will compare the functional groups of the two remaining options: (A) \(\mathrm{CH}_{3}-\mathrm{C}-\mathrm{Cl}\): This molecule is an alkyl halide, which is less reactive than amines in nucleophilic acyl substitution reactions. (D) \(\mathrm{CH}_{3}-\mathrm{CH}_{2}-\mathrm{NH}_{2}\): This molecule is an amine, which is more reactive in nucleophilic acyl substitution reactions. Since the question asks which molecule cannot be used for the esterification of salicylic acid, the answer is: (A) \(\mathrm{CH}_{3}-\mathrm{C}-\mathrm{Cl}\)

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Key Concepts

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

Salicylic Acid
Salicylic acid is a widely studied organic compound due to its significance in pharmaceutical applications, especially as a precursor to aspirin. Structurally characterized by a benzene ring bonded to both a hydroxyl (鈥揙H) and a carboxylic acid group (鈥揅OOH), salicylic acid plays a vital role in organic chemistry reactions, including esterification.

In the context of the Joint Entrance Examination (JEE) for engineering in India, understanding the chemical properties of salicylic acid is crucial. Esterification is one reaction where salicylic acid is treated with alcohols or acyl derivatives to produce ester compounds, which can exhibit a myriad of beneficial pharmacological properties.
Nucleophilic Acyl Substitution
Nucleophilic acyl substitution is a fundamental mechanism in organic chemistry that involves the exchange of a nucleophilic group (a molecule or ion with a penchant for donating an electron pair) with an excellent leaving group in an acyl compound, such as a carboxylic acid derivative.

During this process, nucleophiles attack the electrophilic carbon of the carbonyl group, leading to the substitution and formation of different carboxylic acid derivatives, including esters. Esterification is an example of nucleophilic acyl substitution as observed in the reaction of salicylic acid with alcohols. Students preparing for competitive exams like JEE need to grasp the nuances of this mechanism to solve problems pertaining to organic synthesis.
Organic Chemistry for JEE
Organic chemistry is a significant portion of the JEE syllabus, demanding a strong conceptual understanding and proficiency in mechanisms of reaction and synthesis. JEE aspirants must study various organic reactions, functional groups, and their interconversion.

Esterification reactions, like the conversion of salicylic acid, require a clear understanding of how different moieties react under certain conditions. These principles are critical not only for solving textbook problems but also for applying knowledge in practical scenarios. By mastering concepts such as nucleophilic acyl substitution and the properties of salicylic acid, students can gain a competitive edge in achieving high scores in organic chemistry sections of the JEE.

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Most popular questions from this chapter

Acid amide is not obtained as a product in the reaction: (A) \(\mathrm{Me}_{2} \mathrm{CNOH} \stackrel{\text { Conc. } \mathrm{H}_{2} \mathrm{SO}_{4}}{\longrightarrow}\) (B) \(\mathrm{MeCOOH} \stackrel{\mathrm{N}_{3} \mathrm{H}}{\longrightarrow}\) (C) \(\mathrm{Me}_{2} \mathrm{CO} \frac{\mathrm{N}_{3} \mathrm{H}}{\mathrm{H}_{2} \mathrm{SO}_{4}}\) (D) \(\mathrm{MeCN} \frac{95 \%}{\mathrm{H}_{2} \mathrm{SO}_{4}}{\longrightarrow}\)

There are fundamentally different mechanism possible for following reaction, If \(\mathrm{O}^{18}\) is present once in phenolic \(\mathrm{OH}\) and once in carboxylate group of product on using \(\mathrm{NaOH} / \mathrm{H}_{2} \mathrm{O}^{18}\) then mechanism used by following reaction in both cases are respectively : (A) Ester hydrolysis, Nucleophilic aromatic substitution (B) Nucleophilic aromatic substitution, Base catalysed acyl oxygen cleavage mechanism (C) Electrophilic aromatic substitution, Base catalysed acyl oxygen cleavage mechanism (D) Nucleophilic aromatic substitution, Nucleophilic aromatic substitution

For which value of \(n\) compound, \(\mathrm{HO}_{2} \mathrm{C}-\left(\mathrm{CH}_{2}\right)_{\mathrm{n}}-\mathrm{CO}_{2} \mathrm{H}\) undergo only dehydration on heating? (A) \(n=1\) (B) \(n=2\) (C) \(n=3\) (D) \(n=4\)

\(\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7} / \mathrm{H}^{+}, \Delta\) is a strong oxidising agent which can oxidise \(\mathrm{CH}_{3}-\mathrm{CH}_{2}-\mathrm{OH}\) into \(\mathrm{CH}_{3}-\mathrm{COOH}\) but still \(\mathrm{CH}_{3}-\mathrm{CH}=\mathrm{O}\) is obtained by oxidation of \(\mathrm{CH}_{3}-\mathrm{CH}_{3}-\mathrm{OH}\) by using distillation flask at temperature \(55^{\circ} \mathrm{C}_{\text {with }}\) \(\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}, \mathrm{H}^{+}, \Delta\). It is possible because (A) Acetaldehyde distills out at this temperature and escape further oxidation (B) At this temperature, acetaldehyde is very stable and escape further oxidation (C) Oxidation of \(\mathrm{CH}_{3}-\mathrm{CH}_{2}-\mathrm{OH}\) is slower than oxidation of \(\mathrm{CH}_{3}-\mathrm{CH}=\mathrm{O}\) (D) None

The product obtained on reaction of malonic acid with \(\mathrm{P}_{2} \mathrm{O}_{5} / \Delta\) is
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