Properties of Carboxylic Acids and Their Derivatives Questions in English

(C) The reaction of a nitrile (cyclohexanecarbonitrile) with water in the presence of an acid $(H^+)$ leads to partial hydrolysis,which produces an amide as the major product $A$.
The reaction is as follows:
$R-CN + H_2O \xrightarrow{H^+} R-CONH_2$ (Partial Hydrolysis)
In this case,the nitrile group attached to the cyclohexane ring is converted to an amide group $(-CONH_2)$.
Therefore,product $A$ is cyclohexanecarboxamide.

(C) Sodium bicarbonate $(NaHCO_3)$ is a weak base. It reacts with acids that are stronger than carbonic acid ($H_2CO_3$,$pK_a \approx 6.35$) to liberate $CO_2$ gas.
$1$. $B$ (Benzoic acid,$pK_a \approx 4.2$) is a stronger acid than $H_2CO_3$,so it reacts with $NaHCO_3$ to liberate $CO_2$.
$2$. $C$ (Picric acid,$pK_a \approx 0.38$) is a very strong acid due to the electron-withdrawing effect of three $-NO_2$ groups. It is much stronger than $H_2CO_3$ and thus liberates $CO_2$ with $NaHCO_3$.
$3$. $A$ ($2$,$4$,$6$-triamino phenol) is a very weak acid because the $-NH_2$ groups are electron-donating by resonance,which destabilizes the phenoxide ion. It is weaker than $H_2CO_3$ and does not liberate $CO_2$ with $NaHCO_3$.
Therefore,both $B$ and $C$ will liberate $CO_2$.

(D) The acidic strength depends on the stability of the conjugate base formed after the loss of a proton $(H^+)$.
$1.$ Phenol $(I)$ is much less acidic than carboxylic acids $(II, III, IV)$.
$2.$ Among the carboxylic acids,electron-withdrawing groups $(-NO_2)$ increase acidity,while electron-donating groups $(-CH_3)$ decrease it.
$3.$ Compound $II$ ($p$-nitrobenzoic acid) has a strong $-I$ and $-M$ effect,making it the most acidic.
$4.$ Compound $III$ (benzoic acid) is the reference.
$5.$ Compound $IV$ ($p$-toluic acid) has a $+I$ and $+H$ effect,making it the least acidic among the carboxylic acids.
$6.$ Therefore,the order is $II > III > IV > I$.

(D) $NaBH_4$ (Sodium borohydride) is a selective reducing agent that reduces aldehydes and ketones to their corresponding alcohols but does not reduce esters,carboxylic acids,or amides under normal conditions. In the given substrate,there is a ketone group and an ester group $(-COOCH_3)$. Therefore,$NaBH_4$ will selectively reduce the ketone group to a secondary alcohol $(-CHOH)$ while leaving the ester group unaffected. The final product is a cyclohexane ring with an $-OH$ group,a $-CH_2-COOCH_3$ group,and a $-CH_3$ group.

(A) The reactivity of carboxylic acid derivatives towards nucleophilic acyl substitution (like hydrolysis) depends on the leaving group ability. The better the leaving group,the more reactive the compound.
The leaving group abilities are: $Cl^- > RCOO^- > RO^- > NH_2^-$.
Therefore,the order of reactivity towards hydrolysis is:
$A$ (Acyl chloride) > $B$ (Acid anhydride) > $C$ (Ester) > $D$ (Amide).
Thus,the correct order is $A > B > C > D$.

(B) Maleic anhydride is formed by the dehydration of $cis$-but-$2$-enedioic acid (maleic acid).
When $cis$-but-$2$-enedioic acid is heated,it undergoes intramolecular dehydration to form a cyclic anhydride known as maleic anhydride.
The reaction is as follows:
$cis$-but-$2$-enedioic acid $\xrightarrow{\Delta}$ Maleic anhydride + $H_2O$.
$trans$-but-$2$-enedioic acid (fumaric acid) does not form an anhydride easily upon heating because the carboxylic acid groups are on opposite sides of the double bond,making the formation of a cyclic structure sterically unfavorable.

(A) The given reaction is a nucleophilic acyl substitution reaction. The rate of this reaction depends on the leaving group ability of $Z^-$. $A$ better leaving group makes the reaction faster. The leaving group ability is inversely proportional to the basicity of the group. Among the given options,$Cl^-$ is the weakest base and thus the best leaving group. Therefore,the reaction is fastest when $Z = Cl$.

