Properties of Carboxylic Acids and Their Derivatives Questions in English

(C) The Hell-Volhard-Zelinsky $(HVZ)$ reaction is used for the halogenation of carboxylic acids at the $\alpha$-position.
When carboxylic acids containing $\alpha$-hydrogen atoms are treated with $Br_2$ or $Cl_2$ in the presence of a small amount of red phosphorus,the $\alpha$-hydrogen is replaced by the halogen atom.
The reaction for the preparation of $\alpha$-bromoacetic acid is:
$CH_3COOH + Br_2 \xrightarrow{\text{red } P} CH_2BrCOOH + HBr$
Thus,the correct option is $(C)$.

(A) When succinic acid $(HOOC-CH_2-CH_2-COOH)$ is heated,it undergoes intramolecular dehydration to form succinic anhydride.
The reaction is as follows:
$HOOC-CH_2-CH_2-COOH \xrightarrow{\Delta, -H_2O} \text{Succinic anhydride}$

(B) The reaction between an acyl chloride $(RCOCl)$ and a sodium salt of a carboxylic acid $(R'COONa)$ results in the formation of an acid anhydride.
This is a nucleophilic substitution reaction where the carboxylate ion acts as a nucleophile.
The general reaction is: $RCOCl + R'COONa \xrightarrow{\Delta} RCOOCOR' + NaCl$.
For example: $CH_3COONa + CH_3COCl \xrightarrow{\Delta} (CH_3CO)_2O + NaCl$.

(C) The given reaction is an example of the Perkin reaction.
In the Perkin reaction,an aromatic aldehyde reacts with an acid anhydride (in the presence of the corresponding sodium salt of the acid) to form an $\alpha,\beta$-unsaturated carboxylic acid.
Here,$p$-methoxybenzaldehyde reacts with acetic anhydride $(CH_3CO)_2O$ in the presence of sodium acetate $(CH_3COONa)$ to yield $p$-methoxycinnamic acid.
Therefore,$(X)$ is acetic anhydride,which is $(CH_3CO)_2O$.

(C) The dissociation constant $(K_a)$ of a carboxylic acid is directly proportional to its acidity. The acidity is increased by the electron-withdrawing inductive effect ($-I$ effect) of substituents.
$1$. Comparing the substituents: $F$ has a stronger $-I$ effect than $Br$.
$2$. Comparing the position: The closer the substituent is to the $-COOH$ group,the stronger the $-I$ effect.
$3$. In $CH_3CHFCOOH$ and $CH_3CHBrCOOH$,the halogen is on the $\alpha$-carbon,exerting a strong $-I$ effect.
$4$. In $FCH_2CH_2COOH$ and $BrCH_2CH_2COOH$,the halogen is on the $\beta$-carbon,exerting a weaker $-I$ effect.
$5$. Between $FCH_2CH_2COOH$ and $BrCH_2CH_2COOH$,$Br$ has a weaker $-I$ effect than $F$.
Therefore,$BrCH_2CH_2COOH$ has the weakest $-I$ effect,making it the least acidic and thus having the smallest dissociation constant $(K_a)$.

(B) When vapours of ethanoic acid $(CH_3COOH)$ are passed over $MnO_2$ at $573 \ K$,it undergoes decarboxylation and ketonization to form propanone $(CH_3COCH_3)$.
The chemical equation is:
$2CH_3COOH \xrightarrow{MnO_2, 573 \ K} CH_3COCH_3 + CO_2 + H_2O$

(D) The acidity of carboxylic acids is increased by the presence of electron-withdrawing groups ($-I$ effect) attached to the alpha-carbon.
Chlorine atoms exert a strong $-I$ effect,which stabilizes the carboxylate anion by dispersing the negative charge.
As the number of chlorine atoms increases,the $-I$ effect increases,thereby increasing the acidity.
The order of acidity is: $Cl_3CCOOH > Cl_2CHCOOH > ClCH_2COOH > CH_3COOH$.
Therefore,$Cl_3CCOOH$ (trichloroacetic acid) is the strongest acid.

(B) Ethyl acetate $(CH_3COOCH_2CH_3)$ is a colourless liquid at room temperature with a characteristic pleasant fruity smell. Its boiling point is $77^{\circ}C$.

(B) The reaction of acetamide $(CH_3CONH_2)$ with water $(H_2O)$ is an example of hydrolysis,where the amide bond is cleaved to form acetic acid and ammonia:
$CH_3CONH_2 + H_2O \rightarrow CH_3COOH + NH_3$.

