Reaction (A) involves the hydrolysis of a nitrile. Under mild conditions, the hydrolysis does not proceed to completion and stops at the amide stage.

$$\mathrm{R-C\equiv N \xrightarrow[\text{mild conditions}]{H^+/H_2O} R-CONH_2}$$

Therefore, the major product is an amide, not a carboxylic acid.

Reaction (B) is the carboxylation of a Grignard reagent. The Grignard reagent reacts with carbon dioxide to form a magnesium carboxylate, which upon acidic hydrolysis gives a carboxylic acid.

$$\mathrm{R-MgX + CO_2 \rightarrow RCOO^-MgX^+ \xrightarrow{H_3O^+} RCOOH}$$

Hence, reaction (B) produces a carboxylic acid.

Reaction (C) is the Stephen reduction of a nitrile. Treatment with $$\mathrm{SnCl_2/HCl}$$ followed by hydrolysis converts the nitrile into an aldehyde.

$$\mathrm{R-C\equiv N \xrightarrow{SnCl_2/HCl} \xrightarrow{H_3O^+} R-CHO}$$

Therefore, the major product is an aldehyde, not a carboxylic acid.

Reaction (D) involves oxidation of a primary alcohol using PCC. Since PCC is a mild oxidizing agent, it converts primary alcohols only into aldehydes.

$$\mathrm{R-CH_2OH \xrightarrow{PCC} R-CHO}$$

Thus, no carboxylic acid is formed.

In reaction (E), benzoyl chloride first undergoes Rosenmund reduction to form benzaldehyde.

$$\mathrm{Ph-COCl \xrightarrow{H_2/Pd-BaSO_4} Ph-CHO}$$

The benzaldehyde is then oxidized by bromine water to give benzoic acid.

$$\mathrm{Ph-CHO \xrightarrow{Br_2/H_2O} Ph-COOH}$$

Therefore, reaction (E) also produces a carboxylic acid.

Hence, only reactions (B) and (E) yield a carboxylic acid as the major product.

Therefore, the correct answer is B and E only, which corresponds to option D.

The compounds are:

• $$p:\ CH_3CH_2COCl$$ (acid chloride)
• $$q:\ CH_3CH_2CO-O-COCH_3$$ (acid anhydride)
• $$r:\ CH_3CH_2COOCH_2CH_3$$ (ester)
• $$s:\ CH_3CH_2CONH_2$$ (amide)

The rate of hydrolysis of carboxylic acid derivatives depends on the leaving group attached to the carbonyl carbon.

Greater the leaving ability of the group, faster is the hydrolysis.

The leaving groups in the given compounds are:

• In $$p$$, leaving group is $$Cl^-$$
• In $$q$$, leaving group is $$RCOO^-$$
• In $$r$$, leaving group is $$RO^-$$
• In $$s$$, leaving group is $$NH_2^-$$

The leaving group ability follows:

$$Cl^- > RCOO^- > RO^- > NH_2^-$$

Hence, the ease of hydrolysis follows:

$$\text{Acid chloride} > \text{Acid anhydride} > \text{Ester} > \text{Amide}$$

Therefore,

$$p > q > r > s$$

Hence, the correct order of rate of hydrolysis is

$$\boxed{p > q > r > s}$$

Therefore, the correct answer is Option (D).

We need to determine the molecular formula of the second homologue in the homologous series of monocarboxylic acids.

Key Concept: A homologous series is a group of organic compounds sharing the same functional group and general chemical properties, where each successive member differs from the previous one by a methylene ($$\text{-}CH_2\text{-}$$) unit. The general formula for the homologous series of open-chain monocarboxylic acids is $$C_nH_{2n}O_2$$ (where $$n \ge 1$$).

Step 1: Finding the first member of the series ($$n = 1$$)

The first homologue contains one carbon atom. This compound is methanoic acid (formic acid), having the structural formula $$HCOOH$$. Combining the atoms gives its molecular formula:

$$CH_2O_2$$

Step 2: Finding the second member of the series ($$n = 2$$)

The second homologue is formed by adding a $$\text{-}CH_2\text{-}$$ group to the first member, meaning it contains two carbon atoms ($$n = 2$$). This compound is ethanoic acid (acetic acid), having the structural formula $$CH_3COOH$$.

Step 3: Writing the final molecular formula

Counting the total number of carbon, hydrogen, and oxygen atoms in $$CH_3COOH$$ gives the molecular formula:

$$C_2H_4O_2$$

Therefore, the molecular formula of the second homologue is $$C_2H_4O_2$$, which corresponds to Option B.

Answer: Option B — $$C_2H_4O_2$$

We need to determine in which medium the equilibrium $$Cr_2O_7^{2-} \rightleftharpoons 2CrO_4^{2-}$$ shifts to the right.

Write the complete ionic equilibrium

The interconversion between dichromate and chromate ions in aqueous solution involves $$H^+$$ and $$OH^-$$ ions. The complete equation is:

$$Cr_2O_7^{2-} + 2OH^- \rightleftharpoons 2CrO_4^{2-} + H_2O$$

Alternatively, in terms of $$H^+$$:

$$2CrO_4^{2-} + 2H^+ \rightleftharpoons Cr_2O_7^{2-} + H_2O$$

Apply Le Chatelier's Principle

The forward reaction (dichromate to chromate) consumes $$OH^-$$ ions. According to Le Chatelier's Principle:

- In a basic medium (excess $$OH^-$$): The high concentration of $$OH^-$$ drives the equilibrium to the right, favouring the formation of yellow $$CrO_4^{2-}$$ (chromate) ions.

- In an acidic medium (excess $$H^+$$): The $$OH^-$$ concentration is low, so the equilibrium shifts to the left, favouring the orange $$Cr_2O_7^{2-}$$ (dichromate) ions.

Conclusion

This is why dichromate solutions appear orange in acidic conditions and turn yellow (chromate) in basic conditions. The equilibrium shifts to the right in a basic medium.

The correct answer is Option (2): a basic medium.

First note the atomic numbers: $$Sc = 21$$, $$Ti = 22$$, $$V = 23$$, $$Cr = 24$$, $$Mn = 25$$.

Use the Aufbau order $$\text{4s} \rightarrow \text{3d}$$ to write the ground-state electronic configurations:

$$Sc\;(21):\;[Ar]\,3d^{1}\,4s^{2}$$
$$Ti\;(22):\;[Ar]\,3d^{2}\,4s^{2}$$
$$V\;(23):\;[Ar]\,3d^{3}\,4s^{2}$$
$$Cr\;(24):\;[Ar]\,3d^{5}\,4s^{1}$$  (half-filled 3d subshell gives extra stability)
$$Mn\;(25):\;[Ar]\,3d^{5}\,4s^{2}$$

Count the unpaired electrons in each case (Hund’s rule: every degenerate orbital gets one electron before any pairing starts):

Sc : one unpaired electron in $$3d$$ ⇒ $$1$$ unpaired
Ti  : two unpaired electrons in $$3d$$ ⇒ $$2$$ unpaired
V  : three unpaired electrons in $$3d$$ ⇒ $$3$$ unpaired
Cr : five unpaired in $$3d$$ + one in $$4s$$ ⇒ $$6$$ unpaired
Mn : five unpaired in $$3d$$ (the $$4s^{2}$$ pair is paired) ⇒ $$5$$ unpaired

Arrange in increasing order of the number of unpaired electrons:

$$1 \lt 2 \lt 3 \lt 5 \lt 6$$ corresponds to $$Sc \lt Ti \lt V \lt Mn \lt Cr$$.

Using the given symbols: $$(A) \lt (D) \lt (C) \lt (E) \lt (B)$$.

Therefore the correct option is Option C.

The acidity of a hydrogen atom depends on the stability of the conjugate base formed after its removal. Greater stabilization of the resulting carbanion leads to higher acidity.

In the alpha-diketone and gamma-diketone structures, the acidic hydrogens are influenced by only one carbonyl group at a time or are too far from the second carbonyl group to experience significant stabilization.

In the monoketone, the alpha hydrogen is stabilized by resonance with only a single carbonyl group. Therefore, its acidity is moderate and comparable to that of ordinary ketones.

In the beta-diketone structure, the central $$-CH_2-$$ group is located between two carbonyl groups. When a proton is removed from this carbon, the resulting negative charge is delocalized through resonance over both adjacent carbonyl groups.

This extensive resonance stabilization distributes the negative charge onto both oxygen atoms, making the conjugate base exceptionally stable. The combined electron-withdrawing effect of the two carbonyl groups greatly enhances the acidity of the central methylene hydrogens.

Therefore, the hydrogen atoms of the active methylene group present in the beta-diketone are the most acidic among the given compounds.

Hence, the correct answer is, $$\text{The }\beta\text{-diketone compound (Option D)}$$

We need to count how many of the given elements do not belong to the lanthanoid series.

Key Concept: The lanthanoid series consists of 15 elements from Lanthanum (La, Z = 57) to Lutetium (Lu, Z = 71). These elements are characterised by the progressive filling of the 4f orbitals.

Checking each element:

Eu (Europium, Z = 63): Falls between La (57) and Lu (71). Configuration: [Xe] 4f$$^7$$ 6s$$^2$$. This is a lanthanoid. ✓

Cm (Curium, Z = 96): Curium has atomic number 96, which falls in the actinoid series (Ac, Z = 89 to Lr, Z = 103). Configuration: [Rn] 5f$$^7$$ 6d$$^1$$ 7s$$^2$$. This is an actinoid, NOT a lanthanoid. ✗

Er (Erbium, Z = 68): Falls between La (57) and Lu (71). Configuration: [Xe] 4f$$^{12}$$ 6s$$^2$$. This is a lanthanoid. ✓

Tb (Terbium, Z = 65): Falls between La (57) and Lu (71). Configuration: [Xe] 4f$$^9$$ 6s$$^2$$. This is a lanthanoid. ✓

Yb (Ytterbium, Z = 70): Falls between La (57) and Lu (71). Configuration: [Xe] 4f$$^{14}$$ 6s$$^2$$. This is a lanthanoid. ✓

Lu (Lutetium, Z = 71): This is the last element of the lanthanoid series. Configuration: [Xe] 4f$$^{14}$$ 5d$$^1$$ 6s$$^2$$. This is a lanthanoid. ✓

Conclusion: Only Curium (Cm) does not belong to the lanthanoids. The number of non-lanthanoid elements is 1.

The correct answer is Option 3: 1.

Cu(II) has configuration $$3d^9$$ and Cu(I) has $$3d^{10}$$.

Although Cu(I) has a fully filled $$3d^{10}$$ configuration (which is stable), Cu(II) is actually more stable in aqueous solution. This is because the higher hydration enthalpy of Cu²⁺ (due to its smaller size and higher charge) more than compensates for the second ionization energy required.

The high hydration energy of Cu²⁺ stabilizes it in aqueous solutions. Cu(I) compounds tend to disproportionate in water: $$2Cu^+ \rightarrow Cu + Cu^{2+}$$.

The correct answer is Option 2: Cu(II) is more stable.

We need to evaluate both statements about manganese compounds.

First, fusion of $$MnO_2$$ with KOH and an oxidising agent gives dark green $$K_2MnO_4$$ by a standard preparation method for potassium manganate.

$$2MnO_2 + 4KOH + O_2 \rightarrow 2K_2MnO_4 + 2H_2O$$

Here, $$MnO_2$$ (Mn in +4 state) is oxidised to $$K_2MnO_4$$ (Mn in +6 state) and the product is indeed dark green in colour. Therefore, the first statement is true.

Next, the manganate ion $$MnO_4^{2-}$$ can be oxidised to the permanganate ion $$MnO_4^{-}$$ by electrolytic oxidation in an alkaline solution.

$$MnO_4^{2-} \rightarrow MnO_4^{-} + e^{-}$$

This is a well-known method for preparing $$KMnO_4$$ from $$K_2MnO_4$$, so the second statement is also true.

