JEE Carboxylic Acids Questions

We need to find the mass of 0.1 mole of compound Z.

Identify gas Q.

RMgBr + H₂O → RH + Mg(OH)Br

Gas Q = RH (hydrocarbon)

At STP: 1 g of Q occupies 1.4 dm³ = 1.4 L

Molar volume at STP = 22.4 L/mol

Molar mass of Q = $$\frac{22.4}{1.4} \times 1 = 16$$ g/mol

So Q = CH₄ (methane, molar mass = 16), which means R = CH₃ (methyl group).

Identify compound Z.

CH₃MgBr + CO₂ (dry ice) → CH₃COOMgBr

CH₃COOMgBr + H₃O⁺ → CH₃COOH (acetic acid)

So Z = CH₃COOH, molar mass = 60 g/mol.

Find mass of 0.1 mole of Z.

Mass = 0.1 × 60 = 6 g

Therefore, 0.1 mole of Z weighs 6 g.

To obtain a dicarboxylic acid (an acid containing two -COOH groups) the reaction conditions must be able to introduce one -COOH group on each of two different carbons of the same molecule. The most common laboratory method for doing this in a single step is oxidative cleavage of a carbon-carbon double bond with a strong oxidising agent such as hot, acidic $$KMnO_4$$ (or, equivalently, ozonolysis followed by oxidative work-up). Each carbon of the former $$C=C$$ bond is oxidised up to the carboxylate level, so both ends of the cleaved bond become -COOH groups.

What happens in Option C
  • The substrate given in Option C contains an internal carbon-carbon double bond.
  • The reagent specified is hot, acidic $$KMnO_4$$ (or an equivalent strong oxidising system).
  • Hot $$KMnO_4$$ performs oxidative cleavage: the $$C=C$$ bond is cut and each of the two carbon atoms is converted into a carbonyl that is then further oxidised to a carboxylic acid.
  • Because both carbons of the former double bond stay inside the same molecular framework after the cleavage, the product contains two -COOH groups that remain connected through the rest of the carbon skeleton. Hence a single molecule of a dicarboxylic acid is obtained as the major product.

Why the other options fail
Option A, Option B and Option D either involve oxidation of **only one functional carbon**, or they involve a reaction (such as haloform cleavage or nitrile hydrolysis) that removes part of the carbon skeleton. In every one of those sequences the molecule finally possesses just one -COOH group; the second carbon is either lost from the molecule or remains at a lower oxidation level. Consequently they cannot supply a dicarboxylic acid as the principal product.

Therefore, the only sequence that reliably converts two different carbons of the same molecule into carboxyl groups is the oxidative-cleavage sequence of Option C.

Option C which is: the oxidative cleavage of the double bond with hot, acidic $$KMnO_4$$ (or its equivalent) is the correct choice.

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 — $$C_2H_4O_2$$

$$CH_3CH_2COOH$$ (propionic acid), $$CH_3COOH$$ (acetic acid), $$CH_3CH_2CH_2COOH$$ (butyric acid), $$HCOOH$$ (formic acid).

Recall the factor affecting acidity

The acidic strength of carboxylic acids depends on the stability of the conjugate base ($$RCOO^-$$). Alkyl groups exhibit a +I (electron-donating inductive) effect, which destabilises the carboxylate anion by pushing electron density towards the negatively charged carboxylate group. More alkyl groups (and longer chains) lead to greater +I effect and thus weaker acidity.

Compare the +I effect

$$HCOOH$$: No alkyl group attached — strongest acid in this series.

$$CH_3COOH$$: One methyl group — moderate +I effect.

$$CH_3CH_2COOH$$: Ethyl group — stronger +I effect than methyl.

$$CH_3CH_2CH_2COOH$$: Propyl group — strongest +I effect, weakest acid.

Arrange in decreasing order of acidic strength

$$HCOOH > CH_3COOH > CH_3CH_2COOH > CH_3CH_2CH_2COOH$$

This matches Option 3.

The three compounds are haloacetic acids: $$A = FCH_2COOH$$, $$B = ClCH_2COOH$$ and $$C = BrCH_2COOH$$.

Step 1 : Why the acidity order is $$A \gt B \gt C$$
When an acid $$HA$$ loses $$H^+$$ it forms its conjugate base $$A^-$$. The stronger the electron-withdrawing (-I) effect of the substituent, the better the negative charge on $$A^-$$ is delocalised and the more stable the anion becomes. Fluorine is the most electronegative of the three halogens followed by chlorine and then bromine. Hence the -I effect decreases in the order
$$F \gt Cl \gt Br$$.
Therefore the stability of the corresponding carboxylate anions and hence the acid strength follow the same sequence:
$$FCH_2COOH \; (A) \gt ClCH_2COOH \; (B) \gt BrCH_2COOH \; (C).$$
Thus Assertion A is correct.

Step 2 : Truth value of Reason R
Reason R states that “Fluoro is a stronger electron-withdrawing group than Chloro.” This statement is factually correct because the electronegativity of fluorine (4.0) is higher than that of chlorine (3.0). So Reason R itself is a true statement.

Step 3 : Does Reason R explain Assertion A?
To explain the complete order $$A \gt B \gt C$$ one must compare the -I effects of all three halogens $$F, Cl$$ and $$Br$$. Reason R compares only fluorine with chlorine and says nothing about bromine. Therefore, although R is correct, it is not a sufficient explanation for the full order stated in Assertion A.

Hence, both Assertion A and Reason R are true, but R is NOT the correct (complete) explanation of A.

Option D which is: Both A and R are correct but R is NOT the correct explanation of A.

