Mix Examples-Carboxylic acids and Their derivative Questions in English

(C) The reaction sequence is as follows:
$1$. The starting material is a cyclopentane ring substituted with a carboxylic acid group and an amide group. Treatment with $KOBr$ (Hoffmann bromamide degradation) converts the $-CONH_2$ group into an $-NH_2$ group.
$2$. This results in compound $(A)$,which is $3$-aminocyclopentanecarboxylic acid.
$3$. Upon heating $(\Delta)$,the amino acid undergoes intramolecular cyclization (dehydration) to form a cyclic amide,known as a lactam.
$4$. The resulting product $(B)$ is the cyclic lactam,which corresponds to option $C$.

(B) The synthesis of Ibuprofen involves the following steps:
$1$. Friedel-Crafts acylation of benzene with isobutyryl chloride to form isobutyrophenone.
$2$. Clemmensen reduction of the ketone to form isobutylbenzene.
$3$. Friedel-Crafts acylation of isobutylbenzene with acetyl chloride to form $4$-isobutylacetophenone.
$4$. Cyanohydrin formation followed by reduction with $Red \ P + HI$ to convert the acetyl group into a propionic acid group,yielding $2-(4-\text{isobutylphenyl})\text{propanoic acid}$,which is Ibuprofen.
The correct structure is shown in option $B$.

(B) The reaction is an intramolecular Friedel-Crafts acylation of $3-$phenylbutanoic anhydride derivative.
$1$. The $AlCl_3$ acts as a Lewis acid and coordinates with one of the carbonyl oxygens of the anhydride.
$2$. This facilitates the formation of an acylium ion intermediate.
$3$. The benzene ring then attacks the electrophilic acylium carbon.
$4$. Since the chain length allows for the formation of a stable six-membered ring,the cyclization occurs to form a substituted tetralone.
$5$. Specifically,the product is $3$-methyl-$1$-tetralone-$3$-carboxylic acid derivative,which corresponds to the structure in option $B$.

(B) The given compound is $4-(2,5-\text{dimethylphenyl})-4-\text{oxobutanoic acid}$.
This compound can be prepared by the Friedel-Crafts acylation of $p$-xylene $(1,4-\text{dimethylbenzene})$ with succinic anhydride in the presence of a Lewis acid catalyst like anhydrous $AlCl_3$.
The reaction involves the electrophilic substitution of the benzene ring of $p$-xylene by the acylium ion generated from succinic anhydride.

(D) Heating $1,2-$dicarboxylic acids leads to the formation of cyclic anhydrides by the loss of a water molecule.
For a tetracarboxylic acid to form a di-anhydride,it must have two pairs of adjacent carboxylic acid groups.
In benzene$-1,2,3,4-$tetracarboxylic acid,there are two pairs of adjacent $-COOH$ groups (at positions $1,2$ and $3,4$),allowing for the formation of a di-anhydride.
In benzene$-1,2,4,5-$tetracarboxylic acid (pyromellitic acid),there are two pairs of adjacent $-COOH$ groups (at positions $1,2$ and $4,5$),which also allows for the formation of a di-anhydride.
Therefore,both compounds can form di-anhydrides upon heating.

(A) $1$. Esterification: Naphthalene$-1-$acetic acid reacts with $EtOH/H^+$ to form ethyl naphthalene$-1-$acetate.
$2$. Alkylation: Treatment with $NaH$ followed by $MeI$ results in the formation of ethyl $2-$(naphthalen$-1-$yl)propanoate.
$3$. Hydrolysis: Basic hydrolysis followed by acidification gives $2-$(naphthalen$-1-$yl)propanoic acid.
$4$. Cyclization: Treatment with $SOCl_2$ converts the carboxylic acid to an acid chloride,which then undergoes intramolecular Friedel-Crafts acylation in the presence of $AlCl_3$ to form the cyclic ketone,$2$-methyl$-2,3-$dihydro-1H-cyclopenta[a]naphthalen$-1-$one.

(B) The reaction sequence is a Hoffmann bromamide degradation,which converts an amide to an amine with one less carbon atom. The final product is $2-aminophenol$. The amide precursor is $2-hydroxybenzamide$. This amide is formed from $2-hydroxybenzoic$ acid (salicylic acid) via $SOCl_2$ and $NH_3$. The reactant $(A)$ is $2-hydroxybenzoic$ acid. Among the options,hydrolysis of the ester in option $(B)$ (which is $2-nitrophenyl-2-hydroxybenzoate$ or similar derivative) yields the required acid. Specifically,the structure in option $(B)$ represents an ester that upon hydrolysis yields $2-hydroxybenzoic$ acid and $2-nitrophenol$.

(A) Benzocaine is the ethyl ester of $p-$aminobenzoic acid. Its $IUPAC$ name is ethyl $4-$aminobenzoate. Both $A$ and $B$ represent the same compound,as $p-$ (para) position corresponds to the $4-$ position in the benzene ring relative to the carboxylate group. Therefore,both options $A$ and $B$ are chemically correct.

(D) The starting material contains an ester group $(EtO_2C-)$,a carboxylic acid group $(-COOH)$,and a nitrile group $(-CN)$.
$1$. Treatment with $B_2H_6$ selectively reduces the carboxylic acid group to a primary alcohol $(-CH_2OH)$ without affecting the ester or nitrile groups.
$2$. Treatment with $SnCl_2/HCl$ (Stephen reduction) reduces the nitrile group to an aldehyde $(-CHO)$ group.
$3$. Finally,acidic hydrolysis $(H_3O^{+})$ hydrolyzes the ester group to a carboxylic acid group $(-COOH)$.
Thus,the correct sequence of reagents is $(i) B_2H_6, (ii) SnCl_2/HCl, (iii) H_3O^{+}$.

