Properties of Carboxylic Acids and Their Derivatives Questions in English

(C) $NaBH_4$ (Sodium borohydride) is a selective reducing agent that reduces aldehydes and ketones to their corresponding alcohols,but it does not reduce esters under normal conditions.
In the given molecule,there is a ketone group in the cyclohexane ring and an ester group $(-COOCH_3)$.
$NaBH_4$ will selectively reduce the ketone group to a secondary alcohol $(-CH(OH)-)$ while leaving the ester group intact.
Therefore,the product is a molecule with a cyclohexanol ring attached to a $-CH_2-COOCH_3$ group.
This corresponds to the structure shown in option $C$.

(A) The $pK_{a}$ value is inversely proportional to the acidity of the carboxylic acid. Stronger acids have lower $pK_{a}$ values.
Acidity is determined by the stability of the conjugate base (carboxylate ion).
Electron-withdrawing groups (EWGs) stabilize the carboxylate ion through the inductive effect ($-I$ effect),thereby increasing acidity.
Comparing the substituents:
$1$. $NO_{2}CH_{2}COOH$: The $-NO_{2}$ group is a strong electron-withdrawing group ($-I$ effect),which significantly stabilizes the carboxylate ion.
$2$. $CH_{3}COOH$: The $-CH_{3}$ group is an electron-donating group ($+I$ effect),which destabilizes the carboxylate ion.
$3$. $HCOOH$: Hydrogen has no inductive effect.
$4$. $C_{6}H_{5}COOH$: The phenyl group is electron-withdrawing by resonance but less effective than the $-NO_{2}$ group in this context.
Therefore,$NO_{2}CH_{2}COOH$ is the strongest acid among the given options and has the lowest $pK_{a}$ value.

(B) The acidity of benzoic acid derivatives depends on the electronic effects of the substituents attached to the benzene ring.
$1$. The $-NO_2$ group in $(I)$ is a strong electron-withdrawing group ($-I$ and $-M$ effect),which stabilizes the carboxylate anion,increasing acidity.
$2$. The $-OCH_3$ group in $(II)$ is an electron-donating group ($+M$ effect),which destabilizes the carboxylate anion,decreasing acidity.
$3$. Benzoic acid $(III)$ serves as the reference point.
Therefore,the order of acidity is $4$-Nitrobenzoic acid $(I)$ > Benzoic acid $(III)$ > $4$-Methoxybenzoic acid $(II)$.
The correct option is $B$.

(D) The acidic strength of carboxylic acids depends on the electron-withdrawing effect of substituents. Electron-withdrawing groups $(EWG)$ increase acidity by stabilizing the carboxylate anion,while electron-donating groups decrease it.
$A$. $CH_2FCH_2CH_2COOH < CH_3CHFCH_2COOH$: Correct. The $F$ atom is closer to the $COOH$ group in the second molecule,exerting a stronger $-I$ effect.
$B$. $CH_2ClCOOH < CH_2FCOOH$: Correct. $F$ is more electronegative than $Cl$,so $CH_2FCOOH$ is more acidic.
$C$. $CH_3COOH < CH_2ClCOOH$: Correct. $Cl$ is an $EWG$,making the chloroacetic acid more acidic than acetic acid.
$D$. $HCOOH < C_6H_5COOH$: Incorrect. Formic acid $(HCOOH)$ is more acidic than benzoic acid $(C_6H_5COOH)$ because the phenyl group in benzoic acid is electron-donating by resonance and has a weaker $-I$ effect compared to the hydrogen atom in formic acid. Thus,the correct order is $C_6H_5COOH < HCOOH$.

