Properties of Carboxylic Acids and Their Derivatives Questions in English

(C) The hydrolysis of acyl chloride $(R-COCl)$ with water $(H_2O)$ results in the formation of a carboxylic acid $(R-COOH)$ and hydrogen chloride $(HCl)$.
The chemical reaction is as follows:
$R-COCl + H_2O \rightarrow R-COOH + HCl$
Thus,the correct product obtained is a carboxylic acid.

(C) The reaction of acid chlorides with dialkyl cadmium reagents is a standard method for the preparation of ketones.
The chemical equation is:
$2CH_3COCl + (CH_3)_2Cd \rightarrow 2CH_3COCH_3 + CdCl_2$
Here,$CH_3COCl$ is ethanoyl chloride (also known as acetyl chloride),which reacts with dimethyl cadmium to form propanone (acetone).
Therefore,'$A$' is ethanoyl chloride.

(A) The reaction of benzamide $(C_{6}H_{5}CONH_{2})$ with water in the presence of an acid $(HCl)$ and heat is an acid-catalyzed hydrolysis reaction.
The amide group $(-CONH_{2})$ is hydrolyzed to a carboxylic acid group $(-COOH)$ and ammonium chloride $(NH_{4}Cl)$ is formed as a byproduct.
The chemical equation is:
$C_{6}H_{5}CONH_{2} + H_{2}O + HCl \xrightarrow{\Delta} C_{6}H_{5}COOH + NH_{4}Cl$
Thus,the product $A$ is benzoic acid $(C_{6}H_{5}COOH)$.

(B) Diborane $(B_2H_6)$ is a selective reducing agent.
It is known to reduce carboxylic acids $(-COOH)$ to primary alcohols $(R-CH_2OH)$ rapidly at room temperature.
It does not reduce other functional groups like esters $(-COOR)$,nitro groups $(-NO_2)$,or halides $(-X)$ under these conditions.
Therefore,the correct answer is $-COOH$.

(B) The boiling point of carboxylic acids increases with an increase in the molecular mass due to the increase in the magnitude of van der Waals forces of attraction.
$Valeric \ acid$ $(C_4H_9COOH)$ has the highest molecular mass among the given options ($Formic \ acid$: $HCOOH$,$Acetic \ acid$: $CH_3COOH$,$Butyric \ acid$: $C_3H_7COOH$).
Therefore,$Valeric \ acid$ has the highest boiling point.

(D) The boiling point of a compound depends on the strength of intermolecular forces.
Carboxylic acids $(C_2H_5COOH)$ form stable dimers due to strong intermolecular hydrogen bonding,which is significantly stronger than the hydrogen bonding in alcohols $(n-C_4H_9OH)$ or the dipole-dipole interactions in amines.
Therefore,$C_2H_5COOH$ has the highest boiling point among the given compounds.

(D) The boiling point of carboxylic acids increases with an increase in molar mass due to stronger intermolecular forces (hydrogen bonding and van der Waals forces).
Among the given compounds,the molar masses are:
$1$. Formic acid $(HCOOH)$: $46 \ g/mol$
$2$. Acetic acid $(CH_3COOH)$: $60 \ g/mol$
$3$. Butyric acid $(C_3H_7COOH)$: $88 \ g/mol$
$4$. Valeric acid $(C_4H_9COOH)$: $102 \ g/mol$
Since Formic acid has the lowest molar mass,it has the lowest boiling point.

(C) The reaction of an alkyl halide with a silver salt of a carboxylic acid is a standard method for the preparation of esters. This is a nucleophilic substitution reaction ($S_N2$ mechanism).
The reaction is as follows:
$CH_3CH_2Br + CH_3CH_2COOAg \xrightarrow{\Delta} CH_3CH_2COOCH_2CH_3 + AgBr \downarrow$
Here,$CH_3CH_2COOAg$ is silver propanoate. Thus,the reagent '$A$' is silver propanoate.