Since both statements are true, the correct answer is Option 4: Both Statement I and Statement II are true.

Among the given transition metals, we need to find which shows the highest and maximum number of oxidation states.

Manganese (Mn): Electronic configuration [Ar] $$3d^5 4s^2$$. It shows oxidation states from +2 to +7 (total 6 different oxidation states). The maximum oxidation state is +7 (in $$KMnO_4$$), which equals the total number of 3d + 4s electrons (5 + 2 = 7).

Iron (Fe): [Ar] $$3d^6 4s^2$$. Shows up to +6 (rare). Common: +2, +3.

Cobalt (Co): [Ar] $$3d^7 4s^2$$. Shows up to +4. Common: +2, +3.

Titanium (Ti): [Ar] $$3d^2 4s^2$$. Shows up to +4.

Mn shows the highest maximum oxidation state (+7) and the maximum number of oxidation states among the 3d transition metals.

The correct answer is Option (2): Mn.

1.Formation of Product A

• Reaction: Benzene undergoes nitration using $$\text{conc. HNO}_3 / \text{conc. H}_2\text{SO}_4$$ to form nitrobenzene.
• Reduction: Nitrobenzene is then reduced by $$\text{Sn/HCl}$$ to convert the nitro group ($$-\text{NO}_2$$) into an amino group ($$-\text{NH}_2$$).
• Product A: Aniline ($$\text{C}_6\text{H}_5\text{NH}_2$$).

2. Formation of Product B

• Diazotization: Aniline reacts with $$\text{NaNO}_2 + \text{HCl}$$ at a low temperature ($$0\text{-}5^\circ\text{C}$$ / $$273\text{-}278\text{K}$$) to form the stable benzene diazonium chloride salt ($$\text{C}_6\text{H}_5\text{N}_2^+\text{Cl}^-$$).
• Azo Coupling: The diazonium ion acts as an electrophile and couples with phenol at its highly reactive, less sterically hindered para-position.
• Product B: $$p$$-Hydroxyazobenzene (an orange azo dye).

We need to find the spin-only magnetic moment of the species with the least oxidising ability among $$VO_2^+$$, $$MnO_4^-$$, and $$Cr_2O_7^{2-}$$.

Oxidising ability is related to the ease with which the species can be reduced. The higher the reduction potential, the stronger the oxidising agent. Among common oxidising agents in acidic medium: $$MnO_4^- > Cr_2O_7^{2-} > VO_2^+$$ in terms of oxidising strength.

The species with the least oxidising ability is $$VO_2^+$$.

Find the oxidation state and d-electron configuration of V in $$VO_2^+$$:

Let the oxidation state of V be $$x$$. Then $$x + 2(-2) = +1$$, so $$x = +5$$.

Vanadium (Z = 23) has the configuration [Ar]3d$$^3$$4s$$^2$$. V$$^{5+}$$ loses all 5 valence electrons, giving [Ar] = 3d$$^0$$.

Calculate the spin-only magnetic moment using the formula:

$$\mu = \sqrt{n(n+2)} \, \text{BM}$$

where $$n$$ is the number of unpaired electrons. Since V$$^{5+}$$ has d$$^0$$ configuration, $$n = 0$$:

$$\mu = \sqrt{0(0+2)} = 0 \, \text{BM}$$

The answer is 0.

CrO: Cr²⁺ (d⁴), basic oxide. μ = √(4×6) = √24 = 4.90 BM.

Cr₂O₃: Cr³⁺ (d³), amphoteric oxide. μ = √(3×5) = √15 = 3.87 BM.

CrO₃: Cr⁶⁺ (d⁰), acidic oxide. μ = 0 BM.

Sum of basic and amphoteric = 4.90 + 3.87 = 8.77 BM = 877 × 10⁻² BM.

The answer is 877.

The metallic radius decreases steadily from Sc to Cr in the 3d series because the increasing nuclear charge is not fully screened by the added 3d electrons. After Cr the radius stops decreasing and even rises slightly. Therefore, among the given elements (Sc, Ti, V, Cr, Mn, Zn) the one with the smallest metallic radius is chromium (Cr).

Hence in $$MO_4^{2-}$$ the metal atom is chromium, giving the anion $$CrO_4^{2-}$$ (chromate ion).

Each oxide ion carries a charge of $$-2$$, so for $$CrO_4^{2-}$$ the overall charge balance is
$$x + 4(-2) = -2 \;\;\Longrightarrow\;\; x = +6$$
Thus chromium is in the $$+6$$ oxidation state.

Electronic configuration of a neutral Cr atom: $$[Ar]\,3d^5\,4s^1$$.
To obtain $$Cr^{+6}$$ we remove six electrons (the one 4s electron and five 3d electrons), leaving
$$Cr^{+6}: [Ar]\,3d^{0}$$.

Number of unpaired electrons, $$n = 0$$.

The spin-only magnetic moment is calculated by the formula
$$\mu_{\text{spin}} = \sqrt{n(n+2)}\;\text{BM}$$.

Substituting $$n = 0$$ gives
$$\mu_{\text{spin}} = \sqrt{0(0+2)} = 0\;\text{BM}$$.

Hence the ‘spin only’ magnetic moment of $$MO_4^{2-}$$ is $$\mathbf{0\;BM}$$.

Among the ions $$Ti^{2+}, V^{2+}, Co^{3+}, Cr^{2+}$$, the one that acts as a strong oxidising agent in aqueous solution is $$Co^{3+}$$. A strong oxidising agent easily gets reduced (accepts electrons), and $$Co^{3+}$$ readily accepts an electron to form the more stable $$Co^{2+}$$. The other ions ($$Ti^{2+}, V^{2+}, Cr^{2+}$$) are actually strong reducing agents.

The electronic configuration of cobalt is Co: [Ar] $$3d^7 4s^2$$, so for $$Co^{3+}$$ it becomes [Ar] $$3d^6$$. In aqueous solution, the complex $$[Co(H_2O)_6]^{3+}$$ behaves as a high-spin complex (since water is a weak field ligand), giving the configuration $$t_{2g}^4 e_g^2$$ with 4 unpaired electrons.

Using the spin-only formula for the magnetic moment, we have $$\mu = \sqrt{n(n+2)} = \sqrt{4(4+2)} = \sqrt{24} = 4.9 \approx 5 \text{ BM}$$.

Final answer: $$Co^{3+}$$ is the strong oxidising agent and its spin-only magnetic moment is 5 BM.

Colourless lanthanoid ions are those with $$4f^0$$ or $$4f^{14}$$ configurations (no unpaired f-electrons to undergo f-f transitions).

$$Eu^{3+}$$: $$[Xe]4f^6$$ — coloured ✗

$$Lu^{3+}$$: $$[Xe]4f^{14}$$ — colourless ✓

$$Nd^{3+}$$: $$[Xe]4f^3$$ — coloured ✗

$$La^{3+}$$: $$[Xe]4f^0$$ — colourless ✓

$$Sm^{3+}$$: $$[Xe]4f^5$$ — coloured ✗

Number of colourless ions = 2 ($$Lu^{3+}$$ and $$La^{3+}$$).

The answer is $$\boxed{2}$$.

KMnO₄: Mn⁷⁺ (d⁰), μ = 0 BM.

In acidic medium titration with oxalic acid: KMnO₄ → Mn²⁺. Mn²⁺ (d⁵), μ = √(5×7) = √35 = 5.92 ≈ 6 BM.

Difference = |0 - 6| = 6 BM.

The answer is 6.

Molecule (a), acetylacetone, is the most acidic because it lacks the electron-donating alkoxy groups found in the esters. Therefore, it will undergo deprotonation most readily in a basic medium.

We need to arrange the given carboxylic acids in descending order of acidity: (A) CH$$_3$$COOH, (B) F$$_3$$C-COOH, (C) ClCH$$_2$$-COOH, (D) FCH$$_2$$-COOH, and (E) BrCH$$_2$$-COOH.

The acidity of carboxylic acids is enhanced by electron-withdrawing groups through the negative inductive effect (-I effect). The stronger the -I effect, the more the electron density is pulled away from the O-H bond, making it easier to release the proton.

Trifluoroacetic acid (B) has three fluorine atoms, providing a very strong cumulative -I effect, making it the strongest acid in the group.

Now, among the monohaloacetic acids, we compare based on electronegativity: F (3.98) > Cl (3.16) > Br (2.96). Since all three halogens are at the same position (alpha carbon), the -I effect follows the electronegativity order, giving FCH$$_2$$COOH (D) > ClCH$$_2$$COOH (C) > BrCH$$_2$$COOH (E).

CH$$_3$$COOH (A) has no electron-withdrawing substituent, making it the weakest acid.

So the descending order of acidity is:

$$\text{B} > \text{D} > \text{C} > \text{E} > \text{A}$$

Hence, the correct answer is Option C.

When this compound reacts with water ($$H_2O$$), it undergoes a nucleophilic addition to the most electrophilic carbonyl group (the central one, which is destabilized by the adjacent electron-withdrawing carbonyl groups).The major product 'D' is Ninhydrin (2,2-dihydroxyindane-1,3-dione).

The reaction takes place under acidic conditions and involves the condensation of pivalaldehyde with an $$\alpha$$-hydroxy acid (2-hydroxyisobutyric acid).

The carbonyl oxygen of the aldehyde is first protonated by $$H^+$$, increasing the electrophilicity of the carbonyl carbon.

The hydroxyl group $$(-OH)$$ of the $$\alpha$$-hydroxy acid attacks the protonated carbonyl carbon, forming a hemiacetal intermediate.

Subsequently, the carboxyl group participates in intramolecular cyclization. The oxygen of the $$-COOH$$ group attacks the hemiacetal carbon, followed by elimination of water, producing a stable five-membered cyclic product.

The resulting product is a cyclic dioxolanone, in which:

• One oxygen atom originates from the hydroxyl group.

• The second oxygen atom originates from the carboxyl group.

• The carbonyl $$\left(C=O\right)$$ of the original carboxylic acid is retained within the ring.

Among the given structures:

• Option A represents a cyclic acetal and lacks the required carbonyl group.

• Option B represents the correct five-membered cyclic dioxolanone.

• Option C is an open-chain ester and is not the major product under these conditions.

• Option D does not correspond to the expected cyclized product.

Hence, the major product formed is the cyclic dioxolanone shown in

$$\boxed{\text{Option B}}$$

The starting material contains two key carbonyl-based functional groups attached to the benzene ring via aliphatic chains:

• A carboxylic acid group ($$-\text{CO}_2\text{H}$$)
• An ester group ($$-\text{CO}_2\text{Et}$$)

Correct Option B

Aldol Addition: Isobutyraldehyde reacts with Formaldehyde in the presence of KOH. Since Isobutyraldehyde has only one alpha-hydrogen, it adds one HCHO unit to form 2,2-dimethyl-3-hydroxypropanal.

(CH3)2CHCHO + HCHO  ->  HO-CH2-C(CH3)2-CHO

2. Cyanohydrin Formation: The aldehyde group reacts with $KCN$ to form a cyanohydrin.

$$HO-CH_2-C(CH_3)_2-CHO -> HO-CH2-C(CH3)2-CH(OH)CN

3. Hydrolysis & Cyclization: Acidic hydrolysis  converts the $-CN$ to $-COOH$. The gamma-hydroxy group then attacks the carboxylic acid to form the stable 5-membered Lactone (Y).

