Step 1: Determining the Empirical Formula

First, calculate the percentage of oxygen: $100 - (69.77 + 11.63) = 18.60\%$. [Per the Law of Conservation of Mass].

Empirical formula: $C_5H_{10}O$. Empirical formula mass: $(5 \times 12) + (10 \times 1) + (1 \times 16) = 86$ g/mol. Since the given molecular mass is 86, the Molecular Formula is $C_5H_{10}O$.

Step 2: Analyzing Functional Group Properties

• Negative Tollens' Test: The compound is a ketone, not an aldehyde. [Justification: Tollens' reagent reacts only with aldehydes].
• Sodium Hydrogensulphite Addition: Confirms the presence of a carbonyl group ($>C=O$). [Justification: Characteristic reaction of ketones with low steric hindrance].
• Positive Iodoform Test: Confirms the presence of a methyl ketone group ($CH_3-CO-$). [Justification: The $CH_3-CO-$ group is oxidized to $CHI_3$ (iodoform) and a carboxylate ion by $I_2/OH^-$].

Step 3: Determining Structural Isomers via Oxidation

Vigorous oxidation of a ketone involves the cleavage of the $C-C$ bond adjacent to the carbonyl group. Oxidation of the unknown compound yields ethanoic acid ($CH_3COOH$) and propanoic acid ($CH_3CH_2COOH$).

Given the formula $C_5H_{10}O$ and the methyl ketone requirement, the structure must be $CH_3-CO-CH_2-CH_2-CH_3$ (Pentan-2-one).

Step 4: Verification of Oxidation Products

Cleavage of Pentan-2-one during vigorous oxidation:

$CH_3-CO-CH_2-CH_2-CH_3 \xrightarrow{[O]} CH_3COOH + CH_3CH_2COOH$

The cleavage at the $C_2-C_3$ bond produces two-carbon (ethanoic) and three-carbon (propanoic) fragments, matching the empirical evidence provided. [Popoff’s Rule: Oxidation of unsymmetrical ketones occurs by cleavage of the $C-C$ bond on either side of the carbonyl to produce a mixture of carboxylic acids].

Final Solution: The compound is Pentan-2-one, with the structure $CH_3COCH_2CH_2CH_3$.

Step 1: Aldol Condensation with Acetone (Claisen-Schmidt Reaction)

To convert Benzaldehyde ($C_6H_5CHO$) to a carbon chain suitable for forming $3$-phenylpropan-$1$-ol, we must first increase the carbon count. Benzaldehyde reacts with acetaldehyde (ethanal) via a crossed-aldol condensation in the presence of dilute $NaOH$ to form cinnamaldehyde ($3$-phenylprop-$2$-enal).

Reaction:
$C_6H_5CHO + CH_3CHO \xrightarrow{dil. NaOH, \Delta} C_6H_5CH=CH-CHO + H_2O$

[Justification: Per the Claisen-Schmidt condensation mechanism, the alpha-hydrogen of the aldehyde is abstracted by the base to form an enolate, which attacks the carbonyl carbon of the benzaldehyde followed by dehydration.]

Step 2: Catalytic Hydrogenation

The resulting $3$-phenylprop-$2$-enal contains both a carbon-carbon double bond and a carbonyl group. To achieve $3$-phenylpropan-$1$-ol, we perform a complete catalytic hydrogenation using hydrogen gas in the presence of a metal catalyst like $Ni$, $Pd$, or $Pt$. This reduces both the alkene and the aldehyde functional groups simultaneously.

Reaction:
$C_6H_5CH=CH-CHO + 2H_2 \xrightarrow{Ni, \Delta} C_6H_5CH_2CH_2CH_2OH$

[Justification: Hydrogenation of unsaturated aldehydes over $Ni$ or $Pd$ catalysts results in the saturation of the $\pi$-bonds. The alkene reduces to an alkane, and the aldehyde reduces to a primary alcohol.]

Summary of Transformation

Final Solution: The conversion is achieved by first reacting benzaldehyde with acetaldehyde in the presence of dilute $NaOH$ to obtain $3$-phenylprop-$2$-enal, followed by complete catalytic hydrogenation using $H_2/Ni$ to yield $3$-phenylpropan-$1$-ol.

Step 1: Friedel-Crafts Acylation of Benzene

To introduce an acetyl group ($-COCH_3$) onto the benzene ring, we perform a Friedel-Crafts acylation. Benzene reacts with acetyl chloride in the presence of an anhydrous Lewis acid catalyst, aluminum chloride ($AlCl_3$).