(B) The boiling point of a compound depends on the strength of intermolecular forces.
Carboxylic acids,such as $CH_3COOH$,form stable intermolecular hydrogen-bonded dimers,which significantly increase their boiling points compared to alcohols,aldehydes,and ketones of comparable molecular mass.
Therefore,$Ethanoic \ acid$ has the highest boiling point among the given options.

(C) The reaction of benzoyl chloride $(C_6H_5COCl)$ with water $(H_2O)$ is a hydrolysis reaction.
In this reaction,the chlorine atom $(-Cl)$ of the acid chloride is replaced by a hydroxyl group $(-OH)$ from water.
The chemical equation is: $C_6H_5COCl + H_2O \rightarrow C_6H_5COOH + HCl$.
Here,$C_6H_5COOH$ is benzoic acid,which is the product $B$.

(B) The reaction of ethanoyl chloride $(CH_3COCl)$ with water $(H_2O)$ is a hydrolysis reaction.
The chlorine atom is replaced by a hydroxyl group $(-OH)$ from water,releasing hydrogen chloride $(HCl)$ as a byproduct.
The chemical equation is: $CH_3COCl + H_2O \longrightarrow CH_3COOH + HCl$.
The product $P$ is ethanoic acid $(CH_3COOH)$.

(B) The boiling point of organic compounds depends on the strength of intermolecular forces.
$C_2H_5COOH$ (propanoic acid) can form stable intermolecular hydrogen-bonded dimers,which significantly increases its boiling point compared to alcohols $(C_4H_9OH)$ and amines $(C_4H_9NH_2)$ of similar molecular mass.
Alkanes like $C_2H_5CH(CH_3)_2$ only have weak London dispersion forces.
Therefore,carboxylic acids exhibit the highest boiling point among these functional groups.

(C) The boiling point of carboxylic acids is directly proportional to their molar mass due to the increase in van der Waals forces with increasing chain length.
Among the given options,the molar masses are:
$CH_3COOH$ (Acetic acid) = $60 \ g/mol$
$CH_3CH_2COOH$ (Propionic acid) = $74 \ g/mol$
$CH_3CH_2CH_2COOH$ (Butyric acid) = $88 \ g/mol$
$CH_3CH_2CH_2CH_2COOH$ (Valeric acid) = $102 \ g/mol$
Since Acetic acid has the lowest molar mass,it has the lowest boiling point.

(A) The reaction of the sodium salt of an $\alpha-$chloro carboxylic acid with aqueous sodium nitrite $(NaNO_2)$ leads to the formation of a nitro-substituted intermediate.
Upon subsequent hydrolysis and decarboxylation,this intermediate yields a nitroalkane.
For example,the reaction of sodium $\alpha-$chloroacetate with $NaNO_2$ followed by hydrolysis produces nitromethane $(CH_3NO_2)$.

(A) flavouring agent found in oil of wintergreen is $Methyl \ salicylate$.
It is an organic compound with the formula $C_6H_4(OH)(CO_2CH_3)$.
It is the methyl ester of salicylic acid.
It is a colourless,viscous liquid.
The structure of $Methyl \ salicylate$ is as follows:

(B) Step $1$: The reaction between $CH_3COOH$ and $CH_3CH_2OH$ in the presence of $H^{+}$ is an esterification reaction,which produces ethyl acetate $(A)$ as the product.
$CH_3COOH + CH_3CH_2OH \rightleftharpoons[H^{+}]{H^{+}} CH_3COOCH_2CH_3 (A) + H_2O$
Step $2$: The reduction of an ester $(A)$ using $H_2/Ni, \Delta$ leads to the formation of alcohols.
$CH_3COOCH_2CH_3 + 2H_2 \xrightarrow{Ni, \Delta} CH_3CH_2OH + CH_3CH_2OH$
Thus,the product $B$ is $CH_3CH_2OH$ (ethanol).