(D) is the correct answer.
Acidic strength of benzoic acid derivatives is increased by electron-withdrawing groups $(EWG)$ and decreased by electron-donating groups $(EDG)$.
$4-$Nitrobenzoic acid contains a $-NO_2$ group,which is a strong electron-withdrawing group ($-I$ and $-M$ effect),thereby stabilizing the carboxylate anion and increasing acidity.
$2-$Methoxybenzoic acid and $4-$Methoxybenzoic acid contain a $-OCH_3$ group,which acts as an electron-donating group ($+M$ effect),decreasing the acidic strength compared to benzoic acid.
Therefore,$4-$Nitrobenzoic acid has the maximum acidic strength among the given options.

(D) Step $1$: Reaction of Grignard reagent with $CO_2$.
$C_6H_5CH_2MgBr + CO_2$ $\rightarrow C_6H_5CH_2COOMgBr$ $\xrightarrow{H_3O^+} C_6H_5CH_2COOH$ (Phenylacetic acid,'$X$').
Step $2$: Decarboxylation of the carboxylic acid.
$C_6H_5CH_2COOH \xrightarrow{NaOH + CaO, \Delta} C_6H_5CH_3$ (Toluene,'$Y$').
The final product '$Y$' is $C_6H_5CH_3$.

(A) The $pK_{a}$ value is inversely proportional to the acid strength. Stronger acids have lower $pK_{a}$ values,while weaker acids have higher $pK_{a}$ values.
Electron-withdrawing groups (EWGs) increase the acidity of carboxylic acids by stabilizing the carboxylate anion through the inductive effect.
Comparing the substituents attached to the $CH_{2}COOH$ group:
$1$. $C_{6}H_{5}-$ (Phenyl group): Weakly electron-withdrawing or electron-donating by resonance,making it the least acidic among the options.
$2$. $O_{2}N-$ (Nitro group): Strong electron-withdrawing group ($-I$ and $-M$ effect).
$3$. $F-$ (Fluoro group): Strong electron-withdrawing group ($-I$ effect).
$4$. $NC-$ (Cyano group): Strong electron-withdrawing group ($-I$ effect).
Since $C_{6}H_{5}CH_{2}COOH$ has the weakest electron-withdrawing group,it is the least acidic and therefore has the highest $pK_{a}$ value.

(D) The reaction between benzoyl chloride $(C_6H_5COCl)$ and sodium benzoate $(C_6H_5COONa)$ is a nucleophilic acyl substitution reaction.
In this reaction,the carboxylate oxygen of the sodium benzoate acts as a nucleophile and attacks the carbonyl carbon of the benzoyl chloride,displacing the chloride ion.
This results in the formation of benzoic anhydride $((C_6H_5CO)_2O)$ and sodium chloride $(NaCl)$ as a byproduct.
The reaction is represented as: $C_6H_5COCl + C_6H_5COONa \xrightarrow{\Delta} (C_6H_5CO)_2O + NaCl$.
Therefore,the correct option is $D$.

(D) The acidic strength of carboxylic acids depends on the stability of the carboxylate anion formed after the loss of a proton. Electron-withdrawing groups $(EWG)$ increase acidity,while electron-donating groups $(EDG)$ decrease it.
$A$. The inductive effect of the $-Cl$ atom decreases as its distance from the $-COOH$ group increases. Thus,the order is correct.
$B$. $-NO_2$ is an $EWG$ (increases acidity),and $-CH_3$ is an $EDG$ (decreases acidity) compared to benzoic acid. Thus,the order is correct.
$C$. The number of $-Cl$ atoms increases the electron-withdrawing effect,increasing acidity. Thus,the order is correct.
$D$. Formic acid $(HCOOH)$ is more acidic than acetic acid $(CH_3COOH)$ due to the $+I$ effect of the methyl group. However,benzoic acid $(C_6H_5COOH)$ is more acidic than acetic acid $(CH_3COOH)$ because the phenyl group is electron-withdrawing relative to the methyl group. Therefore,the correct order is $HCOOH > C_6H_5COOH > CH_3COOH$. The given order $HCOOH > CH_3COOH > C_6H_5COOH$ is incorrect.