Water gas is a mixture of carbon monoxide ($$CO$$) and hydrogen ($$H_2$$), produced by passing steam over hot coke:

$$C + H_2O \rightarrow CO + H_2$$

When water gas is passed over a cobalt ($$Co$$) catalyst under appropriate conditions of temperature and pressure, it produces methanoic acid (formic acid).

The balanced reaction is:

$$CO + H_2O \rightarrow HCOOH$$

Here, carbon monoxide reacts with water (present as steam) in the presence of a cobalt catalyst to form methanoic acid ($$HCOOH$$).

Let us verify why the other options are incorrect:

Option A (Methanal / Formaldehyde, $$HCHO$$): Methanal is obtained by catalytic oxidation of methanol using a silver or iron-molybdenum oxide catalyst. It is not a direct product of water gas with a cobalt catalyst.

Option C (Ethanol, $$C_2H_5OH$$): Ethanol can be produced from syngas but requires different catalysts (such as rhodium-based catalysts) and specific conditions. Cobalt catalyst does not directly produce ethanol from water gas.

Option D (Methanol, $$CH_3OH$$): Methanol is produced from syngas ($$CO + 2H_2$$) using a $$ZnO$$-$$Cr_2O_3$$ catalyst at high pressure. The reaction is $$CO + 2H_2 \rightarrow CH_3OH$$. This requires a different catalyst, not cobalt.

Therefore, the correct answer is Option B: Methanoic acid.

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

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 ($$\text{Ar--CH}_2\text{--CH=CH--CH}_2\text{--CH}_2\text{--NH--R}$$).

Explanation

The starting compound is salicylic acid, which contains a phenolic $$-OH$$ group and a carboxylic acid $$-COOH$$ group.

In reaction X, salicylic acid is converted into acetylsalicylic acid (aspirin). During this transformation, the phenolic $$-OH$$ group is converted into an acetate ester $$(-OCOCH_3)$$, while the carboxylic acid group remains unchanged.

This transformation requires acetylation of the phenolic hydroxyl group. Acetic anhydride in the presence of an acid catalyst is commonly used for this purpose.

Therefore,

$$X=(CH_3CO)_2O/H^+$$

In reaction Y, salicylic acid is converted into methyl salicylate. In this transformation, the carboxylic acid group is converted into a methyl ester $$(-COOCH_3)$$, while the phenolic $$-OH$$ group remains unchanged.

This conversion is achieved by Fischer esterification using methanol in the presence of an acid catalyst and heat.

Therefore, $$Y=CH_3OH/H^+,\Delta$$

Hence, $$X=(CH_3CO)_2O/H^+$$ and $$Y=CH_3OH/H^+,\Delta$$

Therefore, the correct answer is $$\text{Option A}$$

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 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.

Explanation: The reaction begins with iodolactonization.

The carboxylic acid is converted into a carboxylate ion by $$NaHCO_3$$.

The alkene reacts with $$I_2$$ to form an iodonium ion intermediate.

The carboxylate ion attacks intramolecularly to form a stable five-membered lactone ring, placing iodine on the more substituted carbon.

In the next step, pyridine and heat promote dehydrohalogenation $$-HI$$ is eliminated to form a double bond.

Elimination occurs preferentially at the $$\alpha$$-carbon adjacent to the carbonyl group because it produces an $$\alpha,\beta$$-unsaturated lactone. The resulting conjugated system is thermodynamically more stable than non-conjugated alternatives.

Therefore, the major product is the conjugated $$\alpha,\beta$$-unsaturated lactone corresponding to Option C.

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.

The reaction sequence (reagents are given in the paper) converts the starting hydrocarbon first into an aldehyde and finally into two oxidised products P and Q. Alkaline $$\mathrm{KMnO_4}$$ (or hot acidic $$\mathrm{K_2Cr_2O_7}$$) is the last reagent mentioned; such strong oxidising conditions convert every benzylic / aliphatic -CH2- group attached to a carbon-carbon multiple bond into a carboxyl group -COOH.

Therefore the two products obtained after this final oxidation step, P and Q, must both be carboxylic acids.

Hence, statement A
Option A : “Compounds P and Q are carboxylic acids.”
is correct.

Let us examine the other three statements one by one.

Statement B
“Compound S decolorises bromine water.”
Compound S is an aldehyde (formed in the middle of the sequence before the final oxidation). Ordinary aldehydes do not possess any C=C or activated aromatic ring that can add Br2 in aqueous medium, so they do not discharge the reddish-brown colour of bromine water. Hence statement B is incorrect.

Statement C
“Compounds P and S react with hydroxylamine to give the corresponding oximes.”
Hydroxylamine, $$\mathrm{NH_2OH}$$, forms oximes only with aldehydes and ketones. It does not react with carboxylic acids under ordinary conditions. Since P is a carboxylic acid, it will not give an oxime. Therefore the statement claiming that both P and S give oximes is false (only S, the aldehyde, would give an oxime).

Statement D
“Compound R reacts with dialkylcadmium to give the corresponding tertiary alcohol.”
Compound R in the sequence is an acyl chloride, $$\mathrm{RCOCl}$$. Dialkylcadmium reagents, $$\mathrm{(R')_2Cd}$$, react with acyl chlorides to yield ketones ($$\mathrm{RCO\,R'}$$), not tertiary alcohols. Therefore statement D is also incorrect.

Only statement A is correct.

Final Answer: Option A which is: Compounds P and Q are carboxylic acids.

• 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.