(A) The reaction of salicylic acid ($2$-hydroxybenzoic acid) with acetic anhydride in the presence of an acid catalyst (like $H_2SO_4$) is an acetylation reaction.
In this reaction,the phenolic $-OH$ group of salicylic acid is acetylated to form $2$-acetoxybenzoic acid,which is commonly known as Aspirin.
The reaction is as follows:
Salicylic acid + Acetic anhydride $\xrightarrow{H^+}$ Aspirin + Acetic acid.
Therefore,$A$ is Aspirin.

(C) The reactivity of carboxylic acids towards esterification is inversely proportional to the steric hindrance around the carbonyl carbon atom.
$CH_3CH_2COOH$ (propanoic acid) has the least steric hindrance among the given options because it is a primary carboxylic acid with the smallest alkyl group attached to the carbonyl carbon.
As the number of alkyl groups or the size of the alkyl groups attached to the $\alpha$-carbon increases,the steric hindrance increases,which decreases the rate of nucleophilic attack by the alcohol.
Therefore,$CH_3CH_2COOH$ is the most reactive.

(C) The reaction between $CH_3COOH$ and $C_2H_5OH$ is an esterification reaction that produces ethyl ethanoate $(CH_3COOC_2H_5)$:
$CH_3COOH + C_2H_5OH \rightarrow CH_3COOC_2H_5 + H_2O$
Ethyl ethanoate can also be prepared by the reaction of acetic anhydride with ethanol:
$(CH_3CO)_2O + C_2H_5OH \rightarrow CH_3COOC_2H_5 + CH_3COOH$
Therefore,the correct option is $C$.

(A) The reaction between a carboxylic acid and an alcohol in the presence of an acid catalyst produces an ester and water. This process is known as esterification.
$RCOOH + R^{\prime}OH \xrightarrow{H^+} RCOOR^{\prime} + H_2O$
Here,$RCOOH$ is the carboxylic acid,$R^{\prime}OH$ is the alcohol,and $RCOOR^{\prime}$ is the ester.

(C) The reaction $C_6H_5COOC_2H_5 \xrightarrow{\text{dil. } H_2SO_4} ?$ represents the acidic hydrolysis of an ester.
In acidic hydrolysis,an ester reacts with water in the presence of an acid catalyst (like $dil. H_2SO_4$) to produce a carboxylic acid and an alcohol.
The general reaction is: $RCOOR' + H_2O \xrightarrow{H^+} RCOOH + R'OH$.
For the given ester,ethyl benzoate $(C_6H_5COOC_2H_5)$,the hydrolysis yields benzoic acid $(C_6H_5COOH)$ and ethanol $(C_2H_5OH)$.

(D) The $Hell-Volhard-Zelinsky$ $(HVZ)$ reaction occurs in carboxylic acids that contain at least one $\alpha-hydrogen$ atom.
In this reaction,the $\alpha-hydrogen$ is replaced by a halogen (chlorine or bromine) in the presence of red phosphorus.
$Butanoic$ $acid$ $(CH_3CH_2CH_2COOH)$,$Propanoic$ $acid$ $(CH_3CH_2COOH)$,and $Ethanoic$ $acid$ $(CH_3COOH)$ all possess $\alpha-hydrogen$ atoms and thus undergo the $HVZ$ reaction.
$Methanoic$ $acid$ $(HCOOH)$,also known as $Formic$ $acid$,does not contain an $\alpha-carbon$ atom,and consequently,it lacks an $\alpha-hydrogen$ atom.
Therefore,$Methanoic$ $acid$ does not undergo the $HVZ$ reaction.

(D) The Hell-Volhard-Zelinsky $(HVZ)$ reaction is a specific reaction used for the $\alpha$-halogenation of carboxylic acids.
In this reaction,carboxylic acids containing an $\alpha$-hydrogen atom react with chlorine or bromine in the presence of a small amount of red phosphorus to form $\alpha$-halo carboxylic acids.
The reaction proceeds as follows:
$CH_{3}CH_{2}COOH$ $\xrightarrow{Br_{2}/P} CH_{3}CH(Br)COOH$ $\xrightarrow{Br_{2}/P} CH_{3}C(Br)_{2}COOH$