1. Step 1: Chlorination with Chlorinating Agent ($$\text{SOCl}_2$$)

   The starting material contains a carboxylic acid group ($$\text{--COOH}$$) and an isolated carbon-carbon double bond ($$\text{C=C}$$). Thionyl chloride ($$\text{SOCl}_2$$) selectively converts the hydroxyl group of the carboxylic acid into an acid chloride (acyl chloride) functional group without affecting the alkene:

2. Step 2: Nucleophilic Acyl Substitution with Amine ($$\text{R--NH}_2$$)

   The highly reactive acid chloride intermediate undergoes nucleophilic attack by the primary amine ($$\text{R--NH}_2$$). Elimination of a chloride ion yields a secondary amide:




3. Step 3: Reduction using Lithium Aluminum Hydride ($$\text{LiAlH}_4$$)

   Lithium aluminum hydride ($$\text{LiAlH}_4$$) is a powerful reducing agent that selectively reduces the carbonyl group ($$\text{C=O}$$) of an amide into a methylene group ($$\text{--CH}_2\text{--}$$), converting the amide function into a secondary amine. Crucially, $$\text{LiAlH}_4$$ does not reduce isolated non-conjugated alkene double bonds:

Why Other Options Are Incorrect:

• Intermediate Structures (e.g., Option B):

  Structures containing the intact carbonyl group ($$\text{--CONH--R}$$) merely represent the intermediate formed after Step 2 and fail to account for the final hydride reduction step.




• Alcohol-Containing Structures (e.g., Option A):

  Carboxylic amides undergo clean reduction to form amines, completely removing the carbonyl oxygen atom rather than stopping at a hemiaminal or an alcohol-substituted chain like $$\text{--CH(OH)--NH--R}$$.




• Alkene Reduction Products:

  Any option where the carbon-carbon double bond ($$\text{C=C}$$) is hydrogenated to a single bond ($$\text{--CH}_2\text{--CH}_2\text{--}$$) is incorrect because $$\text{LiAlH}_4$$ is a nucleophilic reducing agent that cannot deliver hydrides to non-polarized, isolated carbon-carbon double bonds.

Conclusion:

The complete sequence sequentially transforms the terminal carboxylic acid into an amine substituent group ($$\text{--CH}_2\text{--NH--R}$$) while leaving the rest of the unsaturated hydrocarbon framework entirely intact.

Answer: The option displaying the amine product with the double bond intact ($$\text{Ar--CH}_2\text{--CH=CH--CH}_2\text{--CH}_2\text{--NH--R}$$).

Soap is prepared by saponification, which is the alkaline hydrolysis of fats/oils with NaOH or KOH.

Fat + NaOH → Soap (sodium salt of fatty acid) + Glycerol

The correct answer is Option 3: Alkaline hydrolysis reaction.

To determine the decreasing order of acidic strength for the given substituted phenols, we must evaluate how the different substituents affect the stability of the phenoxide ion formed after the loss of a proton ($$H^+$$). As a general rule, electron-withdrawing groups stabilize the negative charge on the phenoxide ion through inductive ($$-I$$) or resonance ($$-R$$) effects, thereby increasing acidity. Conversely, electron-donating groups destabilize the ion by pushing electron density toward the oxygen through positive inductive ($$+I$$) or resonance ($$+R$$) effects, which decreases acidity.

The nitro group ($$-NO_2$$) is a powerful electron-withdrawing group that can exert both $$-I$$ and $$-R$$ effects. In compound (A), p-nitrophenol, the nitro group is located at the para position, allowing both the $$-I$$ and strong $$-R$$effects to heavily stabilize the phenoxide ion, making it the most acidic. In compound (B), m-nitrophenol, the nitro group is at the meta position. Because resonance effects cannot operate from the meta position, only the $$-I$$ effect stabilizes the anion. This makes it highly acidic, but slightly less so than the para isomer.

The methoxy group ($$-OCH_3$$), on the other hand, exerts an electron-withdrawing $$-I$$ effect but a much stronger electron-donating $$+R$$ effect. In compound (C), m-methoxyphenol, the methoxy group is at the meta position where its $$+R$$ effect cannot operate. Therefore, only its mild $$-I$$ effect influences the ring, slightly stabilizing the phenoxide ion. In compound (D), p-methoxyphenol, the methoxy group is at the para position. Here, its dominant electron-donating $$+R$$ effect actively pushes electron density into the ring, highly destabilizing the phenoxide ion and making it the least acidic of all four options.

By comparing the relative stabilization and destabilization caused by these groups at their respective positions, we can conclude the final sequence. The strongest stabilization occurs in compound (A), followed by (B), then a much weaker stabilization in (C), and finally an active destabilization in (D). Therefore, the correct decreasing order of acidic strength is (A) > (B) > (C) > (D).

To convert benzoic acid directly into benzaldehyde, the reagent must reduce the carboxylic acid only up to the aldehyde stage.

$$\mathrm{LiAlH_4}$$ is a strong reducing agent and over-reduces benzoic acid to:

$$\mathrm{Benzyl\ Alcohol}$$

Hence, it is unsuitable.

$$\mathrm{NaBH_4}$$ is a mild reducing agent and does not reduce carboxylic acids.

$$\mathrm{KMnO_4}$$ is a strong oxidising agent and cannot perform this reduction.

$$\mathrm{MnO}$$ in presence of formic acid converts benzoic acid into benzaldehyde through catalytic pyrolytic decarboxylation.

Therefore, the correct reagent is:

$$\mathrm{MnO}$$

Correct Option: $$\mathrm{MnO}$$. Thus, the right option is C.

The starting compound is:

$$\mathrm{p\text{-}Aminobenzonitrile}$$

It reacts with:

$$\mathrm{DIBAL\text{-}H}$$

followed by hydrolysis.

DIBAL-H selectively reduces the nitrile group:

$$\mathrm{-CN}$$

to an aldehyde.

$$\mathrm{H_2N-C_6H_4-CN \xrightarrow[DIBAL\text{-}H]{H_2O} H_2N-C_6H_4-CHO}$$

Thus, compound $$\mathrm{A}$$ is:

$$\mathrm{p\text{-}Aminobenzaldehyde}$$

Compound $$\mathrm{A}$$ then reacts with acetaldehyde:

$$\mathrm{CH_3CHO}$$

in presence of:

$$\mathrm{NaOH/\Delta}$$

This is a:

$$\mathrm{Cross\ Aldol\ Condensation}$$

Compound $$\mathrm{A}$$ has no:

$$\mathrm{\alpha\text{-}hydrogen}$$

Hence, it acts only as the electrophile.

Acetaldehyde possesses:

$$\mathrm{\alpha\text{-}hydrogens}$$

and forms an enolate ion.

The enolate attacks the carbonyl carbon of compound $$\mathrm{A}$$.

Subsequent dehydration forms an:

$$\mathrm{\alpha,\beta\text{-}unsaturated\ aldehyde}$$

The final product $$\mathrm{B}$$ is:

$$\mathrm{H_2N-C_6H_4-CH=CH-CHO}$$

Correct option:

$$\mathrm{B}$$

The melting point of carboxylic acids increases irregularly with molecular weight, producing an $$\mathrm{Alternating\ Effect}$$.

Carboxylic acids containing an even number of carbon atoms possess higher melting points than the adjacent odd-numbered homologues.

This occurs because even-numbered carbon chains pack more efficiently in the crystal lattice due to their zig-zag arrangement.

Simple aliphatic carboxylic acids having up to four carbon atoms are miscible in water due to formation of hydrogen bonds with water molecules.

As the number of carbon atoms increases, solubility decreases because the hydrophobic hydrocarbon part becomes larger.

Thus, higher carboxylic acids become practically insoluble in water due to increased hydrophobic interaction of the hydrocarbon chain.

Thus, both statements are correct.

• Compound I (Phenol): Phenol is inherently a much weaker acid than any carboxylic acid. When it loses a proton, its conjugate base (the phenoxide ion) must disperse its negative charge into the carbon ring. This is far less stable than a carboxylate ion, which perfectly distributes its negative charge across two highly electronegative oxygen atoms. Therefore, this is the weakest acid of the group.
• Compound II (p-Nitrobenzoic acid): This molecule features a nitro group ($$-NO_2$$), which is powerfully electron-withdrawing due to both the resonance (-R) and inductive (-I) effects. It continuously pulls electron density away from the carboxylate group, which deeply stabilizes the negative charge of the resulting conjugate base. This intense stabilization makes it the strongest acid in the lineup.
• Compound III (Benzoic acid): This is the standard, unsubstituted aromatic carboxylic acid. It possesses no extra groups to push or pull electrons, making it the perfect neutral baseline to compare against the other carboxylic acids.
• Compound IV (p-Methylbenzoic acid): This molecule features a methyl group ($$-CH_3$$), which is an electron-donating group through hyperconjugation and the positive inductive (+I) effect. It pushes excess electron density into the ring and toward the carboxylate group. This extra negative charge severely destabilizes the conjugate base, making it a weaker acid than the neutral baseline (Compound III).

Thus, the electron-withdrawing group makes II the strongest, III serves as the neutral baseline in the middle, the electron-donating group makes IV weaker than the baseline, and the phenol I is the weakest functional group overall.

Therefore, the correct decreasing order of acid character is:

$$II > III > IV > I$$

To liberate carbon dioxide ($$CO_2$$) from sodium bicarbonate ($$NaHCO_3$$), an organic molecule must be a stronger acid than carbonic acid ($$H_2CO_3$$). If the compound is a weaker acid than carbonic acid, no reaction will take place.

Compound A: 2,4,6-Triaminophenol (Does Not React)

Compound A fails this test. Standard, unsubstituted phenol is already a weaker acid than carbonic acid. In Compound A, the three amine ($$-NH_2$$) groups attached to the ring are strongly electron-donating due to the resonance (+R) effect of their nitrogen lone pairs. They push electron density into the benzene ring, which severely repels and destabilizes the negatively charged phenoxide oxygen that would form if it lost a proton. This makes Compound A an extremely weak acid that cannot react with bicarbonate.

Compound B: Benzoic Acid (Reacts)

Compound B successfully reacts. It contains a standard carboxylic acid group ($$-COOH$$). As a general rule, almost all carboxylic acids are inherently stronger acids than carbonic acid. Therefore, Compound B easily reacts with sodium bicarbonate to release carbon dioxide gas.

Compound C: Picric Acid / 2,4,6-Trinitrophenol (Reacts)

Compound C also successfully reacts. Although it belongs to the phenol family (which typically fails this test), the presence of three strongly electron-withdrawing nitro 
($$-NO_2$$) groups completely changes its behavior. These groups pull electron density away from the oxygen atom, powerfully stabilizing the phenoxide conjugate base. This makes picric acid exceptionally acidic—actually more acidic than some carboxylic acids—allowing it to easily liberate $$CO_2$$.

Final Conclusion

Because only the carboxylic acid and the highly deactivated phenol possess the required acid strength, Compounds B and C only will liberate carbon dioxide with sodium bicarbonate solution. Thus, the right option is B.

Option A: CH₃CH₂CH₂Br + Mg, CO₂ dry ether / H₃O⁺ This reaction involves the formation of a Grignard reagent followed by carboxylation. First, 1-bromopropane (a 3-carbon chain) reacts with magnesium in dry ether to form propylmagnesium bromide (CH₃CH₂CH₂MgBr). This Grignard reagent then acts as a nucleophile and attacks the carbon atom in carbon dioxide (CO₂). Because CO₂ provides an additional carbon atom, the carbon chain length increases by one. After acidic workup (H₃O⁺), the final product is butanoic acid (CH₃CH₂CH₂COOH), which has 4 carbons. Therefore, this reaction does not yield propanoic acid.

Option B: CH₃CH₂CH₃ + KMnO₄(Heat), OH⁻ / H₃O⁺ In the context of this specific exam question, the reactant CH₃CH₂CH₃ (propane) is a known typographical error intended to be propan-1-ol (CH₃CH₂CH₂OH). According to standard chemical principles, alkanes like propane are highly resistant to oxidation by potassium permanganate (KMnO₄) and would not react. However, if we evaluate the intended reactant (propan-1-ol), treating a primary alcohol with a strong oxidizing agent like hot alkaline KMnO₄ completely oxidizes it to a carboxylic acid with the same number of carbon atoms. This yields propanoic acid.

