Properties of Carboxylic Acids and Their Derivatives Questions in English

(C) The acidity of organic compounds depends on the stability of the conjugate base formed after the loss of a proton $(H^+)$.
$1$. $CH_3CH_2OH$ (Ethanol) and $CH_3CH_2CH_2OH$ (Propan$-1-$ol) are alcohols with $pK_a$ values around $16$.
$2$. $C_6H_5OH$ (Phenol) has a $pK_a$ of approximately $10$,as the phenoxide ion is stabilized by resonance.
$3$. $CH_3COOH$ (Acetic acid) has a $pK_a$ of approximately $4.76$. The carboxylate ion $(CH_3COO^-)$ is significantly more stable than the phenoxide ion because the negative charge is delocalized over two highly electronegative oxygen atoms.
Therefore,$CH_3COOH$ is the most acidic compound among the given options.

(B) The $pK_a$ value is inversely proportional to the acid strength. $A$ weaker acid has a higher $pK_a$ value.
Comparing the acidity of the given acids:
$1$. $FCH_2COOH$ (Strongest due to the strong $-I$ effect of $F$)
$2$. $ClCH_2COOH$ (Stronger than $CH_3COOH$ due to $-I$ effect of $Cl$)
$3$. $HCOOH$ (Stronger than $CH_3COOH$ because $CH_3$ group has a $+I$ effect)
$4$. $CH_3COOH$ (Weakest acid among the given options due to the electron-donating $+I$ effect of the $CH_3$ group).
Since $CH_3COOH$ is the weakest acid,it has the highest $pK_a$ value.

(D) 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 acidity.
Fluorine $(F)$ is more electronegative than Chlorine $(Cl)$,so it exerts a stronger $-I$ effect.
Therefore,$FCH_2COOH$ is the strongest acid among the given options.

(D) The correct option is $(d)$.
Concentrated nitric acid $(HNO_3)$ acts as a strong oxidizing agent.
It oxidizes cane sugar $(C_{12}H_{22}O_{11})$ to oxalic acid $((COOH)_2)$ and water.
The chemical equation for the reaction is:
$C_{12}H_{22}O_{11} + 18HNO_3 \rightarrow 6(COOH)_2 + 9NO_2 + 9H_2O$

(A) In the given reaction,$HCOOH$ (formic acid) undergoes dehydration in the presence of concentrated $H_2SO_4$ to produce carbon monoxide $(CO)$ and water $(H_2O)$.
$HCOOH \xrightarrow{H_2SO_4} CO + H_2O$
Since $H_2SO_4$ removes a water molecule from the reactant,it acts as a dehydrating agent.

(B) Oxalic acid $(COOH)_2$ on heating with concentrated $H_2SO_4$ undergoes dehydration to give a mixture of $CO$ and $CO_2$ gases.
$(COOH)_2 \xrightarrow{\text{conc. } H_2SO_4, \Delta} CO + CO_2 + H_2O$

(B) Carboxylic acids are easily ionized because the resulting carboxylate ion is stabilized by resonance.
In the carboxylate ion,the negative charge is delocalized over the two oxygen atoms due to $\pi$-electron shifting,which significantly stabilizes the conjugate base.
This stabilization makes the release of the $H^{+}$ ion energetically favorable,leading to easy ionization.

(A) The reaction is a nucleophilic acyl substitution reaction.
In this reaction,the leaving group ability of $X^-$ determines the rate of the reaction.
$A$ better leaving group makes the reaction faster.
Leaving group ability is inversely proportional to the basicity of the group.
Since $Cl^-$ is the weakest base among the given options (because its conjugate acid $HCl$ is the strongest acid),it is the best leaving group.
Therefore,the reaction is fastest when $X$ is $Cl$.

(B) The reduction of carboxylic acids with $HI$ in the presence of red phosphorus $(P)$ is a strong reduction method that converts the carboxylic acid group $(-COOH)$ into a methyl group $(-CH_3)$.
Propionic acid is $CH_3-CH_2-COOH$.
Upon reduction,it forms propane $(CH_3-CH_2-CH_3)$.
The reaction is: $CH_3-CH_2-COOH + 6HI \xrightarrow{\text{Red } P} CH_3-CH_2-CH_3 + 2H_2O + 3I_2$.