(B) Malonic acid $(CH_{2}(COOH)_{2})$ undergoes decarboxylation upon heating to produce acetic acid $(CH_{3}COOH)$ and carbon dioxide $(CO_{2})$.
The reaction is as follows:
$CH_{2}(COOH)_{2} \xrightarrow{\Delta} CH_{3}COOH + CO_{2}$

(C) The reaction sequence is as follows:
$1$. The first step is the Hell-Volhard-Zelinsky $(HVZ)$ reaction,where propanoic acid reacts with $Cl_2$ in the presence of red phosphorus to form $\alpha$-chloropropanoic acid $(X)$:
$CH_3-CH_2-COOH \xrightarrow[\text{red } P]{Cl_2} CH_3-CHCl-COOH \text{ (} X \text{)}$
$2$. The second step is a dehydrohalogenation reaction using alcoholic $KOH$,which removes $HCl$ from the $\alpha$-carbon and $\beta$-carbon to form acrylic acid $(Y)$:
$CH_3-CHCl-COOH \xrightarrow{\text{Alc. KOH}} CH_2=CH-COOH \text{ (} Y \text{)}$
Therefore,the end product $Y$ is acrylic acid $(CH_2=CH-COOH)$.

(A) Corrosive sublimate $(HgCl_{2})$ acts as a mild oxidizing agent. It reacts with formic acid $(HCOOH)$ to form a white precipitate of calomel $(Hg_{2}Cl_{2})$,which may appear greyish-black due to the presence of finely divided mercury. Acetic acid $(CH_{3}COOH)$ does not react with $HgCl_{2}$.
$2 HCOOH + 2 HgCl_{2} \longrightarrow Hg_{2}Cl_{2} + 2 CO_{2} + 2 HCl$
$CH_{3}COOH + HgCl_{2} \longrightarrow \text{No reaction}$
Thus,it is used to distinguish between formic acid and acetic acid.

(C) The reaction of benzoic acid with phosphorus pentachloride $(PCl_5)$ is a standard method for the preparation of acid chlorides. The chemical equation is:
$C_6H_5COOH + PCl_5 \rightarrow C_6H_5COCl + POCl_3 + HCl$
Here,$C_6H_5COCl$ is benzoyl chloride,$POCl_3$ is phosphorus oxychloride,and $HCl$ is hydrogen chloride. Thus,the reagent is $PCl_5$.

(A) The conversion of an acid chloride $(C_6H_5COCl)$ to a carboxylic acid $(C_6H_5COOH)$ is a hydrolysis reaction.
Acid chlorides are highly reactive towards nucleophilic acyl substitution.
When treated with water $(H_2O)$,the chloride ion $(Cl^-)$ is replaced by a hydroxyl group $(-OH)$ to form the corresponding carboxylic acid and hydrochloric acid $(HCl)$.

(D) The reaction of acetic anhydride $(CH_3CO)_2O$ with water $(H_2O)$ is a hydrolysis reaction.
In this reaction,the anhydride molecule is cleaved by water to form two molecules of acetic acid.
The chemical equation is: $(CH_3CO)_2O + H_2O \rightarrow 2CH_3COOH$.

(D) Step $1$: Acetic acid $(CH_3COOH)$ reacts with thionyl chloride $(SOCl_2)$ to form acetyl chloride $(CH_3COCl)$ as product $A$.
$CH_3COOH + SOCl_2 \rightarrow CH_3COCl + SO_2 + HCl$
Step $2$: Acetyl chloride $(CH_3COCl)$ reacts with ammonia $(NH_3)$ to form acetamide $(CH_3CONH_2)$ as product $B$.
$CH_3COCl + 2NH_3 \rightarrow CH_3CONH_2 + NH_4Cl$
Therefore,$B$ is acetamide.

(C) Acyl chlorides $(RCOCl)$ react with water (hydrolysis) to produce carboxylic acids $(RCOOH)$ and hydrogen chloride $(HCl)$.
The reaction is as follows:
$RCOCl + H_2O \rightarrow RCOOH + HCl$
Therefore,the correct product is a carboxylic acid.

