Carboxylic Acids MCQ Quiz - Objective Question with Answer for Carboxylic Acids - Download Free PDF

CONCEPT:

Acidity of Aliphatic Carboxylic Acids

• The acidity of carboxylic acids is influenced by the nature of the substituents attached to the carbon chain.
• Electron-withdrawing groups increase acidity by stabilizing the conjugate base (carboxylate ion), while electron-donating groups decrease acidity by destabilizing the conjugate base.
• Formic acid (HCOOH) is the most acidic among simple aliphatic carboxylic acids because it has no alkyl groups, which are electron-donating by nature.
• As the size of the alkyl group increases, the electron-donating inductive effect (-I effect) becomes stronger, making the acid weaker.

EXPLANATION:

• HCOOH (formic acid): No alkyl group, highest acidity.
• CH₃COOH (acetic acid): One methyl group (-CH₃), moderately acidic. (+I)
• (CH₃)₂CHCOOH (isobutyric acid): Two methyl groups attached to the same carbon, further decreases acidity. ( two +I)
• (CH₃)₃CCOOH (pivalic acid): Three methyl groups attached to the same carbon, lowest acidity due to strong electron-donating effect. (three +I)
• The correct order of decreasing acidity is:

  HCOOH > CH₃COOH > (CH₃)₂CHCOOH > (CH₃)₃CCOOH

Therefore, the correct answer is Option 3: HCOOH > CH₃COOH > (CH₃)₂CHCOOH > (CH₃)₃CCOOH.

Explanation:

• Propionic acid has the formula CH₃CH₂COOH, containing 1 carboxyl group.
  
• Oxalic acid has the formula (COOH)₂, containing 2 carboxyl groups.
  
• Malonic acid has the formula CH₂(COOH)₂, containing 2 carboxyl groups.
  
• Citric acid has the formula C₆H₈O₇, which can be structured as HOOC-CH₂-C(OH)(COOH)-CH₂-COOH, containing 3 carboxyl groups.
  

Conclusion:

Therefore, the tricarboxylic acid in the list is Citric acid.

The reaction begins with butanoic acid (C₃H₇COOH) as the reactant. The transformation involves the following steps:

• Step 1: Butanoic acid reacts with phosphorus tribromide (PBr₃), which converts the carboxylic acid (-COOH) group into a corresponding acyl bromide (C₃H₇COBr).

• Step 2: The acyl bromide then reacts with a primary amine (R-NH₂) to form an amide (C₃H₇CONH-R) through nucleophilic substitution.

• Step 3: The amide undergoes reduction using lithium aluminum hydride (LiAlH₄), which reduces the amide to a primary amine (C₃H₇CH₂NH-R).

• Step 4: Finally, the reaction is treated with acid (H₃O⁺) for workup, yielding the final product, which is an alkyl amine (N-alkylbutylamine).

Explanation: 

• In the first step, PBr₃ replaces the hydroxyl group of butanoic acid with a bromine atom, forming butanoyl bromide.

• In the second step, an amine (CH₃-NH₂) acts as a nucleophile, attacking the carbonyl carbon of butanoyl bromide and forming an amide.

• In the third step, LiAlH₄ reduces the amide group (-CONH-CH₃) to an amine group (-CH₂NH-CH₃).

• 

• The final product is an N-methylbutylamine.

Conclusion:

The correct product formed is:

CONCEPT:

Decarboxylation Reaction

• Decarboxylation is a chemical reaction that removes a carboxyl group and releases carbon dioxide (CO₂). Usually, decarboxylation refers to a reaction of carboxylic acids, removing a carbon atom from a carbon chain.
• When a carboxylic acid is heated in the presence of soda lime (a mixture of sodium hydroxide (NaOH) and calcium oxide (CaO)), the carboxyl group (-COOH) is removed as carbon dioxide (CO₂), and the remaining alkyl chain forms a hydrocarbon.

EXPLANATION:

• In the given reaction:

  CH₃CH₂CH₂COONa → CH₃CH₂CH₃ + CO₂ (in the presence of soda lime and heat)

  • The compound CH₃CH₂CH₂COONa (sodium butanoate) undergoes decarboxylation.
  • During the decarboxylation reaction, the -COONa group is removed and replaced by a hydrogen atom.
  • The product formed is propane (CH₃CH₂CH₃).

Therefore, the correct answer is CH₃CH₂CH₃.

CONCEPT:

Oxidation of Aromatic Compounds with Potassium Permanganate (KMnO4)

• When aromatic compounds with side chains (such as alkyl groups) are treated with hot potassium permanganate (KMnO4), the alkyl group undergoes oxidation to a carboxylic acid group (–COOH).
• For a side chain to be oxidized to a carboxyl group (benzoic acid in this case), the compound must have at least one α–H (hydrogen attached to the carbon adjacent to the benzene ring).
• Hot KMnO4 is a strong oxidizing agent that oxidizes the side chain, ultimately converting alkyl groups (such as –CH3, –CH2–) into carboxyl groups (–COOH), leading to the formation of benzoic acid.

