Properties of Carboxylic Acids and Their Derivatives Questions in English

(C) The reaction with $NaHCO_3$ is a characteristic test for carboxylic acids.
Carboxylic acids are stronger acids than carbonic acid $(H_2CO_3)$,which is formed when $NaHCO_3$ reacts with an acid.
Acetic acid $(CH_3COOH)$ is a carboxylic acid and reacts with $NaHCO_3$ to evolve $CO_2$ gas:
$CH_3COOH + NaHCO_3 \rightarrow CH_3COONa + H_2O + CO_2 \uparrow$.
Phenol is a much weaker acid than carbonic acid and does not react with $NaHCO_3$.
$n$-hexanol is an alcohol and is even less acidic than phenol,so it also does not react with $NaHCO_3$.
Therefore,only acetic acid reacts with $NaHCO_3$ to produce $CO_2$.

(B) When benzoic acid $(C_6H_5COOH)$ is treated with lithium aluminium hydride $(LiAlH_4)$,it acts as a strong reducing agent.
It reduces the carboxylic acid group $(-COOH)$ to a primary alcohol group $(-CH_2OH)$.
The reaction is:
$C_6H_5COOH + 4[H] \xrightarrow{LiAlH_4} C_6H_5CH_2OH + H_2O$
Thus,the product obtained is benzyl alcohol.

(C) Acid hydrolysis of an ester involves the reaction with water in the presence of an acid catalyst to produce a carboxylic acid and an alcohol.
For ethyl acetate $(CH_3COOC_2H_5)$:
$CH_3COOC_2H_5 + H_2O \xrightarrow{H^+} CH_3COOH + C_2H_5OH$
Here,$CH_3COOH$ (acetic acid) and $C_2H_5OH$ (ethanol) are two different organic compounds.

(D) Ethane-$1, 2$-dioic acid (oxalic acid) undergoes dehydration in the presence of concentrated $H_2SO_4$ upon heating.
The reaction is as follows:
$(COOH)_2 \xrightarrow[\text{conc. } H_2SO_4]{\Delta} CO + CO_2 + H_2O$
Thus,the products formed are carbon monoxide,carbon dioxide,and water.

(B) The acidity of carboxylic acids depends on the stability of the carboxylate ion formed after the loss of a proton.
Electron-withdrawing groups ($-I$ effect) like $-Cl$ increase acidity by stabilizing the negative charge on the carboxylate ion.
Electron-donating groups ($+I$ effect) like alkyl groups decrease acidity by destabilizing the negative charge.
The $+I$ effect strength follows the order: $-C_2H_5 > -CH_3 > -H$.
Comparing the given acids:
$1$. $ClCH_2COOH$ has a $-Cl$ group ($-I$ effect),making it the most acidic.
$2$. $HCOOH$ has no alkyl group.
$3$. $CH_3COOH$ has a $-CH_3$ group ($+I$ effect).
$4$. $C_2H_5COOH$ has a $-C_2H_5$ group (stronger $+I$ effect than $-CH_3$),making it the least acidic.
Thus,the correct order is $ClCH_2COOH > HCOOH > CH_3COOH > C_2H_5COOH$.

(B) The reaction of carboxylic acids having an $\alpha$-hydrogen with halogen ($Cl_2$ or $Br_2$) in the presence of a small amount of red phosphorus to give $\alpha$-halo carboxylic acids is known as the Hell-Volhard-Zelinsky $(HVZ)$ reaction.

(C) The $-COOH$ group is a strong electron-withdrawing group due to both $-I$ and $-M$ effects.
It deactivates the benzene ring towards electrophilic substitution.
The electron density is decreased at all positions,but the decrease is relatively less at the $m-$position compared to the $o-$ and $p-$positions.
Therefore,the incoming electrophile $(NO_2^+)$ attacks the $m-$position.
The Assertion is correct,but the Reason is incorrect because the carboxyl group decreases,not increases,the electron density.

(A) The decarboxylation of $\beta$-keto carboxylic acids occurs via a cyclic transition state involving the carbonyl oxygen of the ketone and the hydrogen of the carboxylic acid group.
This process releases $CO_2$ and initially forms an enol intermediate.
The enol is unstable and rapidly undergoes tautomerization to form the more stable ketone.
Therefore,both the Assertion and the Reason are correct,and the Reason provides the correct explanation for the decarboxylation process.