First we write the full structure of the given ester. Pent-4-ynoic acid contains five carbon atoms in the main chain and the triple bond starts at carbon-4. Converting the -COOH group into an ethyl ester gives

$$\mathrm{CH_3CH_2C\!\equiv\!C-CO_2CH_2CH_3}$$

Now we examine the nature of each functional group. The molecule has

(i) an ester carbonyl $$-CO_2-$$, and

(ii) a terminal alkyne hydrogen $$\mathrm{\;-C\!\equiv\!C-H}$$ at carbon-5. A terminal alkyne hydrogen is appreciably acidic (pK$$_a\approx25$$) and is removed instantaneously by a Grignard reagent, because every Grignard reagent behaves as a very strong base as well as a nucleophile.

We next recall the two standard facts we shall need.

• Acid-base reaction with a terminal alkyne $$\mathrm{R-C\!\equiv\!C-H + R'MgX \;\longrightarrow\; R-C\!\equiv\!C^{\;-}MgX^{+} + R'H}$$

• Nucleophilic addition of a Grignard reagent to an ester For every mole of an ester, two moles of a Grignard reagent are required to reach the tertiary-alcohol stage:

$$\mathrm{RCOOR' + RMgX \;\xrightarrow{\;(1)\;}\; R-CO-R \;(\text{ketone})}$$

and then

$$\mathrm{R-CO-R + RMgX \;\xrightarrow{\;(2)\;}\; R-C(OMgX)R_2 \;\xrightarrow{\;H_2O\;/\;H^+\;} R-C(OH)R_2}$$

Keeping these two rules in mind, we follow the reaction of ethyl pent-4-yn-oate with methylmagnesium bromide step by step.

Step 1 - deprotonation of the terminal alkyne

$$\mathrm{CH_3CH_2C\!\equiv\!C-H + CH_3MgBr \;\longrightarrow\; CH_3CH_2C\!\equiv\!C^{\;-}MgBr^{+} + CH_4\uparrow}$$

This consumes one mole of $$\mathrm{CH_3MgBr}$$ and produces methane gas. The alkyne is now converted into its magnesium acetylide; the ester function is still intact.

Step 2 - first nucleophilic attack on the ester carbonyl

$$\mathrm{CH_3CH_2C\!\equiv\!C^{\;-}MgBr^{+}\;-CO_2CH_2CH_3 + CH_3MgBr \;\longrightarrow\; CH_3CH_2C\!\equiv\!C-COCH_3 + CH_3CH_2OMgBr}$$

After the first nucleophilic addition and expulsion of the ethoxide ion, a ketone $$\mathrm{RC(=O)CH_3}$$ is formed. A second mole of the Grignard reagent has therefore been used.

Step 3 - second nucleophilic attack on the ketone

$$\mathrm{CH_3CH_2C\!\equiv\!C-COCH_3 + CH_3MgBr \;\longrightarrow\; CH_3CH_2C\!\equiv\!C-C(OMgBr)(CH_3)_2}$$

Hydrolysing the adduct finally gives the alcohol:

$$\mathrm{CH_3CH_2C\!\equiv\!C-C(OH)(CH_3)_2}$$

The carbon bearing the -OH group now carries three alkyl substituents (two $$\mathrm{CH_3}$$ groups and one $$\mathrm{CH_2C\!\equiv\!CCH_2CH_3}$$ group). By definition, such an alcohol is a tertiary (3°) alcohol. Hence Statement I is perfectly correct.

Let us count the number of moles of $$\mathrm{CH_3MgBr}$$ that were consumed overall:

• one mole in Step 1 (acid-base reaction), • one mole in Step 2 (first nucleophilic addition), • one mole in Step 3 (second nucleophilic addition).

Total $$= 3$$ moles of $$\mathrm{CH_3MgBr}$$ per mole of the ester. Statement II, which claims that only two moles are utilised, is therefore incorrect.

We thus conclude that Statement I is true whereas Statement II is false.

Hence, the correct answer is Option 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 starting material in this reaction sequence is a cyanohydrin (specifically, acetone cyanohydrin). To form major product A, the reactant is treated with lithium aluminum hydride (LiAlH₄) followed by acidic workup. LiAlH₄ is a strong reducing agent that specifically targets the nitrile group (-CN) and reduces it to a primary amine (-CH₂NH₂). Because it does not react with tertiary alcohols, the hydroxyl group (-OH) remains completely intact, resulting in an amino alcohol for product A. To form major product B, the reactant is subjected to strongly acidic conditions using hydronium ions and sulfuric acid (H₃O⁺ / H₂SO₄). This triggers two consecutive reactions. First, the acid completely hydrolyzes the nitrile group (-CN) into a carboxylic acid group (-COOH). Immediately following this, the strong dehydrating nature of sulfuric acid causes the elimination of the tertiary hydroxyl group (-OH) along with a hydrogen from an adjacent methyl group. This dehydration forms a stable, conjugated carbon-carbon double bond, yielding an alpha, beta-unsaturated carboxylic acid as product B.

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

The molecular formula of the monobrominated product is $$C_8H_7O_2Br$$.
During bromination one hydrogen atom is replaced by bromine, so the parent compound [P] must be $$C_8H_8O_2$$.

Heating [P] with sodalime (NaOH + CaO) carries out decarboxylation, i.e. $$R-COOH \rightarrow R-H + CO_2$$.
Because the decarboxylation product is toluene ($$C_7H_8$$), [P] must be an aromatic carboxylic acid containing one extra carbon compared to toluene. Therefore [P] is some isomer of $$CH_3-C_6H_4-COOH$$ (methyl-benzoic acid, also called toluic acid).