(C) The strength of an acid is determined by the stability of its conjugate base. Electron-withdrawing groups $(EWG)$ stabilize the carboxylate anion $(RCOO^-)$ through the inductive effect ($-I$ effect),thereby increasing the acidity of the parent carboxylic acid.
Comparing the substituents:
$1$. $NCCH_2-$: Contains a cyano group $(-CN)$,which has a $-I$ effect.
$2$. $O_2NCH_2-$: Contains a nitro group $(-NO_2)$,which has a strong $-I$ effect.
$3$. $F_3C-$: Contains three fluorine atoms. Fluorine is the most electronegative element,and the $-I$ effect of three fluorine atoms is significantly stronger than that of a single $-NO_2$ or $-CN$ group.
$4$. $Cl_3C-$: Contains three chlorine atoms. Chlorine is less electronegative than fluorine,so the $-I$ effect of $Cl_3C-$ is weaker than that of $F_3C-$.
Since fluorine is the most electronegative,$F_3CCOOH$ exhibits the strongest $-I$ effect,making its conjugate base the most stable and the acid the strongest among the given options.

(A) $I.$ Monocarboxylic acids with an even number of carbon atoms exhibit better packing efficiency in the crystal lattice,which results in a higher melting point $(M.P.)$ compared to their odd-numbered counterparts.
$II.$ As the molar mass of monocarboxylic acids increases,the size of the hydrophobic alkyl chain increases,which reduces the interaction with water molecules,thereby decreasing solubility.

(C) The conversion of benzoic acid to benzaldehyde in one step can be achieved by heating benzoic acid with formic acid in the presence of manganese oxide $(MnO)$ as a catalyst at high temperatures. The reaction is as follows:
$C_6H_5COOH + HCOOH \xrightarrow{MnO, \Delta} C_6H_5CHO + CO_2 + H_2O$
Thus,$MnO$ is the correct reagent.

(A) $R-OH + R'-COOH \longrightarrow R'-COOR + H_2O$
This reaction is a nucleophilic acyl substitution reaction where the alcohol acts as a nucleophile.
Electron withdrawing groups $(EWG)$ on the carboxylic acid increase the electrophilicity of the carbonyl carbon by pulling electron density away,which facilitates the attack of the nucleophile,thereby increasing the rate of esterification.

(B) The reaction of $methylenecyclohexane$ with $B_2H_6$ followed by $H_2O_2/NaOH$ is a hydroboration-oxidation reaction,which follows anti-$Markownikoff$ addition of water to the double bond. This results in the formation of a primary alcohol,$cyclohexylmethanol$ $(X)$.
$CrO_3$ in the presence of $H_2SO_4$ is known as Jones reagent,which is a strong oxidizing agent. It oxidizes primary alcohols directly to carboxylic acids. Therefore,$cyclohexylmethanol$ is oxidized to $cyclohexanecarboxylic acid$ $(Y)$.

(C) The reaction between ethyl acetate $(CH_3COOCH_2CH_3)$ and semicarbazide $(NH_2NHCONH_2)$ is a nucleophilic acyl substitution reaction.
$NH_2NHCONH_2$ acts as a nucleophile and attacks the electrophilic carbonyl carbon of the ethyl acetate.
This leads to the elimination of ethanol $(CH_3CH_2OH)$ and the formation of acetyl semicarbazide $(CH_3CONHNHCONH_2)$.

(A) The acidic strength of a compound is directly proportional to the stability of its conjugate base. More stable the conjugate base,stronger is the acid.
$I$: Benzyl alcohol $(C_6H_5CH_2OH)$ forms a conjugate base where the negative charge is on the oxygen atom and is not resonance-stabilized by the benzene ring.
$II$: Benzoic acid $(C_6H_5COOH)$ forms a carboxylate ion $(C_6H_5COO^-)$ which is resonance-stabilized by two oxygen atoms.
$III$: $p$-Cresol $(CH_3-C_6H_4-OH)$ forms a phenoxide ion which is resonance-stabilized,but the electron-donating $+I$ effect of the $-CH_3$ group destabilizes it compared to benzoic acid.
$IV$: Benzenesulfonic acid $(C_6H_5SO_3H)$ forms a sulfonate ion $(C_6H_5SO_3^-)$ where the negative charge is delocalized over three highly electronegative oxygen atoms,making it the most stable conjugate base.
Comparing the stability of conjugate bases:
$I$ (least stable) $< III < II < IV$ (most stable).
Therefore,the order of acidity is $I < III < II < IV$.