EXPLANATION:

• In the given question, we are provided with several compounds. The oxidation ability of KMnO4 depends on the presence of α–H atoms in the side chain of the aromatic compounds.
• The compounds that have at least one α–H (as in the side chain) will react with KMnO4 and undergo oxidation to form benzoic acid.
• Among the given options, the compounds that have α–H in their side chains will yield benzoic acid upon reaction with KMnO4.

There are 5 compounds that will react with KMnO4 to give benzoic acid, which corresponds to option 4.

Therefore, the correct answer is 5 compounds will give benzoic acid with hot KMnO4.

Rosenmund reaction: The Rosenmund reaction is a hydrogenation process where molecular hydrogen reacts with the acetyl chloride in the presence of catalyst – palladium on barium sulfate. 

• The barium sulfate reduces the activity of the palladium due to its low surface area, thereby preventing over-reduction.
• This reaction is used in the preparation of aldehyde from acyl chloride.

Additional Information

• 
• Acetyl chloride is produced in the laboratory by the reaction of acetic acid with chlorodehydrating agents such as PCl₃, PCl₅, SO₂Cl₂, phosgene, or SOCl₂. 

CH3COOH + PCl_{5 ------------>} CH3COCl 

Concept:

Explanation:

• 2-acetoxybenzoic acid is commonly known as aspirin.
• It is prepared by the reaction of acetylation of salicylic acid. It can be achieved by the reaction of salicylic acid with acetic acid in presence of catalyst acids.
• However, the yield is low when acetic acid is used and it can be replaced by acetic anhydride which gives a comparatively much higher yield.
• The acetylation of the phenol group of salicylic acid occurs giving the product Acetylsalicylic acid commonly known as aspirin.
• The reaction is:

Concept:

• Stephen's reaction is named after scientist Henry Stephen who discovered it.
• This reaction is used to synthesize aldehydes from nitriles.
• Gaseous HCl and tin (Sn) are used to carry out the process. Ammonium chloride is important by-product of the reaction.
• The intermediate of the synthesis is an iminium ion.

Explanation:

• When ethane nitrile is treated with Sn/HCl, it undergoes Stephen's reaction to form  Acetaldehyde.
• The reaction is:

• The reaction is more efficient when aromatic nitriles are used rather than aliphatic nitriles.
• The reaction is a redox reaction and electron-withdrawing substituents can improve the rate of the reaction.
• Hence, Stephen's reduction converts ethane nitrile into ethanal or Acetaldehyde.

Correct answer: 2)

Concept:

• Alkaline KMnO₄ is known to be a strong oxidizing agent and hence will first oxidize ethanol.
• KMnO₄ is dark purple in colour. It changes its colour to brown when used in a redox titration.
• An oxidizing agent (also called an oxidant) is a reactant that brings about the oxidation of the other reactant in a chemical reaction and itself undergoes reduction.
• A reducing agent is a reactant that brings about the reduction of the other reactant and itself undergoes oxidation. 

Explanation:

• When a solution of ethanol is heated with potassium permanganate, the pink color of the solution disappears because of the strong oxidizing agent as KMnO₄, which oxidizes ethanol to ethanoic acid by donating nascent oxygen.
• \(CH_{3}CH_{2}-OH\xrightarrow[KMnO_{4}]{Hot, alkaline}CH_{3}COOH\)
• Due to the acidic nature of ethanoic acid, the dark purple colour of KMnO₄ disappears.

Conclusion:

Thus, the Oxidation of ethanol in the presence of hot alkaline KMnO4 yields ethanoic acid.

CONCEPT:

Decarboxylation Reaction

• Decarboxylation is a chemical reaction that removes a carboxyl group and releases carbon dioxide (CO₂). Usually, decarboxylation refers to a reaction of carboxylic acids, removing a carbon atom from a carbon chain.
• When a carboxylic acid is heated in the presence of soda lime (a mixture of sodium hydroxide (NaOH) and calcium oxide (CaO)), the carboxyl group (-COOH) is removed as carbon dioxide (CO₂), and the remaining alkyl chain forms a hydrocarbon.

EXPLANATION:

• In the given reaction:

  CH₃CH₂CH₂COONa → CH₃CH₂CH₃ + CO₂ (in the presence of soda lime and heat)

  • The compound CH₃CH₂CH₂COONa (sodium butanoate) undergoes decarboxylation.
  • During the decarboxylation reaction, the -COONa group is removed and replaced by a hydrogen atom.
  • The product formed is propane (CH₃CH₂CH₃).