Reaction: $C_6H_6 + CH_3COCl \xrightarrow{Anhydrous AlCl_3} C_6H_5COCH_3 + HCl$

[Justification: The $AlCl_3$ polarizes the $CH_3COCl$ bond to generate the electrophile $CH_3CO^+$, which then undergoes electrophilic aromatic substitution on the benzene ring.]

Step 2: Nitration of Acetophenone

The acetyl group ($-COCH_3$) is a deactivating group and a meta-directing group due to the electron-withdrawing nature of the carbonyl oxygen and the resonance effect. Consequently, when acetophenone is treated with a nitrating mixture (concentrated $HNO_3$ and concentrated $H_2SO_4$), the nitro group ($-NO_2^+$) is directed to the meta position.

Reaction: $C_6H_5COCH_3 + conc. HNO_3 \xrightarrow{conc. H_2SO_4, 273-283K} m-NO_2-C_6H_4-COCH_3 + H_2O$

[Justification: The $-COCH_3$ group withdraws electron density from the ortho and para positions via the inductive effect and resonance, making the meta position relatively more electron-rich for electrophilic attack.]

Visual Representation of the Conversion

Final Solution: The conversion of Benzene to m-Nitroacetophenone is achieved in two steps: first, Friedel-Crafts acylation with $CH_3COCl/AlCl_3$ to yield acetophenone, followed by nitration with $conc. HNO_3/conc. H_2SO_4$ to yield m-Nitroacetophenone.

Step 1: Analyzing the Parent Carbonyl Compound

The parent molecule is hexan-3-one. The structural formula for hexan-3-one is determined by its IUPAC name:

• "Hexan" indicates a chain of 6 carbon atoms.
• "-3-one" indicates a ketone functional group ($C=O$) located at the third carbon atom.

The condensed structural formula is: $CH_3-CH_2-C(=O)-CH_2-CH_2-CH_3$. [Per IUPAC nomenclature conventions for ketones].

Step 2: Defining the Ketalization Reaction

The formation of an ethylene ketal occurs through the reaction of a ketone with ethylene glycol ($HO-CH_2-CH_2-OH$) in the presence of an acid catalyst (e.g., $p$-toluenesulfonic acid). This is a nucleophilic addition-elimination reaction. [Per the mechanism of cyclic acetal/ketal formation].

The ketone carbonyl oxygen is replaced by the two oxygen atoms of the ethylene glycol, forming a five-membered dioxolane ring fused to the original carbonyl carbon.

Step 3: Constructing the Chemical Structure

The carbonyl carbon (C3 of the hexane chain) becomes the spiro-center of the dioxolane ring. The remaining alkyl groups attached to the carbonyl carbon ($CH_3CH_2-$ and $-CH_2CH_2CH_3$) remain as substituents on this central carbon.

Step 4: Final IUPAC Interpretation

The structure represents 2-ethyl-2-propyl-1,3-dioxolane. This derivative is the cyclic ketal formed by protecting the ketone functionality of hexan-3-one.

Final Solution: The ethylene ketal of hexan-3-one is 2-ethyl-2-propyl-1,3-dioxolane, featuring a five-membered heterocyclic ring containing two oxygen atoms bonded to the original C3 of the hexane chain.

Step 1: Structural Breakdown of the IUPAC Name

To construct the molecule $p,p'$-Dihydroxybenzophenone, we decompose the nomenclature into its structural components:

• Benzophenone: The parent structure consists of two benzene rings ($C_6H_5-$) attached to a central carbonyl group ($C=O$). The base formula is $(C_6H_5)_2C=O$. [Per IUPAC nomenclature for ketones].
• $p,p'$-Dihydroxy: The prefix "$p$" (para) indicates the substitution occurs at the position $4$ relative to the attachment point of the carbonyl group on each respective benzene ring. The apostrophe ($'$) differentiates the two rings.
• Summary of Substituents: Two hydroxyl ($-OH$) groups are located at the $4$ and $4'$ positions of the benzophenone backbone.

Step 2: Molecular Formula Derivation

Starting from the parent benzophenone ($C_{13}H_{10}O$):

1. Replace one hydrogen atom at the $4$-position of the first ring with an $-OH$ group.
2. Replace one hydrogen atom at the $4'$-position of the second ring with an $-OH$ group.
3. Molecular Formula: $C_{13}H_{10}O_3$.