(B) The correct option is $(B)$.
In the first reaction,$AlCl_3$ acts as a Lewis acid catalyst to facilitate the Fries rearrangement. This reaction involves the migration of the acyl group from the phenolic oxygen to the ortho position of the aryl ring,forming an ortho-hydroxyacetophenone derivative.
In the second reaction,$I_2 / NaOH$ acts as a reagent for the haloform reaction. The methyl ketone group $(-COCH_3)$ attached to the aromatic ring is oxidized to a carboxylic acid group $(-COOH)$ while producing iodoform $(CHI_3)$ as a byproduct.

(A) The correct answer is $A$.
Oxalic acid $(COOH)_2$ undergoes dehydration when heated with concentrated sulfuric acid $(H_2SO_4)$.
The chemical reaction is as follows:
$(COOH)_2 \xrightarrow[H_2SO_4]{\Delta} CO + CO_2 + H_2O$
Here,concentrated $H_2SO_4$ acts as a dehydrating agent,removing a molecule of water from the oxalic acid,resulting in the formation of carbon monoxide $(CO)$,carbon dioxide $(CO_2)$,and water $(H_2O)$.

(A) The correct option is $A$.
When acetic acid $(CH_3COOH)$ reacts with ethanol $(C_2H_5OH)$ in the presence of an acid catalyst like hydrochloric acid $(HCl)$,an esterification reaction occurs.
This reaction produces ethyl acetate $(CH_3COOC_2H_5)$,which is a sweet-smelling compound,along with water $(H_2O)$.
The chemical equation is: $CH_3COOH + C_2H_5OH \xrightarrow{HCl} CH_3COOC_2H_5 + H_2O$.

(B) The correct option is $B$.
When a carboxylic acid reacts with an active metal like sodium,it undergoes a displacement reaction to release hydrogen gas.
The chemical equation for the reaction between acetic acid $(CH_3COOH)$ and sodium metal $(Na)$ is:
$2CH_3COOH + 2Na \longrightarrow 2CH_3COONa + H_2 \uparrow$

(D) The compounds are analyzed for their solubility in cold $NaHCO_3$,cold $NaOH$,and hot $NaOH$:
$1$. $p$-Toluic acid: Dissolves in $NaHCO_3$ and $NaOH$.
$2$. Methyl benzoate: Does not dissolve in cold $NaHCO_3$ or $NaOH$,but undergoes hydrolysis in hot $NaOH$ to form benzoic acid and methanol,thus dissolving.
$3$. $p$-Cresyl acetate: Does not dissolve in cold $NaHCO_3$ or $NaOH$,but undergoes hydrolysis in hot $NaOH$ to form $p$-cresol and acetic acid,thus dissolving.
$4$. $o$-Hydroxyacetophenone: Dissolves in $NaOH$ (phenolic group) but not $NaHCO_3$.
$5$. $3$-Hydroxy$-5-$methylbenzaldehyde: Dissolves in $NaOH$ (phenolic group) but not $NaHCO_3$.
$6$. $1$-Isopropyl$-3-$methoxybenzene: Does not dissolve in any of these.
$7$. Phenyl acetate: Does not dissolve in cold $NaHCO_3$ or $NaOH$,but undergoes hydrolysis in hot $NaOH$ to form phenol and acetic acid,thus dissolving.
Therefore,the compounds that do not dissolve in cold $NaHCO_3$ and $NaOH$ but dissolve in hot $NaOH$ are $2$,$3$,and $7$. The total count is $3$.

(C) $LiBH_4$ is a selective reducing agent that reduces esters to primary alcohols but does not reduce carboxylic acids. In the given molecule,the ester group $(-CO_2Et)$ is reduced to a primary alcohol $(-CH_2OH)$,while the carboxylic acid group $(-CO_2H)$ remains unaffected. Thus,the major product is the one where the ester is converted to an alcohol and the acid remains as it is.