(C) The reaction of two molecules of acetic acid $(CH_3COOH)$ in the presence of a dehydrating agent like phosphorus pentoxide $(P_2O_5)$ and heat $(\Delta)$ results in the removal of one water molecule $(H_2O)$.
This process is known as dehydration.
The reaction is as follows:
$2 CH_3COOH \xrightarrow{\Delta, P_2O_5} (CH_3CO)_2O H_2O$
Here,$(CH_3CO)_2O$ is acetic anhydride.
Therefore,product $A$ is acetic anhydride.

(D) The reaction of carboxylic acids with phosphorus pentachloride $(PCl_5)$ yields acid chlorides,phosphorus oxychloride $(POCl_3)$,and hydrogen chloride $(HCl)$.
The chemical equation is:
$C_6H_5COOH + PCl_5 \xrightarrow{\Delta} C_6H_5COCl + POCl_3 + HCl$
Therefore,the compound $A$ is $PCl_5$.

(C) The reaction of benzoic acid with a mixture of conc. $H_{2}SO_{4}$ and conc. $HNO_{3}$ is an electrophilic aromatic substitution reaction known as nitration.
Benzoic acid contains a $-COOH$ group,which is a deactivating and meta-directing group.
Therefore,the incoming electrophile,the nitronium ion $(NO_{2}^{+})$,attacks the meta-position of the benzene ring.
This results in the formation of $m-$nitrobenzoic acid as the major product.

(D) $1$. The reaction of methyl propanoate $(CH_3CH_2COOCH_3)$ with dilute $NaOH$ followed by heating $(\Delta)$ is a base-catalyzed hydrolysis (saponification) reaction.
$2$. This reaction produces sodium propanoate $(CH_3CH_2COONa)$ and methanol $(CH_3OH)$. Thus,$A$ is $CH_3CH_2COONa$.
$3$. In the second step,the treatment of sodium propanoate $(A)$ with concentrated $HCl$ $(H^+)$ results in the protonation of the carboxylate ion to form the corresponding carboxylic acid.
$4$. The reaction is: $CH_3CH_2COONa + HCl \rightarrow CH_3CH_2COOH + NaCl$.
$5$. Therefore,the product $B$ is propanoic acid $(CH_3CH_2COOH)$.

(D) Anhydrides undergo hydrolysis with water to yield carboxylic acids.
The reaction is as follows:
$(CH_3 CO)_2 O + H_2 O \longrightarrow 2 CH_3 COOH$
Thus,the product obtained is acetic acid $(CH_3 COOH)$.

(A) The hydrolysis of $Methyl$ propanoate $(CH_3CH_2COOCH_3)$ with dil. $NaOH$ yields sodium propanoate $(CH_3CH_2COONa)$ and methanol $(CH_3OH)$.
Subsequent acidification of sodium propanoate with conc. $HCl$ yields propanoic acid $(CH_3CH_2COOH)$.

(D) The alkaline hydrolysis of an ester is known as saponification.
When an ester is heated with aqueous $NaOH$,a sodium salt of the carboxylic acid and an alcohol are formed.
The reaction is represented as:
$R-COOR' + NaOH \rightarrow R-COO^-Na^+ + R'OH$

(B) The reaction of a silver salt of a carboxylic acid with an alkyl halide $(R-X)$ is known as the alkylation of carboxylate ions.
This reaction proceeds via a nucleophilic substitution mechanism $(S_N2)$ where the carboxylate ion $(R'COO^-)$ acts as a nucleophile and attacks the alkyl halide.
The general reaction is: $R'COOAg + R-X \rightarrow R'COOR + AgX \downarrow$.
The product formed is an ester.

(C) The boiling point of a compound depends on the strength of intermolecular forces.
Carboxylic acids $(CH_3-CH_2-COOH)$ form strong intermolecular hydrogen bonds,often existing as dimers in the liquid phase.
This results in significantly higher boiling points compared to alcohols $(CH_3-CH_2-CH_2-CH_2-OH)$,ketones $(CH_3-CH_2-CO-CH_3)$,and aldehydes $(CH_3-CH_2-CH_2-CHO)$ of comparable molecular mass.