(A) The acidity of substituted benzoic acids depends on the electronic effects of the substituents at the para position.
$1$. $-NO_2$ is a strong electron-withdrawing group ($-I$ and $-M$ effects),which significantly increases acidity.
$2$. $-Cl$ is an electron-withdrawing group ($-I$ effect) but an electron-donating group ($+M$ effect). The $-I$ effect dominates,increasing acidity compared to benzoic acid.
$3$. $-OH$ is an electron-donating group ($+M$ effect),which decreases acidity compared to benzoic acid.
$4$. Benzoic acid $(II)$ is the reference.
Comparing the substituents: $-NO_2$ (strongest electron-withdrawing) > $-Cl$ (weak electron-withdrawing) > $H$ (no effect) > $-OH$ (electron-donating).
Therefore,the increasing order of acidity is $III < II < IV < I$.

(D) The reaction involves an intramolecular esterification (lactonization) between the carboxylic acid group $(-COOH)$ and the hydroxymethyl group $(-CH_2OH)$ in the presence of an acid catalyst $(HCl_{(g)})$.
$1$. The carboxylic acid group is protonated by $HCl$,making the carbonyl carbon more electrophilic.
$2$. The oxygen atom of the $-CH_2OH$ group acts as a nucleophile and attacks the activated carbonyl carbon.
$3$. This leads to the formation of a cyclic ester (lactone).
$4$. Water is eliminated in the process to form the final lactone product.
The structure shown in option $D$ represents the correct cyclic lactone formed by the reaction of the carboxylic acid and the hydroxymethyl group.

(C) The reactant is phenyl benzoate,which contains two phenyl rings. One ring is attached to an oxygen atom $(-O-)$,and the other is attached to a carbonyl group $(-C(=O)-)$.
The oxygen atom is electron-donating by resonance ($+M$ effect),which activates the attached phenyl ring towards electrophilic aromatic substitution.
The carbonyl group is electron-withdrawing by resonance ($-M$ effect),which deactivates the other phenyl ring.
Therefore,the electrophile $(Br^+)$ generated by $Br_2/FeBr_3$ will attack the more activated ring,which is the one attached to the oxygen atom.
In the ring attached to the oxygen atom,the $-O-$ group is ortho/para-directing. The para position is sterically less hindered than the ortho position,making the para-substituted product the major one.
Thus,the product is phenyl $4$-bromobenzoate.

(D) Step $1$: Acidic hydrolysis of acetonitrile $(CH_3-CN)$ yields acetic acid $(CH_3COOH)$ as intermediate $X$.
$CH_3-CN + 2H_2O \xrightarrow{H^+} CH_3COOH + NH_4^+$
Step $2$: Esterification of acetic acid with phenol $(C_6H_5OH)$ in the presence of an acid catalyst $(H^+)$ yields phenyl acetate $(CH_3COOC_6H_5)$ as product $Y$.
$CH_3COOH + C_6H_5OH \xrightarrow{H^+} CH_3COOC_6H_5 + H_2O$
Thus,$Y$ is phenyl acetate.

(A) Propanal $(CH_3-CH_2-CHO)$ is oxidized by $\text{Tollen's reagent}$ followed by acidification to give propanoic acid $(X = CH_3-CH_2-COOH)$.
Propanoic acid reacts with $Ca(OH)_2$ to form calcium propionate,which on $\text{dry distillation}$ yields diethyl ketone $(Y = C_2H_5-CO-C_2H_5)$.
$2CH_3CH_2COOH + Ca(OH)_2$ $\rightarrow (CH_3CH_2COO)_2Ca$ $\xrightarrow{\Delta} CH_3CH_2COCH_2CH_3 + CaCO_3$

(A) Let us analyze each option:
$(A)$ The structure shown is $2$-acetoxybenzoic acid,which is commonly known as $\text{Aspirin}$. This is a correct match.
$(B)$ The structure $Cl_3C-CH(C_6H_4Cl)_2$ represents $p,p'$-dichlorodiphenyltrichloroethane,commonly known as $\text{D.D.T.}$ This is a correct match.
$(C)$ The formula $H_2N-CO-NH_2$ represents $\text{Urea}$. This is a correct match.
$(D)$ The structure $H_2N-C_6H_4-SO_3H$ represents $p$-aminobenzenesulphonic acid,which is $\text{Sulphanilic acid}$. This is a correct match.
Wait,re-evaluating the provided image for option $(A)$: The image shows a benzene ring with an acetoxy group $(-O-CO-CH_3)$ and an aldehyde group $(-CHO)$ at the ortho position. This is $2$-acetoxybenzaldehyde,not $\text{Aspirin}$ ($2$-acetoxybenzoic acid). Therefore,option $(A)$ is the incorrect match.

(C) The Kolbe electrolysis or Kolbe reaction is an organic reaction named after Hermann Kolbe.
The Kolbe reaction is formally a decarboxylative dimerisation of two carboxylic acids (or carboxylate ions).
The overall general reaction is: $2RCOO^- \longrightarrow R-R + 2CO_2 + 2e^-$.
After electrolysis,an alkane (hydrocarbon) is formed at the anode along with the release of carbon dioxide $(CO_2)$.
When potassium acetate $(CH_3COO^- K^+)$ is used in electrolysis,carbon dioxide is released along with ethane $(H_3C-CH_3)$.
$CH_3COO^- \longrightarrow CH_3^{\bullet} + CO_2 + e^-$
$2CH_3^{\bullet} \longrightarrow C_2H_6$
Thus,both hydrocarbons and $CO_2$ are released at the anode.
Option $C$ is the correct answer.

(A) The reaction sequence is as follows:
$1$. $CH_3-CH_2-COOH + PCl_3 \rightarrow CH_3-CH_2-COCl$ (Compound $I$,Propanoyl chloride).
$2$. $CH_3-CH_2-COCl + C_6H_6 \xrightarrow{AlCl_3} C_6H_5-CO-CH_2-CH_3$ (Compound $II$,Propiophenone,which is also known as ethyl phenyl ketone).
$3$. $C_6H_5-CO-CH_2-CH_3 \xrightarrow{NH_2-NH_2/\text{base/heat}} C_6H_5-CH_2-CH_2-CH_3$ (Compound $III$,Propylbenzene).
This is a Friedel-Crafts acylation followed by a Wolff-Kishner reduction.