(D) The acidity of carboxylic acids is determined by the stability of the carboxylate anion formed after the loss of a proton. Electron-withdrawing groups ($-I$ effect) increase acidity by stabilizing the negative charge,while electron-donating groups ($+I$ effect) decrease acidity.
$1$. $B$ $(F_3C-COOH)$: Contains three highly electronegative fluorine atoms,exerting a strong $-I$ effect,making it the most acidic.
$2$. $D$ $(FCH_2-COOH)$: Contains one fluorine atom,which has a stronger $-I$ effect than chlorine or bromine.
$3$. $C$ $(ClCH_2-COOH)$: Contains a chlorine atom,which has a weaker $-I$ effect than fluorine but stronger than bromine.
$4$. $E$ $(BrCH_2-COOH)$: Contains a bromine atom,which has the weakest $-I$ effect among the halogens listed.
$5$. $A$ $(CH_3COOH)$: Contains a methyl group,which exerts a $+I$ effect,destabilizing the carboxylate anion and making it the least acidic.
Thus,the descending order of acidity is $B > D > C > E > A$.

(D) The reaction sequence is as follows:
$1$. Treatment of the carboxylic acid with $SOCl_2$ converts it into an acid chloride: $Ph-CH=CH-CH_2-COOH \xrightarrow{SOCl_2} Ph-CH=CH-CH_2-COCl$.
$2$. Reaction with $R-NH_2$ (amine) converts the acid chloride into an amide: $Ph-CH=CH-CH_2-COCl \xrightarrow{R-NH_2} Ph-CH=CH-CH_2-CONHR$.
$3$. Reduction with $LiAlH_4$ followed by $H_3O^+$ reduces the amide group $(-CONHR)$ to an amine group $(-CH_2NHR)$: $Ph-CH=CH-CH_2-CONHR \xrightarrow{LiAlH_4, H_3O^+} Ph-CH=CH-CH_2-CH_2NHR$.
Thus,the final product is $Ph-CH=CH-CH_2-CH_2NHR$.

(A) The given reactant is $N$-methylpyrrolidin$-2-$one,which is a cyclic amide (lactam).
Treatment with $NaOH$ and $\Delta$ (heating) causes the hydrolysis of the amide bond.
The hydroxide ion $(OH^-)$ attacks the carbonyl carbon,leading to the ring-opening of the lactam.
This results in the formation of the carboxylate salt of the corresponding amino acid,$CH_3NH(CH_2)_3COO^-Na^+$.
Subsequent acidification with $H^+$ protonates the carboxylate group to form the carboxylic acid,$CH_3NH(CH_2)_3COOH$.

(A) The reaction sequence is as follows:
$1$. Cyclohexanone $[A]$ reacts with $HCN$ to form a cyanohydrin,which is $1$-hydroxycyclohexanecarbonitrile $[B]$.
$2$. The treatment of $[B]$ with $conc. H_2SO_4$ and heat $(\Delta)$ leads to two simultaneous processes: dehydration of the tertiary alcohol group to form a double bond and hydrolysis of the nitrile group $(-CN)$ to a carboxylic acid group $(-COOH)$.
$3$. The final product $[C]$ is cyclohex-$1$-ene-$1$-carboxylic acid.

(D) $1$. The starting material is propylbenzene. Oxidation with $KMnO_4-KOH$ followed by acidification $(H_3O^+)$ converts the alkyl side chain into a carboxylic acid group,yielding benzoic acid $(B)$.
$2$. Benzoic acid is then subjected to electrophilic aromatic substitution with $Br_2$ in the presence of $FeBr_3$. Since the $-COOH$ group is meta-directing,the product $(C)$ is $m$-bromobenzoic acid.
$3$. The structure of $m$-bromobenzoic acid contains a benzene ring (which has $3$ $\pi$ bonds) and a carbonyl group ($C=O$,which has $1$ $\pi$ bond).
$4$. Total $\pi$ bonds $= 3 + 1 = 4$.

(A) The reaction shown is the Hell-Volhard-Zelinsky $(HVZ)$ reaction.
In this reaction,carboxylic acids having an $\alpha$-hydrogen atom are reacted with $Br_2$ or $Cl_2$ in the presence of a small amount of red phosphorus to form $\alpha$-halo carboxylic acids.
The $\alpha$-carbon is the carbon atom directly attached to the carboxyl group.
In cyclopentanecarboxylic acid,the carbon atom of the cyclopentane ring attached to the $-COOH$ group is the $\alpha$-carbon.
Therefore,the bromine atom replaces the hydrogen atom at the $\alpha$-position,resulting in $1$-bromocyclopentanecarboxylic acid.