(C) The acidic strength of carboxylic acids depends on the stability of the conjugate base (carboxylate ion). Electron-withdrawing groups $(EWG)$ increase acidity by stabilizing the negative charge,while electron-donating groups $(EDG)$ decrease acidity.
$A$. $Cl_3C-COOH > Cl_2CH-COOH > ClCH_2-COOH$: Correct,as the number of $EWG$ $(Cl)$ increases,acidity increases.
$B$. $CH_3CH_2CH(Cl)COOH > CH_3CH(Cl)CH_2COOH > CH_2(Cl)CH_2CH_2COOH$: Correct,as the distance of the $EWG$ from the $-COOH$ group increases,the inductive effect decreases,thus acidity decreases.
$C$. $HCOOH > CH_3COOH > C_6H_5COOH$: This order is incorrect. The correct order is $HCOOH > C_6H_5COOH > CH_3COOH$. $C_6H_5COOH$ is more acidic than $CH_3COOH$ because the phenyl group exerts an electron-withdrawing effect through resonance,whereas the methyl group is electron-donating ($+I$ effect).
$D$. $CH_3COOH > CH_3CH_2COOH > (CH_3)_2CHCOOH$: Correct,as the number of alkyl groups $(EDG)$ increases,the $+I$ effect increases,which destabilizes the carboxylate ion and decreases acidity.

(B) The acidic strength of carboxylic acids depends on the stability of the conjugate base (carboxylate ion).
Electron-withdrawing groups (like $Cl$) increase the acidity by stabilizing the negative charge on the carboxylate ion through the inductive effect ($-I$ effect).
As the number of chlorine atoms increases,the $-I$ effect increases,leading to greater stabilization of the carboxylate ion.
Therefore,the order of acidic strength is $Cl_3CCOOH > Cl_2CHCOOH > ClCH_2COOH > CH_3COOH$.

(D) The acidity of substituted benzoic acids is influenced by the inductive effect $(-I)$ and resonance effect $(-R)$ of the substituents.
$2-$Nitrobenzoic acid is the strongest acid among the given options due to the ortho effect.
The ortho effect is a steric and electronic phenomenon where the presence of a substituent at the ortho position relative to the carboxyl group increases the acidity,regardless of whether the substituent is electron-withdrawing or electron-donating.
Therefore,the correct order of acidity is: $2-$Nitrobenzoic acid $> 3-$Nitrobenzoic acid $> 4-$Nitrobenzoic acid $>$ Benzoic acid.

(D) Sodium benzoate $(C_6H_5COONa)$ is widely used as a food preservative because it inhibits the growth of bacteria,yeast,and mold in acidic conditions.
Therefore,the correct option is $D$.

(C) The $pK_a$ value is inversely proportional to the acidity of the compound. $A$ stronger acid has a lower $pK_a$ value,while a weaker acid has a higher $pK_a$ value.
Acidity depends on the stability of the conjugate base (carboxylate ion). Electron-withdrawing groups $(EWG)$ increase acidity,while electron-donating groups $(EDG)$ decrease acidity.
Comparing the substituents on the benzene ring:
$1$. $-NO_2$ is a strong $EWG$ (increases acidity).
$2$. $-Cl$ is a weak $EWG$ (increases acidity).
$3$. $-H$ is the reference.
$4$. $-CH_3$ is an $EDG$ (decreases acidity).
Since $-CH_3$ is an electron-donating group,it destabilizes the carboxylate ion,making $p$-methylbenzoic acid the weakest acid among the given options.
Therefore,$p$-methylbenzoic acid has the highest $pK_a$ value.

(C) The acidity of carboxylic acids is determined by the stability of the conjugate base (carboxylate ion).
Electron-withdrawing groups (EWGs) stabilize the carboxylate ion through the inductive effect ($-I$ effect),thereby increasing the acidity and the $K_a$ value.
Comparing the substituents:
$1$. $NO_2$ group has a strong $-I$ effect.
$2$. $Br$ group has a moderate $-I$ effect.
$3$. $CCl_3$ group has a very strong $-I$ effect due to three highly electronegative chlorine atoms.
$4$. $CH_3$ group is an electron-donating group ($+I$ effect),which decreases acidity.
Among the given options,$CCl_3COOH$ has the strongest electron-withdrawing effect,making it the strongest acid with the highest $K_a$ value.

