Properties of Carboxylic Acids and Their Derivatives Questions in English

(A) The reaction of salicylic acid with acetic anhydride in the presence of an acid catalyst (or base) is known as acetylation.
This reaction converts the phenolic $-OH$ group of salicylic acid into an acetyl group,resulting in the formation of acetylsalicylic acid,which is commonly known as $Aspirin$.
The chemical equation is:
$C_6H_4(OH)COOH + (CH_3CO)_2O \rightarrow C_6H_4(OCOCH_3)COOH + CH_3COOH$
Thus,the correct option is $A$.

(B) The $pK_a$ value is inversely proportional to the acid strength. $A$ higher $pK_a$ value indicates a weaker acid.
Comparing the given acids:
$FCH_2COOH$ is the strongest due to the strong $-I$ effect of $F$.
$ClCH_2COOH$ is the next strongest due to the $-I$ effect of $Cl$.
$HCOOH$ is stronger than $CH_3COOH$ because the $CH_3$ group in $CH_3COOH$ exerts a $+I$ effect,which destabilizes the carboxylate ion.
Therefore,$CH_3COOH$ is the weakest acid among the given options and has the highest $pK_a$ value.

(A) The reaction of acetamide $(CH_3CONH_2)$ with nitrous acid $(HNO_2)$ results in the formation of acetic acid,nitrogen gas,and water.
The chemical equation is: $CH_3CONH_2 + HNO_2 \rightarrow CH_3COOH + N_2 + H_2O$.

(A) The reaction between a carboxylic acid $(RCOOH)$ and an aromatic amine ($H_2N\phi$,where $\phi = C_6H_5$) is an acylation reaction.
The product formed is an amide,specifically an $N$-substituted amide known as an anilide.
The general structure $RCONH\phi$ represents an anilide.

(C) The reaction between an acid chloride and an amine is an acylation reaction.
$N, N$-Dimethylacetamide has the structure $CH_3CON(CH_3)_2$.
This compound is formed by the reaction of acetyl chloride $(CH_3COCl)$ with dimethylamine $((CH_3)_2NH)$.
The reaction is: $CH_3COCl + (CH_3)_2NH \rightarrow CH_3CON(CH_3)_2 + HCl$.

(C) The reaction involves the conversion of an amide $({C_6}{H_5}CONH_2)$ to a nitrile $({C_6}{H_5}CN)$ using $PCl_5$.
$PCl_5$ acts as a dehydrating agent here,removing a molecule of water $(H_2O)$ from the amide group $(-CONH_2)$ to form the nitrile group $(-CN)$.
Since a water molecule is removed,this process is known as dehydration.

(A) Decarboxylation of $2, 4, 6-$ trinitrobenzoic acid involves the removal of the $-COOH$ group as $CO_2$ upon heating.
The reaction is: $C_6H_2(NO_2)_3COOH \xrightarrow{\Delta} C_6H_3(NO_2)_3 + CO_2$.
The product formed is $2, 4, 6-$ trinitrotoluene $(TNT)$,which is a well-known explosive.

(B) When acetamide $(CH_3CONH_2)$ is heated with phosphorus pentoxide $(P_2O_5)$,it undergoes dehydration to form acetonitrile $(CH_3CN)$.
$CH_3CONH_2 \xrightarrow{P_2O_5, \Delta} CH_3CN + H_2O$

(C) When an ester $(RCOOR')$ reacts with an excess of Grignard reagent $(R''MgX)$,it first forms a ketone,which then reacts further with another molecule of the Grignard reagent to form a tertiary alcohol.
For an ethyl ester $(RCOOCH_2CH_3)$,the reaction with excess $CH_3MgBr$ proceeds as follows:
$1$. $RCOOCH_2CH_3 + CH_3MgBr \rightarrow RCOCH_3 + CH_3CH_2OMgBr$
$2$. $RCOCH_3 + CH_3MgBr \rightarrow R-C(OH)(CH_3)_2$
Assuming the ethyl ester is ethyl acetate $(CH_3COOCH_2CH_3)$:
$CH_3COOCH_2CH_3 + 2CH_3MgBr \rightarrow CH_3-C(OH)(CH_3)_2 + CH_3CH_2OMgBr$
The product is $2$-methylpropan-$2$-ol (tert-butyl alcohol).
However,if the ester is ethyl propionate $(CH_3CH_2COOCH_2CH_3)$:
$CH_3CH_2COOCH_2CH_3 + 2CH_3MgBr \rightarrow CH_3CH_2-C(OH)(CH_3)_2 + CH_3CH_2OMgBr$
The product is $2$-methylbutan-$2$-ol.
Given the options provided,option $C$ represents $2$-methylbutan-$2$-ol,which corresponds to the reaction of ethyl propionate with excess $CH_3MgBr$.

