Class 12 Chemistry Chapter 8 Aldehydes, ketones and Carboxylic Acids

Key Concept: Odor and Volatility

b) Low volatility
[Solution Description] Higher carboxylic acids are practically odorless because they have low volatility. This means that their molecules do not easily evaporate into the gas phase, which reduces the presence of vapor that carries odor.

Key Concept: Reaction Pathway, Test Application

c) Hemiacetal
[Solution Description] The absence of a reaction in Tollen’s test indicates that the compound likely does not contain an aldehyde group. However, the formation of a red precipitate with Benedict’s reagent suggests the presence of a reducing sugar. Reducing sugars often contain free aldehyde groups in equilibrium with their cyclic form or hemiacetal structures, which can reduce Cu$^{2+}$ ions to Cu$_2$O.

Key Concept: Reaction Products

c) Butanal

[Solution Description]

Upon ozonolysis, the double bond in 1-butene breaks and each carbon is converted into a carbonyl group. The reaction sequence involves the formation of intermediate ozonides which are reduced by zinc dust and water to form carbonyl compounds.

The structure of 1-butene is $CH_3-CH_2-CH=CH_2$. When it undergoes ozonolysis:

$$CH_3-CH_2-CH=CH_2 \xrightarrow[\text{Zn/H}_2O]{\text{O}_3} CH_3-CH_2-CHO + HCHO$$

Therefore, the products formed are butanal ($CH_3-CH_2-CHO$) and formaldehyde ($HCHO$).

Thus, the correct answer is Butanal.

Key Concept: Complex Reaction Pathway, Mechanistic Detail Analysis

c) The Grignard reagent forms a tertiary alcohol that is hydrolyzed to yield a ketone.
[Solution Description] In this sequence, the acyl chloride first reacts with the Grignard reagent ($\textit{R-MgX}$) to form a tertiary alcohol through nucleophilic addition to the carbonyl group. This forms an adduct where the oxygen is bound to the magnesium halide. The subsequent hydrolysis step involves protonating the alkoxide to release water and regenerate the carbonyl functional group but as a ketone. Here’s the detailed mechanism:

Step 1: Nucleophilic attack by Grignard reagent
$$
R’\text{-MgX} + ClC(=O)R” \rightarrow R’C(O-MgX)(OH)-R”
$$

Step 2: Hydrolysis of the intermediate
$$
R’C(O-MgX)(OH)-R” + H_3O^+ \rightarrow R’C(=O)R” + Mg^{2+} + X^- + H_2O
$$

The final product is a ketone ($R’C(=O)R”$).

Option C depicts the correct series of transformations leading to the production of the ketone via formation of an alcohol intermediate.

Key Concept: Synthesis Pathways, Mechanistic Insight

d) Acetone, because methanol escapes allowing only monosubstitution

[Solution Description] The reaction of a Grignard reagent with an ester typically goes through two steps leading to a tertiary alcohol. However, in this case, we need to consider the production of a ketone, suggesting that only one equivalent of the Grignard reagent reacts before protonation occurs.

Step 1: Nucleophilic attack by $CH_3^-$ from $CH_3MgBr$ on the carbonyl carbon of $CH_3COOC_2H_5$, forming a tetrahedral intermediate.

Step 2: Collapse of the tetrahedral intermediate expels the ethoxy group ($C_2H_5O^-$), resulting in acetone formation $(CH_3)_2CO$.

This unique behavior denotes that under specific conditions, only one equivalent of Grignard reagent adds to give a ketone rather than further reacting to form a tertiary alcohol.

Key Concept: Reaction Optimization, Reaction Limitations

c) Perform the reaction at lower temperatures to minimize side reactions
[Solution Description] Clemmensen reduction involves using acidic conditions, which are not suitable for substrates with acid-sensitive groups due to possible degradation or side reactions. Thus, mitigating these effects often requires controlling reaction temperature and avoiding prolonged exposure to acids.

Key Concept: Simple Reaction Equation

a) $RCOOH \rightarrow{LiAlH_4} RCH_2OH$
[Solution Description] When carboxylic acids are reduced using $LiAlH_4$, the carboxyl group (-COOH) is converted to a hydroxyl group (-OH), forming a primary alcohol ($RCH_2OH$).

