Properties of Carboxylic Acids and Their Derivatives Questions in English

(B) The reactivity of carboxylic acid derivatives towards nucleophilic acyl substitution depends on the leaving group ability of the group attached to the acyl carbon. The better the leaving group,the more reactive the derivative. The order of leaving group ability is $Cl^- > RCOO^- > RO^- > NH_2^-$. Therefore,the order of reactivity is $R-CO-Cl > R-CO-O-CO-R > R-CO-OR > R-CONH_2$.

(B) The acidic strength of carboxylic acids depends on the stability of the carboxylate anion formed after the loss of a proton.
Electron-withdrawing groups (like $-Cl$) increase acidity by stabilizing the negative charge via the inductive effect ($-I$ effect).
Electron-donating alkyl groups (like $-CH_3$,$-C_2H_5$) decrease acidity by destabilizing the negative charge via the $+I$ effect.
Comparing the compounds:
$1$. $CH_2ClCOOH$: The $-Cl$ atom exerts a strong $-I$ effect,making it the most acidic.
$2$. $HCOOH$: No alkyl group is present,so it is more acidic than the others.
$3$. $CH_3COOH$: The methyl group $(-CH_3)$ has a $+I$ effect,which is weaker than the ethyl group.
$4$. $C_2H_5COOH$: The ethyl group $(-C_2H_5)$ has a stronger $+I$ effect than the methyl group,making it the least acidic.
Thus,the correct order is $CH_2ClCOOH > HCOOH > CH_3COOH > C_2H_5COOH$.

(B) Saponification is the alkaline hydrolysis of an ester. The reaction of ethyl benzoate $(C_6H_5COOC_2H_5)$ with caustic soda $(NaOH)$ proceeds as follows:
$C_6H_5COOC_2H_5 + NaOH \xrightarrow{\Delta} C_6H_5COONa + C_2H_5OH$
Here,ethyl benzoate reacts with sodium hydroxide to form sodium benzoate $(C_6H_5COONa)$ and ethanol $(C_2H_5OH)$.

(B) The oxidation of secondary alcohols to ketones is a standard reaction. $CH_3-CH(OH)-COOH$ (Lactic acid) is a secondary alcohol containing a carboxylic acid group.
Upon oxidation with alkaline $KMnO_4$,the secondary alcoholic group $(-CH(OH)-)$ is oxidized to a ketone group $(-CO-)$.
$CH_3-CH(OH)-COOH + [O] \xrightarrow{KMnO_4} CH_3-CO-COOH + H_2O$.
The product formed is $CH_3-CO-COOH$,which is known as Pyruvic acid.

(D) The reduction of a carboxylic acid $(RCOOH)$ to a primary alcohol $(RCH_2OH)$ requires a strong reducing agent.
$LiAlH_4$ (Lithium aluminium hydride) is a powerful reducing agent capable of reducing carboxylic acids directly to primary alcohols.
Other reagents like $Zn/HCl$ (Clemmensen reduction) reduce carbonyl groups to methylene groups,and $Na/\text{alcohol}$ (Bouveault-Blanc reduction) is typically used for esters.
Therefore,the correct option is $(D)$.

(D) The reaction of carboxylic acids with neutral $FeCl_3$ solution is a characteristic test for the presence of the carboxyl group.
Acetic acid $(CH_3COOH)$ reacts with neutral $FeCl_3$ to form a deep red coloured complex of ferric acetate,$[Fe(OH)_2(CH_3COO)]_n$ or $[Fe(CH_3COO)_3]$.
Other compounds listed,such as acetone $(CH_3COCH_3)$,dimethyl ether $(CH_3OCH_3)$,and ethanol $(CH_3CH_2OH)$,do not give this specific red colouration with neutral $FeCl_3$.

