Carboxylic Acids MCQ Quiz in मल्याळम - Objective Question with Answer for Carboxylic Acids - സൗജന്യ PDF ഡൗൺലോഡ് ചെയ്യുക

Explanation:

• Compound T:
  • T has a carboxylic acid functional group (–COOH), making it soluble in hot aqueous NaOH as it forms a carboxylate salt.
• Formation of U:
  • Reduction of T using LiAlH₄ converts the –COOH group to a primary alcohol (–CH₂OH).
  • U is optically inactive as it doesn't contains a chiral carbon (a carbon bonded to four different groups).
• Formation of W:
  • Oxidation of U with CrO₃/H⁺ forms an aldehyde or ketone. Reaction with excess (CH₃CO)₂O introduces ester groups, resulting in W with molecular formula C₁₀H₁₈O₄.
• Compound V:
  • V is T, which contains a carboxylic acid group that reacts with NaHCO₃ to produce CO₂, causing effervescence.

Conclusion:

Therefore the correct options are 1,3 & 4

Explanation:

It is clear that Bromine is linked with β Carbon. 

So from the options, it is clear that Option 1 is correct.

∴ The common name is β - Bromobutyraldehyde

Explanation:

Acetyl bromide reacts with an excess of  CH3Mgl followed by treatment with a saturated solution of NH4Cl giving tertiary alcohol which in the given option

is: 2-methyl-2-propanol.

CONCEPT:

Grignard Reagent Reaction with Carbon Dioxide

• Grignard reagents (RMgX) are organomagnesium compounds used in organic synthesis to form carbon-carbon bonds.
• When a Grignard reagent reacts with carbon dioxide (dry ice), it forms a carboxylate intermediate.
• This intermediate, upon acid hydrolysis, yields a carboxylic acid.

EXPLANATION:

• In the given reaction sequence:

  Dry ice (CO2" id="MathJax-Element-33-Frame" role="presentation" style="position: relative;" tabindex="0">CO2CO2 $\text{CO}_2$ ) →Dry etherCH3MgBr" id="MathJax-Element-34-Frame" role="presentation" style="position: relative;" tabindex="0">−→−−−−Dry etherCH3MgBr→Dry etherCH3MgBr $\xrightarrow[\text{Dry ether}]{\text{CH}_3\text{MgBr}}$ A" id="MathJax-Element-35-Frame" role="presentation" style="position: relative;" tabindex="0">AA $\mathbf{A}$ →dil. HClH2O" id="MathJax-Element-36-Frame" role="presentation" style="position: relative;" tabindex="0">−→−−−dil. HClH2O→dil. HClH2O $\xrightarrow[\text{dil. HCl}]{\text{H}_2\text{O}}$ B" id="MathJax-Element-37-Frame" role="presentation" style="position: relative;" tabindex="0">BB $\mathbf{B}$

• The Grignard reagent, CH3MgBr" id="MathJax-Element-38-Frame" role="presentation" style="position: relative;" tabindex="0">CH3MgBrCH3MgBr $\text{CH}_3\text{MgBr}$ , reacts with dry ice (CO2" id="MathJax-Element-39-Frame" role="presentation" style="position: relative;" tabindex="0">CO2CO2 $\text{CO}_2$ ) to form the carboxylate intermediate (A" id="MathJax-Element-40-Frame" role="presentation" style="position: relative;" tabindex="0">AA $\mathbf{A}$ ).
  • Intermediate A" id="MathJax-Element-41-Frame" role="presentation" style="position: relative;" tabindex="0">AA $\mathbf{A}$ is CH3COO−MgBr+" id="MathJax-Element-42-Frame" role="presentation" style="position: relative;" tabindex="0">CH3COO−MgBr+CH3COO−MgBr+ $\text{CH}_3\text{COO}^-\text{MgBr}^+$ .
• Upon acid hydrolysis (with dilute HCl and water), the carboxylate intermediate A" id="MathJax-Element-43-Frame" role="presentation" style="position: relative;" tabindex="0">AA $\mathbf{A}$ is converted to the corresponding carboxylic acid.
  • The product B" id="MathJax-Element-44-Frame" role="presentation" style="position: relative;" tabindex="0">BB $\mathbf{B}$ is ethanoic acid (acetic acid), CH3COOH" id="MathJax-Element-45-Frame" role="presentation" style="position: relative;" tabindex="0">CH3COOHCH3COOH $\text{CH}_3\text{COOH}$ .

Therefore, the product B" id="MathJax-Element-46-Frame" role="presentation" style="position: relative;" tabindex="0">BB $\mathbf{B}$ is ethanoic acid, which corresponds to option 2.

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:

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.

Correct answer: 2)

Concept:

• The organic compounds with the functional group \(\rm -NH_2\) are called amines.
• They can be synthesized from carboxylic acids, amides, cyanides, etc by using suitable reagents. 
• Amines are used in the synthesis of dyes, fertilizers, garments industries, etc.

Explanation:

• The named reaction that involves the reaction of carboxylic acids and hydrazoic acid in an acidic medium is called the Schmidt reaction.
• The mechanism of the reaction is shown below:

Conclusion:

Thus, the reaction RCOOH + N3H \(\rm \xrightarrow{conc, \ H_2SO_4}\) RNH2 + CO2 + N2 is called the Schmidt reaction.

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:

Acidity and pKa Values

• The acidity of a compound is inversely related to its pKa value. A lower pKa value indicates a stronger acid.
• The presence of electronegative substituents (such as halogens) near the carboxyl group increases the acidity of carboxylic acids by stabilizing the conjugate base through the inductive effect.

EXPLANATION:

• The electronegativity of the halogens decreases in the order: F > Cl > Br.
• The stronger the electron-withdrawing effect, the more stable the conjugate base, and the stronger the acid (lower pKa).
• Therefore, the order of acidity based on the inductive effect will be:
  • FCH₂COOH (C) > ClCH₂COOH (B) > BrCH₂COOH (A) > HCOOH (D)

Therefore, the correct increasing order of acidity based on pKa values is (D) < (A) < (B) < (C)