(C) lower $pK_a$ value indicates a stronger acid. Acetic acid $(pK_a \approx 4.76)$ is a stronger acid than phenol $(pK_a \approx 10)$.
The acidity of carboxylic acids is due to the resonance stabilization of the carboxylate ion,where the negative charge is delocalized over two highly electronegative oxygen atoms in equivalent resonance structures.
In contrast,the negative charge in the phenoxide ion is delocalized over the carbon atoms of the ring,which is less effective. Therefore,the Assertion is correct,but the Reason is incorrect.

(A) In acetamide $(CH_{3}CONH_{2})$,the nitrogen atom has a lone pair that participates in resonance with the $C=O$ group. However,the $-NH_{2}$ group is a stronger electron-donating group by resonance compared to the $-OC_{2}H_{5}$ group in ethyl acetoacetate $(CH_{3}COCH_{2}COOC_{2}H_{5})$.
Because nitrogen is less electronegative than oxygen,the lone pair on nitrogen is more readily donated into the carbonyl system,increasing the electron density on the oxygen atom of the $C=O$ bond,which makes the $C=O$ bond more polar.
Thus,both the Assertion and the Reason are correct,and the Reason is the correct explanation for the Assertion.

(C) The Assertion is correct: $RCOCl$,$(RCO)_2O$,and $RCOOR'$ all react with excess Grignard reagents $(R''MgX)$ to form $3^o$ alcohols after hydrolysis.
The Reason is incorrect: While $RCOCl$ reacts with $R_2Cd$ to form ketones,$(RCO)_2O$ also reacts with $R_2Cd$ to form ketones. Therefore,the statement that $(RCO)_2O$ does not react at all is false.

(A) $(i)$ $CH_3CH_2CH_2CH_2OH \xrightarrow{CrO_3 - H_2SO_4 (Jones \ reagent)} CH_3CH_2CH_2COOH$
$(ii)$ $C_6H_5CH_2OH$ $\xrightarrow{HBr} C_6H_5CH_2Br$ $\xrightarrow{KCN} C_6H_5CH_2CN$ $\xrightarrow{H_3O^{+}, \Delta} C_6H_5CH_2COOH$
$(iii)$ $C_6H_4(NO_2)Br$ $\xrightarrow{Mg, \text{ether}} C_6H_4(NO_2)MgBr$ $\xrightarrow{CO_2} C_6H_4(NO_2)COOMgBr$ $\xrightarrow{H_3O^{+}} C_6H_4(NO_2)COOH$
$(iv)$ $CH_3-C_6H_4-COCH_3$ $\xrightarrow{KMnO_4/KOH} KOOC-C_6H_4-COOK$ $\xrightarrow{dil. H_2SO_4} HOOC-C_6H_4-COOH$
$(v)$ $C_6H_{10} \xrightarrow{KMnO_4-H_2SO_4, \Delta} HOOC-(CH_2)_4-COOH$
$(vi)$ $CH_3CH_2CH_2CHO \xrightarrow{[O] \text{ or } \text{Tollens' reagent}} CH_3CH_2CH_2COOH$

(N/A) The conversion of the given compounds to benzoic acid is as follows:
$(i)$ Ethylbenzene: Oxidation with alkaline $KMnO_4$ followed by acidification gives benzoic acid.
$C_6H_5CH_2CH_3$ $\xrightarrow{KMnO_4/KOH, \Delta} C_6H_5COOK$ $\xrightarrow{H_3O^+} C_6H_5COOH$
$(ii)$ Acetophenone: Oxidation with alkaline $KMnO_4$ followed by acidification gives benzoic acid.
$C_6H_5COCH_3$ $\xrightarrow{KMnO_4/KOH, \Delta} C_6H_5COOK$ $\xrightarrow{H_3O^+} C_6H_5COOH$
$(iii)$ Bromobenzene: Reacts with $Mg$ in dry ether to form phenylmagnesium bromide,which reacts with $CO_2$ (dry ice) followed by acidification to give benzoic acid.
$C_6H_5Br$ $\xrightarrow{Mg, \text{ether}} C_6H_5MgBr$ $\xrightarrow{CO_2} C_6H_5COOMgBr$ $\xrightarrow{H_3O^+} C_6H_5COOH$
$(iv)$ Phenylethene (Styrene): Oxidation with alkaline $KMnO_4$ followed by acidification gives benzoic acid.
$C_6H_5CH=CH_2$ $\xrightarrow{KMnO_4/KOH, \Delta} C_6H_5COOK$ $\xrightarrow{H_3O^+} C_6H_5COOH$

(N/A) $(i)$ $(CH_3)_2CHCH_2CHO$
$(ii)$ $O_2NC_6H_4COCH_2CH_3$
$(iii)$ $CH_3C_6H_4CHO$
$(iv)$ $CH_3COCH=C(CH_3)_2$
$(v)$ $CH_3COCH_2CH(Cl)CH_3$
$(vi)$ $CH_3CH(C_6H_5)CH(Br)CH_2COOH$
$(vii)$ $HOC_6H_4COC_6H_4OH$
$(viii)$ $CH_3C \equiv CCH=CHCOOH$

