(i) $\text{CH}_3\text{-CO-CH(CH}_3)_2$:

Select the longest continuous carbon chain containing the ketone carbonyl group. The chain has 4 carbons (butanone). Number from left to right to give the carbonyl carbon the lowest possible locant:

There is a methyl substituent at carbon-3.

IUPAC Name: 3-Methylbutan-2-one

(ii) $\text{C}_6\text{H}_5\text{-CH=CH-CHO}$:

The principal functional group is the aldehyde ($\text{-CHO}$), which defines carbon-1 of a 3-carbon parent chain containing a double bond (propenal). A phenyl substituent ($\text{-C}_6\text{H}_5$) is attached at carbon-3.

IUPAC Name: 3-Phenylprop-2-enal (commonly known as cinnamaldehyde)

Correct Option: (a)

Explanation: The reactivity of carbonyl compounds towards nucleophilic addition is governed by two key factors:

1. Steric Hindrance: As the number and size of alkyl groups around the carbonyl carbon increase, the approach of the incoming nucleophile to the carbon is sterically hindered. Formaldehyde ($\text{HCHO}$) has only small hydrogen atoms, presenting minimal hindrance, whereas acetone ($(\text{CH}_3)_2\text{CO}$) has two bulky methyl groups which block the nucleophilic attack.
2. Electronic (Inductive) Effect: Alkyl groups are electron-donating groups ($+I$ inductive effect). They push electron density toward the carbonyl carbon, reducing its positive charge density (electrophilicity). Acetone has two $+I$ groups, acetaldehyde has one, and formaldehyde has none, making formaldehyde's carbonyl carbon highly electrophilic.

Therefore, the increasing reactivity order is:

Aldehydes are more reactive than ketones due to the following two distinct factors:

• Steric Reasons: In aldehydes, there is only one alkyl group attached to the carbonyl carbon, while ketones contain two. The presence of two relatively bulky alkyl groups in ketones sterically hinders the path of the incoming nucleophile trying to attack the carbonyl carbon.
• Electronic Reasons: Alkyl groups are electron-donating by inductive effect ($+I$). In ketones, two alkyl groups release electron density towards the carbonyl carbon, neutralizing its partial positive charge ($\delta^{+}$) and making it less electrophilic. In aldehydes, only one alkyl group is present, so the electrophilic character of the carbonyl carbon is preserved to a much higher degree.

Reason: The primary reason is the **resonance effect** of the benzene ring in benzaldehyde.

The carbonyl group of benzaldehyde is in conjugation with the benzene ring. The lone pairs/electrons of the ring delocalize into the carbonyl group, which can be represented by drawing polar resonance structures where the double bond shifts: $\text{C}=\text{O}$ becomes $\text{C}\text{-}\text{O}^{-}$. This resonance electron-donation increases the electron density on the carbonyl carbon, significantly reducing its electrophilic (electron-deficient) character compared to acetaldehyde, where no such aromatic resonance stabilization exists.

The addition of $\text{HCN}$ to aldehydes/ketones is catalyzed by a base, which accelerates the reaction by generating a stronger nucleophile. The step-by-step mechanism is as follows:

Step 1: Generation of the nucleophile.

Because $\text{HCN}$ is a weak acid, it reacts with the base catalyst ($\text{OH}^{-}$) to form the highly reactive, strongly nucleophilic cyanide ion ($\text{CN}^{-}$):

Step 2: Attack of the nucleophile.

The planar, $sp^2$-hybridized carbonyl carbon is attacked by the nucleophilic cyanide ion from a perpendicular angle. The $\pi$-bond electrons shift entirely to the oxygen atom, forming a tetrahedral alkoxide intermediate:

Step 3: Protonation of the intermediate.

The tetrahedral alkoxide intermediate abstracts a proton from a water molecule to form the final stable product, cyanohydrin, while regenerating the hydroxide base catalyst:

The self-aldol condensation of ethanal to yield but-2-enal on heating proceeds via the following detailed steps:

Step 1: Formation of enolate ion (deprotonation).

The hydroxide ion ($\text{OH}^{-}$) acts as a base and abstracts an acidic $\alpha$-hydrogen atom from the ethanal molecule, generating a resonance-stabilized enolate nucleophile:

Step 2: Nucleophilic attack on another carbonyl molecule.

The nucleophilic enolate ion attacks the electrophilic carbonyl carbon of a second, unreacted ethanal molecule, forming a carbon-carbon single bond and a tetrahedral alkoxide adduct:

Step 3: Protonation of the adduct.