(A) Methanoic acid $(HCOOH)$ contains an aldehydic group $(-CHO)$ in its structure.
Due to the presence of this group,it acts as a reducing agent and reduces Fehling solution.

(A) $LiAlH_4$ is a strong reducing agent that reduces esters to alcohols. The reaction is: $R-COO-R' \xrightarrow{LiAlH_4} R-CH_2OH + R'-OH$. Thus,two units of alcohols are obtained.

(B) $CH_3COOH$ is a carboxylic acid containing an acidic hydrogen atom.
When it reacts with a Grignard reagent $(CH_3MgX)$,the acidic hydrogen is abstracted by the alkyl group of the Grignard reagent.
The reaction is: $CH_3COOH + CH_3MgX \to CH_4 + CH_3COOMgX$.
Here,$CH_4$ (methane) is a hydrocarbon.
Therefore,the correct option is $(B)$.

(B) . Carboxylic acids exhibit stronger hydrogen bonding than alcohols. This is because they form stable dimers through two hydrogen bonds involving the highly electronegative oxygen atoms of the carbonyl and hydroxyl groups.

(D) The acidity of carboxylic acids is influenced by the inductive effect of substituents attached to the alpha-carbon.
Electron-withdrawing groups $(EWG)$ increase the acidity by stabilizing the carboxylate anion through the $-I$ effect.
The strength of the $-I$ effect depends on the electronegativity of the halogen atom: $F > Cl > Br$.
Therefore,the order of acidity is $FCH_2COOH > ClCH_2COOH > BrCH_2COOH > CH_3COOH$.
Thus,$FCH_2COOH$ is the strongest acid.

(D) Formic acid $(HCOOH)$ contains an aldehydic hydrogen atom attached to the carbonyl group,which gives it reducing properties similar to aldehydes.
Therefore,it can reduce Tollen's reagent $([Ag(NH_3)_2]^+)$ to metallic silver $(Ag)$.
The reaction is: $HCOOH + 2[Ag(NH_3)_2]^+ + 3OH^- \rightarrow CO_3^{2-} + 2Ag + 4NH_3 + 2H_2O$.

(B) Oxalic acid $(HOOC-COOH)$ is a dicarboxylic acid that acts as a reducing agent. It can reduce Tollens' reagent (ammoniacal silver nitrate solution) to metallic silver $(Ag)$.
Tartaric acid $(HOOC-CH(OH)-CH(OH)-COOH)$ does not reduce Tollens' reagent.
Therefore,ammoniacal silver nitrate solution can be used to distinguish between them.
Both acids react with sodium bicarbonate to evolve $CO_2$ gas,and both turn blue litmus red,so these cannot be used for distinction.

(B) The reaction of formic acid $(HCOOH)$ with concentrated sulfuric acid $(H_2SO_4)$ is a dehydration reaction.
$HCOOH \xrightarrow{conc. \, H_2SO_4} CO + H_2O$
Concentrated $H_2SO_4$ acts as a dehydrating agent and removes a water molecule from $HCOOH$,resulting in the formation of carbon monoxide $(CO)$ and water $(H_2O)$.

(B) The $-COOH$ group is a strong electron-withdrawing group and is meta-directing in electrophilic aromatic substitution reactions.
Therefore,sulphonation of benzoic acid with concentrated $H_2SO_4$ yields $m-$sulphobenzoic acid as the major product.
The reaction is: $C_6H_5COOH + conc. H_2SO_4 \rightarrow m-HO_3S-C_6H_4-COOH + H_2O$.

(D) The acidity of carboxylic acids is increased by the presence of electron-withdrawing groups (EWGs) attached to the alpha-carbon.
$F$ is more electronegative than $Cl$,so fluorine-substituted acetic acids are stronger than chlorine-substituted ones.
Among the given options,$CHF_2COOH$ contains two highly electronegative $F$ atoms,which exert a strong inductive effect ($-I$ effect) to stabilize the carboxylate anion.
Therefore,$CHF_2COOH$ is the strongest acid.

(C) The silver mirror test (Tollens' test) is given by aldehydes and $\alpha$-hydroxy ketones.
$HCHO$ (formaldehyde),$CH_3CHO$ (acetaldehyde),and $HCOOH$ (formic acid) all contain an aldehydic group or a group that can be oxidized to an aldehyde,thus they give a positive silver mirror test.
$CH_3COOH$ (acetic acid) is a carboxylic acid that does not contain an aldehydic group and cannot be oxidized under these conditions,so it does not give the silver mirror test.