(D) The reaction of an ester $(C_6H_5-COOCH_3)$ with a strong reducing agent like $LiAlH_4$ followed by hydrolysis leads to the reduction of the ester group to a primary alcohol and the release of the corresponding alcohol from the alkoxy part.
$C_6H_5-COOCH_3 + 4[H] \xrightarrow{LiAlH_4} C_6H_5-CH_2OH + CH_3OH$
Thus,the products are benzyl alcohol $(C_6H_5-CH_2-OH)$ and methanol $(CH_3-OH)$.

(D) The reaction between a carboxylic acid and an alcohol in the presence of an acid catalyst is known as esterification.
The general reaction is: $RCOOH + R^{\prime}OH \xrightarrow{H^+} RCOOR^{\prime} + H_2O$.
The product formed is an ester.

(B) $LiAlH_4$ is a strong reducing agent that reduces acid chlorides directly to primary alcohols.
$CH_3COCl \xrightarrow{LiAlH_4} CH_3CH_2OH + HCl$
Therefore,the product formed is ethyl alcohol.

(D) The preparation of an ester from a carboxylic acid and an alcohol is known as Fischer esterification. The reaction is reversible and is represented as: $RCOOH + R'OH \rightleftharpoons RCOOR' + H_2O$.
To shift the equilibrium in the forward direction and increase the yield of the ester,a dehydrating agent is used to remove the water produced.
Concentrated $H_2SO_4$ acts as both a catalyst and a dehydrating agent in this reaction.

(D) 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 $(H_2O)$.
During this mechanism,the $OH^{-}$ group of the carboxylic acid is replaced by the $OR'$ group (alkoxy group) derived from the alcohol.
Therefore,the $OH^{-}$ group of the acid is replaced by the alkoxy group of the alcohol.

(B) Carboxylic acids with low molar mass (up to $4$ carbon atoms) are freely soluble in water.
This solubility is due to the formation of intermolecular $Hydrogen$ bonds between the carboxyl group of the acid and the water molecules.
These interactions allow the acid molecules to disperse uniformly within the water solvent.

(A) Acetamide $(CH_3CONH_2)$ undergoes dehydration in the presence of $P_2O_5$ to form methyl cyanide $(CH_3CN)$.
The reaction is:
$CH_3CONH_2 \xrightarrow{P_2O_5} CH_3C \equiv N + H_2O$
Thus,the correct product is methyl cyanide.

(A) . The reaction of carboxylic acids having an $\alpha$-hydrogen with $Cl_2$ or $Br_2$ in the presence of small amount of red phosphorus to give $\alpha$-halo carboxylic acids is known as the Hell-Volhard-Zelinsky reaction.

(A) The $-COOH$ group present in benzoic acid is a deactivating and meta-directing group.
Therefore,during electrophilic substitution reactions like nitration,the incoming electrophile $(NO_2^+)$ attacks the meta-position.
Thus,the major product formed is $3-$nitrobenzoic acid (also known as $m-$nitrobenzoic acid).

(A) The reduction of carboxylic acids to primary alcohols requires a strong reducing agent.
$LiAlH_4$ (Lithium aluminium hydride) is a powerful reducing agent capable of reducing ethanoic acid $(CH_3COOH)$ to ethanol $(CH_3CH_2OH)$.
The reaction is:
$CH_3COOH + 4[H] \xrightarrow{LiAlH_4} CH_3CH_2OH + H_2O$

(A) The acid strength of substituted benzoic acids is influenced by the electronic effects of the substituents.
$1$. The $-NO_2$ group is a strong electron-withdrawing group ($-I$ and $-M$ effects),which increases the acidity of benzoic acid by stabilizing the carboxylate anion.
$2$. $o-$nitrobenzoic acid exhibits the ortho-effect,which significantly enhances its acidity compared to the meta and para isomers.
$3$. Among the given options,$o-$nitrobenzoic acid has the highest acid strength due to the combined effect of the $-I$ effect and the ortho-effect.
$4$. $p-$nitrophenol is a phenol,which is significantly less acidic than carboxylic acids.
Therefore,the correct order of acidity is $o-$nitrobenzoic acid > $p-$nitrobenzoic acid > $m-$nitrobenzoic acid > $p-$nitrophenol.