Therefore, the correct answer is CH₃CH₂CH₃.

Explanation: -

• Nettle stings or bee-sting have methanoic acid commonly called formic acid (HCOOH).
• The leaves and young stems of this herbaceous plant are fitted with stinging hairs tipped with formic acid and other irritants.
• If touched, these needle-like hairs inject the stinging acid into the skin, triggering a burning, tingling sensation and an itchy rash.

So, the bee sting contains methanoic acid that causes pain and irritation.

The correct answer is Methanoic Acid, option 4.

Additional Information

• Tartaric Acid:
  • It is a white crystalline organic acid found in many fruits including grapes, tamarind, apricots, avocados, and banana.
    It naturally develops in the process of winemaking.
    It is mostly used with baking soda as a flavoring agent to enhance foods.
• Ethanoic Acid:
  • Chemical formula:- CH3COOH.
  • IUPAC name:- Ethanoic acid 
  • Anhydrous Acetic acid is called Glacial Acetic acid.
  • Its crystals are obtained at a temperature slightly lower than 16.7°C. 
• Citric acid:
  • The fruits containing citric acid are termed citrus fruits. eg: lemon, orange, grapefruit, etc.
  • These are sour in taste and are a good source of Vitamin C, thus helping in preventing the disease caused by the deficiency of Vitamin C, i.e., Scurvy.

Concept:

Oxidation reaction:

• Oxidation Reaction refers to a reaction in which either the addition of Oxygen takes place or the removal of Hydrogen takes place.
• It can also be said as the process of loss of one or more electrons by atoms or ions.
• Example is:

Mg + O2 = Mg2O

• Alkyl benzenes and benzyl chloride when subjected to oxidation, yields benzoic acid.
• It can be obtained by a various number of oxidizing agents such as KMnO4, and K2Cr2O7, in presence of acids such as H2SO4.

Explanation:

• Aromatic alkanes when treated with oxidizing agents such as acidified K₂Cr₂O₇, alkaline KMnO₄, or dil. HNO₃, the alkyl side chain gets oxidized to the -COOH group.
• However, the nature of the chain of alkane does not matter.
• The aromatic ring remains intact during the process of oxidation.
• The -COOH group formed is always attached to the aromatic ring and the position of the -COOH group indicates the position of the original side chain in the starting arene.
• The reactions are:

• From the above reaction, it is clear that both toluene and n-butyl benzene gets oxidized to benzoic acid. Thus, the side chain of the arena does not matter.
• If the ring of the arene contains more than one alkyl group, each alkyl group oxidized to -COOH group.
• For example:

Hence, n-Butyl benzene on oxidation gives Benzoic acid.

Mistake Points

• No oxidation takes place when the benzylic carbon is tertiary. 

Concept:

Cannizzaro’s reaction-

• Aldehydes that do not have alpha hydrogen, undergo Cannizaro reaction.
• Self oxidation and reduction take place here in presence of a concentrated aqueous or alcoholic solution of potassium hydroxide.
• Products formed are - an alcohol and a salt of carboxylic acid are produced. 

 Basically,

• Compounds having alpha hydrogen will not give Cannizzaro's reaction
• Compounds not having alpha hydrogen will give Cannizzaro's reaction.

Mechanism:

Explanation:

Let us examine the above structures to check for alpha hydrogens.

1.Formaldehyde has no alpha hydrogen.

Hence, this will show Cannizzaro’s reaction.

2.

Hence, acetaldehyde will not show Cannizzaros Reaction.

3. Benzaldehyde will show Cannizzaro reaction as it has no alpha hydrogen atoms giving products benzyl alcohol and benzoic acid.

4. Formaldehyde will also show Cannizzaro reaction due to the presence of alpha Hydrogen.

Hence, acetaldehyde does not give Cannizzaro reaction.

Additional Information

• All aldehydes can be made to undergo the Cannizaro reaction in the presence of the base Aluminium Ethoxide.
• Under these conditions, the acid and the base combine to form Ester and the reaction is known as Tischenko reaction.

Explanation:

• DIBAL-H is known as diisobutyl Aluminium Hydride.
• It is a reducing agent.
• It converts Carboxylic acid and its derivative into Aldehyde.

So the reactant reagent used to prepare the following aldehyde CH3 - CH=CH-CHO is CH3CH=CH-CN, DIBAL - H.

Concept:

Acid catalyzed to intramolecular esterification reaction.

Intramolecular esterification is if we have a molecule that contains both the carboxylic and hydroxyl groups. In the intramolecular reaction, a cyclic ester is formed. A reaction in which carboxylic acid combines with an alcohol to give a fruity smelling liquid called ester is known as esterification.

The reactant reacts with H₂ So₄ (sulphuric acid) and with chloroform it undergoes intramolecular esterification process to give the final product.