Properties of Carboxylic Acids and Their Derivatives Questions in English

(C) Formic acid contains an aldehyde group $(H-CHO)$ in its structure,which allows it to act as a reducing agent.
Specifically,it reduces Fehling's solution to red precipitate of cuprous oxide $(Cu_2O)$.
Acetic acid $(CH_3COOH)$ lacks this aldehyde group and does not reduce Fehling's solution.

(A) The acidic strength of carboxylic acids is inversely proportional to the $+I$ (inductive) effect of the alkyl group attached to the $-COOH$ group.
As the number of carbon atoms in the alkyl chain increases,the $+I$ effect increases,which destabilizes the carboxylate ion and decreases the acidic strength.
The order is:
$I$ (Methanoic acid,$H-COOH$) > $II$ (Ethanoic acid,$CH_3-COOH$) > $III$ (Propanoic acid,$CH_3CH_2-COOH$) > $IV$ (Butanoic acid,$CH_3CH_2CH_2-COOH$)
Therefore,the correct order is $I > II > III > IV$.

(B) The group $Y$ must be a meta-directing group for the $m-$ isomer to be the major product in an electrophilic aromatic substitution reaction.
$-COOH$ is an electron-withdrawing group and acts as a meta-directing group.
$-NH_2$,$-CH_3$,and $-Cl$ are ortho/para-directing groups.

(A) The acidity of carboxylic acids is increased by the presence of electron-withdrawing groups $(EWG)$ due to the $-I$ (inductive) effect.
The strength of the $-I$ effect for halogens follows the order of their electronegativity:
$F > Cl > Br$.
Therefore,the electron-withdrawing power decreases in the order $F > Cl > Br$,which stabilizes the conjugate base (carboxylate ion) more effectively in the order $FCH_2COO^- > ClCH_2COO^- > BrCH_2COO^-$.
Thus,the correct order of relative acidity is $FCH_2COOH > ClCH_2COOH > BrCH_2COOH$.

(B) The $-COOH$ group attached to the benzene ring is a strong electron-withdrawing group due to its $-I$ and $-M$ effects.
It is a deactivating group and is meta-directing for electrophilic aromatic substitution reactions.
Therefore,when benzoic acid undergoes sulfonation with fuming sulfuric acid ($H_2SO_4$ or $SO_3/H_2SO_4$),the incoming electrophile $(SO_3)$ attacks the meta-position,yielding $m$-sulfobenzoic acid as the major product.

(D) Lactic acid $(CH_3-CH(OH)-COOH)$ undergoes dehydration and cleavage in the presence of dilute $H_2SO_4$ upon heating.
The reaction is: $CH_3-CH(OH)-COOH \xrightarrow[\Delta]{H_2SO_4} CH_3CHO + HCOOH$.
Thus,the products formed are acetaldehyde and formic acid.

(A) The reaction involves the ring-opening of $\beta$-propiolactone by methanol $(\text{CH}_3\text{OH})$.
In the presence of an acid catalyst,the carbonyl oxygen is protonated,making the carbonyl carbon more electrophilic.
Alternatively,the nucleophilic attack of methanol on the carbonyl carbon leads to the cleavage of the acyl-oxygen bond.
This results in the formation of methyl $3$-hydroxypropionate,which is $HO-CH_2-CH_2-COOCH_3$.

(D) The Hell-Volhard-Zelinsky $(HVZ)$ reaction requires the presence of at least one $\alpha$-hydrogen atom in the carboxylic acid.
$HCOOH$ (formic acid) has no $\alpha$-carbon.
$CCl_3COOH$ (trichloroacetic acid) has an $\alpha$-carbon but no $\alpha$-hydrogen.
$C_6H_5COOH$ (benzoic acid) has an $\alpha$-carbon attached to the benzene ring,but it lacks an $\alpha$-hydrogen.
Therefore,none of these compounds can undergo the $HVZ$ reaction.