(A) The reaction of benzoic acid $(C_6H_5COOH)$ with phosphorus pentachloride $(PCl_5)$ is a standard nucleophilic acyl substitution reaction.
The hydroxyl group $(-OH)$ of the carboxylic acid is replaced by a chlorine atom $(-Cl)$,resulting in the formation of benzoyl chloride $(C_6H_5COCl)$.
The reaction is: $C_6H_5COOH + PCl_5 \rightarrow C_6H_5COCl + POCl_3 + HCl$.

(C) Oxalic acid $(HOOC-COOH)$ undergoes dehydration in the presence of concentrated $H_2SO_4$,which acts as a dehydrating agent.
The reaction produces a mixture of carbon monoxide $(CO)$,carbon dioxide $(CO_2)$,and water $(H_2O)$.
The chemical equation is: $HOOC-COOH \xrightarrow{conc. H_2SO_4, \Delta} CO + CO_2 + H_2O$

(A) Salicylic acid ($2$-hydroxybenzoic acid) is more acidic than benzoic acid.
This is primarily due to the stabilization of the conjugate base (salicylate ion) by intramolecular hydrogen bonding between the ortho-hydroxy group and the carboxylate group.
Therefore,the correct option is $A$.

(D) Acetamide $(CH_3CONH_2)$ is an amphoteric compound because it can act as both a weak acid and a weak base.
$1$. As a base: It reacts with strong acids like $HCl$ to form a salt: $CH_3CONH_2 + HCl \to CH_3CONH_3^+ Cl^-$.
$2$. As an acid: It reacts with strong bases or metal oxides like $HgO$ to form metallic derivatives: $2CH_3CONH_2 + HgO \to (CH_3CONH)_2Hg + H_2O$.
Therefore,the correct option is $D$.

(D) The $-COOH$ group present in benzoic acid is a deactivating group and is meta-directing for electrophilic aromatic substitution reactions.
Therefore,when benzoic acid reacts with $Cl_2$ in the presence of a Lewis acid catalyst like $FeCl_3$,the chlorine atom is substituted at the meta-position,resulting in the formation of $m-$chlorobenzoic acid.

(B) 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.
$1$. $CH_3COOH$: No $EWG$,least acidic.
$2$. $CH_2ClCOOH$: Contains one $Cl$ atom ($-I$ effect).
$3$. $CH_2FCOOH$: Contains one $F$ atom (stronger $-I$ effect than $Cl$).
$4$. $CCl_3COOH$: Contains three $Cl$ atoms,providing the strongest $-I$ effect,making it the most acidic.
Thus,the correct order of increasing acid strength is: $CH_3COOH < CH_2ClCOOH < CH_2FCOOH < CCl_3COOH$.

(A) The acidity of carboxylic acids is influenced by the inductive effect of substituents attached to the alpha-carbon.
Electron-withdrawing groups like $Cl$ increase the acidity by stabilizing the carboxylate anion through the $-I$ effect.
As the number of chlorine atoms increases,the $-I$ effect increases,thereby increasing the acidic strength.
The order of acidic strength is: $CH_3COOH < ClCH_2COOH < Cl_2CHCOOH < Cl_3CCOOH$.
Therefore,$CH_3COOH$ is the weakest acid.

(B) The acidity of substituted benzoic acids depends on the electronic effects of the substituents.
$1$. The $-NO_2$ group is a strong electron-withdrawing group ($-I$ and $-M$ effects),which increases the acidity of benzoic acid compared to $PhCOOH$ $(A)$.
$2$. Due to the ortho effect,the ortho-substituted isomer $(B)$ is the most acidic among the nitrobenzoic acids.
$3$. For the meta and para isomers,the $-NO_2$ group at the para position $(C)$ exerts both $-I$ and $-M$ effects,whereas at the meta position $(D)$,it exerts only the $-I$ effect. However,the $-M$ effect is stronger at the para position,making $p-NO_2C_6H_4COOH$ more acidic than $m-NO_2C_6H_4COOH$.
$4$. Therefore,the correct order of acidity is: $o-NO_2C_6H_4COOH (B) > p-NO_2C_6H_4COOH (C) > m-NO_2C_6H_4COOH (D) > PhCOOH (A)$.
Thus,the correct option is $B$.