Key Concept: Comparative Oxidation, Test Specificity

c) Benzaldehyde
[Solution Description] The Fehling’s test is used for distinguishing between aldehydes and ketones. It involves heating the sample with Fehling’s solution, which contains copper(II) ions. Aldehydes reduce Cu$^{2+}$ to Cu$_2$O, resulting in a reddish-brown precipitate, while most ketones do not react due to their resistance to mild oxidizing agents under these conditions. Aromatic aldehydes like benzaldehyde also do not respond to Fehling’s test because they lack the necessary reactivity due to the resonance stabilization from the aromatic ring. Thus, benzaldehyde would not give a positive result.

Key Concept: Simple Product Formation

b) Ammonium salt
[Solution Description] When carboxylic acids react with ammonia, they initially form ammonium salts. These salts then convert to amides upon heating.

Key Concept: Reaction Identification

c) Esterification
[Solution Description] When a carboxylic acid reacts with an alcohol in the presence of an acid catalyst, it forms an ester and water. This process is known as esterification.

Key Concept: Hydrogen Bonding

b) Due to extensive hydrogen bonding
[Solution Description] Carboxylic acids exhibit extensive hydrogen bonding due to their ability to form dimer structures through intermolecular hydrogen bonds. This strong hydrogen bonding increases the boiling point of carboxylic acids compared to other carbonyl-containing compounds like aldehydes and ketones.

Key Concept: Chain Numbering

d) The carboxyl carbon
[Solution Description] In IUPAC nomenclature for carboxylic acids, the carboxyl carbon atom is always numbered as one, regardless of the length or branching of the carbon chain. Therefore, for CH$_3$(CH$_2$)$_2$CH$_2$COOH, the carboxyl carbon is carbon number one.

Reaction Sequence Analysis

d) Reduction with DIBAL-H
[Solution Description] Ethyl acetate, an ester, can be selectively reduced to acetaldehyde using diisobutylaluminium hydride (DIBAL-H). Upon treatment with DIBAL-H, the ester undergoes reduction to form an aldehyde.

Key Concept: Nomenclature, Real-world Applications

c) Propanone; IUPAC Name: Acetone

[Solution Description] Acetone is commonly employed in producing polycarbonate and epoxy resins. Its chemical formula is $$\text{C}_3\text{H}_6\text{O}$$, and its IUPAC name is propan-2-one. Its utility in these industries stems from its ability to dissolve various precursors needed during processing.

Key Concept: Textile Industry

c) Methanoic acid
[Solution Description] Methanoic acid, also known as formic acid, is widely used in the textile industry for its effectiveness in dyeing processes.

Key Concept: Real-world Application, Reaction Pathway

c) 2-Hydroxypropanonitrile (cyanohydrin)
[Solution Description] Propanal undergoes a nucleophilic addition reaction with hydrogen cyanide, forming a cyanohydrin. The mechanism involves the attack of the cyanide ion on the electrophilic carbon of the carbonyl group, followed by protonation to form the cyanohydrin derivative of propanal.

Key Concept: Electrophilic Substitution Mechanism, Synthesis Pathways

b) It deactivates the benzene ring, promoting meta-substitution.
[Solution Description] In Friedel-Crafts acylation, the carbonyl group in acetyl chloride interacts with AlCl$_3$ to form an acylium ion, which serves as the electrophile. Due to the deactivating nature of the carbonyl group, the reaction directs incoming substitutions to the meta position relative to existing substituents on the benzene ring. Thus, the carbonyl group reduces reactivity but provides specific control over the orientation of additional electrophilic substitutions.

Key Concept: Stereochemistry, Reaction Pathways

b) Racemic mixture of both enantiomers
[Solution Description] Sodium borohydride provides hydride ions, which approach the planar carbonyl group from one side leading to racemization. Thus, the resulting alcohol generally forms as a racemic mixture unless the substrate or solvent environment is specifically designed to induce asymmetric induction.