The three possible isomers are:
  • o-methylbenzoic acid (2-toluic acid)
  • m-methylbenzoic acid (3-toluic acid)
  • p-methylbenzoic acid (4-toluic acid)

We now use the bromination data. Electrophilic bromination in $$CCl_4/FeBr_3$$ is governed by the directing effects of the substituents present:

• $$-COOH$$ is a strong deactivating, meta-directing group.
• $$-CH_3$$ is an activating, ortho/para-directing group.

Case 1: o-methylbenzoic acid
  • The ring positions 3, 4, 5, 6 are all chemically nonequivalent and at least two of them are activated for substitution, so a mixture of brominated isomers would be obtained.

Case 2: m-methylbenzoic acid
  • Again several positions (2, 4, 5, 6) differ in reactivity, giving more than one monobromo product.

Case 3: p-methylbenzoic acid
  • Place $$-COOH$$ at position 1 and $$-CH_3$$ at position 4.
  • Both groups favour substitution at positions 3 and 5:
    - For $$-COOH$$ these are its meta positions.
    - For $$-CH_3$$ these are its ortho positions.
  • Positions 3 and 5 are equivalent by symmetry, so only one distinct monobromo isomer can form.

Thus only p-methylbenzoic acid satisfies the statement “produced a single isomer $$C_8H_7O_2Br$$”.

Therefore, the compound [P] is p-methylbenzoic acid (4-methylbenzoic acid).

Option A which is: p-methylbenzoic acid.

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:

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

Electrical conductivity in aqueous solutions of weak acids depends entirely on the concentration of mobile ions produced. At identical concentrations ($$0.1 \text{ M}$$), a stronger acid dissociates more completely, releasing more ions into the solution and exhibiting a higher electrical conductivity. Therefore, the conductivity order perfectly follows the acid dissociation constant ($$K_a$$) order.

Detailed Evaluation of Each Option:

• Formic Acid (A): $$\text{HCOOH}$$

  Formic acid has no electron-donating alkyl groups attached to the carboxyl group. It has the highest dissociation constant ($$K_a \approx 1.8 \times 10^{-4}$$) among the three, making it the strongest electrolyte and the most conductive solution.




• Benzoic Acid (C): $$\text{C}_6\text{H}_5\text{COOH}$$

  The phenyl ring behaves as a weaker electron-donating group via resonance/inductive pathways compared to a normal aliphatic chain when considering overall carboxylic acid strength ($$K_a \approx 6.3 \times 10^{-5}$$). It is weaker than formic acid but significantly stronger than acetic acid.




• Acetic Acid (B): $$\text{CH}_3\text{COOH}$$

  The methyl group ($$\text{--CH}_3$$) exerts a strong electron-donating inductive effect (+I), which destabilizes the carboxylate anion conjugate base. It has the lowest dissociation constant ($$K_a \approx 1.75 \times 10^{-5}$$), making it the weakest acid and the least conductive solution.

Conclusion:

Comparing the ion concentrations produced at equilibrium gives the decreasing order of electrical conductivity as:

$$\mathbf{(A) > (C) > (B)}$$

Answer: Option B — (A) > (C) > (B)

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

All the unsaturated carbons get completely oxidised to -COOH 

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.

We have an aromatic compound ‘A’ whose molecular formula is $$C_7H_6O_2$$. The four structures proposed in the options are:

(i) Furfural (ii) p-Hydroxy-benzaldehyde (iii) Salicylaldehyde     (iv) Benzoic acid

All the options except furfural are benzenoid and possess the same molecular formula as that of ‘A’, so we must look at the given reactions to decide.

First reaction: “A on treating with aqueous ammonia and heating forms compound ‘B’.” Aqueous ammonia converts a carboxylic acid into its amide when the mixture is heated:
$$\text{R-COOH}+NH_3 \;\xrightarrow{\ \Delta\ }\; \text{R-CONH}_2+H_2O$$

Out of the given structures only benzoic acid (Option D) is a carboxylic acid; the two hydroxy-benzaldehydes contain an aldehyde group while furfural is a hetero-aromatic aldehyde. Hence, if ‘A’ were benzoic acid, the reaction with ammonia would give benzamide:

$$C_6H_5COOH + NH_3 \;\xrightarrow{\ \Delta\ }\; C_6H_5CONH_2 + H_2O$$

Let us check the formula of benzamide so obtained. The ring contributes $$C_6H_5$$ and the CONH2 group contributes one carbon, one oxygen, one nitrogen and two hydrogens. Therefore the total is $$C_7H_7NO$$, exactly what would be expected for ‘B’.

Second reaction: “B on reaction with molecular bromine and KOH gives compound ‘C’ having molecular formula $$C_6H_7N$$.” Bromine in alkaline medium converts an amide into an amine with one carbon atom less: this is the Hofmann rearrangement:

$$\text{R-CONH}_2 \;+\; Br_2 \;+\; 4\,KOH \;\longrightarrow\; \text{R-NH}_2 \;+\; K_2CO_3 \;+\; 2\,KBr \;+\; 2\,H_2O$$

Applying the same to benzamide:

$$C_6H_5CONH_2 \xrightarrow[{KOH}]{Br_2}\; C_6H_5NH_2 + CO_2$$

The amine obtained is aniline; its molecular formula is $$C_6H_7N$$ (six carbons from the ring, five ring hydrogens plus two hydrogens on -NH2). This matches perfectly with the formula given for compound ‘C’, confirming our deduction.

Since the whole sequence of reactions fits only when ‘A’ is benzoic acid, we conclude that:

‘A’ ≡ $$C_6H_5COOH$$, i.e., benzoic acid.

Hence, the correct answer is Option D.

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.