(C) The Assertion is correct because the negative charge on the acetate ion $(CH_3COO^-)$ is delocalized over two oxygen atoms through resonance,which stabilizes the ion.
The Reason is incorrect because the acetate ion is a much weaker base than the methoxide ion $(CH_3O^-)$. This is because $CH_3COOH$ is a much stronger acid $(pK_a \approx 4.75)$ than $CH_3OH$ $(pK_a \approx 15.5)$. Since the conjugate base of a stronger acid is always a weaker base,the acetate ion is less basic than the methoxide ion.

(C) The reaction $CH_3COCl + 2NH_3 \rightarrow CH_3CONH_2 + NH_4Cl$ is an example of nucleophilic acyl substitution. The Assertion is correct because $NH_3$ acts as a nucleophile to replace the $Cl$ atom.
However,the Reason is incorrect. While $Cl^{-}$ is indeed a good leaving group,it is a very weak nucleophile compared to $NH_3$. The strength of a nucleophile is generally inversely related to the strength of its conjugate acid. Since $HCl$ is a strong acid,$Cl^{-}$ is a weak base and a weak nucleophile.

(B) When phthalic acid reacts with ammonia $(NH_3)$,it first forms ammonium phthalate.
Upon heating,ammonium phthalate loses two molecules of water to form phthalamide.
On further strong heating,phthalamide loses a molecule of ammonia $(NH_3)$ to form the cyclic imide known as phthalimide.

(D) Carboxylic acids have higher boiling points than aldehydes,ketones,and even alcohols of comparable molecular mass.
This is due to the more extensive association of carboxylic acid molecules through intermolecular $H$-bonding,which often results in the formation of stable dimers in the vapor phase or non-polar solvents.

(D) The given conversion involves the reduction of a carboxylic acid group $(-COOH)$ to a primary alcohol $(-CH_2OH)$ while leaving the ketone $(-COCH_3)$,amide $(-CONH_2)$,and nitrile $(-CN)$ groups unaffected.
$B_2H_6$ (diborane) is a selective electrophilic reducing agent that reduces carboxylic acids to primary alcohols much faster than it reduces other functional groups like ketones,amides,or nitriles under controlled conditions.
$LiAlH_4$ is a strong reducing agent that would reduce all these groups ($-COOH$,$-COCH_3$,$-CONH_2$,$-CN$).
$NaBH_4$ is a milder reducing agent but is generally not effective for reducing carboxylic acids directly.
$H_2/Pd$ would reduce the ketone and potentially the nitrile group.
Therefore,$B_2H_6$ is the most suitable reagent.

(N/A) Ester hydrolysis can be represented as:
$Ester + H_2O \longrightarrow Acid + Alcohol$
The acid produced in the reaction acts as a catalyst and makes the reaction faster.
Substances that act as catalysts in the same reaction in which they are obtained as products are known as autocatalysts.
This phenomenon is called autocatalysis.

(N/A) The resonance structures of the phenoxide ion are shown in the figure.
It can be observed from the resonance structures of the phenoxide ion that in structures $II$,$III$,and $IV$,less electronegative carbon atoms carry a negative charge. Therefore,these three structures contribute negligibly to the resonance stability of the phenoxide ion.
Hence,these structures can be ignored. Only structures $I$ and $V$ carry a negative charge on the more electronegative oxygen atom.
The resonance structures of the carboxylate ion are also shown in the figure.
In the case of the carboxylate ion,resonating structures $I'$ and $II'$ contain a negative charge carried by a more electronegative oxygen atom.
Further,in resonating structures $I'$ and $II'$,the negative charge is delocalized over two oxygen atoms. However,in resonating structures $I$ and $V$ of the phenoxide ion,the negative charge is localized on the same oxygen atom. Therefore,the resonating structures of the carboxylate ion contribute more towards its stability than those of the phenoxide ion. As a result,the carboxylate ion is more resonance-stabilized than the phenoxide ion. Hence,carboxylic acid is a stronger acid than phenol.

(N/A) Sodium salts of carboxylic acids,when heated with soda lime (a mixture of sodium hydroxide and calcium oxide),yield alkanes containing one carbon atom less than the parent carboxylic acid. This process of eliminating carbon dioxide from a carboxylic acid is known as decarboxylation.
$RCOOH \xrightarrow{\text{Soda lime, } \Delta} RH + Na_2CO_3$
Soda lime is a mixture of sodium hydroxide $(NaOH)$ and calcium oxide $(CaO)$. The $NaOH$ in soda lime reacts with the carboxylic acid to form its sodium salt.
Example-$1$: Preparation of methane from ethanoic acid (acetic acid).
$CH_3COONa + NaOH \xrightarrow{CaO, \Delta} CH_4 + Na_2CO_3$
Example-$2$: Preparation of ethane from propanoic acid.
$CH_3CH_2COONa + NaOH \xrightarrow{CaO, \Delta} C_2H_6 + Na_2CO_3$

(N/A) The first three members of aliphatic carboxylic acids ($HCOOH$,$CH_3COOH$,and $CH_3CH_2COOH$) are colourless liquids with a pungent,sharp smell.
Aliphatic carboxylic acids containing $4$ to $9$ carbon atoms are colourless liquids at room temperature with an unpleasant,rancid odour.
Higher carboxylic acids (containing more than $9$ carbon atoms) are wax-like solids and are practically odourless due to their low volatility.