(D) Decarboxylation involves the removal of a carboxyl group $(-COOH)$ as $CO_2$ to form a hydrocarbon.
Cinnamic acid,isophthalic acid,and salicylic acid contain $-COOH$ groups and can undergo decarboxylation to form styrene,benzene,and phenol (which can be further converted) respectively.
Picric acid ($2, 4, 6-$trinitrophenol) does not contain a $-COOH$ group; it is a phenol derivative. Therefore,it cannot undergo decarboxylation to form a hydrocarbon.

(C) The $-COOC_2H_5$ group present in ethyl benzoate is an electron-withdrawing group (deactivating group) that directs incoming electrophiles to the meta position.
Therefore,ethyl benzoate undergoes meta substitution upon monochlorination.
The reaction is:
$C_6H_5COOC_2H_5 + Cl_2 \xrightarrow{FeCl_3} m-Cl-C_6H_4COOC_2H_5 + HCl$

(D) Electrical conductivity in an aqueous solution depends on the concentration of ions present.
Stronger acids dissociate more completely,leading to a higher concentration of ions.
The acidity of carboxylic acids increases with the presence of electron-withdrawing groups due to the negative inductive effect ($-I$ effect).
Among the given options,$0.1\, M$ difluoroacetic acid contains two fluorine atoms,which exert a strong $-I$ effect,significantly increasing the dissociation of the acid compared to acetic acid,chloroacetic acid,and fluoroacetic acid.
Therefore,$0.1\, M$ difluoroacetic acid provides the highest concentration of ions and exhibits the highest electrical conductivity.

(C) The hydrolysis of ethyl acetate is an acid-catalyzed reaction.
$CH_{3}COOC_{2}H_{5} + H_{2}O \xrightarrow{H^{+}} CH_{3}COOH + C_{2}H_{5}OH$
Among the given options,$H_{2}SO_{4}$ is a strong acid that provides $H^{+}$ ions to act as a catalyst for this reaction.
Therefore,the correct option is $(C)$.

(A) The reaction of silver salts of carboxylic acids with bromine in the presence of carbon disulfide $(CS_2)$ is known as the Hunsdiecker reaction.
The general reaction is: $RCOOAg + Br_2 \xrightarrow{CS_2} RBr + AgBr + CO_2$.
For silver acetate $(CH_3COOAg)$,the reaction is: $CH_3COOAg + Br_2 \xrightarrow{CS_2} CH_3Br + AgBr + CO_2$.
Therefore,the main product is methyl bromide $(CH_3Br)$.

(B) The reaction of an ester like ethyl benzoate $(C_6H_5COOC_2H_5)$ with phosphorus pentachloride $(PCl_5)$ involves the cleavage of the $C-O$ bond of the ester group.
The reaction proceeds as follows:
$C_6H_5COOC_2H_5 + PCl_5 \rightarrow C_6H_5COCl + C_2H_5Cl + POCl_3$
Thus,the products formed are benzoyl chloride $(C_6H_5COCl)$,ethyl chloride $(C_2H_5Cl)$,and phosphorus oxychloride $(POCl_3)$.

(A) Grignard reagents $(RMgX)$ are strong nucleophiles and strong bases. They react readily with acidic protons,such as those found in the $ -COOH$ group,to form an alkane and a carboxylate salt. However,a simple $ -CH_3$ group (methyl group),a magnesium atom,or a halogen atom (when part of an alkyl halide) do not possess acidic protons or electrophilic centers that would undergo a characteristic reaction with a Grignard reagent in the same way. Among the given options,the methyl group $( -CH_3)$ is chemically inert towards Grignard reagents.

(C) $LiAlH_4$ is a strong reducing agent that reduces carboxylic acids to primary alcohols.
$CH_3COOH + 4[H] \xrightarrow{LiAlH_4} CH_3CH_2OH + H_2O$
Therefore,acetic acid is converted into ethyl alcohol.

(D) When glycerol is heated with oxalic acid at $110^{\circ}C$,it undergoes decarboxylation to produce formic acid $(HCOOH)$:
$HOOC-COOH \xrightarrow{\text{Glycerol}} HCOOH + CO_2$
Formic acid $(HCOOH)$ acts as a reducing agent and reduces Fehling solution to give a red precipitate of cuprous oxide $(Cu_2O)$:
$HCOOH + 2Cu^{2+} + 5OH^- \to Cu_2O + CO_3^{2-} + 3H_2O$
Thus,the formation of cuprous oxide is due to the presence of $HCOOH$.

