Aldehydes And Ketones MCQ Quiz - Objective Question with Answer for Aldehydes And Ketones - Download Free PDF

CONCEPT:

Reduction of Esters to Aldehydes using DIBAL-H

• DIBAL-H (Diisobutylaluminum hydride, AlH(iBu)₂) is a selective reducing agent that reduces esters to aldehydes without further reducing the aldehyde to an alcohol.
• In the reaction with esters, DIBAL-H (AlH(iBu)₂) reduces the ester (R-C=O-OR') to the aldehyde (R-C=O-H), stopping the reduction at the aldehyde stage.
• This reaction is followed by hydrolysis with water (H₂O) to work up the reaction and form the aldehyde.

EXPLANATION:

• For the given reaction:

  R-C=O-OR' + DIBAL-H → R-C=O-H + ROH

• In this case, DIBAL-H selectively reduces the ester (R-C=O-OR') to an aldehyde (R-C=O-H) and the alcohol group (ROH) is released.

• The correct reagent for the conversion is Option 2 (AlH(iBu)₂, H₂O). DIBAL-H reduces the ester to the aldehyde, and water (H₂O) is used to hydrolyze the intermediate to the final aldehyde product.

Why not the other options:

• Option 1: LiAlH₄ (Lithium aluminium hydride) followed by H₂O:
  • LiAlH₄ is a strong reducing agent that reduces esters to alcohols rather than aldehydes. It would not stop at the aldehyde stage, which makes it unsuitable for this transformation.
• Option 3: NaBH₄ (Sodium borohydride) followed by H₂O:
  • NaBH₄ is a milder reducing agent than LiAlH₄ and typically reduces aldehydes and ketones to alcohols, but it is not reactive enough to reduce esters to aldehydes.
• Option 4: H₂/Pd-BaSO₄ (Hydrogen and palladium on barium sulfate):
  • This is a catalytic hydrogenation method, typically used to reduce alkenes or aromatic compounds. It will reduce aldehydes to alcohols, so it would not be selective enough to stop at the aldehyde stage when reducing esters.

Therefore, the correct answer is AlH(iBu)₂, H₂O, which reduces the ester to an aldehyde selectively.

CONCEPT:

Grignard Reagent Reaction with Nitriles

• Grignard reagents, such as CH₃MgBr, are highly nucleophilic and typically react with electrophilic carbonyl compounds such as aldehydes, ketones, and nitriles.
• When a Grignard reagent like CH₃MgBr reacts with a nitrile group (-C≡N), it adds to the carbon atom of the nitrile, leading to the formation of an imine intermediate.
• Upon subsequent treatment with water (H₂O), the imine intermediate undergoes hydrolysis, resulting in the formation of an amide. In this case, further reduction or reaction with excess Grignard reagent can give a ketone or alcohol, depending on the conditions.

EXPLANATION:​

The Grignard reagent CH₃MgBr reacts with the nitrile group in the compound (benzenecarbonitrile), leading to the formation of an intermediate imine, which is then hydrolyzed to yield an alcohol.

The reaction leads to the following:

• The product formed is 2-phenylethanol (Option 2), where the nitrile group (-C≡N) has been reduced and hydrolyzed to form an alcohol (OH) group attached to the 2nd carbon of the ethyl side chain, with a methyl group attached to the benzene ring.

Therefore, the correct answer is CH₃OH attached to the benzene ring with a CN group at the 2-position.

CONCEPT:

Reactions Producing Benzene

• Benzene is typically produced through reactions such as decarboxylation of aromatic acids, cracking of hydrocarbons, or dehydrohalogenation reactions. These reactions are capable of removing functional groups and producing benzene as the product.

EXPLANATION:

• Let's break down each reaction to determine whether it produces benzene or not:
• Reaction 1: C6H5COONa (Sodium Benzoate) + NaOH (Soda Lime) → C6H6 (Benzene)
  • This is a decarboxylation reaction known as the "Decarboxylation of Sodium Benzoate," which produces benzene as the product.
  • When sodium benzoate reacts with soda lime (a mixture of NaOH and CaO), the carboxyl group is removed, and benzene is formed.
  • 
• Reaction 2: C6H12 (n-hexane) → Benzene (via MoO3 at 773 K, 10–20 atm)
  • This reaction does produce benzene. The catalytic cracking of n-hexane under high pressure and temperature typically results in cyclisation and followed by aromatisation.
    
• Reaction 3: CH≡CH (Ethyne or Acetylene) → Benzene (via red-hot Iron Tube at 873 K)
  • This reaction is known as the "Pyrolysis of Acetylene." When acetylene is passed through a red-hot iron tube, it undergoes cyclotrimerization to form benzene as the product.
    
• Reaction 4: C6H5N2+Cl- (Benzenediazonium Chloride) → Phenol (via H2O warm)
  • This reaction involves the reduction of the benzenediazonium ion to form benzene via water as the reducing agent. This reaction also produces Phenol.
  • 

Conclusion: The correct answer is Option 4.