Key Concept: Oxidation of Primary Alcohols

c) Acetic acid

[Solution Description]

Ethanol ($C_2H_5OH$) can be oxidized by potassium permanganate ($KMnO_4$) to form acetic acid ($CH_3COOH$). Here is the step-by-step process:

1. The ethanol is first oxidized to acetaldehyde ($CH_3CHO$).

2. Acetaldehyde is further oxidized to acetic acid.

The balanced chemical equation for the complete reaction is:

$$C_2H_5OH + 2 KMnO_4 + 3 H_2SO_4 \rightarrow CH_3COOH + 2 MnSO_4 + K_2SO_4 + 4 H_2O$$

Therefore, the final product is acetic acid ($CH_3COOH$).

Key Concept: Reaction Optimization, Real-world Application

c) Polar aprotic solvent, moderate temperature, strong base
[Solution Description]
Cross aldol condensation between acetophenone and formaldehyde can be optimized by considering several factors:

– Solvent: A polar aprotic solvent like DMSO or DMF helps stabilize the charged intermediates, thus favoring higher yields.
– Temperature: Moderately elevated temperatures promote reaction kinetics but should be controlled to prevent side reactions or overreaction leading to polymerization.
– Base Strength: A strong base such as NaOH or KOH in a dilute aqueous solution efficiently generates the enolate ion from acetophenone, which then adds to the formaldehyde.

Thus, using a polar aprotic solvent, moderate temperature, and strong base will optimize the yield of the desired condensation product.

Key Concept: Substituent Influence

b) It directs substitution to the meta position
[Solution Description] The nitro group ($NO_2$) is a strong electron-withdrawing group and directs incoming electrophiles to the meta position due to its -M (mesomeric) effect, which makes ortho and para positions less favorable by destabilizing their corresponding intermediates.

Key Concept: Reaction with Bases

b) Sodium acetate
[Solution Description] Acetic acid ($CH_3COOH$) reacts with sodium hydroxide ($NaOH$) to form a salt and water. The reaction is as follows:

$$
CH_3COOH + NaOH \rightarrow CH_3COONa + H_2O
$$

Therefore, the product formed is sodium acetate.

Key Concept: Reaction Conditions

b) Base
[Solution Description]
Basic hydrolysis of esters, also known as saponification, requires a base such as sodium hydroxide (NaOH). In this reaction, the ester is reacted with a strong base, which converts the ester into a carboxylate ion and an alcohol. Upon acidification of the carboxylate, a carboxylic acid is formed.

Key Concept: Structural Isomers, Complex Naming

c) $$CH_3CH_2CH(CH_3)COCH_3$$
[Solution Description]
The original structure is a pentane derivative with a ketone at the $2^{nd}$ position. To find its structural isomer, rearrange the carbon chain or functional groups but retain the molecular formula. One possible isomer is by moving the ketone to another position while maintaining five carbons total, resulting in 3-methylbutan-2-one.

Key Concept: Complex Reaction Pathway Analysis, Multi-Step Synthesis Design

d) Tertiary Alcohol
[Solution Description] The first step involves a Friedel-Crafts acylation of benzene using an acid chloride and AlCl$_3$, producing an aromatic ketone. In the second step, this ketone reacts with a Grignard reagent to form a tertiary alcohol after hydrolysis. The complete pathway is:

1. Benzene + Acid Chloride $\rightarrow{\text{AlCl}_3}$ Aryl Ketone

2. Aryl Ketone + Grignard Reagent $\rightarrow{\text{hydrolysis}}$ Tertiary Alcohol

This leads us to option d as the correct answer.

Key Concept: Boiling Point Comparison

a) Ethanoic Acid > Methanol >Butanone > Propanal
[Solution Description] Carboxylic acids have the highest boiling points due to extensive hydrogen bonding, followed by alcohols, which also engage in hydrogen bonding but not as extensively. Aldehydes and ketones possess dipole-dipole interactions leading to lower boiling points compared to alcohols and carboxylic acids. Therefore, Ethanoic Acid has the highest boiling point, followed by Methanol, then Butanone, and finally Propanal. Thus, the order is Ethanoic Acid > Methanol > Butanone > Propanal.