(D) In the esterification process,a carboxylic acid reacts with an alcohol in the presence of a small amount of concentrated $H_2SO_4$ to form an ester and water.
Concentrated $H_2SO_4$ acts as a catalyst to increase the rate of the reaction.
Additionally,it acts as a dehydrating agent by removing the water produced during the reaction,which shifts the equilibrium towards the formation of the ester according to Le Chatelier's principle.

(B) The reduction of carboxylic acids to primary alcohols requires a strong reducing agent.
$CH_3COOH \xrightarrow{LiAlH_4 / \text{ether}} CH_3CH_2OH$.
$LiAlH_4$ (Lithium aluminium hydride) in the presence of ether is a strong reducing agent capable of reducing acetic acid to ethanol.
Therefore,the correct option is $(B)$.

(A) The reaction between an alcohol and a carboxylic acid in the presence of an acid catalyst to form an ester and water is known as esterification.
The general reaction is:
$CH_3COOH + CH_3OH \to CH_3COOCH_3 + H_2O$
Therefore,the correct option is $A$.

(B) The boiling point of ethyl alcohol $(C_2H_5OH)$ is approximately $78 \ ^\circ C$.
Formic acid $(HCOOH)$ has a higher boiling point (approximately $100.8 \ ^\circ C$) because it forms a stable dimer through intermolecular hydrogen bonding,as shown below:
$H-C(=O)OH \cdots O=C(H)OH$
This extensive hydrogen bonding increases the effective molecular weight and the energy required to vaporize the substance.
Propane and dimethyl ether have significantly lower boiling points than ethyl alcohol due to weaker intermolecular forces (van der Waals forces).
Therefore,the boiling point of ethyl alcohol is less than that of formic acid.

(B) The reaction between ethyl alcohol $(C_2H_5OH)$ and acetic acid $(CH_3COOH)$ is an esterification reaction.
This reaction occurs in the presence of a concentrated acid catalyst like $H_2SO_4$.
The products formed are ethyl acetate $(CH_3COOC_2H_5)$ and water $(H_2O)$.
The chemical equation is: $C_2H_5OH + CH_3COOH \xrightarrow{H_2SO_4} CH_3COOC_2H_5 + H_2O$.

(B) Esterification is a chemical reaction between a carboxylic acid and an alcohol in the presence of an acid catalyst (like concentrated $H_2SO_4$) to form an ester and water. The general reaction is: $RCOOH + R'OH \rightleftharpoons RCOOR' + H_2O$.

(A) When benzoic acid reacts with ethyl alcohol in the presence of concentrated sulphuric acid,ethyl benzoate is formed. This reaction is known as esterification.
$C_6H_5COOH + C_2H_5OH \xrightarrow{H_2SO_4} C_6H_5COOC_2H_5 + H_2O$

(C) The compound $CH_3C(OH)_3$ contains three hydroxyl $(-OH)$ groups attached to the same carbon atom. Such geminal triols are highly unstable and spontaneously lose a water molecule $(H_2O)$ to form a more stable carboxylic acid.
The reaction is as follows:
$CH_3C(OH)_3 \rightarrow CH_3COOH + H_2O$
Thus,the product obtained is acetic acid $(CH_3COOH)$.

(A) The Bouveault-Blanc reduction is a chemical reaction in which an ester is reduced to a primary alcohol using sodium metal and an alcohol (typically ethanol) as the reducing agent.
The general reaction is: $RCOOR' + 4[H] \xrightarrow{Na/C_2H_5OH} RCH_2OH + R'OH$.
For example: $\underset{\text{Ethyl butyrate}}{C_3H_7COOC_2H_5}$ $\xrightarrow{Na/C_2H_5OH} \underset{\text{Butan-1-ol}}{C_3H_7CH_2OH} + C_2H_5OH$.
Therefore,the correct reagent is $Na/C_2H_5OH$.

(A) The reaction of acetaldehyde $(CH_3CHO)$ with $HCN$ produces a cyanohydrin: $CH_3CHO + HCN \rightarrow CH_3-CH(OH)-CN$.
Upon hydrolysis,the cyano group $(-CN)$ is converted into a carboxylic acid group $(-COOH)$,yielding $CH_3-CH(OH)-COOH$ ($2$-hydroxypropanoic acid).
This molecule contains a chiral carbon atom (the central carbon bonded to $-H$,$-OH$,$-CH_3$,and $-COOH$ groups),which allows it to exhibit Optical isomerism.