(N/A) $(i)$ Carboxylic acids contain the $-COOH$ functional group. In the carboxyl group,the oxygen atom is highly electronegative,creating a polar $C=O$ bond and an $O-H$ bond. This results in a partial positive charge on the hydrogen atom and a partial negative charge on the carbonyl oxygen,leading to the formation of intermolecular hydrogen bonds between carboxylic acid molecules.
$(ii)$ These intermolecular hydrogen bonds are strong enough that they are not completely broken even in the vapour phase. Consequently,most carboxylic acids exist as stable dimers in the vapour phase or in aprotic solvents. This association of molecules increases the effective molecular mass,which significantly raises the boiling points of carboxylic acids compared to alcohols or aldehydes of comparable molecular masses.

(N/A) $(i)$ Simple aliphatic carboxylic acids having up to four carbon atoms ($HCOOH, CH_3COOH, CH_3CH_2COOH$ and $CH_3CH_2CH_2COOH$) are miscible in water due to the formation of hydrogen bonds with water.
$(ii)$ The solubility decreases with an increasing number of carbon atoms.
$(iii)$ Higher carboxylic acids are practically insoluble in water due to the increased hydrophobic interaction of the hydrocarbon part.
$(iv)$ Benzoic acid,the simplest aromatic carboxylic acid,is nearly insoluble in cold water.
$(v)$ Carboxylic acids are also soluble in less polar organic solvents like benzene,ether,alcohol,chloroform,etc.

(N/A) Carboxylic acids act as Bronsted acids because they can donate a proton $(H^+)$. Their acidic nature is proven by the following reactions:
$(i)$ Reaction with metals: Carboxylic acids react with electropositive metals like $Na$ and $K$ to liberate dihydrogen gas,acting as proton donors.
$2 RCOOH + 2 Na \longrightarrow 2 RCOO^- Na^+ + H_2 \uparrow$
$(ii)$ Reaction with carbonates and bicarbonates: Carboxylic acids react with mild bases like aqueous solutions of carbonates and hydrogen carbonates to evolve carbon dioxide gas.
$RCOOH + NaHCO_{3(aq)} \longrightarrow RCOO^- Na^+ + H_2O + CO_2 \uparrow$
$2 RCOOH + Na_2CO_3 \longrightarrow 2 RCOO^- Na^+ + H_2O + CO_2 \uparrow$

(N/A) The acidity of these compounds follows the order: $Carboxylic \ acid > Phenol > Alcohol$.
$1$. $Na$ metal: All three compounds contain an acidic hydrogen atom ($-OH$ group),so they all react with $Na$ to liberate $H_2$ gas.
$2$. $NaOH$: Carboxylic acids and phenols are acidic enough to react with the strong base $NaOH$ to form water-soluble salts. Alcohols are too weakly acidic to react with $NaOH$.
$3$. $NaHCO_3$: Only carboxylic acids are strong enough to react with the weak base $NaHCO_3$ to liberate $CO_2$ gas. Phenols and alcohols are not acidic enough to decompose $NaHCO_3$.

(N/A) $(i)$ Carboxylic acids dissociate in water to give resonance-stabilized carboxylate anions and hydronium ions.
$RCOOH + H_{2}O \rightleftharpoons RCOO^{-} + H_{3}O^{+}$
This carboxylate ion is stable due to its resonance activity.
$(ii)$ If the equilibrium constant for the aqueous acidic solution is $K_{eq}$,then:
$K_{eq} = \frac{[H_{3}O^{+}][RCOO^{-}]}{[RCOOH][H_{2}O]}$
$\therefore K_{eq}[H_{2}O] = \frac{[H_{3}O^{+}][RCOO^{-}]}{[RCOOH]}$
$\therefore K_{a} = \frac{[H_{3}O^{+}][RCOO^{-}]}{[RCOOH]} \quad (\because K_{eq}[H_{2}O] = K_{a})$
Where,$K_{a}$ is the dissociation constant for the acid.
Taking the negative logarithm on both sides:
$-\log K_{a} = -\log \left( \frac{[H_{3}O^{+}][RCOO^{-}]}{[RCOOH]} \right)$
$\therefore pK_{a} = -\log K_{a}$
Relation: $K_{a} \propto \text{acidic strength}$ and $pK_{a} \propto \frac{1}{\text{acidic strength}}$. For convenience,the strength of an acid is generally indicated by its $pK_{a}$ value rather than its $K_{a}$ value.