The alkoxide adduct abstracts a proton from water to form a $\beta$-hydroxyaldehyde (known as an **Aldol**), specifically **3-hydroxybutanal**, while regenerating the hydroxide catalyst:

Step 4: Acid/Base-catalyzed dehydration (heating).

Upon heating, the aldol undergoes spontaneous elimination of a water molecule. The hydrogen on the $\alpha$-carbon is easily lost, yielding a thermodynamically stable conjugated $\alpha,\beta$-unsaturated aldehyde, **but-2-enal** (crotonaldehyde):

Cannizzaro Reaction: Aldehydes which **do not contain any $\alpha$-hydrogen atoms** (such as formaldehyde, $\text{HCHO}$, and benzaldehyde, $\text{C}_6\text{H}_5\text{CHO}$) undergo self-redox (disproportionation) when treated with concentrated alkali ($\ge 50\% \text{ NaOH}$). One molecule is reduced to an alcohol, while the other is oxidized to a carboxylic acid salt.

Step-by-Step Mechanism:

Step 1: Nucleophilic attack of hydroxide ion.

The strongly nucleophilic hydroxide ion attacks the highly electrophilic carbonyl carbon of formaldehyde, forming a tetrahedral dianion/anion intermediate:

Step 2: Hydride transfer (Rate-determining step).

The intermediate anion transfers a hydride ion ($\text{H}^{-}$) directly from its carbon atom to the carbonyl carbon of a second formaldehyde molecule. This hydride transfer is driven by the reforming of the highly stable carbon-oxygen double bond ($\text{C}=\text{O}$):

Step 3: Rapid proton exchange.

Since formic acid is acidic and methoxide is strongly basic, they immediately undergo an instantaneous proton exchange to yield stable sodium formate and methanol:

(i) Distinction between Propanal (Aldehyde) and Propanone (Ketone):

Use **Tollens' Test** (Silver Mirror Test). Propanal will reduce Tollens' reagent, whereas propanone (being a ketone) will not react.

Equation for Propanal:

Observation: A shiny, metallic silver mirror precipitates on the inner walls of the test tube.

(ii) Distinction between Benzaldehyde (Aromatic) and Acetaldehyde (Aliphatic):

Use **Fehling's Test**. Acetaldehyde (an aliphatic aldehyde) reduces Fehling's solution, whereas benzaldehyde (an aromatic aldehyde) lacks the aliphatic structure and fails to react.

Equation for Acetaldehyde:

Observation: The deep blue Fehling's solution turns cloudy and precipitates a bright, red-brick solid of cuprous oxide ($\text{Cu}_2\text{O}$).

Iodoform Test Principle: This test is specific for compounds containing either a **methyl ketone group ($\text{CH}_3\text{CO-}$)** or a methyl carbinol group ($\text{CH}_3\text{CH(OH)-}$). When warmed with iodine ($\text{I}_2$) in sodium hydroxide ($\text{NaOH}$), these compounds undergo halogenation followed by basic cleavage to form a yellow precipitate of iodoform ($\text{CHI}_3$).

Distinguishing Test:

• Pentan-2-one ($\text{CH}_3\text{-CO-CH}_2\text{CH}_2\text{CH}_3$): Contains a methyl ketone group, so it gives a **positive iodoform test**.
• Pentan-3-one ($\text{CH}_3\text{CH}_2\text{-CO-CH}_2\text{CH}_3$): Lacks a methyl ketone group, so it **does not react**.

Balanced Equation for Pentan-2-one:

Observation: A bright yellow crystalline precipitate of iodoform ($\text{CHI}_3$) forms with a characteristic medicinal antiseptic smell.

Correct Option: (a)

Explanation: Carboxylic acids ionize to yield the carboxylate ion. The negative charge is shared equally between **two identical electronegative oxygen atoms** through resonance, resulting in exceptionally high stability. In contrast, when phenol ionizes, the negative charge on the phenoxide ion is delocalized over the aromatic ring, placing the negative charge on less electronegative carbon atoms. Because delocalizing charge over electronegative oxygens is thermodynamically much more stabilizing than delocalizing it over carbons, carboxylic acids are significantly stronger acids than phenols or alcohols.

Increasing Order of Acidity:

Explanation based on Inductive Effect:

The acidic strength of a carboxylic acid depends on the stability of its conjugate base (carboxylate anion). Electron-withdrawing groups ($-I$) pull electron density away from the carboxylate group, dispersing the negative charge and stabilizing the anion, which increases acidity.