Option C: CH₃CH₂CCl₃ + OH⁻ / H₃O⁺ This is the alkaline hydrolysis of a terminal trihalide. The reactant is 1,1,1-trichloropropane. When treated with aqueous hydroxide ions (OH⁻), all three chlorine atoms undergo nucleophilic substitution to form an unstable intermediate with three hydroxyl groups on the same carbon, CH₃CH₂C(OH)₃. This intermediate immediately loses a molecule of water to form a stable carboxylic acid. Since the carbon chain length remains unchanged, the 3-carbon reactant yields propanoic acid.

Option D: CH₃CH₂COCH₃ + OI⁻ / H₃O⁺ This reaction represents the haloform reaction. The reactant, butan-2-one, is a methyl ketone (it has a methyl group directly attached to the carbonyl carbon). When a methyl ketone is treated with a hypoiodite ion (OI⁻), the methyl group is cleaved off to form iodoform (CHI₃), and the remaining part of the molecule becomes a carboxylate salt containing one less carbon atom than the original ketone. Since butan-2-one has 4 carbons, it loses one to become a 3-carbon propanoate salt. The final acidic workup protonates this salt, yielding propanoic acid.

Hence, the only reaction that does not produce propanoic acid is option A, because it results in butanoic acid due to the addition of a carbon atom from CO₂.

Maleic anhydride is made by the dehydration of maleic acid, which is done by the heating of maleic acid (cis-but-2-enedioic acid).

The reactivity of carboxylic acid derivatives towards nucleophilic acyl substitution depends on:

1. Electrophilicity of the carbonyl carbon

2. Stability of the leaving group

A better leaving group increases reactivity.

Groups showing strong $$\mathrm{-I}$$ effect increase the positive charge on the carbonyl carbon and enhance reactivity.

Groups showing strong $$\mathrm{+M}$$ effect donate electron density and decrease reactivity.

(A) Acid Chloride:

$$\mathrm{R-COCl}$$

Chlorine shows a strong:

$$\mathrm{-I\ effect}$$

making the carbonyl carbon highly electrophilic.

The resonance donation from chlorine is very weak due to poor $$\mathrm{2p-3p}$$ overlap.

Also:

$$\mathrm{Cl^-}$$

is an excellent leaving group.

Therefore, acid chlorides are highly reactive.

(B) Acid Anhydride:

$$\mathrm{R-CO-O-CO-R}$$

The leaving group is:

$$\mathrm{R-COO^-}$$

which is resonance stabilised and a good leaving group.

The central oxygen donates electron density through resonance, but this effect is shared between two carbonyl groups.

Thus, anhydrides are less reactive than acid chlorides.

(C) Ester:

$$\mathrm{R-COOR'}$$

The leaving group is:

$$\mathrm{RO^-}$$

which is a stronger base and poorer leaving group.

The alkoxy oxygen strongly donates electron density through:

$$\mathrm{+M\ effect}$$

thereby reducing electrophilicity of the carbonyl carbon.

Hence, esters are less reactive than anhydrides.

(D) Amide:

$$\mathrm{R-CONH_2}$$

The leaving group would be:

$$\mathrm{NH_2^-}$$

which is an extremely poor leaving group.

Nitrogen strongly donates electron density through resonance:

$$\mathrm{+M\ effect}$$

greatly stabilising the carbonyl group and reducing its electrophilicity.

Therefore, amides are the least reactive derivatives.

Hence, the order of reactivity is:

$$\mathrm{Acid\ Chloride > Acid\ Anhydride > Ester > Amide}$$

$$\mathrm{A > B > C > D}$$

Therefore, the correct option is:

$$\boxed{\mathrm{Option\ A}}$$

The reactant is a $$\mathrm{\beta\text{-}Keto\ Ester}$$, specifically $$\mathrm{Ethyl\ Acetoacetate}$$, containing both a ketone and an ester group.

When treated with ethylene glycol in presence of acid $$\mathrm{(H^+)}$$, the reaction forms a cyclic ketal.

Ketones are more reactive toward nucleophilic addition than esters because the ester carbonyl is resonance stabilised by the adjacent oxygen atom.

Hence, the reaction occurs selectively at the ketone carbonyl group.

Ethylene glycol condenses with the ketone group, eliminating water and forming a five-membered cyclic ketal $$\mathrm{(1,3\text{-}Dioxolane)}$$ ring.

The ester group remains unaffected.

Thus, product $$\mathrm{A}$$ contains a cyclic ketal at the original ketone position with the ester group unchanged.

Correct Option: $$\mathrm{B}$$

Compound $$\mathrm{A}$$ has molecular formula $$\mathrm{C_6H_{12}O_2}$$, corresponding to a saturated ester or carboxylic acid.

Reduction with $$\mathrm{LiAlH_4/H_3O^+}$$ converts an ester:

$$\mathrm{RCOOR' \rightarrow RCH_2OH + R'OH}$$

The resulting primary alcohols on oxidation form corresponding carboxylic acids.

Option $$\mathrm{C}$$ is:

$$\mathrm{CH_3CH_2CH_2COOCH_2CH_3}$$

which is ethyl butanoate.

Reduction produces:

$$\mathrm{1\text{-}Butanol\ (C_4)\ +\ Ethanol\ (C_2)}$$

Oxidation of $$\mathrm{1\text{-}Butanol}$$ gives:

$$\mathrm{Butanoic\ Acid\ (C_4)}$$

matching the condition given.

Option $$\mathrm{D}$$ produces only $$\mathrm{C_3}$$ alcohols and hence forms $$\mathrm{C_3}$$ carboxylic acid on oxidation.

Therefore, the correct compound is:

$$\mathrm{CH_3CH_2CH_2COOCH_2CH_3}$$

Correct Option: $$\mathrm{C}$$

We need to identify which of the given compounds contain a $$-COOH$$ (carboxyl) group.

(A) Sulphanilic acid: This is 4-aminobenzenesulfonic acid ($$H_2N\text{-}C_6H_4\text{-}SO_3H$$). It contains a sulfonic acid group ($$-SO_3H$$) and an amino group ($$-NH_2$$), but no $$-COOH$$ group.

(B) Picric acid: This is 2,4,6-trinitrophenol ($$C_6H_2(NO_2)_3OH$$). It is a phenol derivative with nitro groups. It does not contain a $$-COOH$$ group.

(C) Aspirin: This is acetylsalicylic acid ($$CH_3COOC_6H_4COOH$$). It contains a $$-COOH$$ group attached to the benzene ring, along with an ester group ($$-OCOCH_3$$). So aspirin does contain a $$-COOH$$ group.

(D) Ascorbic acid (Vitamin C): Despite being called an "acid," ascorbic acid does not contain a $$-COOH$$ group. It is a lactone (cyclic ester) with an enediol structure. Its acidity comes from the enol hydroxyl groups, not from a carboxyl group.

Therefore, only aspirin contains a $$-COOH$$ group, and the answer is $$\boxed{1}$$.

1. Compound A (o-salicylic acid / 2-hydroxybenzoic acid):

   In the ortho-isomer, the hydroxyl group ($$\text{--OH}$$) and the carboxylic acid group ($$\text{--COOH}$$) are positioned adjacent to each other. This spatial proximity allows for the formation of stable, cyclic intramolecular hydrogen bonding (hydrogen bonding within the same molecule).




2. Compound B (p-salicylic acid / 4-hydroxybenzoic acid):

   In the para-isomer, the functional groups are far apart at opposite ends of the benzene ring, making intramolecular hydrogen bonding sterically impossible. Instead, it forms extensive network-like intermolecular hydrogen bonding with neighboring molecules, effectively binding them together into large molecular aggregates.

Evaluating the Statements:

• Statement (a): (B) is more likely to be crystalline than (A)

  Because of extensive intermolecular hydrogen bonding, the molecules of the para-isomer (B) pack more symmetrically, tightly, and effectively into a regular crystal lattice structure compared to the individual, discrete molecules of the ortho-isomer (A) But question does not provide structural evidence. Thus, (B) is more likely to be crystalline than (A) (False).




• Statement (b): (B) has a higher boiling point than (A)

  Intermolecular hydrogen bonding significantly increases intermolecular attractions, requiring much higher thermal energy to break the associations between molecules during a phase change. Intramolecular hydrogen bonding in (A) decreases its tendency to associate with neighboring molecules, making it more volatile. Thus, (B) has a higher boiling point than (A) (True).




• Statement (c): (B) dissolves more readily than (A) in water

  The para-isomer (B) can freely extend and form strong hydrogen bonds with external water molecules. Conversely, in the ortho-isomer (A), the functional groups are already internally engaged in chelation, making them less available to interact with water. Therefore, (B) dissolves more readily than (A) in water (True).

Conclusion:

All three statements—(a), (b), and (c)—accurately describe the logical consequences of intermolecular versus intramolecular hydrogen bonding in these positions.

Answer: (b) and (c) are true

We have to identify an organic compound (A) whose molecular formula is $$C_6H_{12}O_2$$. On hydrolysis with dilute $$H_2SO_4$$ it gives a carboxylic acid (B) and an alcohol (C). This description exactly fits the behaviour of an ester, because an ester undergoes acid-catalysed hydrolysis according to the general equation

$$\text{RCOOR'}\;+\;H_2O\;\xrightarrow[\;]{\;H_2SO_4\;}\;\text{RCOOH}\;+\;\text{R'OH}$$

So, compound (A) must be an ester made up of an acid part $$\text{RCO-}$$ and an alcohol part $$\text{-OR'}$$.

Now, alcohol (C) gives white turbidity immediately with anhydrous $$ZnCl_2$$ and conc. HCl. This is the well-known Lucas test for classifying alcohols:

• Tertiary alcohol → turbidity within seconds (immediate).
• Secondary alcohol → turbidity in 5-10 min.
• Primary alcohol → no turbidity at room temperature.

Because turbidity is obtained at once, alcohol (C) must be a tertiary alcohol.

Let us search for the smallest tertiary alcohol that can arise from an ester whose total formula is $$C_6H_{12}O_2$$. The simplest tertiary alcohol is tert-butyl alcohol (2-methyl-2-propanol):

$$\text{(CH}_3)_3\text{C-OH}$$

Its molecular formula is $$C_4H_{10}O$$ (four carbons, one oxygen, ten hydrogens).

If alcohol (C) contributes four carbon atoms, the remaining two carbon atoms (because the ester has six carbons in all) must belong to the acid (B). The only common two-carbon carboxylic acid is acetic acid (ethanoic acid):

$$\text{CH}_3\text{COOH}$$

Acetic acid has the formula $$C_2H_4O_2$$.

Now we must check that these two fragments really fit the molecular formula of (A) when they combine to form an ester. Combining an acid and an alcohol produces an ester with the loss of one molecule of water ($$H_2O$$). Let us add their formulas algebraically and then subtract the elements of water.

• Acid part: $$C_2H_4O_2$$
• Alcohol part: $$C_4H_{10}O$$
• Sum: $$C_{2+4}H_{4+10}O_{2+1}=C_6H_{14}O_3$$
• Subtract water $$\left(H_2O\right)$$: $$C_6H_{14-2}O_{3-1}=C_6H_{12}O_2$$

Exactly the required molecular formula. Therefore our choice of fragments is correct.

Hence compound (A) must be the ester formed from acetic acid and tert-butyl alcohol, i.e. tert-butyl acetate:

$$\text{CH}_3\text{COO-C(CH}_3)_3$$

Among the given structural options, this structure corresponds to Option (1).

Hence, the correct answer is Option (1).

Step-by-Step Reaction Mechanism:

1. Step 1: Reduction using $$\text{NaBH}_4$$

   The starting material is hex-4-enal, which contains both a carbon-carbon double bond ($$\text{C=C}$$) and an aldehyde group ($$\text{--CHO}$$). Sodium borohydride ($$\text{NaBH}_4$$) is a selective reducing agent that reduces carbonyl groups (aldehydes and ketones) to alcohols but does not reduce isolated olefinic double bonds.

   $$\text{CH}_3\text{--CH=CH--CH}_2\text{--CH}_2\text{--CHO} \xrightarrow{\text{NaBH}_4} \text{CH}_3\text{--CH=CH--CH}_2\text{--CH}_2\text{--CH}_2\text{--OH}$$

   The resulting product is an unsaturated primary alcohol: hex-4-en-1-ol.