(C) When acetic acid is heated with manganese oxide $(MnO)$ at $300^\circ C$,it undergoes ketonization to form acetone $(CH_3COCH_3)$,carbon dioxide $(CO_2)$,and water $(H_2O)$.
$2CH_3COOH \xrightarrow[300^\circ C]{MnO} CH_3COCH_3 + CO_2 + H_2O$

(C) The acidity of carboxylic acids depends on the stability of the carboxylate ion formed after the loss of a proton.
In $CH_3COOH$,the methyl group $(-CH_3)$ exerts a positive inductive effect ($+I$ effect),which increases the electron density on the carboxylate group,thereby destabilizing the conjugate base and decreasing acidity.
In $HCOOH$,there is only a hydrogen atom attached to the carboxyl group,which does not exert any $+I$ effect.
Therefore,$HCOOH$ is more acidic than $CH_3COOH$.

(C) The reaction of acetic anhydride with ammonia proceeds as follows:
$(CH_3CO)_2O + 2NH_3 \to CH_3CONH_2 + CH_3COONH_4$
In this reaction,one molecule of acetic anhydride reacts with two molecules of ammonia to produce one molecule of acetamide $(CH_3CONH_2)$ and one molecule of ammonium acetate $(CH_3COONH_4)$.

(C) Formic acid $(HCOOH)$ acts as a reducing agent because it contains an aldehydic group $(-CHO)$ in its structure. It reduces Fehling solution to give a red precipitate of cuprous oxide $(Cu_2O)$.
$HCOOH + 2Cu^{2+} + 5OH^- \xrightarrow{\Delta} Cu_2O (\text{red ppt}) + CO_3^{2-} + 3H_2O$
Acetic acid $(CH_3COOH)$ does not contain an aldehydic group and therefore does not reduce Fehling solution.
Both acids react with sodium,form esters with alcohols,and turn blue litmus red.

(D) Formaldehyde $(HCHO)$ and formic acid $(HCOOH)$ can be distinguished using sodium bicarbonate $(NaHCO_3)$.
Formic acid reacts with sodium bicarbonate to evolve carbon dioxide gas,which causes effervescence:
$HCOOH + NaHCO_3 \to HCOONa + H_2O + CO_2 \uparrow$
Formaldehyde does not react with sodium bicarbonate.
Both formaldehyde and formic acid react with Tollen's reagent and Fehling solution,so they cannot be distinguished by these reagents.

(C) Carboxylic acids are stronger acids than carbonic acid $(H_2CO_3)$.
Therefore,they react with sodium bicarbonate $(NaHCO_3)$ to evolve carbon dioxide $(CO_2)$ gas.
Acetic acid $(CH_3COOH)$ is a carboxylic acid,so it reacts as follows:
$CH_3COOH + NaHCO_3 \to CH_3COONa + H_2O + CO_2 \uparrow$
Phenol and $n-$hexanol are much weaker acids than carbonic acid and do not react with $NaHCO_3$ to evolve $CO_2$.

(A) The $pK_a$ value is a measure of the acidity of a substance.
$A$ lower $pK_a$ value corresponds to a higher acid dissociation constant $(K_a)$,which means the acid is stronger.
Comparing the given values:
$HCOOH: 3.77$
$C_6H_5COOH: 4.22$
$CH_3COOH: 4.71$
$CH_3CH_2COOH: 4.88$
Since $3.77$ is the lowest value,$HCOOH$ (formic acid) is the strongest acid.

(B) In the presence of a small amount of iodine catalyst,chlorine reacts with acetic acid to undergo $\alpha$-halogenation,forming monochloroacetic acid $(Cl-CH_2-COOH)$.
This reaction is a variation of the Hell-Volhard-Zelinsky $(HVZ)$ reaction,where the iodine catalyst facilitates the formation of an acyl halide intermediate that undergoes enolization and subsequent halogenation at the $\alpha$-carbon.

(C) When acetic acid $(CH_3COOH)$ is heated with phosphorus pentoxide $(P_2O_5)$,it undergoes dehydration to form acetic anhydride $((CH_3CO)_2O)$.
The reaction is: $2CH_3COOH \xrightarrow{\Delta, P_2O_5} (CH_3CO)_2O + H_2O$.