(N/A) Mineral acids are stronger than carboxylic acids because they completely dissociate in water.
Carboxylic acids are stronger acids than phenols and alcohols because the carboxylate ion is stabilized by resonance,where the negative charge is delocalized over two electronegative oxygen atoms.
Phenols are more acidic than alcohols because the phenoxide ion is stabilized by resonance with the benzene ring.
Alcohols are the least acidic among these as the alkoxide ion is not stabilized by resonance.
The order of acidic strength is: $\text{mineral acid} > \text{carboxylic acid} > \text{phenol} > \text{alcohol}$.

(N/A) The acidity of carboxylic acids is due to the resonance stabilization of the carboxylate ion.
$(i)$ Carboxylic acids are acidic because they can donate a proton to form a carboxylate ion $(RCOO^-)$.
$(a)$ In the carboxylic acid molecule $(RCOOH)$,resonance involves charge separation,which makes the structures less stable.
$(b)$ In the carboxylate ion $(RCOO^-)$,the negative charge is delocalized over two electronegative oxygen atoms through two equivalent resonance structures. This makes the carboxylate ion highly stable.
$(c)$ Since the conjugate base (carboxylate ion) is much more stable than the acid itself,the equilibrium shifts towards the formation of the carboxylate ion,making carboxylic acids acidic.
$(ii)$ Carboxylic acids are more acidic than phenols because:
$(a)$ In the carboxylate ion,the negative charge is delocalized over two electronegative oxygen atoms,which are equivalent. In the phenoxide ion $(C_6H_5O^-)$,the negative charge is delocalized over the carbon atoms of the benzene ring,which are less electronegative than oxygen. Thus,the carboxylate ion is more stable than the phenoxide ion,making carboxylic acids stronger acids than phenols.

(N/A) Substituents may affect the stability of the conjugate base and thus,also affect the acidity of the carboxylic acids.
$(a)$ Effect of substituents on the acidity of carboxylic acids:
$(i)$ Electron withdrawing groups $(EWG)$ increase the acidity of carboxylic acids by stabilizing the conjugate base through delocalization of the negative charge by inductive and/or resonance effects.
$(ii)$ The effect of the following groups in increasing acidity order is:
$Ph < I < Br < Cl < F < CN < NO_{2} < CF_{3}$
$(iii)$ Thus,the following acids are arranged in order of decreasing acidity (based on $pK_{a}$ values):
$CF_{3}COOH > CCl_{3}COOH > CHCl_{2}COOH > NO_{2}CH_{2}COOH > NC-CH_{2}COOH > FCH_{2}COOH > ClCH_{2}COOH > BrCH_{2}COOH > HCOOH > ClCH_{2}CH_{2}COOH > C_{6}H_{5}COOH > C_{6}H_{5}CH_{2}COOH > CH_{3}COOH > CH_{3}CH_{2}COOH$
$(iv)$ Direct attachment of groups such as phenyl or vinyl to the carboxylic acid increases the acidity of the corresponding carboxylic acid,contrary to the decrease expected due to the resonance effect. For example: $CH_{2}=CHCOOH$ is more acidic than $CH_{3}COOH$.

(N/A) The acidic strength of benzoic acid is influenced by the substituents present on the benzene ring:
$1$. Electron-Withdrawing Groups $(EWG)$: Groups like $-NO_{2}$,$-Cl$,etc.,exert an electron-withdrawing effect ($-I$ and/or $-M$ effect),which stabilizes the carboxylate anion by dispersing the negative charge. This increases the acidic strength. For example,the order of acidic strength is: $p-nitrobenzoic \ acid > p-chlorobenzoic \ acid > benzoic \ acid$.
$2$. Electron-Donating Groups $(EDG)$: Groups like $-CH_{3}$,$-OH$,$-OCH_{3}$,etc.,exert an electron-donating effect ($+I$ and/or $+M$ effect),which destabilizes the carboxylate anion by intensifying the negative charge. This decreases the acidic strength. For example,the order of acidic strength is: $benzoic \ acid > p-toluic \ acid > p-hydroxybenzoic \ acid$.