(A) $LiAlH_{4} / \text{ether}$ is a strong reducing agent. It is capable of reducing carboxylic acids to primary alcohols.
The reaction is: $CH_{3}COOH \xrightarrow{LiAlH_{4} / \text{ether}} CH_{3}CH_{2}OH$.

(A) Lower members of aliphatic carboxylic acids are soluble in water. This is due to the formation of hydrogen bonds with water molecules.
Carboxylic acids are polar molecules; they tend to be soluble in water,but as the alkyl chain gets longer,their solubility decreases due to the increasing hydrophobic nature of the carbon chain.

(B) Methanoic acid $(HCOOH)$ contains a hydrogen atom attached to the carbonyl carbon,similar to an aldehyde group $(H-C(=O)-)$.
Therefore,like aldehydes,it reduces Tollen's reagent to metallic silver.
Ethanoic acid $(CH_3COOH)$ is a typical carboxylic acid that lacks this specific $H-C(=O)-$ group and does not reduce Tollen's reagent.
Thus,Tollen's test is used to distinguish between them.

(C) The acidity of carboxylic acids is determined by the stability of the conjugate base (carboxylate ion).
Electron-withdrawing groups ($-I$ effect) stabilize the carboxylate ion by dispersing the negative charge,thereby increasing acidity.
$CH_3CH_2COOH$ (Propanoic acid) has no $-I$ group.
$ClCH_2COOH$ (Chloroacetic acid) has one $-I$ group.
$Cl_2CHCOOH$ (Dichloroacetic acid) has two $-I$ groups.
$Cl_3CCOOH$ (Trichloroacetic acid) has three $-I$ groups.
Since the number of $-I$ groups is highest in Trichloroacetic acid,it is the most acidic.

(C) The acidic strength of substituted benzoic acids is influenced by electronic effects and the ortho effect.
$1$. The $CH_3$ group is an electron-donating group ($+I$ effect),which decreases the acidity of benzoic acid when present at the $meta$ or $para$ positions.
$2$. $p-$toluic acid has a $CH_3$ group at the $para$ position,which destabilizes the carboxylate anion via the $+I$ effect,making it less acidic than benzoic acid.
$3$. $o-$toluic acid experiences the 'ortho effect',which significantly increases its acidity compared to benzoic acid,regardless of the electronic nature of the substituent.
$4$. Therefore,the decreasing order of acidic strength is: $o-$toluic acid $>$ benzoic acid $>$ $p-$toluic acid.

(A) The Hell-Volhard Zelinsky $(HVZ)$ reaction is specific to carboxylic acids that contain at least one $\alpha$-hydrogen atom.
Ethanoic acid $(CH_3COOH)$ contains an $\alpha$-carbon atom attached to three $\alpha$-hydrogen atoms,allowing it to undergo the $HVZ$ reaction.
Methanoic acid $(HCOOH)$ does not have an $\alpha$-carbon atom,and consequently,it lacks $\alpha$-hydrogen atoms,which prevents it from undergoing the $HVZ$ reaction.

(C) The $HVZ$ (Hell-Volhard-Zelinsky) reaction of propanoic acid $(CH_3CH_2COOH)$ with $Cl_2/P$ yields $2-$chloropropanoic acid $(CH_3CHClCOOH)$.
In $2-$chloropropanoic acid,the chlorine atom is an electron-withdrawing group ($-I$ effect).
Electron-withdrawing groups stabilize the carboxylate anion $(RCOO^-)$ by dispersing the negative charge,thereby increasing the acidity of the carboxylic acid.
Therefore,$2-$chloropropanoic acid is a stronger acid than propanoic acid.