(D) The reaction sequence is as follows:
$1$. $CH_3Cl + KCN \rightarrow CH_3CN (A) + KCl$ (Nucleophilic substitution)
$2$. $CH_3CN + H_3O^+ \rightarrow CH_3COOH (B)$ (Acidic hydrolysis of nitrile)
$3$. $CH_3COOH + NH_3 \rightarrow CH_3COONH_4 (C)$ (Formation of ammonium acetate salt)
$4$. $CH_3COONH_4 \xrightarrow{\Delta} CH_3CONH_2 (D) + H_2O$ (Dehydration to form acetamide)
Thus,the final product $D$ is $CH_3CONH_2$.

(A) In benzoic acid,the $-COOH$ group is a deactivating and $meta$-directing group for electrophilic aromatic substitution reactions.
Therefore,the nitration of benzoic acid using a mixture of concentrated $HNO_3$ and concentrated $H_2SO_4$ yields $3$-nitrobenzoic acid as the major product.

(A) In electrophilic aromatic substitution,the orientation of the incoming group depends on the nature of the substituent already present on the benzene ring.
Groups that are electron-withdrawing by inductive or resonance effects (deactivating groups) are meta-directing.
Among the given options,$-COOH$ is an electron-withdrawing group due to the presence of the carbonyl group attached to the benzene ring,which makes it a meta-directing group.
$-Cl$ is ortho/para-directing due to the resonance effect.
$-NH_2$ and $-CH_3$ are ortho/para-directing due to their electron-donating nature.
Therefore,the correct answer is $-COOH$.

(B) Pyrolysis of ethyl acetate $(CH_3COOCH_2CH_3)$ at high temperatures involves a syn-elimination reaction through a cyclic transition state.
This process yields ethene $(CH_2 = CH_2)$ and acetic acid $(CH_3COOH)$.
The reaction is represented as:
$CH_3COOCH_2CH_3 \xrightarrow{\Delta} CH_2 = CH_2 + CH_3COOH$.

(D) The acidity of organic compounds depends on the stability of their conjugate bases.
$1$. $PhSO_3H$ (Benzenesulfonic acid) is a strong acid because the negative charge on the conjugate base is delocalized over three oxygen atoms.
$2$. $PhCO_2H$ (Benzoic acid) is a carboxylic acid,which is more acidic than phenols due to resonance stabilization of the carboxylate ion.
$3$. $PhOH$ (Phenol) is acidic because the phenoxide ion is resonance-stabilized by the benzene ring.
$4$. $PhCH_2OH$ (Benzyl alcohol) is the least acidic as the negative charge on the alkoxide ion is not resonance-stabilized.
Thus,the correct order of decreasing acidity is $PhSO_3H > PhCO_2H > PhOH > PhCH_2OH$.

(A) The acidic strength of carboxylic acids depends on the stability of the conjugate base (carboxylate ion). Electron-withdrawing groups $(EWG)$ increase acidity by stabilizing the negative charge,while electron-donating groups $(EDG)$ decrease acidity by destabilizing it.
$I. CH_3CO_2H$: Reference compound.
$II. MeOCH_2CO_2H$: Contains $-OCH_3$ group,which is an $EWG$ due to the inductive effect ($-I$ effect),making it more acidic than $I$.
$III. CF_3CO_2H$: Contains three fluorine atoms,which are strongly electron-withdrawing ($-I$ effect),making it the most acidic.
$IV. (CH_3)_2CHCO_2H$: Contains two methyl groups,which are electron-donating ($+I$ effect),making it less acidic than $I$.
Comparing the effects: The order of acidity is $IV < I < II < III$.