• $\text{CH}_3\text{COOH}$ is the weakest because the methyl group ($\text{-CH}_3$) is electron-donating ($+I$ inductive effect), which concentrates the negative charge on oxygen, destabilizing the anion.
• Chlorine and fluorine are electron-withdrawing halogens ($-I$ effect). Since fluorine is more electronegative than chlorine, its $-I$ effect is stronger, stabilizing the anion more and making $\text{F-CH}_2\text{-COOH}$ more acidic than $\text{Cl-CH}_2\text{-COOH}$.
• The nitro group ($\text{-NO}_2$) is an exceptionally powerful electron-withdrawing group ($-I$ and $-M$ resonance effects). It stabilizes the carboxylate anion the most, making nitroacetic acid the strongest acid in this group.

Acidity Comparison: **2-Chlorobutanoic acid** is a significantly stronger acid than 3-chlorobutanoic acid.

Scientific Reason:

The inductive effect (electron withdrawal by the electronegative chlorine atom) is highly **distance-dependent**. Its intensity decreases rapidly as the number of intervening single $\sigma$-bonds increases:

• In **2-chlorobutanoic acid**, the chlorine atom is on the $\alpha$-carbon, directly adjacent to the carboxyl group ($\text{-COOH}$). This proximity allows for strong electron withdrawal, effectively stabilizing the carboxylate anion.
• In **3-chlorobutanoic acid**, the chlorine is on the $\beta$-carbon, one bond further away. At this distance, the electron-withdrawing effect is significantly weaker, providing less stabilization to the conjugate base and resulting in lower acidity.

Increasing Order of Acidity:

Justification based on Ring Substituents:

• 4-Nitrobenzoic acid: The nitro group ($-\text{NO}_2$) is a powerful electron-withdrawing group via resonance and induction ($-M, -I$). When located in the para position, it delocalizes the negative charge of the carboxylate anion into the ring, stabilizing the conjugate base and making this the strongest acid.
• Benzoic acid: Serves as the unsubstituted baseline.
• 4-Methylbenzoic acid: The methyl group ($-\text{CH}_3$) is electron-donating through induction ($+I$) and hyperconjugation. This increases the negative charge on the carboxylate group, destabilizing the anion and reducing acidity.
• 4-Methoxybenzoic acid: Although oxygen is electronegative, the methoxy group ($-\text{OCH}_3$) acts as a strong electron donor through resonance ($+M$) when in the para position. This electron donation destabilizes the carboxylate anion the most, making this the weakest acid in the group.

Hell-Volhard-Zelinsky (HVZ) Reaction: Aliphatic carboxylic acids that contain **at least one $\alpha$-hydrogen** react with chlorine ($\text{Cl}_2$) or bromine ($\text{Br}_2$) in the presence of a small amount of red phosphorus to form $\alpha$-halocarboxylic acids. This reaction selectively replaces the hydrogen atom on the $\alpha$-carbon with a halogen.

Equation for Propanoic Acid:

Propanoic acid ($\text{CH}_3\text{CH}_2\text{COOH}$) has two $\alpha$-hydrogens. Reacting it with bromine and red phosphorus replaces one $\alpha$-hydrogen to yield **2-bromopropanoic acid**:

Decarboxylation: The reaction in which carboxylic acids (or their sodium salts) lose a molecule of carbon dioxide ($\text{CO}_2$) to form hydrocarbons. This is typically achieved by heating the sodium salt of a carboxylic acid with soda lime.

Soda Lime: A mixture of sodium hydroxide ($\text{NaOH}$) and calcium oxide ($\text{CaO}$) in a $3:1$ ratio.

Chemical Reaction:

Heating anhydrous sodium acetate ($\text{CH}_3\text{COONa}$) with soda lime yields **methane**:

The product alkane contains one less carbon atom than the starting carboxylic acid salt.