(B) Formic acid $(HCOOH)$ undergoes dehydration when heated with concentrated sulfuric acid $(H_{2}SO_{4})$.
$HCOOH \xrightarrow{\text{conc. } H_{2}SO_{4}} CO + H_{2}O$
The concentrated $H_{2}SO_{4}$ acts as a dehydrating agent,removing a water molecule from the formic acid to produce carbon monoxide $(CO)$ gas.

(D) The reaction sequence is as follows:
$1$. $CH_{3}Br + KCN \rightarrow CH_{3}CN + KBr$ (Product $A$ is $CH_{3}CN$)
$2$. $CH_{3}CN + 2H_{2}O \xrightarrow{H^{+}} CH_{3}COOH + NH_{3}$ (Product $B$ is $CH_{3}COOH$)
$3$. $CH_{3}COOH \xrightarrow{LiAlH_{4}} CH_{3}CH_{2}OH$ (Product $C$ is $CH_{3}CH_{2}OH$)
Thus,the final product $C$ is ethyl alcohol.

(B) $1$. For the formation of $X$: Benzoic acid reacts with $B_2H_6$ followed by hydrolysis $(H_3O^+)$ to undergo reduction,yielding benzyl alcohol $(C_6H_5CH_2OH)$ as $X$.
$2$. For the formation of $Y$: Benzoic acid reacts with $SOCl_2$ to form benzoyl chloride $(C_6H_5COCl)$,which then undergoes Rosenmund reduction using $H_2/Pd-BaSO_4$ to yield benzaldehyde $(C_6H_5CHO)$ as $Y$.
$3$. Therefore,$X$ is benzyl alcohol and $Y$ is benzaldehyde.

(A) $LiAlH_4$ is a strong reducing agent that reduces carboxylic acid to primary alcohol.
$HI / \text{red } P$ is a strong reducing agent that reduces carboxylic acid to the corresponding alkane.
Therefore,for the reaction $CH_3 COOH \xrightarrow{(A)} CH_3 CH_2 OH$,reagent $A$ is $LiAlH_4$.
For the reaction $CH_3 COOH \xrightarrow{(B)} CH_3 CH_3$,reagent $B$ is $HI / \text{red } P$.

(C) The conversion of toluene $(C_6H_5CH_3)$ to benzoic acid $(C_6H_5CO_2H)$ requires strong oxidation,which is achieved using alkaline $KMnO_4$ followed by acidic workup $(H_3O^{+})$.
Thus,$X = (i) \ KMnO_4 / OH^{-}, \Delta, (ii) \ H_3O^{+}$.
The reduction of benzoic acid to benzyl alcohol $(C_6H_5CH_2OH)$ is selectively performed by diborane $(B_2H_6)$ followed by acidic workup $(H_3O^{+})$,as $B_2H_6$ does not reduce the aromatic ring.
Thus,$Y = (i) \ B_2H_6, (ii) \ H_3O^{+}$.
Therefore,the correct option is $C$.

(D) The starting material is $2-cyclohexylethanol$.
$(i)$ Treatment with $CrO_3, H_2SO_4$ (Jones reagent) oxidizes the primary alcohol to a carboxylic acid,yielding $cyclohexylacetic$ acid $(C_6H_{11}CH_2COOH)$.
(ii) The reaction with $Cl_2/Red \ P$ is the Hell-Volhard-Zelinsky $(HVZ)$ reaction,which halogenates the $\alpha$-carbon of the carboxylic acid.
(iii) The product formed is $2-chloro-2-cyclohexylacetic$ acid $(C_6H_{11}CHClCOOH)$.
Thus,the correct option is $D$.