(A) Aspirin (acetylsalicylic acid) is prepared by the acetylation of salicylic acid.
Salicylic acid is chemically known as $o-$hydroxybenzoic acid.
The reaction involves the treatment of salicylic acid with acetic anhydride in the presence of an acid catalyst.

(B) Ethyl butyrate $(C_6H_{12}O_2)$ has the characteristic flavour of pineapple.
It is widely used in the food and beverage industry to impart a pineapple flavour to various products,including alcoholic beverages.

(C) The structures of the given acids are as follows:
$1$. Adipic acid: $HOOC-(CH_2)_4-COOH$ (Dicarboxylic acid)
$2$. Glutaric acid: $HOOC-(CH_2)_3-COOH$ (Dicarboxylic acid)
$3$. Valeric acid: $CH_3-CH_2-CH_2-CH_2-COOH$ (Monocarboxylic acid)
$4$. Malonic acid: $HOOC-CH_2-COOH$ (Dicarboxylic acid)
Therefore,Valeric acid is not a dicarboxylic acid.

(C) The acid strength of carboxylic acids is directly proportional to the $-I$ (inductive) effect of the substituents attached to the alpha-carbon.
As the number of electron-withdrawing groups increases,the $-I$ effect increases,which stabilizes the conjugate base (carboxylate ion) and increases the acidity.
In $Cl_3C-COOH$,there are three chlorine atoms exerting a strong $-I$ effect,whereas the other options have only one halogen atom.
Therefore,$Cl_3C-COOH$ is the strongest acid.

(B) The acidic strength of carboxylic acids is directly proportional to the $-I$ (inductive) effect of the substituent attached to the alpha-carbon.
$Acidic \text{ } strength \propto -I \text{ } effect$.
The order of $-I$ effect for halogens is $F > Cl > Br > I$ due to the difference in electronegativity.
Therefore,the order of acidic strength is $Fluoroacetic \text{ } acid > Chloroacetic \text{ } acid > Bromoacetic \text{ } acid > Iodoacetic \text{ } acid$.
Thus,Fluoroacetic acid is the strongest acid.

(B) The acidity of carboxylic acids is influenced by the presence of electron-withdrawing groups (EWGs) such as the chloro group $(-Cl)$.
These groups exert a $-I$ (negative inductive) effect,which stabilizes the carboxylate anion by dispersing the negative charge.
As the number of electron-withdrawing chlorine atoms increases,the $-I$ effect becomes stronger,leading to greater stabilization of the conjugate base and thus higher acidity.
The order of acidity is: $CCl_3COOH > CHCl_2COOH > CH_2ClCOOH > CH_3COOH$.
Therefore,trichloroacetic acid is the strongest acid among the given options.

(A) The acidic strength of aromatic carboxylic acids is influenced by the electronic effect of the substituent group attached to the benzene ring.
Electron-withdrawing groups $(EWG)$ like $-Cl$,$-CN$,and $-NO_{2}$ stabilize the carboxylate anion through inductive and/or resonance effects,thereby increasing the acidic strength.
Conversely,electron-donating groups $(EDG)$ like $-NH_{2}$ destabilize the carboxylate anion by increasing electron density,which decreases the acidic strength of the aromatic carboxylic acid.
Therefore,$-NH_{2}$ is the group that decreases the acidic strength.

(C) The reaction of benzoyl chloride $(C_6H_5COCl)$ with water $(H_2O)$ is a hydrolysis reaction.
In this reaction,the chlorine atom of the acid chloride is replaced by a hydroxyl group $(-OH)$ from water,resulting in the formation of benzoic acid $(C_6H_5COOH)$ and hydrochloric acid $(HCl)$ as a byproduct.
The chemical equation is:
$C_6H_5COCl + H_2O \rightarrow C_6H_5COOH + HCl$
Therefore,the product is benzoic acid.