(i) Benzoic acid to Benzaldehyde:

1. Convert benzoic acid to benzoyl chloride using thionyl chloride ($\text{SOCl}_2$):
   $$ \text{C}_6\text{H}_5\text{COOH} + \text{SOCl}_2 \rightarrow \text{C}_6\text{H}_5\text{COCl} + \text{SO}_2 + \text{HCl} $$
2. Reduce benzoyl chloride to benzaldehyde using hydrogen gas over a palladium catalyst supported on barium sulfate ($\text{BaSO}_4$), poisoned with sulfur (the **Rosenmund Reduction**):
   $$ \text{C}_6\text{H}_5\text{COCl} + \text{H}_2 \xrightarrow{\text{Pd-BaSO}_4} \text{C}_6\text{H}_5\text{CHO} + \text{HCl} $$

(ii) Ethanol to Ethanoic acid:

This single-step conversion is achieved using a strong oxidizing agent like acidified potassium permanganate ($\text{KMnO}_4$) or alkaline potassium dichromate, which oxidizes the primary alcohol directly to the carboxylic acid:

(i) Propene to Acetone:

1. React propene with water in the presence of an acid catalyst (Markovnikov hydration) to yield propan-2-ol:
   $$ \text{CH}_3\text{-CH}=\text{CH}_2 + \text{H}_2\text{O} \xrightarrow{\text{H}^{+}} \text{CH}_3\text{-CH(OH)-CH}_3 $$
2. Oxidize the secondary alcohol propan-2-ol to acetone using chromium trioxide in acidic medium (Jones reagent):
   $$ \text{CH}_3\text{-CH(OH)-CH}_3 \xrightarrow{\text{CrO}_3 / \text{H}_2\text{SO}_4} \text{CH}_3\text{-CO-CH}_3 \quad \text{(Acetone)} $$

(ii) Benzene to Benzaldehyde:

This conversion can be achieved in a single step using the **Gattermann-Koch reaction**. React benzene with carbon monoxide ($\text{CO}$) and hydrogen chloride ($\text{HCl}$) in the presence of anhydrous aluminium chloride ($\text{AlCl}_3$) and cuprous chloride ($\text{CuCl}$):

This conversion can be carried out in two synthetic steps:

1. Grignard Reaction: React benzaldehyde with phenyl magnesium bromide ($\text{C}_6\text{H}_5\text{MgBr}$), followed by acid hydrolysis, to produce the secondary alcohol **diphenylmethanol**:
   $$ \text{C}_6\text{H}_5\text{-CHO} + \text{C}_6\text{H}_5\text{-MgBr} \rightarrow [(\text{C}_6\text{H}_5)_2\text{CH-OMgBr}] \xrightarrow{\text{H}_3\text{O}^{+}} (\text{C}_6\text{H}_5)_2\text{CH-OH} $$
2. Oxidation: Oxidize diphenylmethanol to the ketone **benzophenone** using a selective oxidizing agent like pyridinium chlorochromate ($\text{PCC}$) or chromic acid:
   $$ (\text{C}_6\text{H}_5)_2\text{CH-OH} \xrightarrow{\text{PCC}} (\text{C}_6\text{H}_5)_2\text{C=O} \quad \text{(Benzophenone)} $$

(i) Reaction of Acetic Acid with $\text{PCl}_5$:

Phosphorus pentachloride replaces the hydroxyl group ($\text{-OH}$) of the carboxylic acid with a chlorine atom, yielding acetyl chloride, phosphorus oxychloride, and hydrogen chloride gas:

(ii) Benzaldehyde with Concentrated Sodium Hydroxide:

Since benzaldehyde lacks $\alpha$-hydrogens, it undergoes the **Cannizzaro Reaction** (disproportionation). One molecule is oxidized to sodium benzoate and the other is reduced to benzyl alcohol:

Let's determine each intermediate step-by-step:

1. Step 1 ($\text{CH}_3\text{CH}_2\text{Br} \xrightarrow{\text{KCN}} \text{A}$): This is a nucleophilic substitution reaction ($S_N2$) where the cyanide ion ($\text{CN}^{-}$) replaces the bromide ion to form a nitrile:
   $$ \text{A} = \text{CH}_3\text{CH}_2\text{CN} \quad \text{(Propanenitrile)} $$
2. Step 2 ($\text{A} \xrightarrow{\text{H}_3\text{O}^{+}} \text{B}$): Complete acid hydrolysis of propanenitrile converts the nitrile group ($\text{-CN}$) into a carboxylic acid group ($\text{-COOH}$):
   $$ \text{B} = \text{CH}_3\text{CH}_2\text{COOH} \quad \text{(Propanoic Acid)} $$
3. Step 3 ($\text{B} \xrightarrow{\text{SOCl}_2} \text{C}$): Thionyl chloride reacts with propanoic acid to replace the hydroxyl group with chlorine, yielding an acid chloride:
   $$ \text{C} = \text{CH}_3\text{CH}_2\text{COCl} \quad \text{(Propanoyl Chloride)} $$

Answer: Intermediate A is propanenitrile, B is propanoic acid, and C is propanoyl chloride.