(N/A) Formation of anhydride: Carboxylic acids on heating with mineral acids such as $H_{2}SO_{4}$ or with $P_{2}O_{5}$ give corresponding anhydride.
$(b)$ Esterification: Carboxylic acids are esterified with alcohols or phenols in the presence of a mineral acid such as concentrated $H_{2}SO_{4}$ or $HCl$ gas as a catalyst.
$RCOOH + R'OH \rightleftharpoons H^{+} \rightleftharpoons RCOOR' + H_{2}O$
In this method,the $C-OH$ bond of the acid and the $O-H$ bond of the alcohol break to form an ester and remove one molecule of water.
$(c)$ Formation of acid chloride: The hydroxyl group $(-OH)$ of carboxylic acids is easily replaced by a chlorine atom on treating with $PCl_{5}$,$PCl_{3}$ or $SOCl_{2}$.
$RCOOH + PCl_{5} \rightarrow RCOCl + POCl_{3} + HCl$
$3RCOOH + PCl_{3} \rightarrow 3RCOCl + H_{3}PO_{3}$
$RCOOH + SOCl_{2} \rightarrow RCOCl + SO_{2} + HCl$
$(d)$ Formation of amide: Carboxylic acids react with ammonia to give ammonium salts which on further heating at high temperature give amides.
$RCOOH + NH_{3}$ $\rightarrow RCOONH_{4}$ $\xrightarrow{\Delta} RCONH_{2} + H_{2}O$

(N/A) $(i)$ Acetic acid reacts with ammonia to form ammonium acetate,which on heating loses water to form acetamide:
$CH_3COOH + NH_3 \rightleftharpoons CH_3COO^-NH_4^+ \xrightarrow{\Delta, -H_2O} CH_3CONH_2$
$(ii)$ Benzoic acid reacts with ammonia to form ammonium benzoate,which on heating loses water to form benzamide:
$C_6H_5COOH + NH_3 \rightleftharpoons C_6H_5COO^-NH_4^+ \xrightarrow{\Delta, -H_2O} C_6H_5CONH_2$
$(iii)$ Phthalic acid reacts with ammonia to form ammonium phthalate,which on heating loses water to form phthalamide,and upon further heating,it loses ammonia to form phthalimide:
$C_6H_4(COOH)_2 + 2NH_3 \rightleftharpoons C_6H_4(COO^-NH_4^+)_2$ $\xrightarrow{\Delta, -2H_2O} C_6H_4(CONH_2)_2$ $\xrightarrow{\Delta, -NH_3} C_6H_4(CO)_2NH$

(N/A) The esterification of carboxylic acids with alcohols is a type of nucleophilic acyl substitution. This reaction proceeds through the following steps:
$(i)$ Protonation of the carbonyl oxygen by the $H^{+}$ acid catalyst to produce the protonated carboxylic acid $(X)$.
$(ii)$ The protonated carbonyl group is activated towards nucleophilic attack by the alcohol $(R'-OH)$,forming a tetrahedral intermediate $(M)$.
$(iii)$ $A$ proton transfer occurs within the tetrahedral intermediate $(M)$ to convert the hydroxyl group into a better leaving group,$-OH_2^{+}$,resulting in intermediate $(Y)$.
$(iv)$ The intermediate $(Y)$ eliminates a water molecule to form the protonated ester $(Z)$.
$(v)$ Finally,the protonated ester $(Z)$ loses a proton to yield the final ester $(P)$.

(N/A) Carboxylic acids having an $\alpha$-hydrogen are halogenated at the $\alpha$-position on treatment with chlorine or bromine in the presence of a small amount of red phosphorus to give $\alpha$-halo-carboxylic acids. This reaction is known as the 'Hell-Volhard-Zelinsky' reaction.
$RCH_{2}COOH \xrightarrow[(ii)\ H_{2}O]{(i)\ X_{2}/\text{Red } P} RCH(X)COOH$
Where,$X = Cl, Br$.
This halogenation reaction continues until all $\alpha$-hydrogens are replaced.
Example: Chlorination of ethanoic acid:
$CH_{3}COOH$ $\xrightarrow{Cl_{2}/\text{Red } P} CH_{2}ClCOOH$ $\xrightarrow{Cl_{2}/\text{Red } P} CHCl_{2}COOH$ $\xrightarrow{Cl_{2}/\text{Red } P} CCl_{3}COOH$
Uses: This reaction is utilized to introduce groups like $-OH, -CN, -NH_{2}$ by first substituting the $\alpha$-position with $Cl$ or $Br$.