Step 3: Visual Representation

Step 4: Structural Analysis and Connectivity

The structure is symmetric. The hydroxyl groups are positioned para to the carbonyl bridge on both rings, resulting in a planar conformation stabilized by the conjugation between the phenyl rings and the carbonyl group. [Per Valence Bond Theory regarding resonance stabilization in aromatic ketones].

Final Solution: The structure of $p,p'$-Dihydroxybenzophenone is defined by a central carbonyl group bonded to two phenyl rings, each substituted with a hydroxyl group at the $4$ and $4'$ positions, represented by the formula $(HO-C_6H_4)_2C=O$.

Step 1: Chemical Characteristics of Phenol and Benzoic Acid

To distinguish between Phenol ($C_6H_5OH$) and Benzoic acid ($C_6H_5COOH$), we must exploit the difference in their acidic strength and functional group reactivity.

• Phenol: A weak acid ($pK_a \approx 9.95$). It does not react with sodium bicarbonate ($NaHCO_3$) to evolve $CO_2$ gas.
• Benzoic acid: A stronger carboxylic acid ($pK_a \approx 4.20$). It reacts with $NaHCO_3$ to evolve $CO_2$ gas with effervescence [Per the Bronsted-Lowry acid-base theory, a stronger acid displaces a weaker acid/carbonic acid from its salts].

Step 2: The Sodium Bicarbonate Test (Primary Distinction)

Add a saturated solution of sodium bicarbonate ($NaHCO_3$) to both samples individually.

Chemical Equation for Benzoic Acid:

$C_6H_5COOH + NaHCO_3 \rightarrow C_6H_5COONa + H_2O + CO_2 \uparrow$

Step 3: Neutral Ferric Chloride Test (Secondary Confirmation)

Both compounds yield distinct colorations with neutral ferric chloride ($FeCl_3$).

• Phenol: Forms a violet-colored complex [Formation of hexaaquairon(III) phenoxide complex].
• Benzoic Acid: Forms a buff-colored (light brown/tan) precipitate of ferric benzoate.

Step 4: Visualization of Chemical Test Principles

Final Solution: The most definitive test is the Sodium Bicarbonate ($NaHCO_3$) test. Benzoic acid reacts to produce brisk effervescence due to the evolution of $CO_2$ gas, whereas Phenol shows no reaction. Alternatively, the neutral $FeCl_3$ test can be used, where Phenol produces a violet coloration and Benzoic acid produces a buff-colored precipitate.

Given Variables & Initial Setup

To distinguish between Benzoic acid ($C_6H_5COOH$) and Ethyl benzoate ($C_6H_5COOC_2H_5$), we must identify a chemical reaction that occurs specifically with the carboxylic acid group ($-COOH$) but does not occur with the ester functional group ($-COOR$).

Step 1: The Sodium Bicarbonate ($NaHCO_3$) Test

Carboxylic acids are sufficiently acidic to react with aqueous sodium bicarbonate, resulting in the evolution of carbon dioxide gas, which manifests as effervescence. Esters, being neutral compounds, do not react with sodium bicarbonate.

Reaction Mechanism:

$C_6H_5COOH + NaHCO_3 \rightarrow C_6H_5COONa + H_2O + CO_2 \uparrow$

• Observation for Benzoic acid: Brisk effervescence is observed due to the liberation of $CO_2$ gas. [Per the Bronsted-Lowry acid-base theory, carboxylic acids act as proton donors to the bicarbonate anion].
• Observation for Ethyl benzoate: No reaction takes place; the mixture remains clear.

Step 2: Comparative Analysis Table

Step 3: Visual Representation of the Test

Step 4: Alternative Testing (Litmus Test)

Benzoic acid will turn blue litmus paper red, indicating the presence of an acidic proton [$H^+$]. Ethyl benzoate is a neutral ester and will not cause a color change in litmus paper. [Based on the Arrhenius definition of acids].

Final Solution: The Sodium Bicarbonate ($NaHCO_3$) test serves as the most effective method to distinguish the two; Benzoic acid will produce brisk effervescence due to $CO_2$ evolution, whereas Ethyl benzoate will show no reaction.

1. Electronic Structure Analysis

To determine the relative acidity, we must examine the stability of the conjugate bases: the carboxylate ion ($RCOO^-$) and the phenoxide ion ($C_6H_5O^-$).