(B) Formic acid $(HCOOH)$ contains an aldehydic group $(-CHO)$ attached to a hydrogen atom.
Due to the presence of this aldehydic group,it acts as a strong reducing agent.
It reduces Tollens' reagent (ammoniacal silver nitrate) to metallic silver $(Ag)$.
Formic acid is actually stronger than acetic acid ($K_a$ of $HCOOH \approx 1.8 \times 10^{-4}$ vs $K_a$ of $CH_3COOH \approx 1.8 \times 10^{-5}$),making option $C$ incorrect.
It is miscible with water due to hydrogen bonding,making option $A$ incorrect.

(B) Formic acid $(HCOOH)$ contains an aldehyde group $(-CHO)$ in its structure,which makes it a strong reducing agent.
It can reduce Tollen's reagent to silver mirror,Fehling's solution,mercuric chloride $(HgCl_2)$,and potassium permanganate $(KMnO_4)$.
Therefore,the statement that it is a reducing agent is true.

(D) The organic liquid is $HCOOH$ (formic acid).
$(i)$ It reacts with sodium carbonate $(Na_2CO_3)$ to liberate carbon dioxide $(CO_2)$: $2HCOOH + Na_2CO_3 \longrightarrow 2HCOONa + H_2O + CO_2$.
$(ii)$ Formic acid acts as a reducing agent and reacts with Tollen's reagent ($Ag_2O$ in $NH_4OH$) to produce a black precipitate of silver $(Ag)$: $HCOOH + Ag_2O \longrightarrow 2Ag + H_2O + CO_2$.

(C) Step $1$: $CH_3-CH_2-COOH$ reacts with $Cl_2$ in the presence of $Fe$ (or $P$) to undergo $\alpha$-halogenation,known as the Hell-Volhard-Zelinsky reaction,to form $X$ $(CH_3-CHCl-COOH)$.
Step $2$: $X$ $(CH_3-CHCl-COOH)$ reacts with alcoholic $KOH$ to undergo dehydrohalogenation (elimination of $HCl$),resulting in the formation of $Y$ $(CH_2=CH-COOH)$,which is acrylic acid.
Reaction sequence: $CH_3-CH_2-COOH$ $\xrightarrow{Cl_2/Fe} CH_3-CHCl-COOH (X)$ $\xrightarrow{\text{Alcoholic } KOH} CH_2=CH-COOH (Y)$.

(B) The Claisen self-condensation reaction requires the presence of at least one $\alpha$-hydrogen atom in the ester molecule.
In $C_6H_5COOC_2H_5$ (ethyl benzoate),the carbon atom adjacent to the carbonyl group is part of the benzene ring and does not have any $\alpha$-hydrogen atom attached to it.
Therefore,it cannot undergo Claisen self-condensation.
Thus,the correct option is $(B)$.

(B) Carboxylic acids are acidic because the carboxylate ion formed after the loss of a proton $(H^+)$ is resonance-stabilized.
In the carboxylate ion,the negative charge is delocalized over two electronegative oxygen atoms,which significantly lowers the energy of the ion and makes the ionization process favorable.

(C) The acidity of carboxylic acids is increased by the presence of electron-withdrawing groups $(EWG)$ due to the inductive effect ($-I$ effect).
Greater the number of $EWG$ and closer they are to the carboxyl group,the stronger the acid.
In $CH_3-CH_2-CCl_2-COOH$,there are two chlorine atoms on the $\alpha$-carbon,which exert a strong $-I$ effect,stabilizing the carboxylate anion significantly.
Therefore,$CH_3-CH_2-CCl_2-COOH$ is the strongest acid among the given options and will be most highly ionised in water.

(A) The alkaline hydrolysis of an ester is known as saponification.
When an ester is heated with an aqueous solution of a strong base like $NaOH$,it undergoes hydrolysis to form the sodium salt of the carboxylic acid and an alcohol.
The general reaction is:
$R-COOR' + NaOH \rightarrow R-COO^-Na^+ + R'OH$
(where $R-COO^-Na^+$ is the sodium carboxylate salt and $R'OH$ is the alcohol).

(C) The reaction sequence is as follows:
$1$. $CH_3COOH + NH_3 \rightarrow CH_3COONH_4$ (Ammonium acetate,$A$)
$2$. $CH_3COONH_4 \xrightarrow[P_2O_5]{\Delta} CH_3CONH_2 + H_2O$ (Acetamide)
$3$. $CH_3CONH_2 \xrightarrow[P_2O_5]{\Delta} CH_3-C \equiv N + H_2O$ (Acetonitrile,$B$)
Thus,the end product $B$ is Acetonitrile.

(B) $LiAlH_4$ is a strong reducing agent that reduces carboxylic acids to primary alcohols. The reaction is represented as: $R-COOH \xrightarrow{LiAlH_4} R-CH_2OH$. For example,$CH_3COOH \xrightarrow{LiAlH_4} CH_3CH_2OH$.