(N/A) $1$. Aromatic carboxylic acids undergo electrophilic substitution reactions.
$2$. The carboxyl group $(-COOH)$ acts as a deactivating and meta-directing group.
$3$. They do not undergo Friedel-Crafts reactions because the carboxyl group is strongly deactivating and the catalyst (Lewis acid,$AlCl_3$) gets bonded to the carboxyl group.
$4$. Examples:
$(i)$ Nitration of benzoic acid: Benzoic acid reacts with conc. $HNO_3$ and conc. $H_2SO_4$ to form $m$-nitrobenzoic acid.
(ii) Bromination of benzoic acid: Benzoic acid reacts with $Br_2$ in the presence of $FeBr_3$ to form $m$-bromobenzoic acid.
(iii) Sulphonation of benzoic acid: Benzoic acid reacts with $H_2SO_4$ (containing $SO_3$) to form $m$-sulphobenzoic acid.

(A) $(i) 2C_6H_5COOH + 2Na \rightarrow 2C_6H_5COO^-Na^+ + H_2 \uparrow$
$(ii) C_6H_5COOH + NaOH \rightarrow C_6H_5COO^-Na^+ + H_2O$
$(iii) C_6H_5COOH + NaHCO_3 \rightarrow C_6H_5COO^-Na^+ + H_2O + CO_2 \uparrow$
$(iv) 2C_6H_5COOH \xrightarrow{P_2O_5/\Delta} (C_6H_5CO)_2O + H_2O$
$(v) C_6H_5COOH + PCl_5 \rightarrow C_6H_5COCl + POCl_3 + HCl$
$(vi) 3C_6H_5COOH + PCl_3 \rightarrow 3C_6H_5COCl + H_3PO_3$
$(vii) C_6H_5COOH + SOCl_2 \rightarrow C_6H_5COCl + SO_2 + HCl$
$(viii) C_6H_5COOH + NH_3 \rightarrow C_6H_5COONH_4 \xrightarrow{\Delta, -H_2O} C_6H_5CONH_2$
$(ix) C_6H_5COOH \xrightarrow{LiAlH_4/\text{ether}} C_6H_5CH_2OH$
$(x) C_6H_5COOH + NaOH/CaO \xrightarrow{\Delta} C_6H_6 + Na_2CO_3$
$(xi) C_6H_5COOH \xrightarrow{\text{conc. } HNO_3/H_2SO_4, \Delta} m\text{-nitrobenzoic acid}$
$(xii) C_6H_5COOH \xrightarrow{Br_2/FeBr_3} m\text{-bromobenzoic acid}$
$(xiii) C_6H_5COOH \xrightarrow{\text{Oleum}, \Delta} m\text{-sulphobenzoic acid}$

(N/A) $(i)$ $A$ compound with the molecular formula $C_2H_4O_2$ that gives effervescence of $CO_2$ gas upon addition of sodium carbonate must possess a $-COOH$ group. Therefore,compound $(A)$ is $CH_3COOH$ (Acetic acid).
$(ii)$ Compound $(P)$ is the sodium salt of the acid,which is $CH_3COONa$ (Sodium ethanoate). When $(P)$ is heated with sodalime $(NaOH + CaO)$,it undergoes decarboxylation to give $CH_4$ gas. Thus,$(Q) = CH_4$ (Methane gas).
The reactions are:
$CH_3COOH + Na_2CO_3 \rightarrow CH_3COONa + CO_2 + H_2O$
$(A) \quad \quad \quad \quad \quad \quad \quad \quad (P)$
$CH_3COONa + NaOH \xrightarrow{\Delta, CaO} CH_4 + Na_2CO_3$
$(P) \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad (Q)$

(N/A) The decreasing order of acidic strength is: $FCH_{2}COOH > ClCH_{2}COOH > C_{6}H_{5}CH_{2}COOH > CH_{3}COOH > CH_{3}CH_{2}OH$.
Reasoning:
$1$. The acidity of carboxylic acids depends on the stability of the carboxylate anion (conjugate base).
$2$. Electron-withdrawing groups $(EWG)$ like $F$ and $Cl$ stabilize the carboxylate anion through the inductive effect ($-I$ effect),thereby increasing acidity. Since $F$ is more electronegative than $Cl$,$FCH_{2}COOH$ is more acidic than $ClCH_{2}COOH$.
$3$. $C_{6}H_{5}CH_{2}COOH$ is more acidic than $CH_{3}COOH$ because the phenyl group exerts a weak electron-withdrawing inductive effect.
$4$. $CH_{3}COOH$ is more acidic than $CH_{3}CH_{2}OH$ because the carboxylate ion is resonance-stabilized,whereas the ethoxide ion is not. Alcohols are much weaker acids than carboxylic acids.