(A) The reaction sequence proceeds as follows:
$1$. $(i) PCC$: Oxidizes primary alcohol to aldehyde $(Cyclopentylmethanol \rightarrow Cyclopentanecarbaldehyde)$.
$2$. $(ii) Tollen's reagent, NaOH$: Oxidizes the aldehyde to a carboxylic acid $(Cyclopentanecarbaldehyde \rightarrow Cyclopentanecarboxylic acid)$.
$3$. $(iii) H^+$: Protonation step.
$4$. $(iv) Br_2, Red P$: This is the Hell-Volhard-Zelinsky $(HVZ)$ reaction,which alpha-halogenates the carboxylic acid $(Cyclopentanecarboxylic acid \rightarrow 1-Bromocyclopentanecarboxylic acid)$.
$5$. $(v) H_2O$: Workup step.
The final product is $1-bromocyclopentanecarboxylic acid$.

(A) The esterification reaction (Fischer esterification) proceeds via the nucleophilic acyl substitution mechanism.
$1$. The carbonyl oxygen of the carboxylic acid is protonated by $H_2SO_4$,making the carbonyl carbon more electrophilic.
$2$. The alcohol $(R'OH)$ acts as a nucleophile and attacks the electrophilic carbonyl carbon,forming a tetrahedral intermediate.
$3$. The structure of this tetrahedral intermediate is $R-C(OH)_2-OR'$,where the oxygen atom derived from the alcohol is protonated (bearing a positive charge).
$4$. Comparing the given options with the mechanism,the structure corresponding to the tetrahedral intermediate is represented in option $A$.

(D) For reaction $(a)$: Benzoic acid contains a $-COOH$ group,which is a deactivating and meta-directing group. Nitration with $Conc. HNO_3 + Conc. H_2SO_4$ yields $m$-nitrobenzoic acid as the major product. Thus,$X$ is $3$-nitrobenzoic acid.
For reaction $(b)$: This is the Hell-Volhard-Zelinsky $(HVZ)$ reaction. Carboxylic acids having an $\alpha$-hydrogen atom are halogenated at the $\alpha$-position on treatment with $Br_2$ in the presence of small amount of red phosphorus. The product $Y$ is an $\alpha$-bromo carboxylic acid,$R-CH(Br)-COOH$.

(A) The conversion of acetic acid $(CH_3COOH)$ to acetyl chloride $(CH_3COCl)$ is achieved using thionyl chloride $(SOCl_2)$. This is a standard reaction for converting carboxylic acids to acid chlorides.
The conversion of acetyl chloride $(CH_3COCl)$ to acetaldehyde $(CH_3CHO)$ is a Rosenmund reduction,which uses hydrogen gas in the presence of palladium catalyst poisoned with barium sulfate $(H_2 / Pd-BaSO_4)$.
Therefore,reagent $A$ is $SOCl_2$ and reagent $B$ is $H_2 / Pd-BaSO_4$.

(A) $1$. The reaction of benzamide $(C_6H_5CONH_2)$ with benzenesulfonyl chloride $(C_6H_5SO_2Cl)$ in the presence of pyridine leads to the dehydration of the amide to form benzonitrile $(C_6H_5CN)$ as intermediate $X$.
$2$. The subsequent hydrolysis of benzonitrile $(C_6H_5CN)$ with $H_3O^+$ yields benzoic acid $(C_6H_5COOH)$.
$3$. The $-COOH$ group is a deactivating and meta-directing group. Therefore,electrophilic aromatic substitution with $Br_2/FeBr_3$ directs the bromine atom to the meta position.
$4$. The final product $Y$ is $3-$bromobenzoic acid.

(C) Step $1$: Identify the products.
Reaction $A$: Oxidation of propylbenzene with $KMnO_4/OH^-$ followed by acid workup gives benzoic acid $(I = C_6H_5COOH)$.
Reaction $B$: Hell-Volhard-Zelinsky reaction of acetic acid gives bromoacetic acid $(II = BrCH_2COOH)$.
Reaction $C$: Grignard reagent formation from bromobenzene followed by reaction with $CO_2$ and acid workup gives benzoic acid $(III = C_6H_5COOH)$.
Note: $I$ and $III$ are both benzoic acid.
Step $2$: Compare acid strength.
$II$ is $BrCH_2COOH$. The electron-withdrawing $-Br$ group increases the acidity of the carboxylic acid compared to benzoic acid.
$I$ and $III$ are both benzoic acid $(C_6H_5COOH)$.
Therefore,the order of acid strength is $II > I = III$.
Given the options,the most appropriate order is $II > I \approx III$.