2. Comparative Stability Analysis

A. Delocalization of Charge: In the carboxylate ion, the negative charge is delocalized over two highly electronegative oxygen atoms [Per the principle of electronegativity]. Because the two oxygen atoms share the negative charge equally, the resonance hybrid is exceptionally stable.

B. Energy Content of Resonating Structures: In the phenoxide ion, while there are more resonating structures (five total), three of these structures place the negative charge on the carbon atoms of the benzene ring. Since carbon is significantly less electronegative than oxygen, these structures are inherently higher in energy and less stable [Per the Electronegativity-Stability Criterion].

3. Visual Representation of Resonating Hybrids

4. Thermodynamic Justification

Acidity is directly proportional to the stability of the conjugate base. By comparing the two ions:

• The carboxylate ion stability is governed by the equivalence of its resonance structures. The negative charge remains on the oxygen atoms, which are more electronegative and thus better able to accommodate the negative charge.
• The phenoxide ion stability is undermined by the non-equivalence of its resonance structures. The delocalization onto carbon atoms involves structures with higher potential energy, making the phenoxide ion less stable than the carboxylate ion.

Consequently, the carboxylate ion is a weaker base (more stable) than the phenoxide ion. By the Brønsted-Lowry Theory, a weaker conjugate base indicates a stronger corresponding parent acid.

Final Solution: Carboxylic acids are stronger acids than phenols because the negative charge in the carboxylate ion is delocalized over two equivalent, highly electronegative oxygen atoms, whereas the negative charge in the phenoxide ion is delocalized onto less electronegative carbon atoms, rendering the phenoxide ion less stable.

Step 1: Structural Expansion and Identification of Principal Functional Group

The given structural formula is: \(CH_{3}CH_{2}COCH(C_{2}H_{5})CH_{2}CH_{2}Cl\).

First, we expand the condensed structural formula to identify the carbon chain and functional groups:

• Principal Functional Group: Ketone (\(>C=O\))
• Substituents: Ethyl group (\(-C_{2}H_{5}\)) and Chloro group (\(-Cl\))

Expanded structure:

      CH3 - CH2 - C(=O) - CH(CH2CH3) - CH2 - CH2Cl

Step 2: Selection of Parent Carbon Chain

According to the IUPAC nomenclature rules, the longest continuous carbon chain containing the principal functional group must be selected [Per IUPAC Rule P-44.1].

• Numbering from left to right: The chain is 7 carbons long (Heptanone).
• Let us map the carbons: \(C^1(H_3) - C^2(H_2) - C^3(=O) - C^4(H) - C^5(H_2) - C^6(H_2) - C^7(H_2Cl)\)
• Note: The ethyl group is attached at carbon 4.

Step 3: Numbering the Parent Chain

Number the chain such that the principal functional group (ketone) receives the lowest possible locant [Per IUPAC Rule P-14.4].

The numbering from left to right is preferred as it assigns the lowest locant to the principal functional group (3 < 5).

Step 4: Assembling the IUPAC Name

Alphabetize the substituents: Chloro (C) comes before Ethyl (E).

• Position of Chloro: 7
• Position of Ethyl: 4
• Ketone location: 3
• Parent chain: Heptane (remove 'e' and add 'one') = Heptan-3-one

Step 5: Final Construction

Combining all elements: 7-chloro-4-ethylheptan-3-one.

Final Solution: 7-chloro-4-ethylheptan-3-one

Step 1: Structural Expansion and Parent Chain Identification

The provided molecular structure is $(CH_{3})_{2}CHCH_{2}CH_{2}CHO$. To identify the IUPAC name, we first expand the condensed structure to visualize the longest continuous carbon chain containing the principal functional group.

Expanded Structure:

Per the IUPAC Rule for Priority of Functional Groups, the aldehyde group ($-CHO$) takes precedence and must be assigned the lowest possible locant, specifically $C-1$.

Step 2: Numbering the Carbon Chain

We number the chain starting from the aldehydic carbon:

• $C-1$: $-CHO$ group carbon
• $C-2$: $CH_{2}$
• $C-3$: $CH_{2}$
• $C-4$: $CH$ (bearing a methyl substituent)
• $C-5$: $CH_{3}$ (terminal carbon of the principal chain)

The longest parent chain consists of $5$ carbon atoms, corresponding to the alkane derivative pentanal. [Per IUPAC nomenclature, the suffix for aldehydes is "-al"].