(C) The reaction of acetic acid with chlorine in the presence of a catalyst like anhydrous $FeCl_3$ (or red phosphorus) leads to the substitution of hydrogen atoms on the $\alpha$-carbon.
$CH_3COOH + 3Cl_2 \xrightarrow{FeCl_3} CCl_3COOH + 3HCl$
This reaction is known as the Hell-Volhard-Zelinsky reaction (or $\alpha$-halogenation).
The final product is $CCl_3COOH$,which is Trichloroacetic acid.

(C) Methanoic acid $(HCOOH)$ is a strong reducing agent because it contains both a carboxylic group and an aldehydic group $(CHO)$ in its structure,which allows it to reduce Tollens' reagent and Fehling's solution.

(C) When two moles of acetic acid $(CH_3COOH)$ are heated with phosphorus pentoxide $(P_2O_5)$,which acts as a dehydrating agent,a molecule of water is removed to form acetic anhydride.
The reaction is: $2CH_3COOH \xrightarrow{P_2O_5, \Delta} (CH_3CO)_2O + H_2O$.
Thus,the product obtained is acetic anhydride.

(B) Low molecular weight carboxylic acids are soluble in water because they can form $H$-bonds with water molecules.

(D) Carboxylic acids exist as dimers in the vapor phase or in aprotic solvents due to the formation of stable intermolecular hydrogen bonds between two carboxylic acid molecules. This structure is represented as follows:
$R-C(=O)OH \dots O=C(OH)-R$
This cyclic structure is stabilized by two hydrogen bonds.

(B) The oxidation of lactic acid $(CH_3-CH(OH)-COOH)$ with alkaline $KMnO_4$ results in the oxidation of the secondary alcohol group to a ketone group.
The reaction is:
$CH_3-CH(OH)-COOH \xrightarrow{KMnO_4} CH_3-CO-COOH$
Here,lactic acid is converted into pyruvic acid $(CH_3-CO-COOH)$.

(D) The conjugate base of a carboxylic acid,i.e.,the carboxylate ion,is resonance stabilized. The negative charge is delocalized over two electronegative oxygen atoms,which makes the carboxylate ion much more stable than the alkoxide or phenoxide ions. This increased stability of the conjugate base shifts the equilibrium towards the formation of $H^+$ ions,thereby increasing the acidity of carboxylic acids. The resonance structure is: $CH_3-COOH \xrightarrow{-H^+} CH_3-COO^- \leftrightarrow CH_3-COO^-$

(C) The acidity of carboxylic acids is increased by the presence of electron-withdrawing groups ($-I$ effect) and decreased by electron-donating groups ($+I$ effect).
In $CH_2ClCOOH$,the chlorine atom exerts a strong $-I$ effect,which stabilizes the carboxylate anion significantly.
Comparing the options:
$CH_3CH_2COOH$ and $CH_3COOH$ have alkyl groups which show $+I$ effect,making them weaker acids.
$CH_2ClCH_2COOH$ has the chlorine atom further away from the carboxyl group,reducing its $-I$ effect compared to $CH_2ClCOOH$.
Therefore,$CH_2ClCOOH$ is the strongest acid among the given options.

(B) The presence of an electron-withdrawing group $(-NO_2)$ on the benzene ring increases the acidity of benzoic acid.
$1$. The ortho-substituted isomer $(II)$ is the most acidic due to the ortho-effect.
$2$. The para-isomer $(III)$ has a strong $-I$ and $-M$ effect,making it more acidic than the meta-isomer.
$3$. The meta-isomer $(IV)$ only exerts a $-I$ effect,making it more acidic than benzoic acid $(I)$ but less acidic than the para-isomer.
$4$. Benzoic acid $(I)$ has no electron-withdrawing substituent.
Therefore,the correct order of acidity is $II > III > IV > I$.