(C) The acidity of carboxylic acids depends on the stability of the carboxylate anion formed after the loss of a proton. Electron-withdrawing groups increase acidity by stabilizing the negative charge,while electron-donating groups decrease acidity.
$1$. $HCOOH$ (Formic acid): The $H$ atom has no inductive effect.
$2$. $CH_3COOH$ (Acetic acid): The $CH_3$ group is electron-donating ($+I$ effect),which destabilizes the carboxylate anion,making it less acidic than formic acid.
$3$. $C_6H_5COOH$ (Benzoic acid): The phenyl group is electron-withdrawing due to its $sp^2$ hybridized carbon atom ($-I$ effect),making it more acidic than acetic acid.
$4$. $C_6H_5CH_2COOH$ (Phenylacetic acid): The $CH_2$ group separates the phenyl ring from the carboxyl group,reducing the electron-withdrawing effect of the phenyl ring compared to benzoic acid.
Comparing the $pK_a$ values: $HCOOH$ $(pK_a \approx 3.75)$ > $C_6H_5COOH$ $(pK_a \approx 4.20)$ > $C_6H_5CH_2COOH$ $(pK_a \approx 4.31)$ > $CH_3COOH$ $(pK_a \approx 4.76)$.
Therefore,$HCOOH$ is the most acidic among the given options.

(D) The acidity of substituted benzoic acids depends on the electronic effects of the substituents. Electron-withdrawing groups $(EWG)$ increase acidity (lower $pK_{a}$),while electron-donating groups $(EDG)$ decrease acidity (higher $pK_{a}$).
$1$. $III$ ($p$-methylbenzoic acid): $-CH_3$ is an $EDG$ ($+I$ and hyperconjugation effect),which decreases acidity. Thus,it has the highest $pK_{a}$.
$2$. $I$ ($p$-iodobenzoic acid): $-I$ is a weak $EWG$ ($-I$ effect),which slightly increases acidity.
$3$. $II$ ($p$-cyanobenzoic acid): $-CN$ is a strong $EWG$ ($-I$ and $-M$ effects),which significantly increases acidity.
$4$. $IV$ ($p$-nitrobenzoic acid): $-NO_2$ is a very strong $EWG$ ($-I$ and $-M$ effects),which increases acidity the most. Thus,it has the lowest $pK_{a}$.
Therefore,the highest $pK_{a}$ is for $III$ and the lowest $pK_{a}$ is for $IV$. The correct option is $D$.

(B) The acidity of substituted benzoic acids is directly proportional to the electron-withdrawing effect ($-I$ and $-M$ effects) of the substituent group.
The substituents present are:
$(A) -CN$ (strong $-I$ and $-M$ effect)
$(B) -F$ (strong $-I$ effect,weak $+M$ effect)
$(C) -NO_2$ (very strong $-I$ and $-M$ effect)
The overall electron-withdrawing power follows the order: $-NO_2 > -CN > -F$.
Therefore,the decreasing order of acidity is $C > A > B$.

(D) The conversion of a primary alcohol $(C_6H_5CH_2CH_2OH)$ to a carboxylic acid $(C_6H_5CH_2COOH)$ requires a strong oxidizing agent like Jones reagent ($CrO_3/H_2SO_4$ in $H_2O/acetone$). Thus,$X=$ Jones reagent.
The conversion of a carboxylic acid to an alkane with one less carbon atom is achieved by heating with soda lime $(NaOH + CaO)$,a process known as decarboxylation. Thus,$Y=NaOH, CaO, \Delta$ and $Z=C_6H_5CH_3$ (toluene).
Therefore,the correct option is $D$.