(A) The acidic strength of carboxylic acids depends on the electron-donating effect of the alkyl group attached to the $-COOH$ group.
Alkyl groups exhibit a $+I$ (positive inductive) effect,which destabilizes the carboxylate anion and decreases the acidic strength.
As the size of the alkyl group increases,the $+I$ effect increases,thereby decreasing the acidic strength.
$HCOOH$ has no alkyl group,so it is the strongest acid.
Comparing the others: $CH_3COOH$ (methyl group) > $CH_3CH_2COOH$ (ethyl group) > $CH_3CH_2CH_2COOH$ (propyl group).
Therefore,the correct decreasing order is: $HCOOH > CH_3COOH > CH_3CH_2COOH > CH_3CH_2CH_2COOH$.

(A) The starting material is methyl $4-$cyanobenzoate. It contains two functional groups: a nitrile $(-CN)$ group and an ester $(-COOCH_3)$ group. Both groups react with Grignard reagents $(CH_3MgBr)$.
$1$. The ester group reacts with two equivalents of $CH_3MgBr$ to form a tertiary alcohol after acidic workup $(H_3O^+)$. The ester $-COOCH_3$ is converted to $-C(OH)(CH_3)_2$.
$2$. The nitrile group reacts with one equivalent of $CH_3MgBr$ to form an imine intermediate,which upon hydrolysis $(H_3O^+)$ yields a ketone. The $-CN$ group is converted to $-COCH_3$.
Therefore,the final product is $4-$($1$-hydroxy$-1-$methylethyl)acetophenone,which corresponds to the structure shown in option $A$.

(A) $1.$ $\text{Hex-3-ynal}$ $(CH_3-CH_2-C \equiv C-CH_2-CHO)$ reacts with $NaBH_4$ to reduce the aldehyde to a primary alcohol $(CH_3-CH_2-C \equiv C-CH_2-CH_2OH)$,followed by $PBr_3$ to convert the alcohol to a bromide. Thus,$I$ is $CH_3-CH_2-C \equiv C-CH_2-CH_2Br$,which corresponds to structure $(D)$.
$2.$ The Grignard reagent formation from $I$ followed by reaction with $CO_2$ and acidic workup yields the carboxylic acid $J$ $(CH_3-CH_2-C \equiv C-CH_2-CH_2-COOH)$. The conversion of $J$ to the acid chloride requires $SOCl_2$ $(K)$. Thus,$J$ corresponds to structure $(B)$ and $K$ is $SOCl_2$. The pair $(J, K)$ is $(B, SOCl_2)$,which matches option $(C)$.
$3.$ The final step is the partial hydrogenation of the alkyne to a cis-alkene using Lindlar's catalyst $(H_2, Pd/BaSO_4, \text{quinoline})$. The product $L$ is a cis-alkene with an aldehyde group,which corresponds to structure $(C)$.
Therefore,the correct sequence for $(1, 2, 3)$ is $(D, C, C)$.

(A) The acidity of carboxylic acids depends on the stability of the carboxylate anion formed after the loss of a proton.
$1$. $2,6$-Dihydroxybenzoic acid $(I)$ is the most acidic because its carboxylate anion is stabilized by intramolecular hydrogen bonding with two $-OH$ groups,which is more effective than the stabilization in salicylic acid $(II)$ which has only one $-OH$ group.
$2$. Salicylic acid $(II)$ is more acidic than $m$-hydroxybenzoic acid $(III)$ due to the ortho-effect and strong intramolecular hydrogen bonding.
$3$. Between $m$-hydroxybenzoic acid $(III)$ and $p$-hydroxybenzoic acid $(IV)$,the $-OH$ group at the $m$-position exerts only a $-I$ effect (electron-withdrawing),whereas at the $p$-position,it exerts a strong $+R$ effect (electron-donating),which destabilizes the carboxylate anion. Therefore,$III$ is more acidic than $IV$.
Thus,the correct order of acidity is $I > II > III > IV$.

