Properties of Carboxylic Acids and Their Derivatives Questions in English

(A) The acidity of carboxylic acids increases with the presence of electron-withdrawing groups due to the $-I$ effect.
The $-NO_2$ group is a strong electron-withdrawing group.
Since the $-I$ effect decreases with distance,the acid where the $-NO_2$ group is closest to the $-COOH$ group (at the $\alpha$-position) will be the most acidic and thus most highly ionized in water.
In $CH_3-CH_2-CH(NO_2)-COOH$,the $-NO_2$ group is at the $\alpha$-position relative to the $-COOH$ group.
Therefore,$CH_3-CH_2-CH(NO_2)-COOH$ is the most highly ionized.

(C) Oxalic acid $(H_2C_2O_4)$ undergoes dehydration when treated with concentrated sulfuric acid $(conc. H_2SO_4)$.
The reaction is as follows:
$H_2C_2O_4 \xrightarrow{conc. H_2SO_4} CO + CO_2 + H_2O$
Concentrated $H_2SO_4$ acts as a dehydrating agent,removing a water molecule from the oxalic acid,resulting in the formation of carbon monoxide $(CO)$ and carbon dioxide $(CO_2)$ gases.

(C) Aspirin (acetylsalicylic acid) is prepared by the acetylation of the phenolic hydroxyl group of salicylic acid.
Salicylic acid reacts with acetic acid $(CH_3COOH)$ or acetic anhydride $((CH_3CO)_2O)$ in the presence of an acid catalyst to form aspirin.
Reaction: $C_6H_4(OH)COOH + CH_3COOH \rightarrow C_6H_4(OCOCH_3)COOH + H_2O$.

(C) Ring strain is inversely proportional to the size of the ring. Smaller rings have higher angle strain due to significant deviation from the ideal tetrahedral bond angle of $109.5^{\circ}$. Among the given options,the $\beta$-lactone is a four-membered ring,which possesses the highest angle strain compared to the five-membered ($\gamma$-lactone) and six-membered ($\delta$-lactone/lactam) rings. Therefore,the $\beta$-lactone is the most highly strained.

(B) The acidic strength of carboxylic acids is directly proportional to the electron-withdrawing effect ($-I$ effect) of the substituents attached to the carbon chain.
$1$. Acetic acid $(CH_3COOH)$ has a $+I$ group $(CH_3)$,which decreases acidity.
$2$. Dichloroacetic acid $(Cl_2CHCOOH)$ has two highly electronegative chlorine atoms,which exert a strong $-I$ effect,significantly stabilizing the carboxylate anion and increasing acidity.
$3$. $4-$Iodocyclohexanecarboxylic acid and $4-$Chlorocyclohexanecarboxylic acid have the electron-withdrawing groups far from the carboxylic acid group,resulting in a much weaker $-I$ effect compared to dichloroacetic acid.
Therefore,dichloroacetic acid is the strongest acid among the given options,meaning it has the highest $K_a$ and lowest $pK_a$ value.

(B) The acid strength is determined by the stability of the conjugate base.
$1.$ In $(x) CH_3SO_3H$,the conjugate base $CH_3SO_3^-$ is highly stabilized by resonance over three oxygen atoms,making it the strongest acid among the three.
$2.$ In $(y) CH_3COOH$,the conjugate base $CH_3COO^-$ is stabilized by resonance over two oxygen atoms,making it a weaker acid than $(x)$.
$3.$ In $(z) CH_3OH$,the conjugate base $CH_3O^-$ has no resonance stabilization,making it the weakest acid.
The order of acid strength is $x > y > z$.
Since $pK_a$ is inversely proportional to acid strength $(pK_a = -\log K_a)$,the increasing order of $pK_a$ values is $x < y < z$.