(A) The acidity of carboxylic acids is determined by the stability of the conjugate base (carboxylate ion).
Electron-withdrawing groups $(EWG)$ like $-Cl$ increase acidity by stabilizing the negative charge on the carboxylate ion through the inductive effect ($-I$ effect).
The strength of the $-I$ effect decreases with distance.
In $CH_3CH_2CH(Cl)CO_2H$,the $-Cl$ group is on the $\alpha$-carbon,which is closer to the $-COOH$ group compared to $ClCH_2CH_2COOH$ (where it is on the $\beta$-carbon).
Therefore,$CH_3CH_2CH(Cl)CO_2H$ is the strongest acid among the given options.

(D) The acidity of carboxylic acids is increased by electron-withdrawing groups $(EWG)$ attached to the ring.
In $p-nitrobenzoic$ acid $(O_2N-C_6H_4-COOH)$,the $-NO_2$ group exerts both a strong $-I$ (inductive) effect and a $-M$ (mesomeric) effect.
These effects stabilize the carboxylate anion by withdrawing electron density,making it the most acidic among the given options.

(B) The acidity of carboxylic acids depends on the stability of the conjugate base (carboxylate ion).
Electron-withdrawing groups ($-I$ effect) stabilize the conjugate base and increase acidity.
The hybridization of the carbon atom attached to the carboxyl group determines the $-I$ effect strength: $sp > sp^2 > sp^3$.
In $(III)$,the carbon is $sp$ hybridized (strongest $-I$ effect).
In $(II)$,the carbon is $sp^2$ hybridized.
In $(I)$,the carbon is $sp^3$ hybridized (weakest $-I$ effect).
Therefore,the decreasing order of acidity is $III > II > I$.

(C) The $pK_a$ value is inversely proportional to the acidic strength of the acid.
Stronger acids have lower $pK_a$ values.
Acidic strength depends on the stability of the conjugate base (carboxylate ion).
Electron-donating groups (like alkyl groups) destabilize the carboxylate ion by increasing the negative charge density,thereby decreasing acidic strength.
$HCOOH$ has no alkyl group attached to the carboxyl group,whereas the others have alkyl groups ($CH_3-$,$CH_3CH_2-$,$(CH_3)_2CH-$) which are electron-donating via the $+I$ effect.
Therefore,$HCOOH$ is the strongest acid among the given options and has the lowest $pK_a$ value.

(C) To determine the acidity,we compare the stability of the conjugate base formed after the loss of a proton $(H^+)$.
$1$. $p$-Nitrophenol is a phenol derivative,which is generally less acidic than carboxylic acids.
$2$. Among the carboxylic acids ($p$-Hydroxybenzoic acid,$o$-Hydroxybenzoic acid,and $p$-Toluic acid),the acidity depends on the inductive and resonance effects of the substituents.
$3$. $o$-Hydroxybenzoic acid (Salicylic acid) exhibits the ortho effect and intramolecular hydrogen bonding,which stabilizes the carboxylate anion significantly.
$4$. Comparing the substituents: $-OH$ at the ortho position in $o$-hydroxybenzoic acid provides strong stabilization compared to the para position in $p$-hydroxybenzoic acid due to the proximity of the hydroxyl group to the carboxylate group.
$5$. Therefore,$o$-Hydroxybenzoic acid is the most acidic among the given options.

(D) In nucleophilic acyl substitution reactions,the rate of reaction depends on the ability of the group $Z$ to act as a leaving group.
The better the leaving group,the faster the reaction.
$A$ good leaving group is the conjugate base of a strong acid,which means it is a weak base.
Comparing the basicity of the leaving groups:
$Cl^-$ is the conjugate base of $HCl$ (a strong acid).
$OCOCH_3^-$ is the conjugate base of $CH_3COOH$ (a weak acid).
$OC_2H_5^-$ is the conjugate base of $C_2H_5OH$ (a very weak acid).
$NH_2^-$ is the conjugate base of $NH_3$ (an extremely weak acid).
Since $Cl^-$ is the weakest base among the given options,it is the best leaving group,making the reaction fastest when $Z = Cl$.

(A) The acidic strength of carboxylic acids is inversely proportional to the $+I$ (inductive) effect of the alkyl group attached to the carboxylate group.
$HCOOH$ has no alkyl group (only $H$),while $CH_3COOH$ has a methyl group ($+I$ effect) and $C_2H_5COOH$ has an ethyl group (stronger $+I$ effect than methyl).
As the $+I$ effect increases,the stability of the conjugate base decreases,thereby decreasing the acidic strength.
Therefore,the correct order is $HCOOH > CH_3COOH > C_2H_5COOH$.