(C) The acidity of carboxylic acids is significantly influenced by substituents on the benzene ring.
$o$-hydroxybenzoic acid (salicylic acid) exhibits the ortho-effect,where the proximity of the hydroxyl group to the carboxyl group stabilizes the conjugate base through intramolecular hydrogen bonding and steric factors,making it more acidic than the other options provided.

(B) Decarboxylation of $\beta$-keto acids occurs readily because the transition state is stabilized by the formation of an enol intermediate,which is further stabilized by the electron-withdrawing effect of the carbonyl group. Among the given options,$2$-oxocyclohexane-$1$-carboxylic acid (option $B$) is a $\beta$-keto acid. The decarboxylation process involves the formation of a cyclic six-membered transition state,which is highly favorable. The electron-withdrawing $-I$ and $-M$ effects of the ketone carbonyl group stabilize the developing negative charge on the $\alpha$-carbon during the transition state,making it the most reactive compound for decarboxylation under mild conditions.

(A) $1$. The reaction of butan$-2-$one $(CH_3-CH_2-CO-CH_3)$ with $CN^-$ (cyanide ion) is a nucleophilic addition reaction,which forms a cyanohydrin intermediate $(G)$: $CH_3-CH_2-C(OH)(CN)-CH_3$.
$2$. Treatment of the cyanohydrin $(G)$ with concentrated $95\% \ H_2SO_4$ and heat leads to two simultaneous processes: hydrolysis of the nitrile $(-CN)$ group to a carboxylic acid $(-COOH)$ group and dehydration of the alcohol $(-OH)$ group to form a double bond.
$3$. The dehydration follows $Saytzeff$ rule,forming the more substituted alkene. The resulting product $H$ is $CH_3-CH=C(CH_3)-COOH$ ($2$-methylbut$-2-$enoic acid).

(C) Compounds that are stronger acids than carbonic acid $(H_2CO_3)$ react with aqueous $NaHCO_3$ to evolve $CO_2$ gas.
$A$. Benzoic acid $(pK_a \approx 4.2)$ is stronger than $H_2CO_3$ $(pK_a \approx 6.35)$.
$C$. $2,4,6$-Trinitrophenol (Picric acid) $(pK_a \approx 0.38)$ is a very strong acid,much stronger than $H_2CO_3$.
$D$. Cyclohexanecarboxylic acid $(pK_a \approx 4.9)$ is stronger than $H_2CO_3$.
Phenol $(B)$ and $4$-Methoxyphenol $(E)$ are weaker acids than $H_2CO_3$ and do not react with $NaHCO_3$ to evolve $CO_2$.
Therefore,compounds $A, C,$ and $D$ produce $CO_2$ with aqueous $NaHCO_3$ solution.

(D) The rate of hydrolysis of carboxylic acid derivatives depends on the leaving group ability of the group attached to the carbonyl carbon.
Better leaving groups make the carbonyl carbon more electrophilic and facilitate the nucleophilic attack by water.
The leaving group ability order is $Cl^- > CH_3COO^- > CH_3CH_2O^- > NH_2^-$.
Therefore,the rate of hydrolysis follows the order $:$
$(p) \ (acid \ chloride) > (q) \ (anhydride) > (r) \ (ester) > (s) \ (amide)$.
Thus,the correct order is $p > q > r > s$.

(C) The acidic strength of carboxylic acids is determined by the stability of the carboxylate anion formed after the loss of a proton. $Electron-donating$ groups ($+I$ effect) destabilize the carboxylate anion by increasing the electron density on the oxygen atom,thereby decreasing the acidic strength. $Electron-withdrawing$ groups ($-I$ effect) stabilize the carboxylate anion,thereby increasing the acidic strength.
The order of $+I$ effect of alkyl groups is: $-(CH_3)_3C > -(CH_3)_2CH > -CH_3 > -H$.
Since the $+I$ effect increases in the order $H < CH_3 < (CH_3)_2CH < (CH_3)_3C$,the acidic strength decreases in the order: $HCOOH > CH_3COOH > (CH_3)_2CHCOOH > (CH_3)_3CCOOH$.