CONCEPT:

• Catalytic Hydrogenation: Aldehydes and ketones undergo hydrogenation in the presence of H₂/Pd-C in ethanol to give corresponding alkanes.
• Acidic Hydrolysis of Esters: Heating esters with dilute acid (H⁺/H₂O/Δ) results in hydrolysis to form carboxylic acids or substituted alkenes depending on stability.
• Wittig Reaction: Reaction between aldehydes/ketones and phosphonium ylides (like (CH₃)₂C=P(C₆H₅)₃) gives alkenes.
• Gilman Reagent Reaction: Organocuprates (R₂CuLi) react with acid chlorides or cyclic ketones to give substituted hydrocarbons upon aqueous work-up.

EXPLANATION:

• A - IV: The compound C₆H₅COCH₃ (acetophenone) undergoes catalytic hydrogenation with H₂/Pd-C in ethanol, reducing the carbonyl to CH₂, forming ethylbenzene (Product IV).
  
• B - III: The ester with two COOCH₃ groups on the aromatic ring undergoes acid hydrolysis and decarboxylation under heat (H⁺/H₂O/Δ), forming isopropylbenzene (cumene) (Product III).
  
• C - II: Benzaldehyde reacts with Wittig reagent (CH₃)₂C=P(C₆H₅)₃, leading to the formation of an alkene with a double bond adjacent to the aromatic ring — product II.
• 
• D - I: Cyclopentanone reacts with Gilman reagent (Li⁺[(CH₃)₂Cu]^–) and then hydrolyzed to form 2-methylcyclopentanol — product I.
• 

Therefore, the correct answer is Option (b): A–IV, B–III, C–II, D–I.

CONCEPT:

Named Organic Reductions and Their Applications

• Each named reduction is associated with specific starting materials and products.
• Some reductions are selective (e.g., partial reduction), others are complete (e.g., carbonyl to alkane).

EXPLANATION:

Column I: Named Reduction	Column II: Reaction/Transformation	Code
A. Rosenmund’s Reduction	Converts acid chloride to aldehyde using H2/Pd/BaSO4	1
B. Birch Reduction	Partial reduction of alkyne to trans-alkene using metal in liquid NH3	3
C. Stephen’s Reduction	Converts nitrile to aldehyde via SnCl2/HCl + H2O	2
D. Clemmensen Reduction	Reduces carbonyl group to alkane under acidic conditions (Zn-Hg/HCl)	4
E. Lindlar’s Catalyst	Partial reduction of alkyne to cis-alkene using poisoned catalyst	6

Therefore, the correct answer is: Option A) A–1, B–3, C–2, D–4, E–6

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:

Reaction of Grignard reagent with CO₂ - 

• It is an important method of preparation of carboxylic acid.
• Grignard reagent behaves as a nucleophile which attacks the substrate CO₂  in presence of dry ether and forms a nucleophilic addition product.
• This nucleophilic addition product is a salt of carboxylic acid which on acidification yields carboxylic acid.

Explanation:

In the given reaction, R-MgX is the Grignard reagent and CO₂ (dry ice) is the substrate.

The reaction is complete in two steps-

1. Attack of Grignard reagent on CO₂ - the nucleophile Grignard reagent attack the CO₂ in presence of dry ether and as a carboxylic acid salt is formed as product name as 'Y'  in the above reaction.
2. Acidification of the carboxylic salt - It gives carboxylic acid as a final product.

It can be represented as -

R^δ--^δ+MgX + O=C=O \(\xrightarrow{dry \hspace{0.1cm}ether}\) RCOO- Mg+X \(\xrightarrow{H_3O^+}\) RCOOH

So, RCOO- Mg+X is obtained as an intermediate product (Y) in the given reaction.

Hence, the correct answer is option 2.

Explanation:

→ Methanal (formaldehyde) cannot undergo Aldol condensation because it has only one carbon atom and does not have an alpha hydrogen atom.

→ Benzaldehyde has an alpha hydrogen atom and can undergo Aldol condensation, but it forms a mixture of products due to the presence of two carbonyl groups that can react.

2,2-Dimethylbutanal does not have an alpha hydrogen atom and cannot undergo Aldol condensation.

→ Phenylacetaldehyde has an alpha hydrogen atom and can undergo Aldol condensation to form an alpha-beta unsaturated aldehyde.

→ Aldol condensation is a reaction between two molecules of aldehydes or ketones that contain alpha hydrogen atoms. The alpha hydrogen atom undergoes deprotonation to form an enolate ion, which then reacts with another molecule of aldehyde or ketone to form a beta-hydroxy aldehyde or ketone. The product undergoes dehydration to form an alpha-beta unsaturated aldehyde or ketone.

→ Therefore, only Phenylacetaldehyde can undergo Aldol condensation among the given options.

Aldehydes which contain at least 2α hydrogen atoms, undergo Aldol condensation reaction Phenylacetaldehyde contains 2α hydrogen atoms and undergoes Aldol condensation reaction in presence of dilute alkali. 