(A) The decreasing order of acidic strength is: $NO_2CH_2COOH > FCH_2COOH > C_6H_5COOH$.
Explanation: The acidity of carboxylic acids depends on the stability of the carboxylate ion (conjugate base) formed after the loss of a proton. Electron-withdrawing groups $(EWG)$ stabilize the carboxylate ion by dispersing the negative charge through the inductive effect ($-I$ effect). The strength of the $-I$ effect follows the order: $-NO_2 > -F$. The nitro group $(-NO_2)$ is a stronger electron-withdrawing group than the fluorine atom $(-F)$,making $NO_2CH_2COOH$ the most acidic. $C_6H_5COOH$ (benzoic acid) is the least acidic among these because the phenyl group is less electron-withdrawing compared to the inductive effect of $-F$ and $-NO_2$ groups attached to the alpha-carbon.

(N/A) The carbonyl carbon in carboxylic acids is less electrophilic than in aldehydes and ketones. This is due to the resonance effect of the $-OH$ group,where the lone pair of electrons on the oxygen atom is delocalized into the carbonyl group.
This resonance structure,as shown below,reduces the partial positive charge on the carbonyl carbon:
$-C(=O)-OH \leftrightarrow -C(-O^-)=O^+-H$
Because the electrophilicity of the carbonyl carbon is significantly reduced,it does not readily undergo nucleophilic addition reactions.

(A) $1$. The reaction of $CH_3-Br$ with $Mg$ in the presence of dry ether forms the Grignard reagent,methylmagnesium bromide,which is $[A] = CH_3MgBr$.
$2$. The reaction of methylmagnesium bromide with $CO_2$ followed by hydrolysis yields ethanoic acid,which is $[B] = CH_3COOH$.
$3$. The esterification of ethanoic acid with methanol $(CH_3OH)$ in the presence of an acid catalyst $(H^+)$ yields methyl ethanoate,which is $[C] = CH_3COOCH_3$.
Therefore,the correct identification is $A = CH_3MgBr, B = CH_3COOH, C = CH_3COOCH_3$.

(A) The acidity of a compound depends on the stability of its conjugate base.
$1$. Carboxylic acids $(R-COOH)$ lose a proton to form a carboxylate ion $(R-COO^-)$. This ion is stabilized by two equivalent resonance structures where the negative charge is delocalized over two highly electronegative oxygen atoms.
$2$. Phenols $(C_6H_5OH)$ lose a proton to form a phenoxide ion $(C_6H_5O^-)$. Although the phenoxide ion is stabilized by resonance,the negative charge is delocalized onto the less electronegative carbon atoms of the benzene ring,making it less stable than the carboxylate ion.
$3$. Alcohols $(R-OH)$ form alkoxide ions $(R-O^-)$ upon losing a proton. These ions have no resonance stabilization,making them the least stable and thus the least acidic.
Therefore,due to the greater stability of the carboxylate ion compared to the phenoxide and alkoxide ions,carboxylic acids are more acidic.

Soaps are sodium or potassium salts of long-chain fatty acids,such as palmitic acid $(C_{15}H_{31}COOH)$,stearic acid $(C_{17}H_{35}COOH)$,and oleic acid $(C_{17}H_{33}COOH)$.
Potassium salts of these fatty acids are known as soft soaps. They are more soluble in water and are used in shaving creams and shampoos.
The reaction for the preparation of potassium stearate (a soft soap) is:
$(C_{17}H_{35}COO)_3C_3H_5 + 3KOH \rightarrow 3C_{17}H_{35}COOK + C_3H_5(OH)_3$
(Glyceryl ester of stearic acid + Potassium hydroxide $\rightarrow$ Potassium stearate + Glycerol)

(N/A) $C_6H_5COOH + NaOH \xrightarrow{CaO, \Delta} C_6H_6 + Na_2CO_3$
Heating benzoic acid with soda lime $(NaOH + CaO)$ leads to decarboxylation,resulting in the formation of benzene and sodium carbonate.

(A) The acidic strength of carboxylic acids depends on the electron-withdrawing effect of substituents attached to the alpha-carbon. The presence of electronegative chlorine atoms increases the acidity by stabilizing the carboxylate anion through the inductive effect ($-I$ effect). As the number of chlorine atoms increases,the $-I$ effect increases,thereby increasing the acidic strength. The order is: $CH_{3}COOH < CH_{2}ClCOOH < CHCl_{2}COOH < CCl_{3}COOH$.