(C) The acidic strength of carboxylic acids depends on the stability of the carboxylate ion formed after the loss of a proton.
Electron-donating groups (like alkyl groups) decrease acidity by destabilizing the carboxylate ion via the $+I$ effect.
Electron-withdrawing groups (like the phenyl group) increase acidity by stabilizing the carboxylate ion via the $-I$ and resonance effects.
Comparing the compounds:
$(I) HCOOH$: No alkyl group attached.
$(II) CH_3COOH$: Contains one methyl group ($+I$ effect).
$(III) CH_3CH_2COOH$: Contains an ethyl group ($+I$ effect,stronger than methyl).
$(IV) C_6H_5COOH$: Contains a phenyl group ($-I$ and resonance effect).
Order of acidity: $HCOOH > C_6H_5COOH > CH_3COOH > CH_3CH_2COOH$.
Thus,the correct order is $(I) > (IV) > (II) > (III)$.

(C) The strength of carboxylic acids is determined by the stability of the conjugate base (carboxylate ion).
Electron-withdrawing groups (EWGs) like $Cl$ increase the acidity by stabilizing the negative charge through the inductive effect ($-I$ effect).
Comparing the given acids:
$1$. $CH_3CH_2COOH$ (Propanoic acid): Contains an electron-donating alkyl group,which decreases acidity.
$2$. $ClCH_2CH_2COOH$ ($3$-Chloropropanoic acid): The $Cl$ atom is at the $\gamma$-position relative to the carboxyl group,so its $-I$ effect is weak.
$3$. $ClCH_2COOH$ (Chloroacetic acid): The $Cl$ atom is at the $\alpha$-position,exerting a strong $-I$ effect,which significantly stabilizes the carboxylate ion.
$4$. $CH_3COOH$ (Acetic acid): Contains a methyl group,which is less electron-donating than the ethyl group in propanoic acid.
Therefore,$ClCH_2COOH$ is the strongest acid among the given options.

(C) The $pK_a$ value is inversely proportional to the acidic strength of the carboxylic acid.
Stronger acids have lower $pK_a$ values,while weaker acids have higher $pK_a$ values.
The acidic strength order is: $CH_2FCOOH > CH_2ClCOOH > HCOOH > CH_3COOH$.
Since $CH_3COOH$ is the weakest acid among the given options due to the $+I$ effect of the methyl group,it has the maximum $pK_a$ value.

(D) The acidic strength of carboxylic acids is directly proportional to the electron-withdrawing effect ($-I$ effect) of the substituents attached to the alpha-carbon.
As the number of chlorine atoms increases,the $-I$ effect increases,which stabilizes the carboxylate anion by dispersing the negative charge.
Therefore,the order of acidic strength is: $CH_3COOH < CH_2ClCOOH < CHCl_2COOH < CCl_3COOH$.
Thus,$CCl_3COOH$ is the strongest acid.

(D) The stability of a carboxylate ion $(R-COO^-)$ is increased by the presence of electron-withdrawing groups (EWGs) attached to the alpha carbon.
These groups stabilize the negative charge through the inductive effect ($-I$ effect) by dispersing the charge density.
Comparing the given options:
$A$: $CH_3-COO^-$ (contains an electron-donating methyl group,which destabilizes the ion).
$B$: $Cl-CH_2-COO^-$ (contains one $Cl$ atom,which has a $-I$ effect).
$C$: $F-CH_2-COO^-$ (contains one $F$ atom,which has a stronger $-I$ effect than $Cl$).
$D$: $F_2CH-COO^-$ (contains two $F$ atoms,which exert a much stronger combined $-I$ effect than a single $F$ or $Cl$ atom).
Since the $-I$ effect is additive and distance-dependent,the presence of two fluorine atoms provides the greatest dispersal of the negative charge,making $F_2CH-COO^-$ the most stable carboxylate ion.

(D) The acidity of substituted benzoic acids is determined by the electronic effects of the substituents.
$1$. The $-NO_2$ group is a strong electron-withdrawing group ($-I$ and $-M$ effect).
$2$. In $(II)$ $o-NO_2C_6H_4COOH$,the ortho effect significantly increases the acidity compared to other isomers.
$3$. In $(III)$ $p-NO_2C_6H_4COOH$,the $-NO_2$ group exerts both $-I$ and $-M$ effects,making it more acidic than $(IV)$ $m-NO_2C_6H_4COOH$,where only the $-I$ effect operates.
$4$. $(I)$ $PhCOOH$ has no electron-withdrawing group,so it is the least acidic.
Therefore,the correct order of acidic strength is $(II) > (III) > (IV) > (I)$.