(B) The $pK_a$ value is inversely proportional to the acid strength.
$HCOOH$ (formic acid) is the strongest acid among the given options because the hydrogen atom attached to the carboxyl group does not exert any electron-donating inductive effect ($+I$ effect),unlike the alkyl groups in other options.
Therefore,$HCOOH$ has the lowest $pK_a$ value.

(A) The $-COOH$ group is a strongly electron-withdrawing group and is meta-directing in electrophilic aromatic substitution reactions.
Therefore,during the nitration of benzoic acid using a mixture of concentrated $HNO_3$ and $H_2SO_4$,the incoming electrophile $(NO_2^+)$ attacks the meta-position.
This results in the formation of $3-$nitrobenzoic acid as the major product.

(A) The decarboxylation of $\beta$-keto acids occurs easily upon heating through a six-membered cyclic transition state. Among the given options,$C_6H_5-CO-CH_2-COOH$ is a $\beta$-keto acid,which facilitates the loss of $CO_2$ via this mechanism.

(D) The reaction is the Hell-Volhard-Zelinsky $(HVZ)$ reaction followed by nucleophilic substitution.
$1$. $CH_3CH_2COOH \xrightarrow{Br_2/P} CH_3CH(Br)COOH$ ($X$ is $2$-bromopropanoic acid).
$2$. $CH_3CH(Br)COOH \xrightarrow{NH_3} CH_3CH(NH_2)COOH$ ($Y$ is $2$-aminopropanoic acid,commonly known as Alanine).

(C) $\beta$-hydroxy carboxylic acids readily lose a water molecule upon heating to form $\alpha, \beta$-unsaturated acids.
$CH_3-CH(OH)-CH_2COOH \xrightarrow{\Delta} CH_3-CH=CH-COOH + H_2O$
This reaction involves the elimination of the hydroxyl group from the $\beta$-carbon and a hydrogen atom from the $\alpha$-carbon,resulting in the formation of a double bond between the $\alpha$ and $\beta$ carbons.

(D) The reaction between propionic acid $(CH_3CH_2COOH)$ and sodium carbonate $(Na_2CO_3)$ or sodium bicarbonate $(NaHCO_3)$ involves the evolution of $CO_2$ gas.
In this acid-base reaction,the $CO_2$ released originates from the carbonate or bicarbonate salt,not from the carboxylic acid group of the organic molecule.
Therefore,the carbon atom in the evolved $CO_2$ comes from the carbonate $(CO_3^{2-})$ or bicarbonate $(HCO_3^-)$ ion.

(C) The acidity of carboxylic acids is influenced by the inductive effect of substituents.
$1$. $\alpha$-halo acids are stronger than $\beta$-halo acids because the electron-withdrawing group is closer to the carboxyl group.
$2$. Fluorine $(F)$ has a stronger electron-withdrawing inductive effect than Bromine $(Br)$.
$3$. Comparing the given options:
- $CH_3CHFCOOH$ ($\alpha$-fluoro acid) is the strongest.
- $CH_3CHBrCOOH$ ($\alpha$-bromo acid) is stronger than $\beta$-substituted acids.
- Between $FCH_2CH_2COOH$ and $BrCH_2CH_2COOH$,both are $\beta$-substituted. Since $F$ is more electronegative than $Br$,$FCH_2CH_2COOH$ is stronger than $BrCH_2CH_2COOH$.
Therefore,$BrCH_2CH_2COOH$ is the weakest acid and will have the lowest dissociation constant $(K_a)$.

(A) Acetic acid $(CH_3COOH)$ reacts with chlorinating agents like $PCl_3$,$PCl_5$,and $SOCl_2$ to form acetyl chloride $(CH_3COCl)$.
$3CH_3COOH + PCl_3 \rightarrow 3CH_3COCl + H_3PO_3$
$CH_3COOH + PCl_5 \rightarrow CH_3COCl + POCl_3 + HCl$
$CH_3COOH + SOCl_2 \rightarrow CH_3COCl + SO_2 + HCl$
Chloroform $(CHCl_3)$ is a solvent and does not act as a chlorinating agent for carboxylic acids to produce acid chlorides.