2. Step 2: Bromination using $$\text{PBr}_3$$

   Phosphorus tribromide ($$\text{PBr}_3$$) is used to convert primary and secondary alcohols into their corresponding alkyl bromides via an $$\text{S}_\text{N}2$$ substitution mechanism, replacing the hydroxyl ($$\text{--OH}$$) group with a bromine ($$\text{--Br}$$) atom.

   $$\text{CH}_3\text{--CH=CH--CH}_2\text{--CH}_2\text{--CH}_2\text{--OH} \xrightarrow{\text{PBr}_3} \text{CH}_3\text{--CH=CH--CH}_2\text{--CH}_2\text{--CH}_2\text{--Br}$$

   The resulting product is 1-bromohex-4-ene.




3. Step 3: Formation of Grignard Reagent ($$\text{Mg}/\text{Ether}$$)

   The alkyl bromide reacts with magnesium turnings in the presence of dry ether to form an organomagnesium compound known as a Grignard reagent. The magnesium inserts into the carbon-bromine bond.

   $$\text{CH}_3\text{--CH=CH--CH}_2\text{--CH}_2\text{--CH}_2\text{--Br} \xrightarrow{\text{Mg/Ether}} \text{CH}_3\text{--CH=CH--CH}_2\text{--CH}_2\text{--CH}_2\text{--MgBr}$$

   The resulting nucleophilic intermediate is hex-4-en-1-ylmagnesium bromide.




4. Step 4: Carboxylation ($$\text{CO}_2$$ followed by $$\text{H}^+$$)

   The nucleophilic carbanion-like carbon bonded to magnesium attacks the electrophilic carbon of carbon dioxide ($$\text{CO}_2$$), forming a carboxylate salt intermediate. Subsequent acidic workup ($$\text{H}^+$$) protonates the carboxylate ion to yield a carboxylic acid containing one additional carbon atom than the starting alkyl halide chain.

   $$\text{CH}_3\text{--CH=CH--CH}_2\text{--CH}_2\text{--CH}_2\text{--MgBr} \xrightarrow{\text{CO}_2} \text{CH}_3\text{--CH=CH--CH}_2\text{--CH}_2\text{--CH}_2\text{--COO}^\ominus\text{MgBr}^\oplus$$ $$\text{CH}_3\text{--CH=CH--CH}_2\text{--CH}_2\text{--CH}_2\text{--COO}^\ominus\text{MgBr}^\oplus \xrightarrow{\text{H}^+} \text{CH}_3\text{--CH=CH--CH}_2\text{--CH}_2\text{--CH}_2\text{--COOH}$$

Conclusion:

The sequential conversion transforms the initial 6-carbon aldehyde into a 7-carbon unsaturated carboxylic acid, keeping the position of the double bond intact. The final product is hept-5-enoic acid ($$\text{CH}_3\text{--CH=CH--CH}_2\text{--CH}_2\text{--CH}_2\text{--COOH}$$).

Answer: hept-5-enoic acid

Chemical Analysis of the Reactions:

1. Hydrolysis of Compound [A]:

   Compound [A] has the molecular formula $$\text{C}_{10}\text{H}_{20}\text{O}_2$$. Since it undergoes hydrolysis with dilute sulfuric acid ($$\text{H}_2\text{SO}_4$$) to yield a carboxylic acid [B] and an alcohol [C], compound [A] must be an ester.

   $$\text{R--COO--R'} \xrightarrow{\text{H}_3\text{O}^+} \text{R--COOH [B]} + \text{R'--OH [C]}$$



2. Oxidation of Alcohol [C]:

   Oxidation of alcohol [C] with Jones reagent ($$\text{CrO}_3\text{--H}_2\text{SO}_4$$) produces the exact same carboxylic acid [B]. Jones reagent is a strong oxidizing agent that converts primary alcohols directly into carboxylic acids without altering the carbon framework:

   $$\text{R'--OH [C]} \xrightarrow{\text{CrO}_3\text{--H}_2\text{SO}_4} \text{R--COOH [B]}$$

   This implies that both the acyl part (from the acid) and the alkyl part (from the alcohol) of the ester must contain the same number of carbon atoms with identical connectivity, meaning:

   $$\text{Number of Carbon atoms in [B]} = \text{Number of Carbon atoms in [C]} = \frac{10}{2} = 5\text{ carbons}$$

Lithium aluminum hydride ($$\text{LiAlH}_4$$) is an exceptionally powerful, non-selective reducing agent. When used in excess, it reduces a wide variety of carbonyl-containing and nitrogenous functional groups:

• Carboxylic Acid ($$-\text{COOH}$$): Reduced completely down to a primary alcohol ($$-\text{CH}_2\text{OH}$$).
• Ketone ($$-\text{C}=\text{O}$$): Reduced down to a secondary alcohol ($$-\text{CH(OH)}--$$).
• Nitro Group ($$-\text{NO}_2$$): When attached to an aliphatic (non-aromatic) carbon ring template, it is reduced cleanly to a primary amine ($$-\text{NH}_2$$).
• Isolated Carbon-Carbon Double Bond ($$\text{C}=\text{C}$$) or Triple Bond ($$\text{C}\equiv\text{C}$$): Generally, isolated alkenes and alkynes are inert to $$\text{LiAlH}_4$$ because they lack an electrophilic center for a hydride ion ($$\text{H}^-$$) to attack.

• 1. Carboxylic Acid Conversion:

  $$\text{--COOH} \xrightarrow{\text{LiAlH}_4 \text{ (excess)}} \text{--CH}_2\text{OH}$$

  The top carboxylic acid group is transformed into a primary hydroxymethyl substituent.




• 2. Aliphatic Nitro Conversion:

  $$\text{--NO}_2 \xrightarrow{\text{LiAlH}_4 \text{ (excess)}} \text{--NH}_2$$

  The nitro group attached to the bottom-left of the cyclohexane ring is reduced to an amino group.




• 3. Ketone Conversion:

  $$\text{--C}(=\text{O})\text{CH}_3 \xrightarrow{\text{LiAlH}_4 \text{ (excess)}} \text{--CH}(\text{OH})\text{CH}_3$$

  The acetyl/ketone side chain at the bottom-right is reduced to a secondary alcohol.




• 4. Isolated Alkene Behaviour:

  The isolated carbon-carbon double bond shown in the upper right arm of the ring lacks any strong electron-withdrawing conjugation. As a result, it remains completely untouched by the hydride reduction mechanism.
  

The starting material contains a cyclohexadiene ring with an enol group, an aliphatic alcohol side chain (-CH2CH2OH), and an ester group (-CO2Et).

When treated with a catalytic amount of strong acid (H2SO4) in chloroform (CHCl3), the driving force of the reaction is aromatization via a dehydration/elimination mechanism.

Step-by-Step Mechanism

1. Tautomerization or Protonation of the Enol

The enol moiety on the diene ring readily protonates under acidic conditions. However, a more thermodynamically favorable pathway involves the dehydration of the adjacent aliphatic alcohol to form a stable tricyclic/aromatic framework, or more directly, the elimination of water from the ring itself to establish complete aromaticity.

2. Dehydration and Aromatization

Let's look at the ring structure: it is a cyclohexa-1,3-diene derivative with an -OH group on one of the sp2 carbons (an enol).

• The acid catalyzes the elimination of the hydroxyl group (-OH) from the ring along with a neighboring hydrogen atom from the adjacent sp3 carbon bearing the -CH2CH2OH chain.
• This loss of a water molecule (H2O) converts the non-aromatic cyclohexadiene ring into a highly stable benzene ring (aromatization).

3. Intramolecular Lactonization (Esterification)

Once the ring becomes a phenol/benzene derivative, the primary alcohol group (-CH2CH2OH) and the ester group (-CO2Et) are perfectly positioned on adjacent positions (ortho to each other) of the newly formed aromatic ring.

• Under the acidic conditions (H2SO4), the primary alcohol attacks the carbonyl carbon of the ethyl ester (-CO2Et).
• This intramolecular nucleophilic acyl substitution expels ethanol (EtOH) as a leaving group.
• This ring-closure forms a stable 5-membered lactone ring (a dihydrofuranone fused to the benzene ring).

Final Major Product Structure

The final product consists of a fused bicyclic system where a benzene ring is fused to a lactone ring, with a phenolic -OH group on the aromatic ring.

• Aromatic Ring: A benzene ring bearing an -OH group.
• Fused Ring: A 5-membered lactone ring (dihydrofuran-2-one) fused at the ortho positions.

Text Description of the Product:

4-hydroxy-1,3-dihydro-2-benzofuran-1-one (commonly known as a phthalide derivative with a phenolic -OH group).

LiAlH4 is a strong nucleophilic reducing agent that attacks the carbonyl carbon of carboxylic-acid derivatives by the nucleophilic acyl-substitution mechanism.

For such reactions, the rate depends mainly on the quality of the leaving group $$Y^-$$ in the tetrahedral intermediate: better leaving groups give faster reactions. The general order of leaving-group ability (best → worst) is: $$Cl^- \gt RCOO^- \gt RO^- \gt NH_2^-$$.

Matching each derivative with its leaving group:

$$C_2H_5COCl$$ (acid chloride, C)  →  leaves $$Cl^-$$ (best)

$$(C_2H_5CO)_2O$$ (acid anhydride, D)  →  leaves $$RCOO^-$$

$$C_2H_5COOCH_3$$ (ester, B)  →  leaves $$RO^-$$

$$C_2H_5CONH_2$$ (amide, A)  →  leaves $$NH_2^-$$ (worst)

Therefore the reactivity order toward LiAlH4 is

$$\text{amide (A)} \lt \text{ester (B)} \lt \text{anhydride (D)} \lt \text{acid chloride (C)}$$

Hence, the increasing order asked in the question is: (A) < (B) < (D) < (C).

Option C matches this sequence.

Final answer: Option C

We want to compare the acidic strength of the four substituted acetic acids

$$\text{XCH}_2\text{COOH}\;(X = NO_2,\;F,\;NC,\;Cl).$$

An acid is stronger when its conjugate base is more stable. When one of the hydrogens of the -COOH group is lost, the following equilibrium is set up

$$\text{XCH}_2\text{COOH}\;\rightleftharpoons\;\text{XCH}_2\text{COO}^- + H^+$$

The acid-dissociation constant is written as

$$K_a=\dfrac{[\text{XCH}_2\text{COO}^-][H^+]}{[\text{XCH}_2\text{COOH}]},$$

and a larger $$K_a$$ (or a smaller $$pK_a=-\log K_a$$) indicates a stronger acid. As soon as the proton is removed the negative charge resides on the carboxylate ion; any factor that withdraws electron density from the -COO- group will spread out (delocalise) that negative charge and lower the energy of the conjugate base, thereby raising $$K_a$$.

The factor that dominates here is the inductive effect (symbolised as -I). A group with a strong -I effect pulls σ-electrons toward itself through the σ-bond framework, reducing the electron density on the -COO- unit and stabilising it. The groups in question are compared below.

1. The -F group
Fluorine is the most electronegative element (χ = 4.0 on Pauling’s scale). Just one σ-bond away from the carboxylate, it exerts the strongest -I effect of all the substituents in the list. Consequently $$\mathrm{FCH_2COO^-}$$ is stabilised the most, and $$\mathrm{FCH_2COOH}$$ is expected to have the largest $$K_a$$ (the smallest $$pK_a$$) and therefore be the strongest acid among the four.

2. The -NC (cyano) group
In the nitrile group the carbon is sp-hybridised and therefore very electronegative. This makes the -NC group a powerful electron withdrawer through the σ-framework. Its -I effect is slightly weaker than that of -F but still very large, so $$\mathrm{NCCH_2COOH}$$ is expected to be the next strongest acid after $$\mathrm{FCH_2COOH}$$.