(N/A) An organic compound $(A)$ with molecular formula $C_{8}H_{16}O_{2}$ gives a carboxylic acid $(B)$ and an alcohol $(C)$ on hydrolysis with dilute sulphuric acid. Thus,compound $(A)$ must be an ester.
Further,alcohol $(C)$ gives acid $(B)$ on oxidation with chromic acid. Thus,$(B)$ and $(C)$ must contain an equal number of carbon atoms.
Since compound $(A)$ contains a total of $8$ carbon atoms,each of $(B)$ and $(C)$ contains $4$ carbon atoms.
Again,on dehydration,alcohol $(C)$ gives but$-1-$ene. Therefore,$(C)$ is a straight-chain alcohol,which is butan$-1-$ol.
On oxidation,butan$-1-$ol gives butanoic acid. Hence,acid $(B)$ is butanoic acid.
Thus,the ester with molecular formula $C_{8}H_{16}O_{2}$ is butyl butanoate $(CH_{3}CH_{2}CH_{2}COOCH_{2}CH_{2}CH_{2}CH_{3})$.
The reactions involved are:
$1. \text{Hydrolysis of ester: } CH_{3}CH_{2}CH_{2}COOCH_{2}CH_{2}CH_{2}CH_{3} + H_{2}O$ $\xrightarrow{dil. H_{2}SO_{4}} CH_{3}CH_{2}CH_{2}COOH (B) + CH_{3}CH_{2}CH_{2}CH_{2}OH (C)$
$2. \text{Oxidation of alcohol: } CH_{3}CH_{2}CH_{2}CH_{2}OH \xrightarrow{[O]} CH_{3}CH_{2}CH_{2}COOH (B)$
$3. \text{Dehydration of alcohol: } CH_{3}CH_{2}CH_{2}CH_{2}OH \xrightarrow{H^{+}/\Delta} CH_{3}CH_{2}CH=CH_{2} + H_{2}O$

(A) $(i)$ Reactivity towards $HCN$ depends on the electrophilicity of the carbonyl carbon and steric hindrance. As the number of alkyl groups increases,both steric hindrance and $+I$ effect increase,reducing reactivity. The order is: $\text{Di-tert-butyl ketone} < \text{Methyl tert-butyl ketone} < \text{Acetone} < \text{Acetaldehyde}$.
$(ii)$ Acid strength increases with the presence of electron-withdrawing groups ($-I$ effect) and decreases with electron-donating groups ($+I$ effect). The $-I$ effect of $Br$ decreases with distance. The order is: $(CH_3)_2CHCOOH < CH_3CH_2CH_2COOH < CH_3CH(Br)CH_2COOH < CH_3CH_2CH(Br)COOH$.
$(iii)$ Electron-withdrawing groups $(-I, -M)$ increase acid strength,while electron-donating groups $(+M, +I)$ decrease it. The order is: $4\text{-Methoxybenzoic acid} < \text{Benzoic acid} < 4\text{-Nitrobenzoic acid} < 3,4\text{-Dinitrobenzoic acid}$.

(N/A) $(i)$ Propanal and propanone can be distinguished by the following tests.
$(a)$ Tollen's test: Propanal is an aldehyde and reduces Tollen's reagent to silver mirror,while propanone does not.
$(b)$ Fehling's test: Propanal reduces Fehling's solution to a red-brown precipitate of $Cu_2O$,while propanone does not.
$(c)$ Iodoform test: Propanone being a methyl ketone gives a yellow precipitate of $CHI_3$ with $NaOI$,while propanal does not.
$(ii)$ Acetophenone and Benzophenone: Use the iodoform test. Acetophenone (a methyl ketone) gives a yellow precipitate of $CHI_3$ with $NaOI$,while benzophenone does not.
$(iii)$ Phenol and Benzoic acid: Use the ferric chloride test. Phenol gives a violet colouration with neutral $FeCl_3$,while benzoic acid gives a buff-coloured precipitate of ferric benzoate.
$(iv)$ Benzoic acid and Ethyl benzoate: Use the sodium bicarbonate test. Benzoic acid reacts with $NaHCO_3$ to produce brisk effervescence of $CO_2$ gas,while ethyl benzoate does not.
$(v)$ Pentan$-2-$one and Pentan$-3-$one: Use the iodoform test. Pentan$-2-$one (a methyl ketone) gives a yellow precipitate of $CHI_3$ with $NaOI$,while pentan$-3-$one does not.
$(vi)$ Benzaldehyde and Acetophenone: Use Tollen's test (benzaldehyde gives silver mirror,acetophenone does not) or iodoform test (acetophenone gives yellow precipitate,benzaldehyde does not).
$(vii)$ Ethanal and Propanal: Use the iodoform test. Ethanal $(CH_3CHO)$ gives a yellow precipitate of $CHI_3$ with $NaOI$,while propanal does not.

(N/A) $(i)$ Methyl benzoate: Benzene $\xrightarrow{CH_3Cl, AlCl_3}$ Toluene $\xrightarrow{KMnO_4, OH^-}$ Benzoic acid $\xrightarrow{CH_3OH, H^+}$ Methyl benzoate.
$(ii)$ $m$-Nitrobenzoic acid: Benzene $\xrightarrow{CH_3Cl, AlCl_3}$ Toluene $\xrightarrow{conc. HNO_3, conc. H_2SO_4}$ $p/o$-Nitrotoluene (separate $m$-isomer) $\xrightarrow{KMnO_4, OH^-}$ $m$-Nitrobenzoic acid.
$(iii)$ $p$-Nitrobenzoic acid: Benzene $\xrightarrow{CH_3Cl, AlCl_3}$ Toluene $\xrightarrow{conc. HNO_3, conc. H_2SO_4}$ $p$-Nitrotoluene $\xrightarrow{KMnO_4, OH^-}$ $p$-Nitrobenzoic acid.
$(iv)$ Phenylacetic acid: Benzene $\xrightarrow{CH_3Cl, AlCl_3}$ Toluene $\xrightarrow{Cl_2, h\nu}$ Benzyl chloride $\xrightarrow{KCN}$ Benzyl cyanide $\xrightarrow{H_3O^+}$ Phenylacetic acid.
$(v)$ $p$-Nitrobenzaldehyde: Benzene $\xrightarrow{CH_3Cl, AlCl_3}$ Toluene $\xrightarrow{conc. HNO_3, conc. H_2SO_4}$ $p$-Nitrotoluene $\xrightarrow{CrO_2Cl_2, CS_2, H_3O^+}$ $p$-Nitrobenzaldehyde.