(A) $a)$. $CH_3COOH + NH_3 \rightarrow CH_3COONH_4$. Acetic acid reacts with ammonia to form ammonium acetate,not acetamide.
$b)$. $CH_3COCl + 2NH_3 \rightarrow CH_3CONH_2 + NH_4Cl$. Acetyl chloride reacts with ammonia to form acetamide.
$c)$. $(CH_3CO)_2O + 2NH_3 \rightarrow CH_3CONH_2 + CH_3COONH_4$. Acetic anhydride reacts with ammonia to form acetamide.
$d)$. $HCOOCH_3 + NH_3 \rightarrow HCONH_2 + CH_3OH$. Methyl formate reacts with ammonia to form formamide,not acetamide.

(B) The reaction is: $C_6H_5CHO + (CH_3CO)_2O \xrightarrow{CH_3COONa} C_6H_5CH=CHCOOH + CH_3COOH$.
This reaction is known as the Perkin reaction,where an aromatic aldehyde reacts with an acid anhydride in the presence of the sodium salt of the corresponding acid to form an $\alpha,\beta$-unsaturated carboxylic acid.

(D) The substitution of an alkoxy group of an ester by an alcohol is called trans-esterification.
It occurs in the presence of either an acid or a base catalyst.

(B) Aliphatic aldehydes react with Fehling's solution to give a reddish-brown precipitate of $Cu_2O$,while aromatic aldehydes and most carboxylic acids do not react.
However,formic acid $(HCOOH)$ contains an aldehydic group $(-CHO)$ attached to a hydrogen atom,which makes it a reducing agent.
Therefore,formic acid gives a positive Fehling's test.
The reaction is: $HCOOH + 4OH^{-} + 2Cu^{2+} \rightarrow CO_2 + 3H_2O + Cu_2O$ (red precipitate).

(A) Among the given compounds,the reactivity towards nucleophilic attack depends on the electrophilicity of the carbonyl carbon.
$MeCOCl$ (acid chloride) is the most reactive because the $Cl$ atom exerts a strong $-I$ effect and a weak $+R$ effect,making the carbonyl carbon highly electron-deficient.
The general order of reactivity towards nucleophilic acyl substitution is: $MeCOCl > MeCOOCOMe > MeCOOMe > MeCHO$.
Therefore,$MeCOCl$ is the most susceptible to nucleophilic attack.

(D) The compound known as oil of winter green is $Methyl \ salicylate$.
It is an ester formed by the reaction of salicylic acid with methanol.
Its chemical formula is $C_8H_8O_3$.

(C) Vinegar is a dilute solution of acetic acid $(CH_3COOH)$.
It is produced by the fermentation of ethanol,which can be derived from cane sugar.
Vinegar typically contains $8-10\%$ acetic acid by volume.

(A) . The main component of vinegar is acetic acid $(CH_3COOH)$,which is produced by the fermentation of sugarcane juice. The term vinegar is derived from the Latin word $acetum$,meaning vinegar.

(A) The acidity of carboxylic acids depends on the stability of the conjugate base (carboxylate ion) formed after the loss of a proton.
Electron-withdrawing groups ($-I$ effect) stabilize the carboxylate anion by dispersing the negative charge,thereby increasing the acidity of the parent acid.
In $ClCH_2COOH$,the chlorine atom exerts a strong $-I$ effect,which significantly stabilizes the carboxylate ion.
In $C_6H_5COOH$,the phenyl group has a weaker effect compared to the chlorine atom.
In $CH_3CH_2COOH$ and $CD_3COOH$,the alkyl groups exert a $+I$ effect,which destabilizes the carboxylate ion and decreases acidity.
Therefore,$ClCH_2COOH$ is the most acidic among the given options.

(D) The order of reactivity of acid derivatives towards nucleophilic acyl substitution reactions is: $RCOCl > (RCO)_2O > RCOOR' > RCONH_2$.
This order is determined by the leaving group ability; the reactivity decreases as the basicity of the leaving group increases.
The leaving groups are $Cl^-$,$RCOO^-$,$RO^-$,and $NH_2^-$.
Since $Cl^-$ is the weakest base,it is the best leaving group,making $Acetyl \ chloride$ $(RCOCl)$ the most reactive.

(A) The reaction of formic acid $(HCOOH)$ with phosphorus pentachloride $(PCl_5)$ yields formyl chloride $(HCOCl)$,phosphorus oxychloride $(POCl_3)$,and hydrogen chloride $(HCl)$.
$HCOOH + PCl_5 \to HCOCl + POCl_3 + HCl$
Note: Formyl chloride is unstable and decomposes into $CO$ and $HCl$.