The acidic strength of a carboxylic acid depends on how stable its conjugate base (carboxylate anion) is. Greater stability of the anion means stronger acidity because the equilibrium $$\ce{RCOOH \rightleftharpoons RCOO^- + H^+}$$ lies further to the right.

Stability of the anion is increased by electron-withdrawing groups (EWG, $$-I$$ effect) and decreased by electron-donating groups (EDG, $$+I$$ or $$+R$$ effects). Therefore:

• A strong EWG (-NO2) makes the anion most stable → strongest acid.
• A halogen (-Cl) has a strong $$-I$$ effect but a weak $$+R$$ effect; net result: more acidic than the unsubstituted compound.
• No substituent (-H) is taken as the reference, i.e. benzoic acid itself.
• A methyl group (-CH3) is an EDG due to its $$+I$$ effect; it destabilises the anion → weakest acid.

Assigning the given acids:
  I : p-nitro­benzoic acid  (strong EWG)
  II: benzoic acid  (no substituent)
  III: p-methyl­benzoic acid  (EDG)
  IV: p-chloro­benzoic acid  (halogen EWG)

Accordingly, the stability of their conjugate bases (and hence their acidity) follows the order

$$\text{p-Me-C}_6\text{H}_4\text{COOH} \; \lt \; \text{C}_6\text{H}_5\text{COOH} \; \lt \; \text{p-Cl-C}_6\text{H}_4\text{COOH} \; \lt \; \text{p-NO}_2\text{-C}_6\text{H}_4\text{COOH}$$

That is, $$\text{III} \lt \text{II} \lt \text{IV} \lt \text{I}$$.

Hence the correct increasing order of acidity is Option A: III < II < IV < I.

The starting compound is a primary alcohol. PCC (pyridinium chlorochromate) is a mild oxidising agent that stops at the aldehyde stage.

Step-1 (action of PCC)
$$C_6H_5CH_2CH_2OH \xrightarrow[\text{dichloromethane}]{\text{PCC}} C_6H_5CH_2CHO \;$$ $$-(1)$$
Thus compound $$A$$ is $$C_6H_5CH_2CHO$$ (phenyl-acetaldehyde).

Step-2 (action of hot, alkaline $$KMnO_4$$ followed by acidification)
Alkaline $$KMnO_4$$ is a strong oxidising agent. Any side chain attached to an aromatic ring that still contains at least one benzylic hydrogen is oxidised completely to a carboxyl group $$\bigl(-COOH\bigr)$$.

In $$A$$ the benzylic carbon is the one bearing the aldehydic $$-CHO$$ group in $$C_6H_5CH_2\underline{CHO}$$. It possesses benzylic hydrogens, so the entire -CH₂CHO side chain is shortened and converted to -COOH:

$$C_6H_5CH_2CHO \xrightarrow[\Delta]{KMnO_4/OH^-} C_6H_5COO^-K^+ \xrightarrow{H^+} C_6H_5COOH \;$$ $$-(2)$$

Hence compound $$B$$ is benzoic acid, $$C_6H_5COOH$$.

Among the given structures, benzoic acid corresponds to Option D.

Therefore, the major product $$B$$ is obtained as benzoic acid.
Answer: Option D which is: benzoic acid $$\bigl(C_6H_5COOH\bigr)$$.

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.

The given compound contains a carboxylic acid and a primary alcohol group positioned suitably for an intra-molecular reaction.

In the presence of $$HCL(g)/CCl_4$$, a strongly acidic and non-aqueous medium, the carboxylic acid group gets activated, facilitating intramolecular esterification reaction with the neighboring alcohol group, resulting in the formation of a five-membered lactone ring.

The amide and the phenolic groups remain unaffected under the given conditions. 

Thus, the major product 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.

The structure shown consists of a benzene ring with a carboxylic acid group ($$-COOH$$) and an acetoxy group ($$-OCOCH_3$$) at the ortho position. It's common name is Aspirin.

Primary Medicinal Uses of Aspirin:

Analgesic: To reduce minor to moderate pain (e.g., headaches, toothaches).

Antipyretic: To reduce fever by acting on the hypothalamus.

Anti-inflammatory: To decrease swelling and redness associated with conditions like arthritis.

Anti-platelet: In low doses, it prevents blood clotting, reducing the risk of heart attacks.

Hence, option (D) matches correctly.

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

For every carboxylic acid, its acidic strength depends mainly on the nature of the groups attached to the -COOH moiety.
(i) Electron-withdrawing (-I/-M) groups stabilise the conjugate base $$\left(RCOO^{-}\right)$$ and therefore increase acidity.
(ii) Electron-releasing (+I/+M) groups do the opposite and decrease acidity.

The four reactions given in the paper convert different aldehydes/ketones into four major carboxylic-acid products. From the reagents supplied (Cl2/OH-, HVZ conditions, Tollens oxidation, etc.) the structures of those products are:

Case A: $$CH_{3}COOH \xrightarrow[\text{red P}]{Cl_{2}} CH_{2}ClCOOH$$ (chloro-acetic acid)

Case B: $$CH_{3}CH_{2}CHO \xrightarrow[\text{Tollens}]{[Ag(NH_{3})_{2}]^{+}} CH_{3}CH_{2}COOH$$ (propionic acid)

Case C: $$CH_{3}COCH_{3} \xrightarrow[OH^{-}]{3Cl_{2}} CHCl_{3}+ CCl_{3}COO^{-} \xrightarrow{H^{+}} CCl_{3}COOH$$ (trichloro-acetic acid)

Case D: $$CH_{3}CHO \xrightarrow[\text{Tollens}]{[Ag(NH_{3})_{2}]^{+}} CH_{3}COOH$$ (acetic acid)