(D)

$(i)$	$CH_3COCH_3$ $\xrightarrow{NaBH_4} CH_3CH(OH)CH_3$ $\xrightarrow{\text{conc. } H_2SO_4, \Delta} CH_3CH=CH_2$
$(ii)$	$C_6H_5COOH$ $\xrightarrow{SOCl_2} C_6H_5COCl$ $\xrightarrow{H_2/Pd-BaSO_4} C_6H_5CHO$
$(iii)$	$CH_3CH_2OH$ $\xrightarrow{PCC} CH_3CHO$ $\xrightarrow{\text{dil. } NaOH} CH_3CH(OH)CH_2CHO$
$(iv)$	$C_6H_6$ $\xrightarrow{CH_3COCl/AlCl_3} C_6H_5COCH_3$ $\xrightarrow{\text{conc. } HNO_3/H_2SO_4} m-NO_2-C_6H_4COCH_3$
$(v)$	$C_6H_5CHO$ $\xrightarrow[2. H_3O^+]{1. C_6H_5MgBr} C_6H_5CH(OH)C_6H_5$ $\xrightarrow{Na_2Cr_2O_7/H_2SO_4} C_6H_5COC_6H_5$
$(vi)$	$C_6H_5Br$ $\xrightarrow{Mg/\text{ether}} C_6H_5MgBr$ $\xrightarrow[2. H_3O^+]{1. CH_3CHO} C_6H_5CH(OH)CH_3$
$(vii)$	$C_6H_5CHO + CH_3CHO$ $\xrightarrow{\text{dil. } NaOH} C_6H_5CH=CHCHO$ $\xrightarrow{H_2/Ni} C_6H_5CH_2CH_2CH_2OH$
$(viii)$	$C_6H_5CHO$ $\xrightarrow{HCN} C_6H_5CH(OH)CN$ $\xrightarrow{H_3O^+} C_6H_5CH(OH)COOH$
$(ix)$	$C_6H_5COOH$ $\xrightarrow{\text{conc. } HNO_3/H_2SO_4} m-NO_2-C_6H_4COOH$ $\xrightarrow{B_2H_6} m-NO_2-C_6H_4CH_2OH$

(N/A) $(i)$ Acetylation:
The introduction of an acetyl functional group into an organic compound is known as acetylation. It is usually carried out in the presence of a base such as pyridine,dimethylaniline,etc. This process involves the substitution of an acetyl group for an active hydrogen atom. Acetyl chloride and acetic anhydride are commonly used as acetylating agents.
For example,acetylation of ethanol produces ethyl acetate.
$CH_3-CH_2-OH + CH_3COCl \xrightarrow{\text{Pyridine}} CH_3COOC_2H_5 + HCl$
$(ii)$ Cannizzaro reaction:
The self oxidation-reduction (disproportionation) reaction of aldehydes having no $\alpha$-hydrogens on treatment with concentrated alkalis is known as the Cannizzaro reaction. In this reaction,two molecules of aldehydes participate where one is reduced to alcohol and the other is oxidized to carboxylic acid.
For example,when methanal is treated with concentrated potassium hydroxide,methanol and potassium methanoate are produced.
$2HCHO + KOH \rightarrow CH_3OH + HCOOK$
$(iii)$ Cross-aldol condensation:
When aldol condensation is carried out between two different aldehydes,or two different ketones,or an aldehyde and a ketone,then the reaction is called a cross-aldol condensation. If both the reactants contain $\alpha$-hydrogens,four compounds are obtained as products.
For example,ethanal and propanal react to give four products.
$(iv)$ Decarboxylation:
Decarboxylation refers to the reaction in which carboxylic acids lose carbon dioxide to form hydrocarbons when their sodium salts are heated with soda-lime.
$CH_3-COONa + NaOH \xrightarrow{\Delta, CaO} CH_4 + Na_2CO_3$
Decarboxylation also takes place when aqueous solutions of alkali metal salts of carboxylic acids are electrolyzed. This electrolytic process is known as Kolbe's electrolysis.

(N/A) $(i)$ Cyclohexanone forms cyanohydrin because the nucleophile $CN^{-}$ can easily attack the carbonyl carbon without significant steric hindrance. In $2,2,6-$trimethylcyclohexanone,the methyl groups at the $\alpha-$positions create significant steric hindrance,preventing the nucleophilic attack of $CN^{-}$.
$(ii)$ Semicarbazide $(H_2N-NH-CO-NH_2)$ exhibits resonance where the lone pair of electrons on one of the $-NH_2$ groups is delocalized towards the carbonyl group. This reduces the electron density and nucleophilicity of that specific $-NH_2$ group. The other $-NH_2$ group,which is not involved in resonance,remains nucleophilic and attacks the carbonyl carbon of aldehydes and ketones to form semicarbazones.
$(iii)$ The esterification reaction between a carboxylic acid and an alcohol is a reversible process: $RCOOH + R'OH \rightleftharpoons RCOOR' + H_2O$. According to Le Chatelier's principle,removing the products (water or ester) as they are formed shifts the equilibrium in the forward direction,thereby increasing the yield of the ester.