Step 3: Identification of Substituents and Locants

With the parent chain established as a $5$-carbon chain (pentane derivative), we identify the substituent group attached to the chain:

• A methyl group ($CH_{3}-$) is located at the $C-4$ position.

Step 4: Assembly of the Systematic Name

Following the IUPAC format: (Locant)-(Substituent Name)(Parent Alkane Name)-(Suffix)

Combining these elements results in: $4\text{-methylpentanal}$.

Final Solution: 4-methylpentanal

Step 1: Theoretical Basis of Nucleophilic Addition

The reaction of carbonyl compounds with $HCN$ is a classic nucleophilic addition reaction. The rate-determining step is the nucleophilic attack of the $CN^-$ ion on the electrophilic carbonyl carbon atom. The reactivity of the carbonyl compound is governed by two primary factors:

• Electronic Factors: The presence of electron-donating groups (like alkyl groups) decreases the positive charge density on the carbonyl carbon through the $+I$ (inductive) effect, thereby reducing its electrophilicity [Per the Inductive Effect Theory].
• Steric Factors: The presence of bulky alkyl groups around the carbonyl carbon creates steric hindrance, which obstructs the approach of the $CN^-$ nucleophile to the reaction center [Per Transition State Theory].

Step 2: Analysis of the Given Compounds

Step 3: Sequential Ordering Based on Reactivity

By comparing the substituents on the carbonyl group:

1. Acetaldehyde: Being the least sterically hindered ($H$ and $CH_3$), it is the most reactive toward the $CN^-$ nucleophile.
2. Acetone: Contains two methyl groups; more hindered than acetaldehyde but less than the others.
3. Methyl tert-butyl ketone: The tert-butyl group is significantly larger than a methyl group, drastically reducing the accessibility of the carbonyl carbon.
4. Di-tert-butyl ketone: Contains two bulky tert-butyl groups, which provide extreme steric crowding, making it the least reactive compound in the series.

Step 4: Visualizing Steric Hindrance

Final Solution: The increasing order of reactivity towards $HCN$ is: Di-tert-butyl ketone < Methyl tert-butyl ketone < Acetone < Acetaldehyde.

Step 1: Analyzing the Molecular Transformation

The goal is to convert Ethanal ($CH_3CHO$) into But-2-enoic acid ($CH_3-CH=CH-COOH$). The transformation requires a carbon-carbon bond formation (increasing the carbon chain from $C_2$ to $C_4$) and the oxidation of an aldehyde functional group to a carboxylic acid group.

Step 2: Aldol Condensation (Carbon Chain Extension)

Ethanal contains $\alpha$-hydrogens, which makes it susceptible to Aldol condensation in the presence of a dilute alkali. Two molecules of ethanal condense to form 3-hydroxybutanal ($CH_3-CH(OH)-CH_2-CHO$).

Reaction: $2CH_3CHO \xrightarrow{Dil. NaOH} CH_3-CH(OH)-CH_2-CHO$ [Mechanism: Nucleophilic addition of enolate ion to carbonyl group].

Step 3: Dehydration to Crotonaldehyde

The 3-hydroxybutanal is unstable upon heating; it undergoes dehydration (elimination of water) to produce but-2-enal (crotonaldehyde). This step is facilitated by the stability of the conjugated system formed ($C=C-C=O$).

Reaction: $CH_3-CH(OH)-CH_2-CHO \xrightarrow{\Delta, -H_2O} CH_3-CH=CH-CHO$ [Mechanism: E1cB elimination].

Step 4: Oxidation to But-2-enoic Acid

The final step involves the selective oxidation of the aldehyde group to a carboxylic acid group. Tollen’s reagent ($Ag(NH_3)_2^+$) is chosen because it specifically oxidizes aldehydes without affecting the carbon-carbon double bond.

Reaction: $CH_3-CH=CH-CHO \xrightarrow{[Ag(NH_3)_2]^+OH^-} CH_3-CH=CH-COOH$ [Justification: Tollen's test oxidizes $R-CHO$ to $R-COOH$ while preserving alkenyl unsaturation].

Summary of Synthetic Route

Visual Representation of Molecular Skeleton

Final Solution: The conversion of ethanal to but-2-enoic acid is achieved via an aldol condensation followed by dehydration to crotonaldehyde, and a final selective oxidation using Tollen's reagent ($CH_3-CH=CH-COOH$).

Step 1: Structural Analysis of Phenylacetaldehyde

The chemical structure of Phenylacetaldehyde is $C_6H_5-CH_2-CHO$. To determine its reactivity toward Aldol condensation or the Cannizzaro reaction, we must analyze the presence of $\alpha$-hydrogens.