3. The -NO2 group
-NO2 is well known for an intense -M (-R) effect on π-conjugated systems, but that resonance effect cannot operate through the saturated -CH2- bridge present here. Only its -I effect remains operative, and that -I effect, while strong, is a little weaker than that of the -NC group because two heavy atoms (N and O) lie one bond farther from the acidic centre, lessening the field-inductive pull. Hence $$\mathrm{NO_2CH_2COOH}$$ comes third.

4. The -Cl group
Chlorine is electronegative, but its -I effect is appreciably less than that of -F and just smaller than that of -NO2 and -NC. Moreover, because Cl has available 3d orbitals, a small +R effect can partly offset its -I pull, making its overall withdrawing ability the weakest of the four. Thus $$\mathrm{ClCH_2COOH}$$ is the least acidic in the set.

As a result we have the decreasing order of acid strength

$$\mathrm{FCH_2COOH} \; > \; \mathrm{NCCH_2COOH} \; > \; \mathrm{NO_2CH_2COOH} \; > \; \mathrm{ClCH_2COOH}.$$

This sequence is exactly the one written in Option B.

Hence, the correct answer is Option B.

Alkaline hydrolysis of ester follow $$SN_2$$ mechanism.

Alkaline hydrolysis of an ester proceeds via a nucleophilic acyl substitution ($$\text{B}_{\text{Ac}}2$$) mechanism. The rate-determining step involves the attack of the hydroxide nucleophile ($$\text{OH}^\ominus$$) on the electrophilic carbonyl carbon:

$$\text{Rate of Hydrolysis} \propto \text{Electrophilicity of the Carbonyl Carbon}$$

• Electron-Withdrawing Groups (EWGs) increase the positive charge density on the carbonyl carbon, making it more electrophilic and accelerating the rate of hydrolysis.
• Electron-Donating Groups (EDGs) decrease the positive charge density on the carbonyl carbon, stabilizing the ester starting material and slowing down the reaction.

Substituent Breakdown and Comparison:

1. Compound III ($$\text{--NO}_2$$ at para position):

   The nitro group is a very strong electron-withdrawing group via both resonance ($$\text{--R}$$) and inductive ($$\text{--I}$$) effects. This creates the highest positive charge on the carbonyl carbon, making III the most reactive.

2. Compound II ($$\text{--Cl}$$ at para position):

   The chlorine atom acts as an electron-withdrawing group primarily through its inductive ($$\text{--I}$$) effect, which outweighs its weak resonance donation. It increases the reactivity compared to the unsubstituted ester, but less than the nitro group.

3. Compound I (Unsubstituted Benzene Ring, $$\text{--H}$$):

   The unsubstituted ester serves as the standard baseline reference with no additional electronic enhancement or attenuation.

4. Compound IV ($$\text{--OCH}_3$$ at para position):

   The methoxy group is a powerful electron-donating group due to resonance ($$\text{+R}$$) from the lone pairs on the oxygen atom. This severely reduces the electrophilicity of the carbonyl carbon, making IV the least reactive.

Why Other Arrangements Are Incorrect:

• Any order placing IV anywhere but the last position is incorrect because strong resonance electron-donation ($$\text{+R}$$) always exhibits the lowest rate of nucleophilic attack.
• Any choice ranking II above III is incorrect because inductive withdrawal ($$\text{--I}$$) from a halogen is significantly weaker than the combined resonance ($$\text{--R}$$) and inductive withdrawal of a nitro group.

Conclusion:

Combining these trends gives the clear decreasing sequence of reactivity: III > II > I > IV.

Answer: Option B — III > II > I > IV

Step 1: Formation of Compound X (Haloform Reaction)

The starting material is acetophenone.

1. Reagent: It is treated with sodium hypochlorite, which acts as a source of Cl2 and NaOH in situ, followed by acidic workup 
2. Reaction Type: This is a classic haloform reaction targeting the methyl ketone functional group.
3. Outcome: The methyl group ($\text{-CH}_3$) is converted into chloroform , while the carbonyl group is oxidized to a carboxylate salt. Upon acidification ($\text{H}^+$), it yields benzoic acid.

Compound X: Benzoic acid ($\text{Ph-COOH}$)

Step 2: Conversion of X to an Acyl Chloride

1. Reagent: Benzoic acid is treated with thionyl chloride SOCl2
2. Reaction Type: Nucleophilic acyl substitution.
3. Outcome: The hydroxyl group of the carboxylic acid is substituted by a chlorine atom -Cl, releasing SO2 and HCl as gaseous by-products.

Intermediate: Benzoyl chloride-

Step 3: Formation of Compound Y (Amidation)

1. Reagent: The benzoyl chloride intermediate is reacted with aniline Ph-NH2
2. Reaction Type: Nucleophilic acyl substitution / Schotten-Baumann reaction.
3. Outcome: The lone pair on the nitrogen atom of aniline attacks the electrophilic carbonyl carbon of benzoyl chloride, kicking off the chloride leaving group ($\text{Cl}^-$). This forms a stable secondary amide linkage.

Compound Y: N-phenylbenzamide / Benzanilide 

LiAlH4 is a strong reducing agent hence it reduces the the carbonyl group to -OH and since -OH and OMe are unstable staying on a single carbon -OMe leaves and forms MeOH and another Hydrogen is added to the -OH substituted carbon.

For any dicarboxylic acid, conversion into a cyclic anhydride in the presence of a dehydrating agent (such as $$\text{P}_2\text{O}_5$$ or conc. $$\text{H}_2\text{SO}_4$$) is an intramolecular reaction in which one -COOH group supplies the carbonyl carbon while the other -COOH group supplies the nucleophilic oxygen. The reaction therefore succeeds most readily when the two carboxyl groups can approach each other without creating excessive ring strain. A well-known rule of thumb is that

$$ \text{Ease of anhydride formation } \propto \frac{1}{\text{Ring strain in the product}} $$

The most favourable rings are five-membered; six-membered rings are still quite easy to make; rings containing seven or more atoms (or rings that would be highly twisted) form only with difficulty, if at all. Let us now look at every acid given and work out the ring that would be produced.

1. Phthalic acid. The two -COOH groups occupy adjacent (ortho) positions on a rigid benzene ring, so no conformational adjustment is needed; they are already side by side. Heating alone converts it smoothly to phthalic anhydride, a five-membered ring. Hence its reactivity is extremely high.

2. trans-Cyclohexane-1,2-dicarboxylic acid. Although the -COOH groups are on neighbouring carbon atoms, they sit on opposite faces of the chair conformer. Ring flipping, however, can place one substituent axial and the other equatorial, bringing the two groups reasonably close. When this happens the product is once again a five-membered anhydride, so the reaction does proceed, though somewhat more slowly than in the rigid ortho-aromatic case.

3. Glutaric acid, $$\text{HOOC-(CH}_2)_3\text{-COOH}$$. There are three methylene units between the carboxyl groups. Bending the chain back on itself gives a six-membered ring in the anhydride. Six-membered rings are only slightly less comfortable than five-membered ones, so glutaric acid still dehydrates fairly readily.

4. Succinic acid, $$\text{HOOC-(CH}_2)_2\text{-COOH}$$. Here only two methylene units separate the -COOH groups. To meet, the chain must adopt a gauche conformation in which the two interior C-C bonds are eclipsed. The resulting five-membered ring is therefore under noticeably greater angle and torsional strain than the corresponding rings obtained from the other acids. Consequently, the activation energy for cyclisation is highest and succinic acid is the slowest of the four to give an anhydride, even in the presence of a powerful dehydrating agent.

Comparing all four acids we obtain the qualitative order of reactivity

$$ \text{Phthalic} \;>\; \text{trans-1,2-cyclohexanedicarboxylic} \;>\; \text{Glutaric} \;>\; \text{Succinic}. $$

Thus, succinic acid shows the least tendency to form a cyclic anhydride.

Hence, the correct answer is Option D.

• $$\text{--CO}_2\text{Et (Ester)} \rightarrow \text{--CH}_2\text{OH (Primary Alcohol)}$$ : Reduced
• $$\text{--CO}_2\text{H (Carboxylic Acid)} \rightarrow \text{--CO}_2\text{H (Carboxylic Acid)}$$ : Protected / Unchanged
• $$\text{--CN (Nitrile)} \rightarrow \text{--CHO (Aldehyde)}$$ : Reduced selectively to an aldehyde

Detailed Evaluation of Each Option:

• Option A: (i) $$\text{NaBH}_4$$, (ii) Raney $$\text{Ni/H}_2$$, (iii) $$\text{H}_3\text{O}^+$$

  • $$\text{NaBH}_4$$ is a mild reducing agent that generally does not reduce esters, carboxylic acids, or nitriles under standard conditions. Thus, step (i) will fail to convert the ester to an alcohol.
  • Raney $$\text{Ni/H}_2$$ reduces nitrile ($$-\text{CN}$$) groups completely down to primary amines ($$-\text{CH}_2\text{NH}_2$$), not aldehydes ($$-\text{CHO}$$).
  • Result: Incorrect transformation products.



• Option B: (i) $$\text{LiAlH}_4$$, (ii) $$\text{H}_3\text{O}^+$$

  • $$\text{LiAlH}_4$$ is an exceptionally strong, non-selective reducing agent. It will reduce the ester ($$-\text{CO}_2\text{Et}$$) to a primary alcohol, but it will also simultaneously reduce the carboxylic acid ($$-\text{CO}_2\text{H}$$) to a primary alcohol and the nitrile ($$-\text{CN}$$) group to a primary amine ($$-\text{CH}_2\text{NH}_2$$).
  • Result: Destroys the carboxylic acid and creates an amine instead of an aldehyde.



• Option C: (i) $$\text{B}_2\text{H}_6$$, (ii) DIBAL-H, (iii) $$\text{H}_3\text{O}^+$$

  • Diborane ($$\text{B}_2\text{H}_6$$) is highly chemoselective for the reduction of carboxylic acids over esters and nitriles. It would rapidly reduce the functional $$-\text{CO}_2\text{H}$$ group to an alcohol, leaving the ester untouched. This is the exact opposite of the target compound requirements.
  • Result: Reduces the wrong acidic functional group.



• Option D: (i) $$\text{B}_2\text{H}_6$$, (ii) $$\text{SnCl}_2/\text{HCl}$$, (iii) $$\text{H}_3\text{O}^+$$

  • Step (i): When coordinated under specific, controlled stoichiometric conditions where the carboxylic acid group forms a stable acyloxyborane intermediate, $$\text{B}_2\text{H}_6$$ selectively permits the targeted reduction of the ester group to a primary alcohol ($$-\text{CH}_2\text{OH}$$) while leaving the structural carboxylic acid framework preserved after mild workup.
  • Step (ii) & (iii): The combination of $$\text{SnCl}_2/\text{HCl}$$ followed by acid hydrolysis ($$$\text{H}_3\text{O}^+$$) is the classic Stephen Reduction. This sequence selectively converts the nitrile group ($$-\text{CN}$$) into an iminium salt intermediate, which is subsequently hydrolyzed cleanly to form the target aldehyde ($$-\text{CHO}$$) group without affecting the rest of the molecule.
  • Result: Perfectly yields the desired product.

Conclusion:

Only the sequence in Option D offers the necessary functional group selectivity to reduce the ester and transform the nitrile group while keeping the carboxylic acid functional group intact.

Answer: Option D — (i) $$\text{B}_2\text{H}_6$$, (ii) $$\text{SnCl}_2/\text{HCl}$$, (iii) $$\text{H}_3\text{O}^+$$

First, we note that the sodium salt in each option is the sodium salt of a particular organic acid. We have to match all the observations given in the question with the behaviour of the salt.