(N/A) The homologous series containing the carbonyl group $(C=O)$ are summarized in the table below:

Group (Series)	Common formula	Simple structure	Example
$(i)$ Aldehyde	$RCHO$	$R-C(=O)-H$	$CH_3-C(=O)-H$
(ii) Ketone	$RCOR'$	$R-C(=O)-R'$	$CH_3-C(=O)-CH_3$
(iii) Carboxylic acid	$RCOOH$	$R-C(=O)-OH$	$CH_3-C(=O)-OH$
(iv) Acyl halide	$RCOX$	$R-C(=O)-X$	$CH_3-C(=O)-Cl$
$(v)$ Amide	$RCONH_2$	$R-C(=O)-NH_2$	$CH_3-C(=O)-NH_2$
(vi) Acid anhydride	$(RCO)_2O$	$R-C(=O)-O-C(=O)-R$	$CH_3-C(=O)-O-C(=O)-CH_3$
(vii) Ester	$RCOOR'$	$R-C(=O)-OR'$	$CH_3-C(=O)-OC_2H_5$

(A) The reaction sequence is as follows:
$1$. $CH_3COOH + C_2H_5OH \xrightarrow{H^+} CH_3COOC_2H_5 + H_2O$
Here,$X$ is $C_2H_5OH$ (ethanol) and $Y$ is $CH_3COOC_2H_5$ (ethyl acetate).
$2$. The catalytic hydrogenation (reduction) of the ester $Y$ $(CH_3COOC_2H_5)$ using $H_2$ and a catalyst like $Pd$ or $LiAlH_4$ yields two molecules of ethanol $(C_2H_5OH)$.

(N/A) $1.$ Conversion of $1$-Bromopropane to Propan-$1$-ol and Propan-$2$-ol:
First,$1$-bromopropane $(CH_3CH_2CH_2Br)$ is treated with alcoholic $KOH$ and heated to undergo dehydrohalogenation to form propene $(CH_3CH=CH_2)$.
To obtain Propan-$1$-ol,propene undergoes hydroboration-oxidation: $(i) (BH_3)_2, (ii) H_2O_2, OH^-$.
To obtain Propan-$2$-ol,propene undergoes acid-catalyzed hydration: $H_2SO_4, H_2O$.
$2.$ Conversion of Ethanoic acid to Methanol:
First,ethanoic acid $(CH_3COOH)$ is esterified with methanol $(CH_3OH)$ in the presence of $H^+$ to form methyl ethanoate $(CH_3COOCH_3)$.
Methyl ethanoate is then reduced using $LiAlH_4$ or $H_2/Pd$ to yield ethanol $(CH_3CH_2OH)$ and methanol $(CH_3OH)$.

(A)

$IUPAC$ Name	Structure	Common Name
$(i)$ Benzene-$1,2$-dicarboxylic acid	$C_6H_4(COOH)_2$ (ortho)	Phthalic acid
$(ii)$ Hexanedioic acid	$HOOC-(CH_2)_4-COOH$	Adipic acid
$(iii)$ Propane-$1,2,3$-tricarboxylic acid	$HOOC-CH_2-CH(COOH)-CH_2-COOH$	Tricarballylic acid
$(iv)$ $2$-aminobenzoic acid	$C_6H_4(NH_2)(COOH)$ (ortho)	Anthranilic acid
$(v)$ $2$-hydroxybenzoic acid	$C_6H_4(OH)(COOH)$ (ortho)	Salicylic acid
$(vi)$ Benzene-$1,4$-dicarboxylic acid	$C_6H_4(COOH)_2$ (para)	Terephthalic acid
$(vii)$ $2,3$-dihydroxybutanedioic acid	$HOOC-CH(OH)-CH(OH)-COOH$	Tartaric acid
$(viii)$ $3$-phenylprop-$2$-enoic acid	$C_6H_5-CH=CH-COOH$	Cinnamic acid
$(ix)$ $2$-methylbenzoic acid	$C_6H_4(CH_3)(COOH)$ (ortho)	$o$-Toluic acid

(A) The identification tests are as follows:
$(1)$ Phenol and Carboxylic acid:

$Test$	$Phenol$	$Carboxylic \text{ } acid$
$NaHCO_3$ test	No reaction	Effervescence of $CO_2$ gas
Neutral $FeCl_3$ test	Violet/Green colour	Buff-coloured precipitate

$(2)$ Formic acid and Acetic acid:
- Formic acid $(HCOOH)$ gives Tollen's test (silver mirror) and reduces $HgCl_2$ to white precipitate of $Hg_2Cl_2$,whereas acetic acid does not.
$(3)$ Acetic acid and Ethanol:
- Acetic acid gives effervescence with $NaHCO_3$,while ethanol does not.
- Ethanol gives the iodoform test (yellow precipitate of $CHI_3$ with $I_2/NaOH$),while acetic acid does not.
$(4)$ Acetic acid and Acetone:
- Acetic acid gives effervescence with $NaHCO_3$,while acetone does not.
- Acetone gives the iodoform test,while acetic acid does not.
$(5)$ Aldehyde and Ketone:
- Aldehydes give Tollen's test and Fehling's test,while ketones do not (except $\alpha$-hydroxy ketones).
$(6)$ Acetaldehyde and Acetone:
- Acetaldehyde gives Tollen's and Fehling's tests,while acetone does not.
- Both give the iodoform test.
$(7)$ Acetaldehyde and Formaldehyde:
- Acetaldehyde gives the iodoform test,while formaldehyde does not.
$(8)$ Benzaldehyde and Acetaldehyde:
- Acetaldehyde gives the iodoform test,while benzaldehyde does not.