(C) Carboxylic acids $(RCOOH)$ react with diazomethane $(CH_2N_2)$ to form methyl esters $(RCOOCH_3)$ with the evolution of nitrogen gas $(N_2)$.
The reaction involves the transfer of a $CH_2$ group from diazomethane to the carboxylic acid,resulting in esterification.
Option $C$ is correct.

(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 as follows:
$2CH_3COOH + P_2O_5 \rightarrow (CH_3CO)_2O + H_2O$
Thus,the correct product is acetic anhydride.

(A) The reaction of acetic acid $(CH_3COOH)$ with chlorinating agents like $SOCl_2$,$PCl_3$,and $PCl_5$ produces acetyl chloride $(CH_3COCl)$.
$CH_3COOH + SOCl_2 \to CH_3COCl + SO_2 + HCl$
$3CH_3COOH + PCl_3 \to 3CH_3COCl + H_3PO_3$
$CH_3COOH + PCl_5 \to CH_3COCl + POCl_3 + HCl$
Chloroform $(CHCl_3)$ does not act as a chlorinating agent for carboxylic acids,hence no reaction occurs.

(A) In the esterification reaction,a carboxylic acid $(RCOOH)$ reacts with an alcohol $(R'OH)$ in the presence of an acid catalyst to form an ester $(RCOOR')$ and water $(H_2O)$.
According to the mechanism of esterification,the $OH$ group is removed from the carboxylic acid,and the $H$ atom is removed from the alcohol.
Therefore,the $OH^-$ ion (or the $OH$ part) required to form $H_2O$ comes from the carboxylic acid.

(C) The reaction between ethyl alcohol $(C_2H_5OH)$ and acetic acid $(CH_3COOH)$ in the presence of concentrated $H_2SO_4$ is given by:
$C_2H_5OH + CH_3COOH \xrightarrow{Conc. H_2SO_4} CH_3COOC_2H_5 + H_2O$
This reaction produces an ester (ethyl acetate),which has a characteristic fruity smell.
Therefore,this reaction is known as esterification.

(B) The reaction of carboxylic acids with alcohols in the presence of an acid catalyst like $Dry \ HCl$ is known as Fischer esterification.
The reaction is: $RCOOH + C_2H_5OH \xrightarrow{Dry \ HCl} RCOOC_2H_5 + H_2O$.
Thus,$Dry \ HCl + C_2H_5OH$ is the correct reagent.

(C) The reaction of acetyl chloride $(CH_3COCl)$ with sodium acetate $(CH_3COONa)$ yields acetic anhydride.
The chemical equation is:
$CH_3COCl + CH_3COONa \to (CH_3CO)_2O + NaCl$

(A) The hydrolysis of acetamide $(CH_3CONH_2)$ in the presence of an acid or base catalyst yields acetic acid $(CH_3COOH)$ and ammonia $(NH_3)$.
The reaction is as follows:
$CH_3CONH_2 + H_2O \xrightarrow{\text{Hydrolysis}} CH_3COOH + NH_3$
Therefore,the correct product is acetic acid.

(C) The reaction between sodium acetate $(CH_3COONa)$ and acetyl chloride $(CH_3COCl)$ is a nucleophilic acyl substitution reaction.
The reaction proceeds as follows:
$CH_3COONa + CH_3COCl \to NaCl + CH_3COOCOCH_3$
The product formed is acetic anhydride $(CH_3COOCOCH_3)$.

(A) When ammonium acetate $(CH_3COONH_4)$ is heated with acetic acid $(CH_3COOH)$ at $110\,^{\circ}C$,it undergoes dehydration to form acetamide $(CH_3CONH_2)$ and water $(H_2O)$.
The reaction is: $CH_3COONH_4 + CH_3COOH \xrightarrow{110\,^{\circ}C} CH_3CONH_2 + 2H_2O$.

(B) The reaction of a silver salt of a fatty acid with an alkyl halide is known as the $Borodin-Hunsdiecker$ reaction or simply an esterification reaction.
The general chemical equation is: $RCOOAg + R'X \to RCOOR' + AgX$
Here,$RCOOAg$ is the silver salt of the fatty acid,$R'X$ is the alkyl halide,and the product $RCOOR'$ is an $Ester$.