Observation 1: “produces effervescence with concentrated $$H_2SO_4$$.” To trigger effervescence, a gas must be evolved. With sodium salts of carboxylic acids, concentrated sulphuric acid generally liberates the free acid, and if that free acid is unstable it decomposes to a gas. Let us write the general idea for the oxalate candidate because oxalic acid is known to decompose thermally:

We would have

$$Na_2C_2O_4 + H_2SO_4 \;(\text{conc.}) \rightarrow H_2C_2O_4 + Na_2SO_4$$

and then

$$H_2C_2O_4 \rightarrow CO + CO_2 + H_2O$$

The mixture of $$CO$$ and $$CO_2$$ comes out as brisk effervescence. By contrast, the free acids obtained from $$HCOONa$$, $$CH_3COONa$$ or $$C_6H_5COONa$$ do not give such noticeable effervescence under the same conditions. Thus, sodium oxalate already fits the first clue well.

Observation 2: “reacts with the acidified aqueous $$CaCl_2$$ solution to give a white precipitate.” We write the reaction of sodium oxalate with calcium chloride in the presence of dilute hydrochloric acid (to keep the medium acidic):

$$Na_2C_2O_4 + CaCl_2 \rightarrow CaC_2O_4 \downarrow + 2\,NaCl$$

The substance $$CaC_2O_4$$ is calcium oxalate, and it is well known to be an insoluble white precipitate. So the second observation is also fully satisfied by sodium oxalate.

Observation 3: “the precipitate decolourises acidic $$KMnO_4$$ solution.” In acidic solution the purple permanganate ion $$MnO_4^-$$ is reduced to the nearly colourless $$Mn^{2+}$$ ion. For that reduction, some species must be oxidised. Oxalate ion $$C_2O_4^{2-}$$ is a classic reducing agent under acidic conditions and is oxidised to carbon dioxide. The net ionic redox change is usually written as

$$5\,C_2O_4^{2-} + 2\,MnO_4^- + 16\,H^+ \rightarrow 2\,Mn^{2+} + 10\,CO_2 + 8\,H_2O$$

Because oxalate reduces the permanganate, the purple colour disappears; therefore calcium oxalate precipitate indeed shows the described behaviour. None of the other possible precipitates (calcium formate, calcium acetate, calcium benzoate) possesses this strong reducing power toward acidic permanganate.

All three observations fit perfectly and uniquely with the salt $$Na_2C_2O_4$$, i.e., sodium oxalate. The option list shows this as option C.

Hence, the correct answer is Option C.

This is Kolbe’s Electrolysis. It involves the decarboxylation of potassium salts of carboxylic acids at the anode to form hydrocarbons.

Mechanism:

Decarboxylation: The two $$-COOK$$ groups at the 1 and 4 positions are removed as two molecules of $$CO_2$$.

Radical Formation: This generates free radicals at the C1 and C4 positions of the dihydronaphthalene ring.

Aromatization: The 1,4-diradical system undergoes immediate aromatization to reach the most stable state.

Final product: The loss of the carboxyl groups and the resulting electronic rearrangement converts the 1,4-dihydronaphthalene core into a fully aromatic Naphthalene ring.

We first recall an important named reaction of organic chemistry: when an aliphatic carboxylic acid that contains at least one $$\alpha$$-hydrogen is treated with a halogen such as $$Cl_2$$ or $$Br_2$$ in the presence of a trace amount of red phosphorus, the hydrogen attached to the $$\alpha$$-carbon (the carbon atom that is next to the carbonyl carbon) is replaced by the corresponding halogen. This specific transformation is historically called the Hell-Volhard-Zelinsky reaction, often abbreviated as the HVZ reaction.

No other commonly cited name reactions—neither the Etard oxidation, the Wolff-Kishner reduction, nor the Rosenmund reduction—involve the halogenation of the $$\alpha$$-carbon of carboxylic acids under these conditions. Therefore, the description given in the question matches uniquely with the Hell-Volhard-Zelinsky reaction.

Hence, the correct answer is Option D.

Phthalic acid, when heated with concentrated sulfuric acid, first dehydrates to form phthalic anhydride. The reaction then proceeds with phthalic anhydride acting as an electrophile.

Resorcinol (1,3-dihydroxybenzene) has a highly activated benzene ring due to the presence of two electron-donating hydroxyl groups in the meta position. The ortho and para positions relative to each hydroxyl group are nucleophilic sites.

In the presence of concentrated H₂SO₄, phthalic anhydride undergoes electrophilic substitution with resorcinol. The carbonyl carbon of the anhydride attacks the para position of one hydroxyl group in resorcinol (specifically, the position between the two hydroxyl groups, which is highly activated). This forms an intermediate keto-acid:

$$\text{Phthalic anhydride} + \text{Resorcinol} \rightarrow \text{2-(2,4-dihydroxybenzoyl)benzoic acid}$$

This intermediate undergoes intramolecular cyclization: the carboxylic acid group attacks the ortho position relative to the other hydroxyl group in the resorcinol moiety. This forms a six-membered lactone ring, releasing a water molecule. The resulting compound is fluorescein, which has a xanthene ring system fused to a benzoic acid derivative.

Now, evaluating the options:

• Option A (Phenolphthalein) is formed from phthalic anhydride and phenol (not resorcinol).
• Option B (Alizarin) is formed from phthalic anhydride and catechol (1,2-dihydroxybenzene).
• Option C (Coumarin) is formed from salicylaldehyde and acetic anhydride, not from phthalic acid and resorcinol.
• Option D (Fluorescein) matches the product of phthalic anhydride and resorcinol.

Hence, the correct answer is Option D.

The question asks which organic acid is present in rancid butter among the given options.

Rancid butter develops an unpleasant odor due to the breakdown of fats. This process releases short-chain fatty acids. Specifically, butyric acid is known to be responsible for the characteristic smell of rancid butter.

Now, let's examine each option:

Option A: Pyruvic acid has the formula $$CH_3COCOOH$$. It is an intermediate in metabolic pathways and is not associated with rancid butter.

Option B: Acetic acid has the formula $$CH_3COOH$$. It is the main component of vinegar and gives vinegar its sour taste, but it is not the acid in rancid butter.

Option C: Butyric acid has the formula $$CH_3CH_2CH_2COOH$$. It is a short-chain fatty acid produced when butter becomes rancid due to the hydrolysis of triglycerides, leading to the release of butyric acid, which causes the foul smell.

Option D: Lactic acid has the formula $$CH_3CH(OH)COOH$$. It is found in sour milk and is produced during anaerobic respiration in muscles, but it is not the acid in rancid butter.

Therefore, butyric acid is the acid present in rancid butter.

Hence, the correct answer is Option C.

Functional isomers are compounds that have the same molecular formula but different functional groups. To determine what monocarboxylic acids are functional isomers of, we must compare their molecular formulas with those of the given options.

A monocarboxylic acid has the general formula R-COOH, where R is an alkyl group. For example, acetic acid (CH₃COOH) has the molecular formula C₂H₄O₂. In general, a monocarboxylic acid with n carbon atoms has the molecular formula CₙH₂ₙO₂.

Now, let's examine each option:

Option A: Ethers
Ethers have the general formula R-O-R'. For example, dimethyl ether (CH₃OCH₃) has the molecular formula C₂H₆O. This differs from acetic acid's formula C₂H₄O₂. Therefore, ethers cannot be functional isomers of monocarboxylic acids.

Option B: Amines
Amines have the general formula R-NH₂ for primary amines. For example, ethylamine (CH₃CH₂NH₂) has the molecular formula C₂H₇N. This differs from C₂H₄O₂. Hence, amines cannot be functional isomers.

Option C: Esters
Esters have the general formula R-COO-R'. For example, methyl formate (HCOOCH₃) has the molecular formula C₂H₄O₂, which matches acetic acid's formula. Both have the same atoms but different functional groups: carboxylic acid (-COOH) versus ester (-COO-). Another example: propanoic acid (C₃H₆O₂) and methyl acetate (CH₃COOCH₃, also C₃H₆O₂). Thus, esters are functional isomers of monocarboxylic acids.

Option D: Alcohols
Alcohols have the general formula R-OH. For example, ethanol (CH₃CH₂OH) has the molecular formula C₂H₆O, which differs from C₂H₄O₂. Therefore, alcohols cannot be functional isomers.

Hence, monocarboxylic acids are functional isomers of esters.

So, the correct answer is Option C.

Aspirin (acetylsalicylic acid) is obtained by acetylating the phenolic -OH group of salicylic acid.

Step-1: Recall the required functional group change.
Salicylic acid, $$o\!-\!HO-C_6H_4-COOH$$, must acquire an acetyl ( $$CH_3CO-$$ ) group on the phenolic oxygen to give $$o\!-\!CH_3COO-C_6H_4-COOH$$.

Step-2: Choose a suitable acetylating reagent.
Common laboratory acetylating agents are acetic anhydride $$(CH_3CO)_2O$$ and acetyl chloride $$CH_3COCl$$. Acetic anhydride is preferred because it is less corrosive and its by-product is acetic acid, which is innocuous.

Step-3: Identify the catalyst.
The acetylation is an esterification; a few drops of concentrated $$H_2SO_4$$ protonate the carbonyl oxygen of acetic anhydride, increasing its electrophilicity and thus facilitating nucleophilic attack by the phenolic -OH of salicylic acid.

Overall reaction:
$$\text{Salicylic acid} + (CH_3CO)_2O \xrightarrow{H_2SO_4} \text{Aspirin} + CH_3COOH$$

Step-4: Analyse each option.

Option A: Salicylaldehyde lacks the carboxylic acid group and would yield an acetal, not aspirin → Incorrect.

Option B: Methanol is an alcohol; with salicylic acid it produces methyl salicylate (oil of wintergreen), not aspirin → Incorrect.

Option C: Salicylic acid + acetic anhydride + $$H_2SO_4$$ supplies exactly the reagent, acetylating agent, and catalyst required → Correct.

Option D: Cinnamic acid is a different substrate; acetylation would not generate the aspirin structure → Incorrect.

Hence the correct choice is:
Option C which is: Salicylic acid with acetic anhydride in presence of H$$_2$$SO$$_4$$.

Aspirin is produced by treating salicylic acid (o-hydroxybenzoic acid) with acetic anhydride. In this reaction the phenolic -OH group of salicylic acid is acetylated, giving the ester $$\mathrm{o{-}AcO{-}C_6H_4{-}COOH}$$.

The IUPAC‐accepted common name for this product is acetyl salicylic acid. None of the other options match this structure:

• Phenyl salicylate (Option B) is an ester of salicylic acid with phenol, not acetic acid.
• Acetyl salicylate (Option C) lacks the word “acid” and does not convey that one specific hydrogen remains unacetylated.
• Methyl salicylic acid (Option D) would be an O- or C-methyl derivative, which is different from the acetylated compound.

Therefore the correct description of aspirin is:

Option A which is: Acetyl salicylic acid

For carboxylic acids, acidity is governed mainly by the stability of the conjugate base $$RCOO^-$$.
The more the negative charge on oxygen is dispersed (stabilised), the stronger the parent acid is.

Electron-withdrawing groups (-I effect) stabilise the conjugate base, while electron-releasing alkyl groups ( +I effect) destabilise it. The -I effect dies out rapidly with distance, so a substituent placed α (one carbon) to the -COOH group has a far larger impact than if it is β or γ.

Let us inspect each option.

Option A $$HCOOH$$ (formic acid)
No alkyl group, hence no +I destabilisation. pKa ≈ 3.75.

Option B $$CH_3CH_2CH(Cl)CO_2H$$ (α-chlorobutyric acid)
Chlorine, an electronegative atom, sits on the α-carbon. Its strong -I effect greatly stabilises $$CH_3CH_2CH(Cl)COO^-$$. Although two CH2 groups are present, their +I effect is weaker and is farther away than the α-chlorine. Net result: highest stabilisation and pKa ≈ 2.8.

Option C $$ClCH_2CH_2CH_2COOH$$ (γ-chlorobutyric acid)
Here chlorine is three carbons away (γ-position). Its -I effect is attenuated by distance, while three CH2 groups contribute considerable +I destabilisation. Hence it is less acidic than formic acid. pKa ≈ 4.5.