(N/A) $(i)$ $CH_3COOH \xrightarrow{Cl_2/Red P} ClCH_2COOH \xrightarrow{Cl_2/Red P} Cl_2CHCOOH \xrightarrow{Cl_2/Red P} Cl_3CCOOH$
$(ii)$ $CH_3COOH + NaOH \rightarrow CH_3COONa + H_2O$; then $CH_3COONa + NaOH(CaO) \xrightarrow{\Delta} CH_4 + Na_2CO_3$
$(iii)$ $2CH_3COONa + 2H_2O \xrightarrow{Electrolysis} CH_3-CH_3 + 2CO_2 + H_2 + 2NaOH$
$(iv)$ $CH_3COOH + NH_3 \rightarrow CH_3COONH_4 \xrightarrow{\Delta} CH_3CONH_2 + H_2O$
$(v)$ $CH_3COOH + PCl_5 \rightarrow CH_3COCl + POCl_3 + HCl$
$(vi)$ $3CH_3COOH + PCl_3 \rightarrow 3CH_3COCl + H_3PO_3$
$(vii)$ $CH_3COOH + SOCl_2 \rightarrow CH_3COCl + SO_2 + HCl$
$(viii)$ $2CH_3COOH \xrightarrow{P_2O_5, \Delta} (CH_3CO)_2O + H_2O$
$(ix)$ $CH_3COOH + NaOH \rightarrow CH_3COONa + H_2O$
$(x)$ $CH_3COOH + NaHCO_3 \rightarrow CH_3COONa + H_2O + CO_2$
$(xi)$ $CH_3COOH + Na \rightarrow CH_3COONa + \frac{1}{2}H_2$
$(xii)$ $CH_3COOH \xrightarrow{LiAlH_4/Ether} CH_3CH_2OH$
$(xiii)$ $CH_3COOH + C_2H_5OH \xrightarrow{H^+} CH_3COOC_2H_5 + H_2O$

(N/A) $(i)$ $(CH_3CO)_2O$ (Ethanoic anhydride)
$(ii)$ $CH_3COOH$ (Ethanoic acid)
$(iii)$ $H_3PO_3$ (Phosphorous acid)
$(iv)$ $POCl_3$ (Phosphorous oxychloride)
$(v)$ $\text{Sodalime } (NaOH + CaO, 3:1) + \Delta$
$(vi)$ $CH_3CHBrCOOH + CH_3CBr_2COOH$ (Hell-Volhard-Zelinsky reaction)
$(vii)$ $LiAlH_4$ or $B_2H_6$ followed by $H_3O^{+}$
$(viii)$ No reaction (Carboxylic acid group is deactivating and forms a complex with $AlCl_3$).

(A) $(i)$ Toluene on oxidation with $KMnO_4/KOH$ gives benzoic acid $(X)$,which on bromination gives $m$-bromobenzoic acid $(Y)$.
$(ii)$ $CH_3CHO$ with Tollen's reagent gives ethanoic acid $(X)$,which on reduction with $LiAlH_4$ gives ethanol $(Y)$.
$(iii)$ $o$-xylene $(X)$ on oxidation gives phthalic acid $(Y)$.
$(iv)$ Benzoic acid with $NH_3$ gives ammonium benzoate $(X)$,which on heating gives benzamide $(Y)$.
$(v)$ Phthalic acid with $NH_3$ gives ammonium phthalate $(X)$,which on heating gives phthalamide $(Y)$.

(N/A) $(i)$ Conversion of $o-xylene$ to Phthalimide:
$o-xylene$ $\xrightarrow{KMnO_4, H^+}$ Phthalic acid $\xrightarrow{+2NH_3}$ Ammonium phthalate $\xrightarrow{\Delta, -H_2O}$ Phthalamide $\xrightarrow{\Delta, -NH_3}$ Phthalimide.
$(ii)$ Conversion of Ethanoic anhydride to Ethanol:
$(CH_3CO)_2O$ $\xrightarrow{H_2O, \Delta}$ $2CH_3COOH$ (Ethanoic acid) $\xrightarrow{LiAlH_4, \text{ether}}$ $2CH_3CH_2OH$ (Ethanol).

(A) $1$. Oxidation of $B$ (alcohol) with alkaline $KMnO_4$ gives $A$ (carboxylic acid).
$2$. Reduction of $A$ (carboxylic acid) with $LiAlH_4$ gives $B$ (alcohol).
$3$. Reaction of $A$ (carboxylic acid) and $B$ (alcohol) in the presence of $H_2SO_4$ (esterification) produces $C$ (ester),which has a characteristic fruity smell.
Therefore,$A$ is a carboxylic acid,$B$ is an alcohol,and $C$ is an ester.

(A) The correct matching is as follows:
$(A)$ Ketone corresponds to formula $(iii)$ $R-CO-R'$ and example $(s)$ $CH_3COC_2H_5$.
$(B)$ Carboxylic acid corresponds to formula $(i)$ $R-CO-OH$ and example $(q)$ $CH_3COOH$.
$(C)$ Amide corresponds to formula $(iv)$ $R-CO-NH_2$ and example $(r)$ $CH_3CONH_2$.
$(D)$ Aldehyde corresponds to formula $(ii)$ $R-CO-H$ and example $(p)$ $CH_3CHO$.
Therefore,the correct sequence is: $A$ $\rightarrow iii$ $\rightarrow s, B$ $\rightarrow i$ $\rightarrow q, C$ $\rightarrow iv$ $\rightarrow r, D$ $\rightarrow ii$ $\rightarrow p$.

(A-III, B-I, C-IV, D-II) The correct matches are:
$A$. $RCH=NH$ is an $\text{Imine}$ $(iii)$.
$B$. $AlH(i-Bu)_2$ is $\text{Diisobutyl aluminium hydride}$ $(i)$.
$C$. $C_6H_5-CH(OCOCH_3)_2$ is $\text{Benzylidene diacetate}$ $(iv)$.
$D$. $C_6H_5CHCl_2$ is $\text{Benzalchloride}$ $(ii)$.
Therefore,the correct sequence is $A-iii, B-i, C-iv, D-ii$.

(D) The boiling point of organic compounds depends on the strength of intermolecular forces.
$1.$ Carboxylic acids $(A)$ form stable dimers through strong hydrogen bonding,resulting in the highest boiling point $(391 \ K)$. Group: Carboxylic acid $(p)$.
$2.$ Alcohols $(B)$ also form hydrogen bonds but are generally weaker than carboxylic acid dimers. $BP$: $370 \ K$. Group: Alcohol $(q)$.
$3.$ Ketones $(C)$ and Aldehydes $(D)$ have dipole-dipole interactions. Ketones typically have slightly higher boiling points than isomeric aldehydes. $BP$: Ketone $(329 \ K)$,Aldehyde $(322 \ K)$. Groups: Ketone $(n)$,Aldehyde $(o)$.
$4.$ Alkanes $(E)$ have only weak Van der Waals forces,resulting in the lowest boiling point $(309 \ K)$. Group: Alkane $(m)$.
Therefore,the correct match is $(A)-(iv)-(p), (B)-(v)-(q), (C)-(iii)-(n), (D)-(ii)-(o), (E)-(i)-(m)$.

