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

CONCEPT:

Cinnamaldehyde Reactions and Aldol Condensation

• Cinnamaldehyde (C6H5CH=CHCHO) is an α,β-unsaturated aldehyde that exhibits both electrophilic reactivity and the potential for aldol condensation.
• The aldehyde group (–CHO) in cinnamaldehyde can react with Tollen’s reagent (Ag(NH3)2+), as it contains the aldehyde functional group, which is easily oxidized to a carboxylic acid.
• The α,β-unsaturation in cinnamaldehyde makes it reactive for aldol condensation, especially under basic conditions, due to the enolate formation from the α-hydrogen of the aldehyde.

EXPLANATION:

• Statement I: Cinnamaldehyde contains an aldehyde group, which will react with Tollen’s reagent to give a silver mirror reaction. So, Statement I is correct.
• Statement II: No, cinnamaldehyde does not readily undergo self-aldol condensation. While it contains a carbonyl group (C=O) and a carbon-carbon double bond (C=C), it lacks an alpha-hydrogen, which is necessary for the formation of an enolate ion, a crucial intermediate in the aldol condensation reaction. So, Statement II is also Incorrect.

Therefore, the correct answer is Statement I is correct but Statement II is incorrect

CONCEPT:

Formation of Aldehyde from Alcohol

• The conversion of alcohol to aldehyde is an oxidation process.
• Primary alcohols are oxidized to aldehydes using mild oxidizing agents.
• Strong oxidizing agents may further oxidize aldehydes to carboxylic acids, so a selective reagent is required for controlled oxidation.

EXPLANATION:

• 
• Among the options, PCC in CH₂Cl₂ (dichloromethane) is the correct reagent for the formation of aldehyde from alcohol.

Therefore, the correct answer is Option 3: PCC/CH₂Cl₂.

CONCEPT:

pKa and Acidity

• pKa is the negative logarithm of the acid dissociation constant (Ka) of a compound. It is a measure of the strength of an acid.
• A lower pKa value indicates a stronger acid, as the acid dissociates more easily in solution.
• A higher pKa value indicates a weaker acid, as the acid dissociates less easily.
• The presence of electron-withdrawing groups (such as fluorine atoms) increases the acidity of a molecule by stabilizing the conjugate base, thus lowering the pKa value.

EXPLANATION:

• Acetic acid (CH₃COOH) has no electron-withdrawing groups attached to the carbon adjacent to the carboxylic group. Therefore, it has the highest pKa value, as it is the weakest acid.
• Fluoroacetic acid (FCH₂COOH) has one fluorine atom attached to the carbon adjacent to the carboxylic group. This fluorine atom is an electron-withdrawing group, which increases the acidity (lowers the pKa value).
• Difluoroacetic acid (F₂CHCOOH) has two fluorine atoms, further increasing the acidity compared to fluoroacetic acid.
• Trifluoroacetic acid (F₃CCOOH) has three fluorine atoms, making it the most acidic compound among the given options (lowest pKa value).
• Since acetic acid lacks electron-withdrawing groups, it has the highest pKa value among the options.

Therefore, the correct answer is Acetic acid.

CONCEPT:

The C-O Bond in Carboxylic Acid

• Carboxylic acids contain a carboxyl group (-COOH), which includes a C=O (carbonyl) bond and a C-O (single bond).
• The C-O bond in carboxylic acids is a result of sp²-sp³ orbital overlap.
• The oxygen of the hydroxyl group (-OH) is sp³ hybridized, while the carbon of the carboxyl group is sp² hybridized due to the presence of the double bond with the oxygen atom in the carbonyl group.
• 

EXPLANATION:

• The bond length of the C-O bond in carboxylic acid is influenced by resonance.
• In carboxylic acids, resonance occurs between the carbonyl group (C=O) and the hydroxyl group (C-OH):
  O=C-OH ↔ O-C=O^-
• This resonance gives partial double-bond character to the C-O bond, making it shorter than a typical sp²-sp³ single bond.
• Therefore, the C-O bond in carboxylic acid is sp²-sp³ and shorter than expected due to resonance.

Therefore, the correct answer is (sp²-sp³ shorter).

EXPLANATION:

Common Nitrogen-containing Reagents

• Certain organic reagents containing nitrogen groups are widely used in qualitative and quantitative analysis, especially in carbonyl compound identification.
• Each of these has a distinct structure and chemical name:
  • Hydroxylamine: NH₂–OH
  • Hydrazine: NH₂–NH₂
  • Semicarbazide: NH₂–NH–CO–NH₂

• a. NH₂–OH → Hydroxylamine → Match: iii ✅
• b. NH₂–NH₂ → Hydrazine → Match: ii ✅
• c. NH₂–NH–CO–NH₂ → Semicarbazide → Match: i ✅

Therefore, the correct match is: a - iii, b - ii , c - i

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