Now compare their electron-withdrawing or electron-donating effects:

• C Trichloroacetic acid, $$CCl_{3}COOH$$: three highly electronegative Cl atoms exert a very strong -I effect, greatly stabilising $$CCl_{3}COO^{-}$$. Hence it is the strongest acid of the set.
• A Chloroacetic acid, $$CH_{2}ClCOOH$$: contains one Cl atom. Its -I effect is strong, but weaker than that of three Cl atoms, so it is less acidic than C but more acidic than those lacking Cl altogether.
• D Acetic acid, $$CH_{3}COOH$$: the adjacent methyl group has a weak +I effect, giving an acidity close to that of a simple aliphatic acid.
• B Propionic acid, $$CH_{3}CH_{2}COOH$$: the ethyl group exerts a slightly stronger +I effect than a methyl group, so this conjugate base is the least stabilised and B is the weakest acid.

Therefore the acidic strength decreases in the order

$$CCl_{3}COOH \; (C) \; \gt \; CH_{2}ClCOOH \; (A) \; \gt \; CH_{3}COOH \; (D) \; \gt \; CH_{3}CH_{2}COOH \; (B)$$

The correct option is
Option D which is: C > A > D > B

The sequence of reagents tells us step-by-step what functional group has to be present in each compound.

Step 1 : A → B by $$NaOH/H_3O^+$$
Alkaline hydrolysis followed by acidification converts an amide, nitrile or ester into a carboxylic acid. Because ‘A’ gives a positive Lassaigne’s test for nitrogen, the functional group in A must contain nitrogen. Among the common acid derivatives only amides and nitriles satisfy this.

Molar-mass check for A
Given molar mass of A = 121 g mol$$^{-1}$$. An aromatic amide of formula $$C_6H_5CONH_2$$ (benzamide) has
$$7\;C \;(7\times 12)=84,\; 7\;H=7,\; 1\;N=14,\; 1\;O=16$$
Total $$=84+7+14+16=121\;g\,mol^{-1}$$, exactly the required value. Hence $$A = C_6H_5CONH_2$$ (benzamide).

Step 2 : B gives effervescence with $$NaHCO_3$$
Only acids that are stronger than $$H_2CO_3$$ liberate $$CO_2$$ from $$NaHCO_3$$. The product B obtained after hydrolysis must therefore be a carboxylic acid: $$C_6H_5CONH_2 \xrightarrow{NaOH/H_3O^+} C_6H_5COOH$$ (benzoic acid). Thus $$B = C_6H_5COOH$$.

Step 3 : B → C by $$EtOH/H_2SO_4,\,\Delta$$
Concentrated $$H_2SO_4$$ and hot ethanol carry out Fischer esterification, converting a carboxylic acid into its ethyl ester, which typically has a pleasant fruity odour. $$C_6H_5COOH + HOCH_2CH_3 \xrightarrow{H_2SO_4,\;\Delta} C_6H_5COOCH_2CH_3 + H_2O$$ Hence $$C = C_6H_5COOCH_2CH_3$$ (ethyl benzoate), noted for its sweet fruity smell.

Therefore
A : benzamide $$\left(C_6H_5CONH_2\right)$$
B : benzoic acid $$\left(C_6H_5COOH\right)$$
C : ethyl benzoate $$\left(C_6H_5COOCH_2CH_3\right)$$.

Option A which is: A = benzamide, B = benzoic acid, C = ethyl benzoate.

We need to identify the type of soap/detergent formed when stearic acid reacts with polyethylene glycol.

Detergents are classified into three types based on the nature of their hydrophilic head: (i) anionic detergents carry a negative charge, for example sodium alkyl sulphates and sodium alkylbenzene sulphonates; (ii) cationic detergents carry a positive charge, such as quaternary ammonium salts like cetyltrimethylammonium bromide; and (iii) non-ionic detergents have a hydrophilic part with no charge and are formed by the reaction of a long-chain fatty acid with polyethylene glycol.

Stearic acid $$CH_3(CH_2)_{16}COOH$$ is a long-chain fatty acid, and when it reacts with polyethylene glycol $$HO(CH_2CH_2O)_nH$$ it forms an ester:

This product has a long hydrocarbon chain serving as the hydrophobic tail from stearic acid, and a polyethylene glycol chain serving as the hydrophilic head which is polar due to multiple ether linkages but carries no ionic charge.

Since the hydrophilic part does not carry any charge (neither positive nor negative), this is a non-ionic detergent.

The product formed is a non-ionic detergent; it is not a soap, which requires a metal salt of a fatty acid, nor is it anionic or cationic.

Hence, the correct answer is Option D: Non-ionic detergent.

Looking at the transformation, only the carboxylic acid group ($$\text{--HO}_2\text{C}$$) is converted into a primary alcohol ($$\text{--HOH}_2\text{C}$$), while all other reducible functional groups (amide, ketone, and nitrile) remain completely intact. This requires a highly chemoselective reducing agent.

Diborane ($$\text{B}_2\text{H}_6$$) is uniquely suited for this task. It is an electrophilic reducing agent that reacts exceptionally fast with electron-rich carboxylic acids via a mechanism that involves initial coordination with the carbonyl oxygen followed by hydride delivery. The relative rate of reduction by diborane follows the order:

$$\text{--COOH} > \text{Olefin} > \text{--COCH}_3 >> \text{--COOR} \sim \text{--CN} \sim \text{--CONH}_2$$

Under controlled conditions, $$\text{B}_2\text{H}_6$$ selectively reduces the carboxylic acid group while leaving ketones, nitriles, and amides untouched.