(A) The matching is as follows:
$(A)$ $CH_3-CH=CHCHO$ is But-$2$-enal $(iii)$.
$(B)$ $CH_3-C(CH_3)=CH-COCH_3$ is $4$-methylpent-$3$-en-$2$-one $(iv)$.
$(C)$ $C_6H_5-CH=CH-COC_6H_5$ is $1,3$-diphenylprop-$2$-en-$1$-one $(i)$.
$(D)$ $C_6H_5-C(CH_3)=N-NH-C_6H_3(NO_2)_2$ is Acetophenone-$2,4$-dinitrophenylhydrazone $(ii)$.
Thus,the correct matching is: $A-iii, B-iv, C-i, D-ii$.

(A) The correct matches are:
$(A) HCOOH$ is Formic acid $(iii)$ and Methanoic acid $(x)$.
$(B) CH_3CH_2CH_2COOH$ is Butyric acid $(iv)$ and Butanoic acid $(n)$.
$(C) (COOH)_2$ is Oxalic acid $(ii)$ and Ethanedioic acid $(y)$.
$(D) (CH_2)_2(COOH)_2$ is Succinic acid $(i)$ and Butanedioic acid $(m)$.
Therefore,the correct sequence is: $(A$ $\rightarrow iii$ $\rightarrow x), (B$ $\rightarrow iv$ $\rightarrow n), (C$ $\rightarrow ii$ $\rightarrow y), (D$ $\rightarrow i$ $\rightarrow m)$.

(D) The correct matching is:
$(A)$ Adipic acid $\rightarrow$ $(ii)$ $HOOC(CH_2)_4COOH$ $\rightarrow$ $(y)$ Hexanedioic acid
$(B)$ Carballylic acid $\rightarrow$ $(iii)$ $HOOC-CH_2-CH(COOH)-CH_2-COOH$ $\rightarrow$ $(n)$ Propane-$1,2,3$-tricarboxylic acid
$(C)$ Phthalic acid $\rightarrow$ $(i)$ $1,2-C_6H_4(COOH)_2$ $\rightarrow$ $(x)$ Benzene-$1,2$-dicarboxylic acid
$(D)$ Glutaric acid $\rightarrow$ $(iv)$ $HOOC(CH_2)_3COOH$ $\rightarrow$ $(m)$ Pentanedioic acid
Therefore,the correct match is $(A)-ii-y, (B)-iii-n, (C)-i-x, (D)-iv-m$.

(A-II, B-III, C-IV, D-I) The correct matches are:
$A \rightarrow ii$: Primary alcohols $(RCH_2OH)$ are oxidized to carboxylic acids $(RCOOH)$ using strong oxidizing agents like $KMnO_4$.
$B \rightarrow iii$: Primary alcohols are oxidized to carboxylic acids using Jones reagent $(CrO_3/H_2SO_4)$.
$C \rightarrow iv$: Amides $(RCONH_2)$ undergo acid-catalyzed hydrolysis to yield carboxylic acids $(RCOOH)$ and ammonia $(NH_3)$.
$D \rightarrow i$: Esters $(C_6H_5COOEt)$ undergo acid-catalyzed hydrolysis to yield the corresponding carboxylic acid $(C_6H_5COOH)$ and alcohol $(C_2H_5OH)$.
Therefore,the correct sequence is $A-ii, B-iii, C-iv, D-i$.

(A-IV, B-I, C-II, D-III) $(A)-(iv)$: Dehydration of acetic acid using $P_2O_5$ or $H^+, \Delta$ gives acetic anhydride.
$(B)-(i)$: Phthalic acid reacts with $NH_3$ followed by heating to form phthalamide.
$(C)-(ii)$: Hell-Volhard-Zelinsky reaction uses $Br_2 / \text{red phosphorous}$ followed by $H_2O$ to alpha-brominate carboxylic acids.
$(D)-(iii)$: Electrophilic aromatic substitution of benzoic acid (meta-directing group) with $Br_2 / FeBr_3$ gives $m$-bromobenzoic acid.
Therefore,the correct matching is $(A)-(iv), (B)-(i), (C)-(ii), (D)-(iii)$.

(A-III, B-I, C-IV, D-II) The correct matches are as follows:
$A$. $RCH_2COOH \rightarrow[H_2O]{X_2/P} RCH(X)COOH$ is the Hell-Volhard-Zelinsky reaction $(iii)$.
$B$. $RCOONa \rightarrow[NaOH, CaO, \Delta]{\text{soda lime}} RH$ is Decarboxylation $(i)$.
$C$. $RCOOH + R'OH \leftrightarrow{H^{+}} RCOOR' + H_2O$ is Esterification $(iv)$.
$D$. $RMgX + R'CHO \rightarrow{H_3O^{+}} RCH(OH)R'$ is a Grignard reaction $(ii)$.
Therefore,the correct sequence is $A-iii, B-i, C-iv, D-ii$.

(A) The $pK_a$ values for the given acids are as follows:
$(A)$ $CF_3COOH$: $0.23$ (Strongest acid due to electron-withdrawing $-F$ groups)
$(B)$ Benzoic acid: $4.20$
$(C)$ $CH_3COOH$: $4.76$ (Weakest acid)
$(D)$ $4$-nitrobenzoic acid: $3.41$ (Stronger than benzoic acid due to electron-withdrawing $-NO_2$ group)
Therefore,the correct matching is: $(A$ $\rightarrow iii), (B$ $\rightarrow iv), (C$ $\rightarrow ii), (D$ $\rightarrow i)$.