Key Concept: Electrolysis Outcome

c) Ethane
[Solution Description]
During Kolbe electrolysis, alkali metal salts of carboxylic acids undergo decarboxylation, resulting in hydrocarbons that have twice the number of carbon atoms present in the alkyl group of the acid. For sodium acetate ($CH_3COONa$), the reaction proceeds as follows:

$$
2CH_3COONa + 2H_2O \rightarrow C_2H_6 + 2CO_2 + H_2 + 2NaOH
$$

This shows that ethane ($C_2H_6$) is produced.

Key Concept: Product Stability

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

[Solution Description]

The assertion states that the reaction of aldehydes with ammonia derivatives forms stable imine products. This is true as the process typically results in imines which are relatively stable due to their ability to form double-bonded structures comparable to carbonyls.

The reason provided indicates that imine formation involves a rapid dehydration step, which stabilizes the final product. This explanation is accurate since during the reaction, water is eliminated, and the resulting cationic intermediate undergoes deprotonation to give the neutral imine. Therefore, both the assertion and the reason are correct, and the reason provides an appropriate explanation for the assertion.

Hence, both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

Key Concept: Reaction Conditions

b) Red phosphorus
[Solution Description] For the halogenation of carboxylic acids at the $\alpha$-position, the necessary reagent is red phosphorus along with halogens like chlorine or bromine. This combination facilitates the formation of an acyl halide intermediate followed by the substitution reaction.

Key Concept: Aromatic Uses

c) Benzaldehyde

[Solution Description] Benzaldehyde is widely used in the perfumery industry due to its characteristic almond aroma. It is often employed as a flavoring agent and fragrance in various products.

Key Concept: Real-World Application, Reaction Optimization

c) High temperature, acidic medium
[Solution Description] Industrial processes prioritize efficiency, cost-effectiveness, and environmental impact. For the oxidation of aldehydes to carboxylic acids, potassium dichromate in an acidic medium is often used effectively.

To optimize yields and minimize waste, it is crucial to consider the reactivity and selectivity of the oxidizing agent. Potassium dichromate ($\text{K}_2\text{Cr}_2\text{O}_7$) is not only powerful but also provides consistent results:

$$
3RCHO + \text{K}_2\text{Cr}_2\text{O}_7 + 4\text{H}_2\text{SO}_4 \rightarrow 3RCOOH + \text{Cr}_2\text{(SO}_4\text{)}_3 + \text{K}_2\text{SO}_4 + 4\text{H}_2\text{O}
$$

The use of an acidic medium prevents side reactions and ensures complete conversion, making high-temperature acidic conditions suitable for maximizing yield and reducing waste generation.

Key Concept: Flavor Identification

b) Vanillin
[Solution Description] Vanillin is the primary component responsible for the vanilla flavor. It is used extensively in flavoring agents due to its pleasant aroma.

Key Concept: Real-world Application, Comparative Electronegativity

c) It creates an electrophilic site that can be attacked by nucleophiles.
[Solution Description] The polarity of the carbonyl group arises from the electronegativity difference between carbon and oxygen, creating a dipole moment where the carbon is electrophilic and the oxygen is nucleophilic. This allows the carbonyl group to interact with various biological molecules, facilitating reactions such as nucleophilic addition or reduction in metabolic pathways.

Key Concept: Basic Reaction

c) Cyanohydrin
[Solution Description] When a ketone undergoes reaction with hydrogen cyanide (HCN), the nucleophile ($CN^-$) attacks the electrophilic carbon atom of the carbonyl group, resulting in the formation of a cyanohydrin.

Key Concept: Multi-step Reactions, Reaction Optimization

b) Both Assertion and Reason are true, but Reason is NOT the correct explanation of Assertion.

[Solution Description]

In Wolff-Kishner reduction, using a high-boiling solvent such as ethylene glycol allows the reaction to reach higher temperatures without boiling off the solvent. This high temperature helps in achieving full conversion of carbonyl compounds to hydrocarbons by facilitating the elimination of nitrogen gas formed during the hydrazone formation process. Hence, the assertion is true.