Why Other Options Are Incorrect:

• Option B — $$\text{NaBH}_4$$ (Sodium Borohydride):

  Sodium borohydride is too weak of a reducing agent to reduce carboxylic acids. Instead, it would selectively reduce the ketone ($$\text{--COCH}_3$$) group into a secondary alcohol, leaving the carboxylic acid untouched.




• Option C — $$\text{LiAlH}_4$$ (Lithium Aluminum Hydride):

  Lithium aluminum hydride is an extremely powerful, unselective nucleophilic reducing agent. It would reduce all of the functional groups present: the carboxylic acid to an alcohol, the ketone to a secondary alcohol, the nitrile to a primary amine, and the amide to an amine.




• Option D — $$\text{H}_2/\text{Pd}$$ (Catalytic Hydrogenation):

  Catalytic hydrogenation under normal laboratory conditions does not reduce carboxylic acids, amides, or nitriles. It is typically used for reducing carbon-carbon double/triple bonds, nitro groups, or aldehydes/ketones, making it unreactive or completely non-selective for the desired conversion.

Conclusion:

Diborane ($$\text{B}_2\text{H}_6$$) is the only reagent capable of performing the selective reduction of a carboxylic acid group in the presence of ketone, nitrile, and amide functionalities.

Answer: Option A — $$\text{B}_2\text{H}_6$$

The two reactions given in the question belong to the standard conversion of an aromatic 1,2-dicarboxylic acid into its cyclic anhydride and back to the acid.

Case 1: Formation of the anhydride
Heating any 1,2-dicarboxylic acid (for example, phthalic acid) with a strong dehydrating agent such as $$P_2O_5$$ or concentrated $$H_2SO_4$$ removes one molecule of water intramolecularly and gives the corresponding cyclic anhydride.

The anhydride possesses two C=O bonds - one on each side of the bridging oxygen:
$$\text{O=C-O-C=O}$$

Hence, in the major product $$\mathbf{X}$$ (the anhydride) the total number of carbonyl groups is
$$n_{C=O}(X)=2$$

Case 2: Hydrolysis of the anhydride
When the anhydride $$\mathbf{X}$$ is treated with water (acidic, basic, or even neutral), the ring opens to regenerate the original 1,2-dicarboxylic acid.

Each -COOH group contains one carbonyl bond, so the dicarboxylic acid $$\mathbf{Y}$$ again has two C=O bonds:

$$n_{C=O}(Y)=2$$

Sum of carbonyl groups
$$n_{C=O}(X)+n_{C=O}(Y)=2+2=4$$

Therefore, the required sum is 4.

Final answer: 4.00

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 test with aqueous $$NaHCO_{3}$$ is an acid-base reaction. It is positive (effervescence of $$CO_{2}$$) only for those organic compounds whose acidity is stronger than that of carbonic acid $$\bigl($$ $$pK_{a}\,\approx\,6.37$$ $$\bigr)$$. In the syllabus this essentially restricts the list to:

• Carboxylic acids $$\left( \,\ce{R-COOH}\, \right)$$
• Sulfonic acids $$\left( \,\ce{R-SO_{3}H}\, \right)$$ (rare in JEE questions)
Phenols, enols, alcohols, aldehydes and ketones are all too weakly acidic to react with $$NaHCO_{3}$$.

Now inspect the five compounds given in the question:

Thus only compounds A, C and D satisfy the condition.

Option A which is: A, C and D Only

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).

Effervescence of $$CO_2$$ with aqueous $$NaHCO_3$$ is observed only when the organic compound can donate a proton more acidic than the first proton of carbonic acid $$H_2CO_3$$.

The equilibrium that decides the test is
$$\text{HA} + NaHCO_3 \;\rightleftharpoons\; NaA + H_2CO_3 \;\rightleftharpoons\; NaA + CO_2\uparrow + H_2O$$
For the reaction to proceed to the right, the acid strength order must be $$\text{p}K_a(\text{HA}) \lt \text{p}K_a(H_2CO_3)\;( \approx 6.4 ).$$

Now examine each option:

Thus the compound least likely to give effervescence of $$CO_2$$ with aqueous $$NaHCO_3$$ is the phenol in Option D.

Final answer: Option D which is: p-Cresol (or any phenolic compound shown in Option D).

We need to find the mass of benzoic acid that produces $$CO_2$$ occupying 11.2 L at STP.

Benzoic acid reacts with aqueous sodium bicarbonate: $$C_6H_5COOH + NaHCO_3 \rightarrow C_6H_5COONa + H_2O + CO_2 \uparrow$$ The stoichiometry shows that 1 mole of benzoic acid produces 1 mole of $$CO_2$$.

At STP (Standard Temperature and Pressure: 0 degC, 1 atm), 1 mole of any ideal gas occupies 22.4 L. $$\text{Moles of } CO_2 = \frac{11.2 \text{ L}}{22.4 \text{ L/mol}} = 0.5 \text{ mol}$$

From the 1:1 stoichiometry: $$\text{Moles of benzoic acid} = \text{Moles of } CO_2 = 0.5 \text{ mol}$$

The molar mass of benzoic acid ($$C_6H_5COOH = C_7H_6O_2$$) is $$(7 \times 12) + (6 \times 1) + (2 \times 16) = 84 + 6 + 32 = 122 \text{ g/mol}$$

$$X = 0.5 \times 122 = 61 \text{ g}$$

The answer is 61 g.