(B) The $pK_{a}$ value is inversely proportional to the acidic strength of the carboxylic acid.
Acidic strength increases with the presence of electron-withdrawing groups $(EWG)$ and decreases with electron-donating groups $(EDG)$.
Comparing the compounds:
$(S) \ C_{2}H_{5}COOH$ has an ethyl group $(EDG)$,which decreases acidity,making it the weakest acid.
$(P) \ CH_{3}COOH$ has a methyl group $(EDG)$,which is less donating than the ethyl group,making it stronger than $(S)$.
$(Q) \ FCH_{2}COOH$ and $(R) \ ClCH_{2}COOH$ have halogen atoms $(EWG)$. Since $F$ is more electronegative than $Cl$,$(Q)$ is a stronger acid than $(R)$.
Thus,the order of acidic strength is: $(S) < (P) < (R) < (Q)$.
Since $pK_{a} = -\log(K_{a})$,the order of $pK_{a}$ is the reverse of the acidic strength order: $(Q) < (R) < (P) < (S)$.

(C) Statement-$I$ is correct: Carboxylic acids form stable dimers in the vapour phase or in aprotic solvents (like benzene) due to intermolecular hydrogen bonding between two carboxylic acid molecules.
Statement-$II$ is correct: As the length of the hydrocarbon chain increases in carboxylic acids,the hydrophobic (water-repelling) nature of the alkyl group dominates,making higher carboxylic acids practically insoluble in water.

(A) The $pK_a$ value is inversely proportional to the acid strength $(K_a)$.
Stronger acids have lower $pK_a$ values.
Acid strength depends on the stability of the conjugate base (carboxylate ion).
Electron-withdrawing groups $(EWG)$ increase acidity by stabilizing the negative charge on the carboxylate group.
The hybridization of the carbon atom attached to the $-COOH$ group affects acidity: $sp > sp^2 > sp^3$.
$1$. $HC \equiv C - COOH$: The carbon attached to $-COOH$ is $sp$ hybridized,which is highly electronegative and exerts a strong $-I$ effect.
$2$. $CH_2 = CH - COOH$: The carbon is $sp^2$ hybridized,exerting a moderate $-I$ effect.
$3$. $CH_3 - CH_2 - COOH$ and $CH_3 - CH_2 - CH_2 - COOH$: These have alkyl groups which exert a $+I$ effect,decreasing acidity.
Therefore,$HC \equiv C - COOH$ is the strongest acid and has the lowest $pK_a$ value.

(A) Step $1$: Reaction of Grignard reagent with $CO_2$.
$C_6H_5CH_2MgBr + CO_2 \rightarrow C_6H_5CH_2COOMgBr$.
Upon hydrolysis with $H_3O^+$,it forms phenylacetic acid $(C_6H_5CH_2COOH)$,which is '$X$'.
Step $2$: Decarboxylation of '$X$'.
$C_6H_5CH_2COOH + NaOH + CaO \xrightarrow{\Delta} C_6H_5CH_3 + Na_2CO_3$.
The product '$Y$' is toluene $(C_6H_5CH_3)$.

(A) The Assertion is True because both aldehydes $(R-CHO)$ and acid derivatives ($R-CO-Z$,where $Z = -Cl, -OR, -NH_2$) contain the carbonyl group $(>C=O)$.
The Reason is also True. Aldehydes undergo nucleophilic addition reactions because the carbonyl carbon is electrophilic and the addition product is stable.
However,in acid derivatives,the lone pair on the substituent $(Z)$ participates in resonance with the carbonyl group,which reduces the electrophilicity of the carbonyl carbon. Consequently,they undergo nucleophilic acyl substitution rather than simple nucleophilic addition.
Thus,the Reason correctly explains why acid derivatives do not exhibit the same nucleophilic addition reactions as aldehydes.

(A) The reaction sequence is as follows:
$1$. $CH_3COOH + SOCl_2 \xrightarrow{\Delta} CH_3COCl + SO_2 + HCl$
In the first step,acetic acid reacts with thionyl chloride $(SOCl_2)$ to form acetyl chloride $(CH_3COCl)$.
$2$. $CH_3COCl + CH_3OH \rightarrow CH_3COOCH_3 + HCl$
In the second step,acetyl chloride reacts with methanol $(CH_3OH)$ to form methyl acetate $(CH_3COOCH_3)$ as the major organic product.