While the reason provided describes the process of Clemmensen reduction accurately, it does not explain why the choice of a high-boiling solvent is critical for Wolff-Kishner reduction. Consequently, both statements are individually true; however, the reason is not the correct explanation of the assertion.

Thus, the correct answer is option b).

Key Concept: Solubility and Hydrogen Bonding

d) Hydrogen bonding with water
[Solution Description] Lower aldehydes are soluble in water primarily due to their ability to form hydrogen bonds with water molecules. The carbonyl group (C=O) in aldehydes can engage in hydrogen bonding with water, allowing them to be miscible.

Key Concept: Solubility in Organic Solvents

b) Ethanol
[Solution Description] Benzoic acid is nearly insoluble in cold water but is soluble in less polar organic solvents like benzene, ether, alcohol, and chloroform. Therefore, it will be more soluble in ethanol among the given options.

Key Concept: Basic Reaction

c) Carboxylic acids
[Solution Description] The Hell-Volhard-Zelinsky Reaction primarily involves carboxylic acids as the main reactant. This reaction converts carboxylic acids to $\alpha$-halo carboxylic acids by replacing the $\alpha$-hydrogen with a halogen atom.

Key Concept: Carboxylic Acid IUPAC Names

c) Hexanoic acid
[Solution Description] The given structure corresponds to a linear chain of six carbon atoms, including the carboxyl group at the end. For carboxylic acids, the IUPAC name replaces –e in alkanes with –oic acid. Therefore, the IUPAC name is Hexanoic acid.

Key Concept: Simple Chemical Tests

d) Tollens’ test

[Solution Description]

Acetone is a ketone whereas propanal is an aldehyde. A simple test to differentiate them is Tollens’ test. Aldehydes like propanal will reduce Tollens’ reagent to form a silver mirror, whereas ketones like acetone do not.

Key Concept: Multiplicative Prefixes

b) Propanedioic acid

[Solution Description] The compound $$\text{HOOC-CH}_2-\text{COOH}$$ has two carboxyl groups, hence it is a dicarboxylic acid. According to IUPAC nomenclature rules, the base name for a three-carbon chain is ‘prop’. Since there are two carboxyl groups, we use the prefix ‘di’ and replace ‘-e’ of propane with ‘-oic acid’, resulting in propanedioic acid.

Key Concept: Comparative Reactivity, Resonance Structures and Polarity

b) Formaldehyde, due to lesser steric hindrance
[Solution Description] Formaldehyde is more reactive than acetaldehyde in nucleophilic addition reactions. This is because formaldehyde has less steric hindrance around the carbonyl carbon due to the absence of alkyl groups attached to it, thereby increasing its electrophilicity compared to acetaldehyde, which has a methyl group that provides some electron donating effect, reducing its reactivity.

Key Concept: Basic Concept

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

[Solution Description]

The hydration of ethyne ($C_2H_2$) with water ($H_2O$) in the presence of catalysts $H_2SO_4$ (sulfuric acid) and $HgSO_4$ (mercuric sulfate) results in the formation of an intermediate enol, which tautomerizes to form acetaldehyde, a type of aldehyde. The process is specific to ethyne for producing acetaldehyde, while other alkynes typically form ketones. Thus, both assertion and reason are true, and the reason correctly explains the assertion as it provides the conditions necessary for this reaction to occur.

Key Concept: Multi-Step Synthesis, Advanced Mechanism

b) Oxidation, Controlled Dehydrogenation
[Solution Description] Oxidizing methanol ($CH_3OH$) initially produces formaldehyde ($HCHO$). To retain this intermediate without proceeding to formic acid, dehydrogenation can be used to stabilize formaldehyde. This requires careful control to avoid excessive oxidation. Thus, the correct sequence is oxidation followed by controlled dehydrogenation.

Key Concept: Reaction Pathway, Real-World Application

b) Both Assertion and Reason are true, but Reason is NOT the correct explanation of Assertion.

[Solution Description]

Let’s analyze each statement:

Assertion: In an aldol condensation, the enolate ion acts as a nucleophile that attacks the carbonyl carbon of another molecule, leading to the formation of a β-hydroxy carbonyl compound. This is indeed true because the enolate ion provides the necessary electrons for the nucleophilic addition step.