(D) The reaction between ethyl alcohol $(CH_3CH_2OH)$ and acetyl chloride $(CH_3COCl)$ is an esterification reaction.
The reaction is: $CH_3COCl + CH_3CH_2OH \to CH_3COOCH_2CH_3 + HCl$.
The product formed is ethyl acetate.

(D) Acetamide $(CH_3CONH_2)$ is formed when compounds containing an acetyl group $(CH_3CO-)$ react with ammonia $(NH_3)$.
$1$. Acetic acid $(CH_3COOH)$ reacts with $NH_3$ to form ammonium acetate,which on heating gives acetamide.
$2$. Acetyl chloride $(CH_3COCl)$ reacts with $NH_3$ to form acetamide.
$3$. Methyl acetate $(CH_3COOCH_3)$ reacts with $NH_3$ to form acetamide.
$4$. Methyl propionate $(CH_3CH_2COOCH_3)$ contains a propionyl group $(CH_3CH_2CO-)$ instead of an acetyl group. Therefore,it reacts with $NH_3$ to produce propionamide $(CH_3CH_2CONH_2)$ and methanol $(CH_3OH)$:
$CH_3CH_2COOCH_3 + NH_3 \to CH_3CH_2CONH_2 + CH_3OH$.

(B) The reaction of a carboxylic acid with an alcohol in the presence of an acid catalyst (like dry $HCl$) is known as Fischer esterification.
$RCOOH + C_2H_5OH \xrightarrow{Dry \ HCl} RCOOC_2H_5 + H_2O$
Here,the carboxylic acid reacts with ethanol in the presence of dry $HCl$ to form an ester.

(C) $1$. Toluene on oxidation gives benzoic acid $(A)$.
$2$. Benzoic acid reacts with $SOCl_2$ to form benzoyl chloride $(B)$.
$3$. Benzoyl chloride reacts with $NaN_3$ to form benzoyl azide $(C)$.
$4$. Upon heating,benzoyl azide undergoes Curtius rearrangement to form phenyl isocyanate $(D)$.

(C) The hydrolysis of acetamide $(CH_3CONH_2)$ in the presence of a boiling mineral acid (like $HCl$) results in the formation of acetic acid and ammonium chloride.
The reaction is as follows:
$CH_3CONH_2 + H_2O + HCl \xrightarrow{\Delta} CH_3COOH + NH_4Cl$
Therefore,the product obtained is acetic acid.

(A) $LiAlH_4$ is a strong reducing agent that reduces esters to primary alcohols. The reaction is as follows:
$R-COOR' + 4[H] \xrightarrow{LiAlH_4} R-CH_2OH + R'OH$
In this reaction,the ester is cleaved to produce two alcohol molecules: one derived from the acyl part $(R-CH_2OH)$ and one from the alkoxy part $(R'OH)$.

(C) The reaction of acetyl chloride with sodium acetate yields acetic anhydride.
$CH_3COCl + CH_3COONa \rightarrow (CH_3CO)_2O + NaCl$
Thus,$CH_3COONa$ is the correct reagent.

(B) Saponification of an ester like ethyl benzoate $(C_6H_5COOC_2H_5)$ with a strong base like caustic soda $(NaOH)$ results in the formation of a sodium salt of the carboxylic acid and an alcohol.
The reaction is: $C_6H_5COOC_2H_5 + NaOH \rightarrow C_6H_5COONa + C_2H_5OH$.
Here,$C_6H_5COONa$ is sodium benzoate and $C_2H_5OH$ is ethanol.

(C) The reaction of an alcohol with a carboxylic acid in the presence of an acid catalyst produces an ester. Esters are known for their characteristic sweet,fruity smell. The reaction is: $C_2H_5OH + CH_3COOH \rightarrow CH_3COOC_2H_5 + H_2O$.