(D) The reaction is known as transesterification or alcoholysis.
In this reaction,the alkoxy group of the ester is replaced by the alkoxy group of the alcohol in the presence of an acid catalyst.
The reaction is represented as: $RCOOR' + R''OH \overset{H^{\oplus}}{\longleftrightarrow} RCOOR'' + R'OH$.
Therefore,the product formed is $RCOOR''$.

(B) $RCOOH + PCl_5 \to RCOCl + POCl_3 + HCl$
$RCOCl + KCN \to RCOCN + KCl$
$RCOCN + 2H_2O \xrightarrow{H^+} RCOCOOH + NH_3$
$RCOCOOH \xrightarrow{Zn(Hg)/HCl} RCH_2COOH$
The final product obtained is $RCH_2COOH$.

(B) $PCl_5$,$SOCl_2$,and $PCl_3$ are standard chlorinating agents used to convert carboxylic acids into acid chlorides $(RCOCl)$.
$CH_3COOH + PCl_5 \to CH_3COCl + POCl_3 + HCl$
$CH_3COOH + SOCl_2 \to CH_3COCl + SO_2 + HCl$
$3CH_3COOH + PCl_3 \to 3CH_3COCl + H_3PO_3$
However,$Cl_2$ (chlorine gas) in the presence of red phosphorus (Hell-Volhard-Zelinsky reaction) reacts with carboxylic acids to substitute an $\alpha$-hydrogen atom,resulting in $\alpha$-chloro carboxylic acid,not an acid chloride.
$CH_3COOH + Cl_2 \xrightarrow{\text{Red } P} CH_2ClCOOH + HCl$

(A) The reaction involves the alkaline hydrolysis (saponification) of an ester followed by acidification.
$C_6H_5COOC_2H_5 + KOH \text{ (alc)} \to C_6H_5COOK + C_2H_5OH$
Upon adding $HCl$,the potassium salt of the acid is converted into the corresponding carboxylic acid:
$C_6H_5COOK + HCl \to C_6H_5COOH (\text{white solid}) + KCl$
Since $C_6H_5COOH$ (benzoic acid) is a white solid that precipitates upon acidification,the starting compound is ethyl benzoate.

(C) The reaction of alcohols and carboxylic acids with $PCl_5$ (phosphorus pentachloride) results in the substitution of the hydroxyl group $(-OH)$ or the carboxyl group $(-COOH)$ with a chlorine atom $(-Cl)$.
For alcohols: $R-OH + PCl_5 \to R-Cl + POCl_3 + HCl$
For carboxylic acids: $R-COOH + PCl_5 \to R-COCl + POCl_3 + HCl$
Therefore,the correct reagent is phosphorus pentachloride.

(D) $Benedict's$ solution is a mild oxidizing agent that reacts with aldehydes and $\alpha$-hydroxy ketones (like glucose) to form a red precipitate of $Cu_2O$.
Acetic anhydride is a derivative of a carboxylic acid and does not contain a free aldehyde group,therefore it does not reduce $Benedict's$ solution.

(A) When acetic acid $(CH_3COOH)$ reacts with phosphorus pentachloride $(PCl_5)$,the hydroxyl group $(-OH)$ of the carboxylic acid is replaced by a chlorine atom.
The reaction is:
$CH_3COOH + PCl_5 \to CH_3COCl + POCl_3 + HCl$
The products formed are acetyl chloride $(CH_3COCl)$,phosphoryl chloride $(POCl_3)$,and hydrogen chloride $(HCl)$.