(C) The stability of a carboxylate ion $(R-COO^-)$ is increased by electron-withdrawing groups $(EWG)$ and decreased by electron-donating groups $(EDG)$ attached to the benzene ring.
$1.$ The $-NO_2$ group is a strong electron-withdrawing group ($-I$ and $-M$ effect).
$2.$ The $-CN$ group is also an electron-withdrawing group ($-I$ and $-M$ effect).
$3.$ The $-CH_3$ group is an electron-donating group ($+I$ and hyperconjugation).
Comparing the options:
- Option $A$ has one $-NO_2$ group.
- Option $B$ has one $-CN$ group.
- Option $C$ has two $-NO_2$ groups and one $-CN$ group,all of which are strong electron-withdrawing groups. This combination significantly disperses the negative charge on the carboxylate group,making it the most stable.
- Option $D$ has a $-CH_3$ group,which destabilizes the carboxylate ion.
Therefore,the ion with the most electron-withdrawing groups is the most stable.

(C) The starting material is a polyhydroxy compound containing a tertiary alcohol,a secondary alcohol,and a primary alcohol. Acidic $KMnO_4$ is a strong oxidizing agent.
$1$. The primary alcohol group $(-CH_2OH)$ is oxidized to a carboxylic acid $(-COOH)$.
$2$. The secondary alcohol group is oxidized to a ketone $(C=O)$.
$3$. The tertiary alcohol group is generally resistant to oxidation under these conditions.
Therefore,the primary alcohol becomes a carboxylic acid,and the secondary alcohol becomes a ketone,resulting in the final product.

(D) The given reaction is the reduction of an ester $(CH_3(CH_2)_9COOC_2H_5)$ to an aldehyde $(CH_3(CH_2)_9CHO)$.
$LiAlH_4$ and $NaBH_4$ are strong reducing agents that typically reduce esters to primary alcohols.
$Pd/H_2$ is a catalyst for hydrogenation,which would reduce the ester to an alcohol or reduce other functional groups if present.
$DIBAL-H$ (Diisobutylaluminium hydride) is a selective reducing agent that reduces esters to aldehydes at low temperatures $(-78 \ ^\circ C)$.
Therefore,the correct reducing agent is $DIBAL-H$.

(B) The boiling point of a compound depends on the strength of intermolecular forces.
$1$. $CH_3COCH_3$ (Propanone) and $CH_3CH_2CHO$ (Propanal) exhibit dipole-dipole interactions.
$2$. $CH_3CH_2CH_2OH$ (Propan-$1$-ol) exhibits hydrogen bonding.
$3$. $CH_3COOH$ (Ethanoic acid) exhibits strong intermolecular hydrogen bonding,forming stable dimers in the liquid phase.
Due to the presence of strong hydrogen-bonded dimers,$CH_3COOH$ has the highest boiling point $(118 \ ^\circ C)$ compared to propan-$1$-ol $(97 \ ^\circ C)$,propanone $(56 \ ^\circ C)$,and propanal $(49 \ ^\circ C)$.

(A) $LiAlH_4$ is a strong reducing agent.
It reduces esters $(R-COOR')$ to primary alcohols.
The ester bond is cleaved to form two alcohols: the alcohol derived from the acyl part $(R-CH_2OH)$ and the alcohol derived from the alkoxy part $(R'-OH)$.
Therefore,the reaction is: $R-COOR' \xrightarrow{LiAlH_4} R-CH_2OH + R'-OH$.
Thus,$A = R-CH_2OH$ and $B = R'-OH$.

(D) Carboxylic acids $(R-COOH)$ are reduced to primary alcohols $(R-CH_2OH)$ using strong reducing agents like lithium aluminum hydride $(LiAlH_4)$.
The reaction is: $R-COOH \xrightarrow[\text{Reduction}]{LiAlH_4} R-CH_2OH$.

(C) $Methyl$ propanoate is $CH_3CH_2COOCH_3$.
On alkaline hydrolysis with dil $NaOH$,it undergoes saponification to form sodium propanoate and methanol:
$CH_3CH_2COOCH_3 + NaOH \rightarrow CH_3CH_2COONa + CH_3OH$
Upon acidification with conc. $HCl$,the sodium salt of the carboxylic acid (sodium propanoate) is converted into propanoic acid:
$CH_3CH_2COONa + HCl \rightarrow CH_3CH_2COOH + NaCl$
Thus,the final product is propanoic acid,which corresponds to option $C$.