• The carbonyl group ($-CHO$) is attached to a carbon atom, which is designated as the $\alpha$-carbon.
• In $C_6H_5-CH_2-CHO$, the $\alpha$-carbon is bonded to two hydrogen atoms ($-\alpha-H$).
• [Per the definition of Aldol Condensation]: Compounds containing at least one $\alpha$-hydrogen atom undergo aldol condensation in the presence of dilute alkali.
• [Per the definition of Cannizzaro Reaction]: Compounds lacking $\alpha$-hydrogen atoms undergo self-oxidation and reduction (disproportionation) in the presence of concentrated alkali.

Step 2: Determination of Reaction Type

Since Phenylacetaldehyde possesses $\alpha$-hydrogens, it undergoes Aldol Condensation. It does not undergo the Cannizzaro reaction.

Step 3: Mechanism and Product Formation (Aldol Condensation)

In the presence of dilute $NaOH$, the $\alpha$-hydrogen is abstracted to form an enolate, which then attacks another molecule of Phenylacetaldehyde. The product is 3-hydroxy-2,4-diphenylbutanal.

Step 4: Summary Table of Reactivity

Step 5: Structural Formula of the Product

The product formed via Aldol condensation is 3-hydroxy-2,4-diphenylbutanal. The structural evolution is as follows:

$2 \cdot C_6H_5CH_2CHO \xrightarrow{dil. NaOH} C_6H_5CH_2-CH(OH)-CH(C_6H_5)-CHO$

Final Solution: Phenylacetaldehyde undergoes Aldol condensation due to the presence of $\alpha$-hydrogens. It does not undergo the Cannizzaro reaction. The product of the Aldol condensation is 3-hydroxy-2,4-diphenylbutanal ($C_6H_5CH_2CH(OH)CH(C_6H_5)CHO$).

Step 1: Chemical Structure Identification

The reactant is cyclohexanecarbaldehyde, which consists of a cyclohexane ring ($C_6H_{11}-$) attached to an aldehyde functional group ($-CHO$). The reagent is semicarbazide ($NH_2CONHNH_2$) in the presence of a weak acid ($H^+$ catalyst).

Step 2: Mechanism of Nucleophilic Addition-Elimination

The reaction between an aldehyde and semicarbazide proceeds via a nucleophilic addition-elimination mechanism (specifically, the formation of a semicarbazone). The weak acid is required to protonate the carbonyl oxygen, making the carbonyl carbon more electrophilic, while also facilitating the loss of water in the final step.

• Nucleophilic Attack: The primary amino group ($-NH_2$) of the semicarbazide, specifically the terminal nitrogen of the $-NH_2$ group (not the nitrogen attached to the amide), acts as the nucleophile. [Per the Nucleophilic Addition Mechanism to Carbonyls].
• Proton Transfer: A proton transfer occurs to form a carbinolamine intermediate ($-NH-CH(OH)-$).
• Elimination: Under acidic conditions, the hydroxyl group ($-OH$) is protonated to form $-OH_2^+$, which acts as a good leaving group. A molecule of water ($H_2O$) is eliminated, forming the carbon-nitrogen double bond ($C=N$).

Step 3: Chemical Reaction Equation

The general reaction is: $R-CHO + H_2N-NH-CO-NH_2 \xrightarrow{H^+} R-CH=N-NH-CO-NH_2 + H_2O$

Substituting cyclohexanecarbaldehyde ($R = C_6H_{11}$):

$C_6H_{11}CHO + H_2N-NH-CONH_2 \xrightarrow{H^+} C_6H_{11}CH=N-NH-CONH_2 + H_2O$

Step 4: Visual Representation of Product Formation

Step 5: Structural Summary

Final Solution: The reaction of cyclohexanecarbaldehyde with semicarbazide in the presence of a weak acid yields cyclohexanecarbaldehyde semicarbazone ($C_6H_{11}CH=NNHCONH_2$) and water ($H_2O$).

1. Analysis of the Molecular Structure

The chemical formula provided is $PhCOPh$. This structure consists of a carbonyl group ($C=O$) bonded to two phenyl groups ($C_6H_5-$). In structural terms, the carbon of the carbonyl group acts as the central link between two benzene rings.