(C) The acidity of carboxylic acids is determined by the stability of the conjugate base (carboxylate ion).
Electron-withdrawing groups ($-I$ effect) stabilize the carboxylate ion and increase acidity,while electron-donating groups (like the methyl group) decrease it.
The strength of the $-I$ effect follows the order: $F > Cl > Br > I$.
Therefore,the acidity order is $FCH_2COOH > ClCH_2COOH > BrCH_2COOH > CH_3COOH$.

(C) $Methyl \ salicylate$ is commonly known as oil of wintergreen. It is used in medicinal preparations,particularly in ointments and balms,for its analgesic and anti-inflammatory properties to relieve muscle and joint pain.

(A) The reaction of salicylic acid $(C_7H_6O_3)$ with acetic anhydride $((CH_3CO)_2O)$ in the presence of an acid catalyst is known as acetylation.
This reaction produces acetylsalicylic acid,which is commonly known as $Aspirin$ and acetic acid as a byproduct.
The chemical equation is: $C_6H_4(OH)COOH + (CH_3CO)_2O \rightarrow C_6H_4(OCOCH_3)COOH + CH_3COOH$.

(C) The reaction between ethanol and acetic acid in the presence of an acid catalyst $(H^+)$ is an esterification reaction:
$C_2H_5OH + CH_3COOH \xrightarrow{H^+} CH_3COOC_2H_5 + H_2O$
The product formed is ethyl acetate,which is an ester.
Esters are known for their characteristic fruity,sweet smell.

(A) Carboxylic acids are reduced to primary alcohols using strong reducing agents like lithium aluminium hydride $(LiAlH_4)$.
The reaction is: $CH_3COOH + 4[H] \xrightarrow{LiAlH_4} CH_3CH_2OH + H_2O$.

(A) The reaction between an alcohol and a carboxylic acid produces an ester. Therefore,this reaction is called esterification.
$RCOOH + R'OH \to RCOOR' + H_2O$
$\text{Carboxylic acid} + \text{Alcohol} \to \text{Ester} + \text{Water}$

(D) Carboxylic acids are reduced to primary alcohols by strong reducing agents like lithium aluminium hydride $(LiAlH_4)$.
The reaction is:
$CH_3CH_2COOH \xrightarrow{LiAlH_4} CH_3CH_2CH_2OH$
Thus,$LiAlH_4$ is the correct reagent.

(B) The boiling point depends on the strength of intermolecular forces,primarily hydrogen bonding and dipole-dipole interactions.
$III$ $(CH_3CH_2CH_2COOH)$ is a carboxylic acid,which forms strong intermolecular hydrogen bonds (dimers),leading to the highest boiling point.
$I$ $(CH_3CH_2CH_2CH_2OH)$ is an alcohol,which also forms intermolecular hydrogen bonds,but they are generally weaker than those in carboxylic acids.
$II$ $(CH_3CH_2CH_2CHO)$ is an aldehyde,which exhibits only dipole-dipole interactions and lacks hydrogen bonding,resulting in the lowest boiling point.
Therefore,the correct order is $III > I > II$.

(B) Carboxylic acids,such as acetic acid $(CH_3COOH)$,are stronger acids than carbonic acid $(H_2CO_3)$.
Therefore,they react with sodium bicarbonate $(NaHCO_3)$ to evolve $CO_2$ gas.
Phenol and $n$-hexanol are much weaker acids than carbonic acid and do not react with $NaHCO_3$ to release $CO_2$.

(B) The starting material is $1,2-bis(hydroxymethyl)benzene$ (also known as $o-xylene$ glycol).
$1$. Vigorous oxidation of $1,2-bis(hydroxymethyl)benzene$ using a strong oxidizing agent (like $KMnO_4$) converts the two $-CH_2OH$ groups into $-COOH$ groups,forming $X$,which is Phthalic acid $(benzene-1,2-dicarboxylic \ acid)$.
$2$. Upon dry heating,Phthalic acid undergoes dehydration to form $Z$,which is Phthalic anhydride.
The reaction sequence is:
$1,2-bis(hydroxymethyl)benzene$ $\xrightarrow[\text{Oxidation}]{} \text{Phthalic acid} (X)$ $\xrightarrow[\text{Heating}]{} \text{Phthalic anhydride} (Z) + H_2O$.