(B) The reaction of ethyl chloroformate $(ClCOOC_2H_5)$ with methyl magnesium iodide $(CH_3MgI)$ proceeds via a nucleophilic acyl substitution mechanism.
$Cl-CO-OC_2H_5 + CH_3MgI \rightarrow CH_3-CO-OC_2H_5 + Mg(I)Cl$
In this reaction,the methyl group from the Grignard reagent replaces the chlorine atom of the ethyl chloroformate,resulting in the formation of ethyl acetate $(CH_3COOC_2H_5)$.

(B) The reaction between the silver salt of a fatty acid $(RCOOAg)$ and an alkyl halide $(R'X)$ is known as the nucleophilic substitution reaction,which yields an ester as the product.
The chemical equation is: $RCOOAg + R'X \to RCOOR' + AgX$.

(A) The reaction involves the alkaline hydrolysis (saponification) of an ester.
When an ester is boiled with alcoholic $KOH$,it forms the potassium salt of the corresponding carboxylic acid and an alcohol.
Upon acidification with $HCl$,the potassium salt is converted into the free carboxylic acid.
If the resulting carboxylic acid is a solid at room temperature and insoluble in water,it will precipitate as a white solid.
Among the given options,$Ethyl \ benzoate$ $(C_6H_5COOC_2H_5)$ undergoes hydrolysis to form potassium benzoate $(C_6H_5COOK)$ and ethanol.
Potassium benzoate,upon acidification with $HCl$,yields benzoic acid $(C_6H_5COOH)$,which is a white crystalline solid that is sparingly soluble in water and thus precipitates out.
The reaction is:
$C_6H_5COOC_2H_5 + KOH \rightarrow C_6H_5COOK + C_2H_5OH$
$C_6H_5COOK + HCl \rightarrow C_6H_5COOH(s) + KCl$

(C) The compound $(CH_3)_2NCOCH_3$ is an $N,N$-dimethylacetamide.
Acidic hydrolysis (refluxing with acid) of an amide involves the cleavage of the $C-N$ bond.
$(CH_3)_2NCOCH_3 + H_2O \xrightarrow{H^+} CH_3COOH + (CH_3)_2NH$.
Thus,the products are acetic acid and dimethylamine.

(B) The hydrolysis of $CH_3COCl$ produces $HCl$ and $CH_3COOH$.
$CH_3COCl + H_2O \rightarrow CH_3COOH + HCl$.
Since $HCl$ is a strong acid,it dissociates completely in water to produce a high concentration of $H^+$ ions,resulting in the lowest $pH$ among the given options.

(B) Ethyl acetate $(CH_3COOC_2H_5)$ reacts with $CH_3MgI$ in two steps.
Step $1$: Nucleophilic attack of $CH_3^-$ on the carbonyl carbon of ethyl acetate forms an unstable intermediate,which eliminates ethoxide to form acetone $(CH_3COCH_3)$.
Step $2$: Since $CH_3MgI$ is in excess,it reacts with the formed acetone to produce a magnesium alkoxide intermediate,which upon hydrolysis yields tert-butyl alcohol ($2$-methylpropan-$2$-ol).
Reaction: $CH_3COOC_2H_5 + CH_3MgI \rightarrow CH_3COCH_3 + Mg(OC_2H_5)I$.
Then,$CH_3COCH_3 + CH_3MgI \rightarrow (CH_3)_3COMgI$.
Finally,$(CH_3)_3COMgI + H_2O/H^+ \rightarrow (CH_3)_3COH + Mg(OH)I$.