CONCEPT:

Aldol Reaction

• The Aldol reaction involves the reaction of aldehydes or ketones with α-hydrogen atoms in the presence of a base to form β-hydroxy aldehydes or β-hydroxy ketones (aldols), which can further dehydrate to form α,β-unsaturated carbonyl compounds.
• The acidity of α-hydrogens is crucial because the base abstracts these protons to form enolate ions, which are the reactive intermediates that attack the carbonyl group of another molecule, leading to the Aldol product.

Explanation:-

1. Statement I: Acidity of α-hydrogens of aldehydes and ketones is responsible for Aldol reaction.
   • This statement is correct because the Aldol reaction relies on the formation of enolate ions, which is facilitated by the acidity of the α-hydrogens.
2. Statement II: Reaction between benzaldehyde and ethanal will NOT give Cross – Aldol product.
   • Benzaldehyde does not have α-hydrogens, so it cannot form enolate ions. However, ethanal (acetaldehyde) can form an enolate ion due to the presence of α-hydrogens.
   • In a cross-Aldol reaction, ethanal can act as the enolate ion to react with the carbonyl carbon of benzaldehyde, leading to a cross-Aldol product.
   • This statement is incorrect because benzaldehyde can indeed react with ethanal under Aldol conditions to give a cross-Aldol product.

• Reviewing the accuracy of each statement:
  • Statement I: Correct
  • Statement II: Incorrect

Aldehyde and ketones having acidic

α-hydrogen show aldol reaction

The correct answer is Statement I is correct but Statement II is incorrect.

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:

Grignard reagent usually attacks on > C = O group as:

The question is related to above reaction only with the condition that the consumption of RMgX will be more than 1 equivalent in some of the given cases.

Among the given compounds B, i.e.

Contain additional groups which can give active hydrogen’s. Grignard reagents produces alkanes whenever come in contact with any group or compound which can give active hydrogen as:

These reactions are equivalent to acid-base reactions. So, in both of these compounds more than one equivalent will be required to form Grignard products. Remember these compounds will give 2 type of products as:

(i) from the > C = O group

(ii) from the group which release active hydrogen.

The additional reactions involved are:

Explanation:-

NaBH₄ can reduce aldehyde and ketone functional groups into primary and secondary alcohol respectively.

Explanation: -

 The compound with the -al suffix is Propanal.

Hence, the correct option is (1).

Explanation:

Iodoform test: This test is used to distinguish between ethanal and propanal. In this test, a yellow precipitate of iodoform (CHI3) is formed when a compound containing a methyl ketone or a secondary alcohol with a methyl group is treated with an iodine and sodium hydroxide (NaOH) solution.
The reactions for ethanal and propanal are as follows:

Ethanal + I₂ + NaOH → CHI₃ + Na₂CO₃ + H₂O

Propanal + 3I₂ + 4NaOH → CHI₃ + 3NaI + 2Na₂CO₃ + 2H₂O

In the Iodoform test, the yellow precipitate of iodoform is formed only when the compound contains a methyl ketone or secondary alcohol with a methyl group.

Therefore, the test can be used to distinguish between ethanal (which is an aldehyde and does not have a methyl group) and propanal (which is an aldehyde and has a methyl group).

CH₃CHO \(\underset{\mathrm{NaOH}}{\stackrel{\mathrm{I}_2}{\longrightarrow}} \underset{\substack{\text { Yellow } \\ \text { Precipitate }}}{\mathrm{CH}_3}\) \(+\mathrm{HCOO}^{\ominus} \mathrm{Na}^{\oplus}\)

CH₃CH₂CHO \(\underset{\mathrm{NaOH}}{\stackrel{\mathrm{I}_2}{\longrightarrow}}\) No characteristic change

Ethanal gives positive iodoform test. 

Explanation:-

The reaction involves the reduction of a carbonyl compound (specifically a ketone) using the Clemmensen reduction method. The Clemmensen reduction is used to reduce ketones (or aldehydes) to alkanes. The reagents used are zinc amalgam (Zn-Hg) and hydrochloric acid (HCl).

The given reactant is:

\(\text{4-methylacetophenone} \ (C_6H_5COCH_3)\)

The Clemmensen reduction reduces the carbonyl group (C=O) to a methylene group (CH2). 

Here is the reaction:

\(\text{C}_6\text{H}_5\text{COCH}_3 \xrightarrow{\text{Zn-Hg, HCl}} \text{C}_6\text{H}_5\text{CH}_2\text{CH}_3\)

The carbonyl group (C=O) in 4-methylacetophenone is reduced to an ethyl group (CH2).

So, the product of the reaction is:

\(\text{4-ethyl toluene (p-ethyltoluene)}\)

1. The first option shows a compound with two hydroxyl groups attached to the benzene ring, which is incorrect.
2. The second option shows the starting material, which is incorrect.
3. The third option shows a compound with a hydroxyl group attached to the benzene ring, which is incorrect.
4. The fourth option shows 4-ethyl toluene (p-ethyltoluene), which is the correct product.

Thus, the correct answer is 4