2. Determination of Common Name

According to the nomenclature of ketones, when two identical alkyl or aryl groups are attached to the carbonyl group, the common name is formed by naming the groups followed by the suffix "-ketone".

• The two groups are Phenyl ($Ph-$).
• Combining the prefix and suffix yields: Benzophenone. [Per IUPAC and common nomenclature standards for diaryl ketones].

3. Determination of IUPAC Name

The IUPAC system dictates specific rules for ketones involving aromatic rings:

• Parent Chain Selection: The carbonyl group is part of a one-carbon chain attached to two phenyl substituents. Therefore, the parent alkane is methane.
• Suffix Assignment: The ketone functionality is indicated by the suffix "-one". Thus, the parent structure is "methanone".
• Substituent Identification: Two phenyl rings are attached to the central carbon atom.
• Synthesis: By prefixing the substituents to the parent alkane name, we arrive at: Diphenylmethanone.

4. Summary of Nomenclature

Final Solution: The IUPAC name is Diphenylmethanone and the common name is Benzophenone.

Step 1: Conversion of Benzaldehyde to Mandelic Acid Nitrile (Cyanohydrin)

To convert Benzaldehyde ($C_6H_5CHO$) into $\alpha$-Hydroxyphenylacetic acid (also known as Mandelic acid), the first step involves a nucleophilic addition reaction with hydrogen cyanide ($HCN$).

Reaction: $C_6H_5CHO + HCN \rightarrow C_6H_5CH(OH)CN$

Mechanism: The nucleophilic cyanide ion ($CN^-$), generated from $NaCN/HCl$, attacks the electrophilic carbonyl carbon of the benzaldehyde. The carbonyl oxygen is subsequently protonated to form a cyanohydrin [Per Nucleophilic Addition Mechanism].

Step 2: Acidic Hydrolysis of Cyanohydrin

The resulting mandelonitrile ($C_6H_5CH(OH)CN$) is subjected to complete acid-catalyzed hydrolysis. This converts the nitrile ($-CN$) functional group into a carboxylic acid ($-COOH$) group.

Reaction: $C_6H_5CH(OH)CN + 2H_2O + H^+ \rightarrow C_6H_5CH(OH)COOH + NH_4^+$

Justification: Nitriles undergo hydrolysis in the presence of strong aqueous acids (e.g., $H_2SO_4$ or $HCl$) to form amides as intermediates, which are further hydrolyzed to carboxylic acids [Per Nitrile Hydrolysis Protocol].

Final Solution: The conversion is achieved in two steps: first, nucleophilic addition of HCN to Benzaldehyde to form mandelonitrile, followed by the acid-catalyzed hydrolysis of the nitrile group to yield $\alpha$-Hydroxyphenylacetic acid ($C_6H_5CH(OH)COOH$).

Step 1: Identifying the Parent Alkane and Functional Group

According to the IUPAC nomenclature rules, the suffix "-al" indicates the presence of an aldehyde functional group ($-CHO$). The parent chain is identified by the prefix "butan-", which denotes a carbon chain length of $n=4$ atoms. [Per IUPAC Nomenclature of Organic Compounds].

Step 2: Determining the Carbon Skeleton and Substituents

The name "3-Methylbutanal" implies the following structure:

• Parent Chain: Butanal ($C_1-C_2-C_3-C_4$).
• Functional Group Position: In aldehydes, the carbonyl carbon is designated as $C_1$.
• Substituent: A methyl group ($-CH_3$) is attached at the $C_3$ position.

Step 3: Constructing the Structural Formula

The numbering proceeds as follows:

1. $C_1$: Carbonyl carbon of the aldehyde group ($CHO$).
2. $C_2$: $CH_2$ group.
3. $C_3$: $CH$ group bonded to a $CH_3$ substituent.
4. $C_4$: $CH_3$ terminal group.

Visual Representation

Chemical Composition Table

Final Solution: The structural formula for 3-Methylbutanal is $CH_3-CH(CH_3)-CH_2-CHO$.

Theoretical Basis: Fischer Esterification Equilibrium

The preparation of an ester from a carboxylic acid and an alcohol is a reversible chemical process known as Fischer Esterification. The chemical equation for the reaction is as follows:

$RCOOH + R'OH \xrightleftharpoons[H^+]{} RCOOR' + H_2O$

This reaction exists in a state of dynamic equilibrium where the rate of the forward reaction (esterification) is equal to the rate of the backward reaction (hydrolysis of the ester) [Per the Law of Mass Action].