(B) The given reaction is the $Perkin$ reaction.
$Perkin$ reaction is a condensation reaction in which an aromatic aldehyde is heated with an aliphatic acid anhydride in the presence of the sodium salt of the corresponding acid to form an $\alpha, \beta-$unsaturated acid.
$C_6H_5CHO + (CH_3CO)_2O \xrightarrow{CH_3COONa} C_6H_5CH=CHCOOH + CH_3COOH$.
Here,$C_6H_5CHO$ is benzaldehyde and $(CH_3CO)_2O$ is acetic anhydride.
The product $(A)$ formed is $C_6H_5CH=CHCOOH$,which is cinnamic acid.

(A) The starting material $C_8H_6O_4$ is phthalic acid.
When phthalic acid is heated $(\Delta)$,it undergoes dehydration to form phthalic anhydride,which is compound $X$.
Phthalic anhydride then reacts with $NH_3$ to form phthalamic acid.
Therefore,compound $X$ is phthalic anhydride.

(A) The quick vinegar process involves the oxidation of ethanol using the bacteria $Acetobacter \ aceti$.
This process is carried out at a temperature of approximately $300 \ K$.
The chemical reaction is: $CH_3CH_2OH + O_2 \xrightarrow{Acetobacter, 300 \ K} CH_3COOH + H_2O$.

(D) Formic acid $(HCOOH)$ acts as both a carboxylic acid and an aldehyde due to the presence of the aldehydic group $(-CHO)$ in its structure.
$(A)$ Tollen's reagent: $HCOOH + Ag_2O \to CO_2 + H_2O + 2Ag$ (Silver mirror is formed).
$(B)$ Mercuric chloride: $HCOOH + 2HgCl_2 \to 2CO_2 + 4HCl + 2Hg$ (Black precipitate of $Hg$ is formed).
$(C)$ $KMnO_4$: $5HCOOH + 2KMnO_4 + 3H_2SO_4 \to 5CO_2 + K_2SO_4 + 2MnSO_4 + 8H_2O$ (Decolorization of $KMnO_4$ occurs).
Since formic acid reduces all of these,the correct option is $(D)$.

(D) The dimerisation of carboxylic acids in non-aqueous solvents like benzene is due to the formation of $2$ intermolecular hydrogen bonds between two carboxylic acid molecules,resulting in a stable cyclic structure.

(A) The acidity of carboxylic acids is determined by the stability of the conjugate base (carboxylate ion).
Electron-withdrawing groups $(EWG)$ stabilize the carboxylate ion through the inductive effect ($-I$ effect),thereby increasing acidity.
The strength of the $-I$ effect depends on the electronegativity of the substituent.
Fluorine $(F)$ is more electronegative than Chlorine $(Cl)$ and Bromine $(Br)$.
Therefore,the $-I$ effect follows the order: $CF_3 > CCl_3 > CBr_3 > CH_3$.
Thus,$CF_3COOH$ is the strongest acid among the given options.

(D) Aspirin (acetylsalicylic acid) is prepared by the acetylation of the phenolic $-OH$ group of salicylic acid using acetic anhydride in the presence of an acid catalyst.
The reaction is as follows:
$C_6H_4(OH)COOH + (CH_3CO)_2O \rightarrow C_6H_4(OCOCH_3)COOH + CH_3COOH$
Therefore,the correct option is $(d)$.

(C) The reduction of oxalic acid $(HOOC-COOH)$ with zinc and $H_2SO_4$ is a specific chemical reaction that yields glycollic acid $(CH_2(OH)-COOH)$.
The chemical equation for the reaction is:
$HOOC-COOH + 4[H] \xrightarrow{Zn / H_2SO_4} CH_2(OH)-COOH + H_2O$

(A) Carboxylic acids $(RCOOH)$ react with ammonia $(NH_3)$ to form ammonium salts $(RCOONH_4)$,which upon heating lose a water molecule to form amides $(RCONH_2)$.
$RCOOH + NH_3 \to RCOONH_4$
$RCOONH_4 \xrightarrow{\Delta} RCONH_2 + H_2O$
Therefore,carboxylic acids are the substances that yield amides upon heating with $NH_3$.