(B) $CH_3COOC_2H_5$ reacts with two equivalents of $C_2H_5MgBr$.
The first equivalent converts the ester to a ketone $(CH_3COCH_2CH_3)$.
The second equivalent reacts with the ketone to form a tertiary alcohol,$3-methylpentan-3-ol$.
Reaction: $CH_3COOC_2H_5 + 2C_2H_5MgBr \xrightarrow{H_2O} CH_3-C(OH)(C_2H_5)_2 + C_2H_5OH$.

(A) Vinegar is defined as a $6 - 10 \%$ aqueous solution of acetic acid $(CH_3COOH)$. Therefore,the correct option is $A$.

(A) Esters are known for their characteristic pleasant,fruity odors. Many esters are used in the perfume and flavoring industries due to this property,for example,benzyl acetate.

(B) When $2-$hydroxybenzoic acid (salicylic acid) is distilled with zinc dust,the phenolic $-OH$ group is removed by reduction,resulting in the formation of benzoic acid $(C_6H_5COOH)$.

(C) Step $I$: Ethyl acetate $(CH_3CO_2C_2H_5)$ undergoes Claisen condensation in the presence of sodium ethoxide $(NaOC_2H_5)$ to form ethyl acetoacetate $(CH_3COCH_2COOC_2H_5)$,which is compound $A$.
Step $II$: Ethyl acetoacetate,when heated in the presence of an acid,undergoes decarboxylation and dehydration/rearrangement to form the cyclic compound shown in the provided image (a derivative of cyclobutanone or similar cyclic structure as per the reaction mechanism provided in the image).

(B) The $pK_a$ value is inversely proportional to the acid dissociation constant $(K_a)$,i.e.,$pK_a = -\log(K_a)$.
Therefore,the acid with the highest $K_a$ value will have the lowest $pK_a$ value.
Among the given carboxylic acids,formic acid $(HCOOH)$ is the strongest acid because the hydrogen atom attached to the carboxyl group exerts no electron-donating inductive effect ($+I$ effect),whereas alkyl groups in other options ($CH_3-$,$CH_3CH_2-$,$(CH_3)_2CH-$) exert a $+I$ effect,which destabilizes the carboxylate anion and decreases acidity.
Thus,$HCOOH$ has the highest $K_a$ and the lowest $pK_a$ value.

(C) The reaction of a carboxylic acid with soda lime $(NaOH + CaO)$ is known as decarboxylation.
In this process,the carboxylic acid loses a molecule of $CO_2$ to form an alkane with one carbon atom less than the parent acid.
To obtain ethane ($C_2H_6$,which has $2$ carbon atoms),the starting carboxylic acid must have $3$ carbon atoms.
Therefore,$X$ is propanoic acid $(C_2H_5COOH)$.
Reaction: $C_2H_5COOH + NaOH \xrightarrow{CaO} C_2H_6 + Na_2CO_3$.

(A) The hydrolysis of acid derivatives depends on the leaving group ability of the group attached to the acyl group $(RCO-)$.
The leaving group ability follows the order: $Cl^- > RCOO^- > RO^- > NH_2^-$.
Therefore,the reactivity towards nucleophilic acyl substitution (hydrolysis) follows the order:
$RCOCl > (RCO)_2O > RCOOR > RCONH_2$.
Thus,the correct order is $I > IV > II > III$.

(D) Carboxylic acids undergo dehydration to form acid anhydrides when heated with strong dehydrating agents such as phosphorus pentoxide $(P_2O_5)$.
$2R-COOH \xrightarrow{P_2O_5, \Delta} (RCO)_2O + H_2O$

(D) The acidity of a proton depends on the stability of the conjugate base formed after the removal of the proton.
In $(CH_3CO)_3CH$,the central carbon is attached to three electron-withdrawing acetyl groups $(-COCH_3)$.
Upon removal of the proton,the resulting carbanion is stabilized by resonance across three carbonyl groups,making it the most stable conjugate base among the given options.
Therefore,$(CH_3CO)_3CH$ has the most acidic proton.