(C) $1$. The reaction of styrene $(C_6H_5-CH=CH_2)$ with alkaline $KMnO_4$ followed by acidic workup $(H_3O^+)$ is an oxidative cleavage reaction that converts the vinyl group into a carboxylic acid group,yielding benzoic acid $(C_6H_5COOH)$ as product $X$.
$2$. Benzoic acid contains a $-COOH$ group attached to the benzene ring. The $-COOH$ group is a strongly deactivating group and is meta-directing for electrophilic aromatic substitution reactions.
$3$. Therefore,the bromination of benzoic acid $(X)$ using $Br_2 / FeBr_3$ will result in the electrophilic substitution of the bromine atom at the meta-position,yielding $m$-bromobenzoic acid as product $Y$.

(A) The acidic strength depends on the stability of the conjugate base formed after the loss of a proton.
$(i)$ is $4$-ethylcyclohexanecarboxylic acid (aliphatic).
$(ii)$ is $4$-methoxybenzoic acid (aromatic with $+R$ effect of $-OCH_3$ group).
$(iii)$ is $4$-nitrobenzoic acid (aromatic with $-R$ effect of $-NO_2$ group).
$1$. Aromatic carboxylic acids are generally stronger than aliphatic carboxylic acids due to the resonance stabilization of the carboxylate anion by the benzene ring.
$2$. The $-NO_2$ group in $(iii)$ exerts a strong $-R$ (electron-withdrawing) effect,which stabilizes the carboxylate anion and significantly increases acidity.
$3$. The $-OCH_3$ group in $(ii)$ exerts a $+R$ (electron-donating) effect,which destabilizes the carboxylate anion and decreases acidity compared to benzoic acid,but it is still more acidic than the aliphatic acid $(i)$ due to the aromatic ring.
Therefore,the order of acidic strength is $(iii) > (ii) > (i)$.

(A) The reaction sequence is as follows:
$1$. $CH_3COOH$ reacts with $P_2O_5$ under heating $(\Delta)$ to form acetic anhydride $(X)$ via dehydration.
$2$. Acetic anhydride $(X)$ then reacts with salicylic acid $(Y)$ in the presence of an acid catalyst $(H^ )$ to undergo acetylation of the phenolic $-OH$ group.
$3$. This reaction produces $2-\text{acetoxybenzoic acid}$,commonly known as aspirin $(B)$,which acts as an analgesic drug.
Therefore,$A = P_2O_5, \Delta$ and $B = \text{2-acetoxybenzoic acid}$.

(D) The reaction of benzoic acid $(C_6H_5COOH)$ with $Br_2$ in the presence of a Lewis acid catalyst like $FeBr_3$ is an electrophilic aromatic substitution reaction.
The $-COOH$ group attached to the benzene ring is a strongly deactivating group and is meta-directing.
Therefore,the electrophile $(Br^+)$ will attack the meta-position of the benzene ring,resulting in the formation of $3-$bromo benzoic acid as the major product.

(D) The reaction $X = CH_3Cl / AlCl_3$ is not feasible for benzoic acid because the $-COOH$ group is a strongly deactivating group,which makes the benzene ring unreactive towards electrophilic aromatic substitution. Furthermore,the $-COOH$ group coordinates with the Lewis acid $AlCl_3$ to form a complex,which further deactivates the ring and prevents the Friedel-Crafts alkylation reaction from occurring.

(C) The conversion of a carboxylic acid into an acyl chloride is best achieved using thionyl chloride $(SOCl_2)$.
The reaction is given by: $CH_3CH_2COOH + SOCl_2 \rightarrow CH_3CH_2COCl + SO_2 + HCl$.
This method is preferred because the by-products ($SO_2$ and $HCl$) are gases,which escape the reaction mixture,driving the reaction to completion.