(D) is $o$-hydroxybenzoic acid (salicylic acid),which exhibits intramolecular $H$-bonding.
$(B)$ is $p$-hydroxybenzoic acid,which exhibits intermolecular $H$-bonding.
$(a)$ Due to intermolecular $H$-bonding,$(B)$ molecules associate to form a lattice structure,making it more likely to be crystalline than $(A)$,which exists as discrete molecules due to intramolecular $H$-bonding. Thus,$(a)$ is true.
$(b)$ Intermolecular $H$-bonding in $(B)$ leads to higher molecular association,requiring more energy to break,thus $(B)$ has a higher boiling point than $(A)$. Thus,$(b)$ is true.
$(c)$ $(B)$ can form more extensive $H$-bonding with water molecules compared to $(A)$ due to the availability of its functional groups,making it more soluble in water. Thus,$(c)$ is true.
Therefore,all three statements are correct.

(D) The acidity of a hydrogen atom depends on the stability of the conjugate base formed after the removal of the proton.
In trimethyl methanetricarboxylate,the central carbon is attached to three electron-withdrawing ester groups $(-COOCH_3)$.
These groups stabilize the resulting carbanion through strong inductive $(-I)$ and resonance $(-R)$ effects.
Since there are three such groups,the negative charge on the carbanion is highly delocalized,making the hydrogen atom significantly more acidic compared to the other options.

(D) The acidity of $\alpha$-hydrogen depends on the stability of the resulting carbanion. The more stable the carbanion,the more acidic the $\alpha$-hydrogen.
$(B)$ $Ph-CO-CH_2-CO-Ph$ is the most acidic because the carbanion is stabilized by resonance with two carbonyl groups and two phenyl rings.
$(A)$ $CH_3-CO-CH_3$ (acetone) has a carbanion stabilized by one carbonyl group.
$(C)$ $CH_3-CO-OCH_3$ (methyl acetate) has a carbanion that is less stable than $(A)$ due to cross-conjugation of the lone pair on oxygen with the carbonyl group,which reduces the electron-withdrawing effect of the carbonyl.
$(D)$ $CH_3-CO-N(CH_3)_2$ ($N$,$N$-dimethylacetamide) is the least acidic because the nitrogen lone pair participates in strong resonance (cross-conjugation) with the carbonyl group,making the carbonyl carbon less electron-withdrawing.
Therefore,the increasing order of acidity is $(D) < (C) < (A) < (B)$.

(B) The stability of alcohol derivatives in an aqueous base depends on their susceptibility to hydrolysis.
Esters,represented by the structure $RO-CO-CH_3$,undergo base-catalyzed hydrolysis (saponification) in an aqueous base.
The mechanism involves the nucleophilic attack of the $OH^-$ ion on the carbonyl carbon of the ester,followed by the elimination of the alkoxide ion $(RO^-)$,which then abstracts a proton from the resulting carboxylic acid to form an alcohol and a carboxylate ion.
Other derivatives like ethers $(RO-CH_2-C_6H_5)$ and acetals/ketals $(RO-C_5H_9O)$ are generally stable in basic media.

(A) Phthalic acid $(C_6H_4(COOH)_2)$ on heating loses a water molecule to form phthalic anhydride. When heated strongly with ammonia,it forms phthalimide,which is an acid imide. Succinic acid also behaves similarly,but phthalic acid is the standard example for this reaction in the context of aromatic dicarboxylic acids.

(A) The ease of nucleophilic substitution depends upon the nature of the leaving group. When the leaving tendency of a group in a compound is high,the compound is more reactive towards nucleophilic substitution.
The nucleophilic acyl substitution proceeds in two steps as shown in the mechanism:
$1$. Nucleophilic attack on the carbonyl carbon to form a tetrahedral intermediate.
$2$. Elimination of the leaving group $(A^-)$ to regenerate the carbonyl group.
The order of leaving tendency is $Cl^{-} > RCOO^{-} > RO^{-} > NH_{2}^{-}$.
Therefore,the order of reactivity of acyl compounds is:
$RCOCl$ (Acyl chloride) $ > (RCO)_2O$ (Acid anhydride) $ > RCOOR$ (Ester) $ > RCONH_2$ (Amide).

(A) The conversion of $3-$vinylbenzaldehyde to $3-(2-$bromoethyl$)$benzoic acid involves two main transformations:
$1$. Oxidation of the aldehyde group $(-CHO)$ to a carboxylic acid group $(-COOH)$ using the iodoform test reagents $(I_2/NaOH)$ which selectively oxidizes the methyl ketone or aldehyde group while leaving the alkene intact.
$2$. Anti-Markovnikov addition of $HBr$ to the vinyl group $(-CH=CH_2)$ using $HBr/R_2O_2$ (peroxide effect) to yield the $2-$bromoethyl substituent.
Thus,the correct sequence of reagents is $I_2/NaOH$ followed by $HBr/R_2O_2$.