Given sequence uses:
NH₃/Δ → amide formation, Br₂/KOH → Hofmann bromamide degradation, CHCl₃/KOH → Carbylamine reaction (isocyanide formation)

A: Carboxylic acid (cyclohexyl-COOH)
NH₃/Δ → cyclohexyl-CONH₂ (amide formation)
Br₂/KOH → cyclohexyl-NH₂ (Hofmann degradation, one C less)
CHCl₃/KOH → cyclohexyl-NC (Carbylamine)
Gives required product → correct

B: Aldehyde (cyclohexyl-CH₂-CHO)
NH₃ does not convert aldehyde to amide under these conditions
No proper substrate for Hofmann reaction
Fails

C: Ketone (cyclohexyl-CO-CH₃)
NH₃ forms imine, not amide
Hofmann reaction needs amide → not possible
Fails

D: Amide (cyclohexyl-CONH₂)
Already amide, so NH₃/Δ does nothing
Br₂/KOH → cyclohexyl-NH₂
CHCl₃/KOH → cyclohexyl-NC
But initial step is unnecessary and does not match sequence properly
Not correct as per given pathway

Final answer: A

To determine the major product P, we analyze each step of the reaction sequence starting from propanenitrile $$CH_3CH_2CN$$.

Treatment of a nitrile with aqueous base followed by acidification results in complete hydrolysis of the nitrile group to a carboxylic acid. Thus,

$$CH_3CH_2CN \xrightarrow{OH^-/H_2O,\ \Delta} CH_3CH_2COO^- \xrightarrow{H_3O^+} CH_3CH_2COOH.$$

Therefore, the intermediate formed is propanoic acid.

The next step involves treatment with $$Cl_2$$ and red phosphorus followed by hydrolysis, which is the Hell-Volhard-Zelinsky (HVZ) reaction. This reaction selectively replaces an $$\alpha$$-hydrogen of a carboxylic acid with a chlorine atom. Since propanoic acid contains two $$\alpha$$-hydrogens on the carbon adjacent to the carboxyl group, one of them is substituted by chlorine to give

$$CH_3CH_2COOH \xrightarrow{Cl_2,\ \text{Red P}} CH_3CHClCOOH.$$

Hence, the major product P is 2-chloropropanoic acid with the structure

$$CH_3CHClCOOH.$$

Therefore, the correct answer is 2-chloropropanoic acid, corresponding to option A.

$$\text{CH}_3\text{Cl} / \text{Anhydrous AlCl}_3$$ (Friedel-Crafts Alkylation):

Benzene reacts with methyl chloride in the presence of a Lewis acid to form Toluene ($$\text{C}_6\text{H}_5\text{CH}_3$$).

$$\text{Cl}_2 / \text{FeCl}_3$$ (Electrophilic Aromatic Substitution): The methyl ($$-\text{CH}_3$$) group on toluene is an ortho/para-directing group. Due to steric hindrance at the ortho-position, the para-isomer is formed as the major product. This gives $$p$$-chlorotoluene.

$$\text{K}_2\text{Cr}_2\text{O}_7 / \text{H}_2\text{SO}_4$$ (Oxidation):

Potassium dichromate in an acidic medium is a strong oxidizing agent. It oxidizes the benzylic alkyl group ($$-\text{CH}_3$$) completely into a carboxylic acid group ($$-\text{COOH}$$). The chlorine atom remains unaffected.

Thus, the major product ($$X$$) is $$p$$-chlorobenzoic acid ($$\text{Cl}-\text{C}_6\text{H}_4-\text{COOH}$$).

The molecular formula of $$p$$-chlorobenzoic acid is $$\text{C}_7\text{H}_5\text{O}_2\text{Cl}$$.

Using the given atomic masses:

• $$\text{C} = 12 \times 7 = 84$$
• $$\text{H} = 1 \times 5 = 5$$
• $$\text{O} = 16 \times 2 = 32$$
• $$\text{Cl} = 35.5 \times 1 = 35.5$$

$$\text{Molar Mass of } X = 84 + 5 + 32 + 35.5 = 157.5\text{ g mol}^{-1}$$

Carboxylic acids react with sodium bicarbonate ($$\text{NaHCO}_3$$) to liberate carbon dioxide ($$\text{CO}_2$$) gas according to the following balanced equation:

$$\text{R-COOH} + \text{NaHCO}_3 \rightarrow \text{R-COONa} + \text{H}_2\text{O} + \text{CO}_2\uparrow$$

From the stoichiometry of the reaction:

$$\text{1 mole of } \text{R-COOH} \text{ liberates } \text{1 mole of } \text{CO}_2\text{ gas}$$

• Given volume of $$\text{CO}_2$$ liberated = $$11.2\text{ dm}^3$$
• Moles of $$\text{CO}_2$$ = $$\frac{11.2}{22.4} = 0.5\text{ moles}$$

Since the molar ratio between $$p$$-chlorobenzoic acid and $$\text{CO}_2$$ is $$1:1$$:

$$\text{Moles of major product } (X) = 0.5\text{ moles}$$

Now, find the mass $$P$$:

$$\text{Mass } P = \text{Moles} \times \text{Molar Mass}$$ 

$$P = 0.5 \times 157.5 = 78.75\text{ g}$$

Step 1: NaOEt
Reaction name: Elimination (E2 dehydrohalogenation)

• Removes HBr from the side chain
• Forms an alkene (styrene type: Ph-CH=CH₂)

Step 2: B₂H₆ / H₂O₂
Reaction name: Hydroboration-oxidation

• Adds OH in anti-Markovnikov manner
• Converts alkene to primary alcohol (Ph-CH₂-CH₂-OH)

Step 3: Jones reagent (CrO₃ / H⁺)
Reaction name: Oxidation of primary alcohol

• Converts -CH₂OH → -COOH
• Forms Ph-CH₂-COOH