(C) Amides undergo alkaline hydrolysis when boiled with caustic soda $(NaOH)$ to produce the corresponding carboxylate salt and ammonia gas $(NH_3)$.
$CH_3CONH_2 + NaOH \xrightarrow{\text{Boil}} CH_3COONa + NH_3 \uparrow$
Therefore,$Acetamide$ is the correct substance.

(A) Formic acid $(HCOOH)$ contains an aldehyde group $(-CHO)$ attached directly to a hydrogen atom.
Due to the presence of this $-CHO$ group,it acts as a reducing agent and gives positive tests for aldehydes,such as the Tollen's reagent test and Fehling's solution test.
Other carboxylic acids do not contain this $-CHO$ group and therefore do not show these tests.

(C) The acid strength of a compound is determined by the stability of its conjugate base. The more stable the conjugate base,the stronger the acid.
$1$. $RCOOH$ (Carboxylic acid) is the strongest acid among the given options because its conjugate base,the carboxylate ion $(RCOO^-)$,is resonance-stabilized.
$2$. $HOH$ $(H_2O)$ is a stronger acid than $ROH$ (alcohol) because the alkyl group in $ROH$ is electron-donating ($+I$ effect),which destabilizes the alkoxide ion $(RO^-)$.
$3$. $HC \equiv CH$ (terminal alkyne) is the weakest acid among these because the $sp$-hybridized carbon is less electronegative than oxygen,and the negative charge on the carbon atom is less stable compared to the negative charge on oxygen.
Therefore,the correct order of acid strength is: $RCOOH > HOH > ROH > HC \equiv CH$.

(A) The rate of nucleophilic acyl substitution reaction depends on the leaving group ability and the electrophilicity of the carbonyl carbon.
$1$. Anhydrides $(RCO-O-COR)$ are highly reactive because the carboxylate ion $(RCOO^-)$ is a good leaving group.
$2$. Esters $(RCOOR')$ are less reactive than anhydrides because the alkoxide ion $(RO^-)$ is a poorer leaving group compared to the carboxylate ion.
$3$. Ethers $(R-O-R')$ do not contain a carbonyl group and are generally inert towards nucleophilic attack by ammonia under standard conditions.
Therefore,the decreasing order of reactivity is: $Anhydrides > Esters > Ethers$.

(B) The acidity of carboxylic acids depends on 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.
$1$. $CH_3CH_2COOH$ (Propionic acid): Contains an electron-donating alkyl group ($+I$ effect),which decreases acidity.
$2$. $CH_3COOH$ (Acetic acid): Contains a methyl group ($+I$ effect),which is less electron-donating than the ethyl group in propionic acid.
$3$. $HCOOH$ (Formic acid): No alkyl group attached to the carboxyl group.
$4$. $ClCH_2COOH$ (Chloroacetic acid): Contains a chlorine atom,which is a strong electron-withdrawing group ($-I$ effect). This significantly stabilizes the carboxylate ion,making it the most acidic among the given options.

(A) The reaction of a carboxylic acid with $PCl_5$ produces an acyl chloride (acid chloride).
The chemical equation is: $CH_3COOH + PCl_5 \to CH_3COCl + POCl_3 + HCl$
Therefore,the correct option is $A$.

(C) . The reaction between an alcohol $(C_2H_5OH)$ and a carboxylic acid $(CH_3COOH)$ in the presence of an acid catalyst is known as esterification.
The product formed is an ester $(CH_3COOC_2H_5)$,which is characterized by a pleasant,fruity smell.

(A) The acidity of $\alpha$-halo substituted carboxylic acids depends on the electron-withdrawing inductive effect ($-I$ effect) of the halogen atom.
As the electronegativity of the halogen increases,the $-I$ effect increases,which stabilizes the carboxylate anion and increases the acidity.
The electronegativity order of halogens is $F > Cl > Br$.
Therefore,the correct order of acid strength is $FCH_2COOH > ClCH_2COOH > BrCH_2COOH$.

(A) $HCOOH$ (methanoic acid) acts as a reducing agent because it contains an aldehydic group $(-CHO)$ in its structure.
It reduces ammoniacal silver nitrate solution (Tollen's reagent) to metallic silver,forming a silver mirror.
$CH_3COOH$ (ethanoic acid) does not contain an aldehydic group and therefore does not reduce Tollen's reagent.
Thus,Tollen's reagent can be used to distinguish between them.