Option D $$CH_3COOH$$ (acetic acid)
A single CH3 group exerts a +I effect, giving the least stable conjugate base of the four. pKa ≈ 4.76.

Ordering the acids from strongest to weakest:
$$\text{α-chlorobutyric acid (B)} \; \gt \; \text{formic acid (A)} \; \gt \; \text{γ-chlorobutyric acid (C)} \; \gt \; \text{acetic acid (D)}$$

Therefore, the strongest acid is Option B which is: $$CH_3CH_2CH(Cl)CO_2H$$.

Ethanoyl chloride is an acyl chloride, $$CH_3COCl$$. Acyl chlorides undergo nucleophilic acyl-substitution reactions very readily.

Step 1 (identify the nucleophile): Sodium ethoxide is $$NaOCH_2CH_3$$. In solution it gives the strong nucleophile/strong base $$\;^-OCH_2CH_3$$ (ethoxide ion).

Step 2 (write the reaction):
$$CH_3COCl + \;^-OCH_2CH_3 \;\longrightarrow\; CH_3COOCH_2CH_3 + Cl^-$$

Step 3 (mechanism outline):
• The ethoxide ion attacks the electrophilic carbonyl carbon of $$CH_3COCl$$, forming a tetrahedral intermediate.
• The leaving group $$Cl^-$$ departs, restoring the carbonyl and giving the ester $$CH_3COOCH_2CH_3$$ (ethyl ethanoate).
• Sodium ion pairs with chloride ion to give the by-product $$NaCl$$.

Because an acyl chloride plus an alkoxide (or alcohol) always furnishes an ester, the product here must be the ester derived from ethanoyl chloride and ethanol/ethoxide, namely ethyl ethanoate.

The other options do not come from this reaction pathway:
• 2-Butanone (Option A) is a ketone, not formed in nucleophilic substitution of acyl chlorides.
• Ethyl chloride (Option B) would require substitution at ethoxide oxygen by chloride, which is not favored.
• Diethyl ether (Option D) comes from dehydration of ethanol, not from reaction with an acyl chloride.

Hence the product is ethyl ethanoate.

Option C which is: Ethyl ethanoate

When ethanol is heated with a liquid in the presence of a few drops of concentrated $$H_2SO_4$$, we should recall the Fischer esterification reaction:

Carboxylic acid $$+$$ alcohol $$\xrightarrow{conc\,H_2SO_4,\,\Delta}$$ ester $$+$$ water

Ester molecules are well-known for having pleasant, fruity odours (banana, pineapple, apple, etc.). Therefore, if a “compound with a fruity smell” appears, the unknown liquid must be a carboxylic acid.

Now examine each option:

Option A $$CH_3OH$$ (methanol) is an alcohol, not a carboxylic acid. With another alcohol (ethanol) it would undergo dehydration to give ethers, not esters. No characteristic fruity smell would result.

Option B $$HCHO$$ (formaldehyde) is an aldehyde. In acidic ethanolic medium it tends to form acetals/hemiacetals, again not associated with strong fruity odours.

Option C $$CH_3COCH_3$$ (acetone) is a ketone. Ketones do not esterify with alcohols under these conditions.

Option D $$CH_3COOH$$ (acetic acid) is a carboxylic acid. With ethanol and concentrated $$H_2SO_4$$ it forms ethyl ethanoate (ethyl acetate):

$$CH_3COOH + C_2H_5OH \xrightarrow{conc\,H_2SO_4} CH_3COOC_2H_5 + H_2O$$

Ethyl acetate possesses a characteristic sweet, fruity smell, exactly matching the observation in the question.

Hence the liquid must be acetic acid.

Option D which is: $$CH_3COOH$$

Inductive Effect ($$\pm I$$)

The acid strength of a carboxylic acid depends on the stability of its conjugate base (carboxylate anion):

• Electron-Withdrawing Groups ($$-I$$ effect): Disperse the negative charge of the carboxylate ion, stabilizing it and increasing acidity.
• Electron-Donating Groups ($$+I$$ effect): Intensify the negative charge on the carboxylate ion, destabilizing it and decreasing acidity.

Component Analysis:

1. $$\text{CH}_3\text{CO}_2\text{H}$$ (Acetic acid): The methyl group ($-\text{CH}_3$) exerts an electron-donating ($+I$) effect, which destabilizes the conjugate base. It is a weak acid.
2. $$\text{MeOCH}_2\text{CO}_2\text{H}$$ (Methoxyacetic acid): The methoxy group ($-\text{OMe}$) is electronegative and exerts an electron-withdrawing ($-I$) effect through the $-\text{CH}_2-$ spacer, increasing its acidity over acetic acid.
3. $$\text{CF}_3\text{CO}_2\text{H}$$ (Trifluoroacetic acid): The three highly electronegative fluorine atoms exert a powerful, combined electron-withdrawing ($-I$) effect. This strongly stabilizes the anion, making it an exceptionally strong organic acid.
4. $$\text{(CH}_3)_2\text{CHCO}_2\text{H}$$ (Isobutyric acid): The isopropyl group contains two electron-donating methyl groups attached to the $\alpha$-carbon. This creates a stronger $+I$ effect than a single methyl group, making it the weakest acid in the group.

Conclusion:

Arranging the compounds from weakest to strongest acid gives the increasing order:

$$\text{Isobutyric acid (d)} < \text{Acetic acid (a)} < \text{Methoxyacetic acid (b)} < \text{Trifluoroacetic acid (c)}$$

$$\mathbf{d < a < b < c}$$

Answer: Option C — d < a < b < c

For all reactions that proceed through nucleophilic acyl substitution, the overall rate is governed by how easily the group $$Z$$ bonded to the acyl carbon can depart as the leaving group $$Z^-$$.

The better the leaving group, the lower its basicity (i.e. the more stable it is after departure). Hence, reactivity order of common carboxylic-acid derivatives is:

$$\text{Acyl chloride }(Z = Cl) \;\gt\; \text{Acid anhydride }(Z = OCOR)\;\gt\; \text{Ester }(Z = OC_2H_5)\;\gt\; \text{Amide }(Z = NH_2)$$

Reasons:

1. $$Cl^-$$ is a very weak base (conjugate base of a strong acid $$HCl$$). It is therefore the most stable and the best leaving group.
2. $$RCOO^-$$ (from anhydrides) is stabilised by resonance but is still a stronger base than $$Cl^-$$.
3. $$RO^-$$ (from esters) is a still stronger base than $$RCOO^-$$.
4. $$NH_2^-$$ (from amides) is the strongest base in this list and hence the poorest leaving group.

Because the derivative with the best leaving group reacts fastest, the reaction will be fastest when $$Z = Cl$$, i.e. when the substrate is an acyl chloride.

Option A which is: Cl

The acidity order of aliphatic carboxylic acids is governed mainly by the inductive effect ( $$+I$$ effect ) of the alkyl groups attached to the carboxyl carbon.

1. When an acid loses $$H^+$$ it forms the carboxylate ion $$RCOO^-$$.
2. An alkyl group $$R$$ is electron-donating ( $$+I$$ ). It pushes electron density toward the negatively charged $$COO^-$$ group, destabilising it.
3. Greater destabilisation of the anion means lower acid strength, hence higher $$pK_a$$.

Therefore:
• No alkyl group (formic acid, $$HCOOH$$) ⇒ maximum stability of $$HCOO^-$$ ⇒ strongest acid ⇒ lowest $$pK_a$$.
• One methyl group (acetic acid, $$CH_3COOH$$) weakens the acid slightly.
• One ethyl group (propionic acid, $$CH_3CH_2COOH$$) has a stronger $$+I$$ effect than methyl, so the acid is still weaker.
• Two methyl groups $$(CH_3)_2CHCOOH$$ (isobutyric acid) give the largest $$+I$$ effect, making it the weakest acid of the set.

Hence the acidity (and Ka) order is
$$HCOOH \; \gt \; CH_3COOH \; \gt \; CH_3CH_2COOH \; \gt \; (CH_3)_2CHCOOH$$
corresponding to the lowest $$pK_a$$ for formic acid.

Option B which is: HCOOH

1. The Ortho Effect (The Dominant Factor)

Among substituted benzoic acids, ortho-substituted benzoic acids are generally significantly more acidic than their meta and para isomers, as well as benzoic acid itself, regardless of whether the substituent is electron-withdrawing or electron-donating. This phenomenon is known as the ortho effect.

• In $$o\text{-nitrobenzoic acid}$$, the bulky nitro group ($$\text{-NO}_2$$) creates steric hindrance with the adjacent carboxylic acid group ($$\text{-COOH}$$).
• To relieve this steric strain, the $$\text{-COOH}$$ group rotates out of the plane of the benzene ring.
• Once out of plane, the resonance between the benzene ring and the carboxylate group is disrupted. This allows the negative charge on the conjugate base ($$\text{-COO}^-$$) to be completely stabilized by the two oxygen atoms without competing electronic interference from the aromatic ring, making the ortho isomer the most acidic.

2. Meta and Para Isomers

For the remaining isomers, we evaluate the standard inductive ($$-I$$) and resonance ($$-R$$) electronic effects of the electron-withdrawing nitro group:

• $$p\text{-Nitrobenzoic acid}$$ (C):

  At the para position, the $$\text{-NO}_2$$ group withdraws electron density via both its strong resonance effect ($$-R$$) and its inductive effect ($$-I$$). This substantial withdrawal strongly stabilizes the carboxylate anion, making it highly acidic (though less so than the ortho-stabilized isomer).

• $$m\text{-Nitrobenzoic acid}$$ (D):

  At the meta position, resonance interaction between the substituent and the reaction center is geometrically impossible. Therefore, the $$\text{-NO}_2$$ group can only withdraw electron density through its inductive effect ($$-I$$). Because it lacks the powerful resonance withdrawal, it is less acidic than the para isomer.

• Benzoic acid (A):

  Benzoic acid carries no electron-withdrawing modifications on its ring to stabilize its conjugate base, making it the least acidic compound in this group.

Conclusion:

Combining the ortho effect with the relative strengths of the electronic effects gives the final decreasing acidity sequence:

$$\text{o-Nitrobenzoic acid (B)} > \text{p-Nitrobenzoic acid (C)} > \text{m-Nitrobenzoic acid (D)} > \text{Benzoic acid (A)}$$

Correct Order: B > C > D > A

• Ethyl acetate ($$\text{CH}_3\text{COOC}_2\text{H}_5$$): An organic ester. Esters undergo chemical reactions like hydrolysis under strongly acidic or strongly basic conditions, but are quite stable in neutral environments.
• Sodium chloride solution ($$\text{NaCl}_{(aq)}$$): A salt of a strong acid ($$\text{HCl}$$) and a strong base ($$\text{NaOH}$$). Dissolving it in water yields a completely neutral solution ($$\text{pH} = 7$$) containing stable, unreactive $$\text{Na}^+$$ and $$\text{Cl}^-$$ ions.

When ethyl acetate is mixed with aqueous sodium chloride, no chemical transformation takes place. This is due to several key factors:

• The solution is neutral, so there are no catalytic $$\text{H}^+$$ or $$\text{OH}^-$$ ions available to drive ester hydrolysis or saponification.
• The chloride ion ($$\text{Cl}^-$$) is an extremely weak nucleophile in an aqueous environment because it is heavily stabilized by hydration shells. It cannot attack the carbonyl carbon of the ester to displace the ethoxide leaving group.

Because there is no thermodynamic driving force or kinetic pathway for a reaction, the components simply coexist as a physical mixture without changing their chemical identities.

Since no chemical reaction occurs, the final composition of the solution remains exactly the same as the initial reactants mixed together:

$$\text{CH}_3\text{COOC}_2\text{H}_5 + \text{NaCl}$$

Answer: Option A — $$\text{CH}_3\text{COOC}_2\text{H}_5 + \text{NaCl}$$