(B) The reaction where phenyl esters undergo rearrangement in the presence of anhydrous $AlCl_3$ to yield $o-$ and $p-$ hydroxy ketones is known as the Fries rearrangement.
The reaction is as follows:
$Phenyl \ acetate \xrightarrow{Anhyd. AlCl_3, \Delta} o-Hydroxyacetophenone + p-Hydroxyacetophenone$

(D) Methyl salicylate is commonly known as oil of wintergreen. Its chemical structure is $C_6H_4(OH)(COOCH_3)$.

(B) The reaction of salicylic acid with methyl alcohol in the presence of $H_2SO_4$ (esterification) produces methyl salicylate,which is known as oil of wintergreen.

(C) The reaction is a Knoevenagel condensation followed by decarboxylation.
$CH_3CHO + CH_2(COOH)_2$ $\xrightarrow{\Delta} CH_3CH=C(COOH)_2$ $\xrightarrow{-\text{CO}_2} CH_3CH=CHCOOH$ (Crotonic acid).
Thus,the product $X$ is $CH_3CH=CHCOOH$.

(C) The reaction of acetaldehyde with $HCN$ forms a cyanohydrin,which upon hydrolysis yields lactic acid $(2-hydroxypropanoic \ acid)$.
$CH_3CHO + HCN \to CH_3CH(OH)CN \xrightarrow{H_3O^+} CH_3CH(OH)COOH$.
Lactic acid contains a chiral carbon atom (the central carbon bonded to $-H, -OH, -CH_3, -COOH$ groups).
Due to the presence of this chiral center,lactic acid exists as two non-superimposable mirror images,thus exhibiting enantiomerism.

(D) Phenols are less acidic than carboxylic acids and do not react with $NaHCO_3$ to evolve $CO_2$ gas.
Carboxylic acids react with $NaHCO_3$ to form water-soluble sodium carboxylate salts.
Therefore,$NaHCO_3$ can be used to extract the carboxylic acid $(B)$ from the mixture,leaving the phenol $(A)$ in the organic phase.

(C) The reaction of acetic acid with $P_2O_5$ (a dehydrating agent) leads to the formation of ethanoic anhydride.
$2CH_3COOH \xrightarrow{P_2O_5, \Delta} (CH_3CO)_2O + H_2O$
Here,$CH_3COOH$ is acetic acid and $(CH_3CO)_2O$ is ethanoic anhydride.

(B) The reaction of carboxylic acids having an $\alpha$-hydrogen with bromine or chlorine in the presence of a small amount of red phosphorus to produce $\alpha$-halo carboxylic acids is known as the $Hell-Volhard-Zelinsky$ $(HVZ)$ reaction.

(D) Pure acetic acid has a melting point of $16.6\,^{\circ}C$.
At temperatures below this,it forms ice-like crystals,which is why it is known as glacial acetic acid.

(D) Oxalic acid undergoes dehydration in the presence of concentrated $H_2SO_4$ to produce carbon monoxide,carbon dioxide,and water.
The chemical reaction is as follows:
$(COOH)_2 \xrightarrow{conc. H_2SO_4} CO + CO_2 + H_2O$

(C) The reaction of acetic acid $(CH_3COOH)$ with phosphorus pentachloride $(PCl_5)$ yields acetyl chloride $(CH_3COCl)$ along with phosphorus oxychloride $(POCl_3)$ and hydrogen chloride $(HCl)$.
The chemical equation is:
$CH_3COOH + PCl_5 \to CH_3COCl + POCl_3 + HCl$

(C) Decarboxylation of carboxylic acids with soda lime $(NaOH + CaO)$ removes one carbon atom from the chain as $Na_2CO_3$.
To obtain ethane $(C_2H_6)$,we need a carboxylic acid with three carbon atoms.
Propanoic acid $(CH_3CH_2COOH)$ undergoes decarboxylation to form ethane $(CH_3CH_3)$:
$CH_3CH_2COOH + NaOH \xrightarrow{CaO} CH_3CH_3 + Na_2CO_3$.