Reason: While aldol reactions do produce α,β-unsaturated carbonyl compounds, this occurs through dehydration of the initially formed β-hydroxy carbonyl compounds, not through direct nucleophilic attack by water. Therefore, while the reason mentions a true fact about aldol reactions leading to α,β-unsaturated carbonyls, it incorrectly attributes the mechanism.

Since both statements are true on their own, but the reason does not correctly explain the assertion, the correct answer is:

Key Concept: Industrial Applications

b) Methanoic acid

[Solution Description] Methanoic acid, also known as formic acid, is used extensively in the rubber, textile, dyeing, leather, and electroplating industries. The acidic properties help in coagulating latex in rubber production and act as a mordant in dyeing processes.

Key Concept: Aromatic Uses

b) Perfume Industry
[Solution Description] Benzaldehyde is primarily used in the perfumery industry because it has a characteristic almond-like scent which makes it desirable for fragrance production.

Key Concept: Odour Change with Size

b) Less pungent and more fragrant

[Solution Description] Lower aldehydes typically have sharp pungent odors due to their volatile nature. As the molecular size and complexity increase, the volatility diminishes, resulting in less pungent and often more pleasant or fragrant odors.

Key Concept: Basic Esterification

b) It activates the carbonyl group towards nucleophilic addition.
[Solution Description] Protonation of the carbonyl oxygen enhances the electrophilicity of the carbonyl carbon, making it more susceptible to nucleophilic attack by the alcohol.

Key Concept: Reaction Conditions

b) Sodalime
[Solution Description] The decarboxylation of sodium salts of carboxylic acids requires sodalime, which consists of $\text{NaOH}$ and $\text{CaO}$ in a 3:1 ratio, as an essential component.

Key Concept: Solubility Trends, Real-world Application

c) They inhibit microbial growth due to acidity.

[Solution Description]

Carboxylic acids are commonly used in the food industry due to their acidity, which inhibits microbial growth, thus acting as a preservative. Their solubility in water and ability to interact with other molecules ensures they mix well within food products. Although their physical properties like high boiling points influence some applications, their primary role in food preservation comes from acidity.

Therefore, the correct answer involves understanding both their chemical and practical uses, focusing on the acidity and antimicrobial properties.

Key Concept: Boiling Point Comparison

c) Methanal has a lower boiling point than ethanol.
[Solution Description]
Methanal (an aldehyde) does not form significant hydrogen bonds, whereas ethanol (an alcohol) does. Hydrogen bonding in ethanol leads to higher boiling points compared to methanal.

Key Concept: Electronegativity

c) Oxygen is more electronegative than carbon
[Solution Description] The polarization of the carbon-oxygen bond in the carbonyl group occurs due to the difference in electronegativity between carbon and oxygen. Oxygen is more electronegative than carbon, which causes the shared electrons in the bond to be drawn closer to the oxygen atom, creating a partial negative charge on oxygen and a partial positive charge on carbon.

Key Concept: Reaction Optimization, Real-World Application

c) Use stoichiometric DIBAL-H at low temperature
[Solution Description] The reduction of esters to aldehydes using DIBAL-H requires precise control over the reaction conditions to prevent further reduction to alcohols. To achieve selective reduction to the aldehyde stage, it is crucial to use stoichiometric amounts of DIBAL-H and conduct the reaction at low temperatures. This minimizes the reactivity of the hydride ion, thereby preventing over-reduction.

Key Concept: Reaction Optimization, Reaction Variants

b) Low temperature and moderate concentration of NaOH.
[Solution Description] In a mixture of formaldehyde and benzaldehyde, the crossed Cannizzaro reaction occurs where one molecule of formaldehyde is oxidized while benzaldehyde is reduced. Formaldehyde preferentially undergoes oxidation due to its smaller size and higher reactivity. To maximize benzyl alcohol production, controlling the concentration of the base and temperature helps minimize over-oxidation or undesirable side reactions. Using a moderate concentration of sodium hydroxide at low temperatures optimizes the primary reaction pathway.