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Consider the following acids: (1) \(\mathrm{HCN}\) (2) \(\mathrm{HCOOH}\) (3) \(\mathrm{CH}_{3} \mathrm{COOH}\) (4) \(\mathrm{Cl}-\mathrm{CH}_{2}-\mathrm{COOH}\) The acid strengths of these acids are such that (a) \(4>3>1>2\) (b) \(1>3>2>4\) (c) \(2>3>4>1\) (d) \(4>2>3>1\)

Short Answer

Expert verified
The correct order is: (d) 4 > 2 > 3 > 1.

Step by step solution

01

Identify Acid Type

Consider the type of each of the given acids. 1. HCN - hydrocyanic acid 2. HCOOH - formic acid 3. CH₃COOH - acetic acid 4. Cl-CH₂-COOH - chloroacetic acid Among these, HCN is a hydrogen cyanide and is relatively weak as it is not a carboxylic acid like the others.
02

Analyze Electron-Withdrawing Effects

Examine the electronegativity and structure of each compound. The presence of electronegative elements, like chlorine, enhances acidity due to electron-withdrawing effects. - Cl-CH₂-COOH has a chlorine atom, which strongly withdraws electrons, making the acid stronger. Thus, (4) is the strongest. - HCOOH has no additional electronegative substituents affecting acidity beyond its carboxylic group. - CH₃COOH is similar to HCOOH but has a methyl group which offers a slightly electron-donating effect. - HCN, being not a conventional carboxylic acid, is weaker.
03

Ranking the Acidity

From the above reasoning, the expected order based on acidity strength due to electron-withdrawing effects is: 4 (Cl-CH₂-COOH) > 2 (HCOOH) > 3 (CH₃COOH) > 1 (HCN). This matches the option 'd' in the question statement.

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Key Concepts

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

Electron-Withdrawing Effects
In the realm of chemistry, understanding electron-withdrawing effects is key to evaluating acidity. Electron-withdrawing groups are atoms or molecules attached to a chemical compound that are highly electronegative. These groups can pull electron density away from other parts of the molecule.
In acids, this shift in electron density towards the electron-withdrawing group makes the hydrogen atoms more positive, making it easier for them to be released as protons (\(H^+\)). Therefore, acids with strong electron-withdrawing groups tend to be stronger because they stabilize the negative charge of the conjugate base.
Taking chloroacetic acid (\(\text{Cl-CH}_2\text{COOH}\)), for instance, the chlorine atom is a potent electron-withdrawing group. This makes it much stronger than formic acid or acetic acid, which lack this robust electronegative element.
Carboxylic Acids
Carboxylic acids are a fascinating group of organic compounds characterized by the presence of a carboxyl group (\(\text{-COOH}\)). This functional group is crucial because it significantly contributes to the acidic properties of these compounds.
The acidity of carboxylic acids results from the stability of their conjugate base, the carboxylate ion (\(\text{RCOO}^-\)). The resonance stabilization offered by the carboxylate ion explains why carboxylic acids are generally stronger than other types of acids like hydrocyanic acid (\(\text{HCN}\)), which lacks this resonance stabilization.
It's particularly interesting to observe the role of different substituents on the carbon chain of carboxylic acids. Electronegativity and electron-withdrawing ability of substituents can greatly influence their acidity, as seen in chloroacetic acid compared to just acetic acid.
Electronegativity
Electronegativity is a fantastic concept central to chemistry, reflecting an atom's ability to attract electrons in a chemical bond. In the context of acidity, the presence of highly electronegative atoms positively influences acid strength as they stabilize the conjugate base.
For example, the chlorine atom in chloroacetic acid is much more electronegative than hydrogen in acetic acid. This pulls electron density away from the molecule while making it easier for the molecule to donate a proton.
The more electronegative the substituent attached to the carboxylic acid, the stronger the acid tends to be. This explains why chloroacetic acid is stronger than regular acetic acid; the chlorine substitution intensifies the electronegativity and hence, the acidity.
Acidity Ranking
Acidity ranking involves comparing and ordering acids from the strongest to the weakest based on specific characteristics. One critical factor is the electrophilic nature of the acid molecules—acids more likely to donate protons are deemed stronger.
In this context, understanding the structure and substituents attached to acids becomes crucial. For instance, chloroacetic acid outranks other acids like formic acid, acetic acid, and hydrocyanic acid because of its electron-withdrawing chlorine atom.
The ranking usually follows the path: acids with stronger electron-withdrawing groups and greater resonance stabilization are at the forefront. Hence, the logical order from strongest to weakest based on the exercise is:
  • Cloroacetic Acid (Cl-CH₂-COOH)
  • Formic Acid (HCOOH)
  • Acetic Acid (CH₃COOH)
  • Hydrocyanic Acid (HCN)
This sequence highlights the critical factors influencing the acidity such as electronegativity, molecular structure, and resonance effects.

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Most popular questions from this chapter

In the following sequence of reactions: \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH} \stackrel{\mathrm{KMnO}_{4}}{\longrightarrow}(\mathrm{a}) \stackrel{\mathrm{SOCl}_{2}, \mathrm{NH}_{3}}{\longrightarrow}\) (b) \(\mathrm{Br}_{2}+\mathrm{NaOH}\) (c) the end product (c) is (a) Acetone (b) Ethylamine (c) Acetic acid (d) Methyl amine

\(\mathrm{CH}_{3} \mathrm{COCl}\) reacts with (a) \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{2}\) (b) Salicylic acid (c) \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{OH}\) (d) all of these

Among the following mixtures, dipole-dipole as the major interaction, is present in (a) benzene and ethanol (b) acetonitrile and acetone (c) \(\mathrm{KCl}\) and water (d) benzene and carbon tetrachloride

Here the strongest acid is (a) \(\mathrm{CH}_{3}-\mathrm{COOH}\) (b) \(\mathrm{CH}_{3} \mathrm{CH}_{2}-\mathrm{COOH}\) (c) \(\mathrm{Cl}-\mathrm{CH}_{2}-\mathrm{COOH}\) (d) \(\mathrm{Br}-\mathrm{CH}_{2}-\mathrm{COOH}\)

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) \(\mathrm{CH}_{3}-\mathrm{C}=\mathrm{NNHCONH}_{2}\)

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