(D) $(d)$ Carboxylic acids are more acidic than phenols and alcohols because the carboxylate ion, which is the conjugate base of a carboxylic acid, is stabilized by resonance. In the carboxylate ion, the negative charge is delocalized over two electronegative oxygen atoms, making it significantly more stable than the phenoxide or alkoxide ions.

(D) The reaction between propionic acid and sodium bicarbonate is an acid-base reaction:
$CH_3CH_2COOH_{(aq)} + NaHCO_{3(aq)} \to CH_3CH_2COONa_{(aq)} + H_2O_{(l)} + CO_{2(g)}$
In this reaction,the carboxylic acid $(R-COOH)$ acts as a proton donor,and the bicarbonate ion $(HCO_3^-)$ acts as a base.
The $HCO_3^-$ ion accepts a proton to form carbonic acid $(H_2CO_3)$,which then decomposes into $H_2O$ and $CO_2$.
Therefore,the carbon atom in the evolved $CO_2$ gas originates from the bicarbonate ion $(NaHCO_3)$.
Thus,the correct option is $(D)$.

(B) The boiling point depends on the intermolecular forces present in the compounds.
$CH_3CH_2CH_2COOH$ $(3)$ contains a carboxylic acid group,which forms strong intermolecular hydrogen bonds,often existing as dimers,leading to the highest boiling point.
$CH_3CH_2CH_2CH_2OH$ $(1)$ contains a hydroxyl group,which also forms intermolecular hydrogen bonds,but they are generally weaker than those in carboxylic acids.
$CH_3CH_2CH_2CHO$ $(2)$ contains an aldehyde group,which is polar but cannot form intermolecular hydrogen bonds,resulting in the lowest boiling point among the three.
Therefore,the correct order of boiling points is $3 > 1 > 2$.

(D) The iodoform test is given by compounds containing either a methyl ketone group $(CH_3-CO-)$ or a methyl carbinol group $(CH_3-CH(OH)-)$.
$1.$ Acetone $(CH_3-CO-CH_3)$ and Ethyl alcohol $(CH_3-CH_2-OH)$ both give a positive iodoform test.
$2.$ Acetic acid $(CH_3-COOH)$ does not give the iodoform test because the hydroxyl group attached to the carbonyl carbon prevents the reaction.

(B) When acetamide $(CH_3CONH_2)$ is heated in the presence of a dehydrating agent like $P_2O_5$,it undergoes dehydration to form acetonitrile $(CH_3CN)$.
Reaction: $CH_3CONH_2 \xrightarrow{P_2O_5} CH_3-C\equiv N + H_2O$
Therefore,the correct option is $B$.

(A) The compound $(CH_3)_2NCOCH_3$ is an $N,N$-dimethylacetamide,which is an amide.
Acid-catalyzed hydrolysis (refluxing with acid) of an amide breaks the $C-N$ bond to yield a carboxylic acid and an amine (or its salt).
The reaction is: $(CH_3)_2NCOCH_3 + H_2O \xrightarrow{H^+} (CH_3)_2NH + CH_3COOH$.
Thus,the products are dimethylamine $(CH_3)_2NH$ and acetic acid $CH_3COOH$.

(D) When benzamide $(C_6H_5CONH_2)$ is treated with a dehydrating agent like phosphorus oxychloride $(POCl_3)$,it undergoes dehydration to form benzonitrile $(C_6H_5CN)$.
The reaction is as follows:
$C_6H_5CONH_2 \xrightarrow{POCl_3, \Delta} C_6H_5CN + H_2O$
Therefore,the correct option is $(D)$.

(C) $Methyl \ salicylate$ is commonly known as oil of wintergreen.
It is used as a medicine,specifically in topical ointments for pain relief due to its analgesic and anti-inflammatory properties.