Step 1: Application of Le Chatelier’s Principle

According to Le Chatelier’s Principle, if a dynamic equilibrium is disturbed by changing the conditions (concentration, pressure, or temperature), the position of equilibrium shifts to counteract the change.

In this system, the equilibrium constant ($K_c$) is defined as:

$K_c = \frac{[RCOOR'][H_2O]}{[RCOOH][R'OH]}$

If the concentration of the products ($RCOOR'$ or $H_2O$) is reduced, the quotient of the expression becomes smaller than $K_c$. To restore equilibrium, the system must increase the concentration of the products by favoring the forward reaction [By the Principle of Dynamic Equilibrium].

Step 2: Mechanistic Necessity for Product Removal

By removing the water or the ester as soon as it is formed, the chemical system is prevented from reaching a state of equilibrium where significant amounts of reactants remain. The removal acts as a persistent stress on the system, forcing the reaction to continuously produce more ester to satisfy the equilibrium constant. This is a common industrial technique to maximize the Percentage Yield of the desired product.

Visual Representation of Equilibrium Displacement

Conclusion

The removal of products effectively prevents the reverse hydrolysis reaction ($RCOOR' + H_2O \rightarrow RCOOH + R'OH$) from occurring. Consequently, the forward reaction proceeds toward completion, significantly increasing the conversion efficiency of the alcohol and acid into the ester.

Final Solution: The preparation of esters is a reversible equilibrium process. By removing the water or ester as it forms, the concentration of the products is decreased, which—in accordance with Le Chatelier’s Principle—shifts the equilibrium toward the forward direction, thereby increasing the yield of the ester.

Step 1: Structural Breakdown of the IUPAC Name

The name p,p’-dihydroxybenzophenone provides the following chemical information:

• Benzophenone: The parent structure consists of two phenyl rings linked by a carbonyl group ($C=O$). This is represented as $Ph-CO-Ph$.
• p,p’-Dihydroxy: The prefix "p" stands for para, indicating the position of the hydroxyl ($-OH$) groups relative to the carbonyl bridge. In a monosubstituted benzene, positions 2 and 6 are ortho, 3 and 5 are meta, and 4 is para.
• Logic: We position the hydroxyl groups at the 4 and 4' positions of the two phenyl rings, located opposite to the carbonyl attachment point on each ring.

Step 2: Molecular Geometry and Connectivity

The molecular formula is derived from the connectivity: $(HO-C_6H_4)_2C=O$.

Step 3: Visual Representation

Step 4: Theoretical Justification

The structure follows the principles of valence bond theory. Each benzene ring maintains its aromaticity ($4n+2$ Huckel's rule). The carbonyl carbon is $sp^2$ hybridized, resulting in a trigonal planar geometry, which dictates the $120^\circ$ bond angles around the central carbon atom [Per VSEPR Theory]. The para-substitution ensures maximal resonance stabilization and spatial distance between the two hydroxyl groups.

Final Solution: The structure of $p,p’$-dihydroxybenzophenone is composed of two $4$-hydroxyphenyl groups attached to a central carbonyl carbon atom, represented by the formula $(HO-C_6H_4)_2CO$.

Step 1: Structural Analysis and Functional Group Identification

The given molecular structure is $CH_{3}-CH=CH-CHO$. To identify the IUPAC name, we must perform the following analysis:

• Principal Functional Group: The molecule contains an aldehyde group ($-CHO$). Per IUPAC nomenclature rules, the aldehyde carbon must be assigned the lowest possible locant, specifically $C-1$. [Per IUPAC Blue Book P-66.1.1.1.1]
• Carbon Chain Length: Counting the carbons starting from the aldehyde group:
  • $C-1$: $-CHO$ (Aldehyde carbon)
  • $C-2$: $-CH=$
  • $C-3$: $=CH-$
  • $C-4$: $-CH_{3}$
• Parent Alkane: The longest chain consists of 4 carbon atoms, corresponding to the alkane "butane".
• Unsaturation: There is a carbon-carbon double bond between $C-2$ and $C-3$.

Step 2: IUPAC Nomenclature Derivation

We assemble the name based on the identified components:

Step 3: Constructing the Final IUPAC Name

By combining the components, we get but-2-enal. [Per IUPAC rule P-44.4.1: The principal functional group suffix must be added to the unsaturated parent chain name].

Visual Representation of the Molecule

Final Solution: The IUPAC name of $CH_{3}CH=CHCHO$ is But-2-enal.