(C) The solubility of carboxylic acids in $H_2O$ depends on the size of the hydrophobic hydrocarbon chain.
Smaller carboxylic acids are more soluble because they can form stronger hydrogen bonds with water relative to their non-polar part.
Comparing the given acids:
$1$. Malonic acid $(HOOC-CH_2-COOH)$ has $3$ carbons.
$2$. Succinic acid $(HOOC-CH_2-CH_2-COOH)$ has $4$ carbons.
$3$. Phthalic acid $(C_6H_4(COOH)_2)$ has $8$ carbons.
$4$. Salicylic acid $(C_6H_4(OH)(COOH))$ has $7$ carbons.
Since malonic acid has the smallest hydrocarbon chain,it exhibits the maximum solubility in water.

(A) The acidic strength 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.
In option $(A)$,$CH_3CO_2H$ has no electron-withdrawing group,while $CH_2FCO_2H$ has a highly electronegative $F$ atom,which exerts a strong $-I$ effect.
Therefore,$CH_2FCO_2H$ is significantly stronger than $CH_3CO_2H$.
In other options,the second acid is weaker than the first because the $-I$ effect decreases as the distance of the substituent from the $-CO_2H$ group increases or as the electronegativity of the substituent decreases.

(B) $pK_a$ is inversely proportional to the acidity of the compound $(pK_a \propto 1/K_a)$.
Electron-withdrawing groups $(EWG)$ like $-Cl$ increase acidity (decrease $pK_a$) through the $-I$ effect.
The closer the $EWG$ is to the $-COOH$ group,the stronger the effect.
Electron-donating groups $(EDG)$ like alkyl groups increase $pK_a$ (decrease acidity) through the $+I$ effect.
Among the given options,$CH_3-CH_2-COOH$ has no electron-withdrawing groups,making it the weakest acid and thus having the highest $pK_a$ value.

(A) The hydrogen labeled as $a$ is the most acidic because it is part of a carboxylic acid group $(-COOH)$.
Upon losing a proton,the resulting carboxylate anion $(R-COO^-)$ is stabilized by resonance,where the negative charge is delocalized over two equivalent oxygen atoms,making it significantly more stable than the anions formed by removing hydrogens from the other positions (alcohol $-OH$,thiol $-SH$,or $\alpha$-hydrogen to a carbonyl).

(B) The acidity of a carbon acid depends on the stability of the conjugate base formed after the removal of a proton.
In $CH_2(CO_2Et)_2$ and $CH_3COCH_2COOC_2H_5$,the conjugate base is stabilized by resonance with two carbonyl groups.
However,in bicyclic systems,the acidity is significantly influenced by the geometry.
In bicyclo[$2.2$.$1$]heptane$-2,3-$dione,the two carbonyl groups are adjacent to each other. The removal of a proton from the bridgehead or adjacent carbon creates a carbanion that is stabilized by the inductive and resonance effects of the two adjacent carbonyl groups.
Specifically,the enolate formed from bicyclo[$2.2$.$1$]heptane$-2,3-$dione is highly stabilized due to the proximity of the two carbonyl groups,making it a stronger carbon acid compared to the others listed.

(C) The acidic strength is determined by the stability of the conjugate base.
$(ii)$ $p$-methoxybenzoic acid: The $-OCH_3$ group exerts a strong $+M$ (mesomeric) effect,which destabilizes the carboxylate anion,making it the least acidic.
$(i)$ benzoic acid: This is the reference compound.
$(iii)$ $o$-methoxybenzoic acid: Due to the ortho effect (or Steric Inhibition of Resonance,$SIR$),the $-COOH$ group is forced out of the plane of the benzene ring,which reduces the resonance and increases the acidity significantly compared to benzoic acid.
Therefore,the increasing order of acidic strength is $(ii) < (i) < (iii)$.

(C) The acidic strength is determined by the stability of the conjugate base formed after the loss of a proton $(H^+)$.
$1$. $Ph-CO_2H$ $(C)$: The conjugate base is a carboxylate ion $(Ph-COO^-)$,which is stabilized by resonance with the benzene ring and the two oxygen atoms.
$2$. $Ph-OH$ $(A)$: The conjugate base is a phenoxide ion $(Ph-O^-)$,which is stabilized by resonance with the benzene ring.
$3$. $Ph-CH_2-NH_3^+$ $(D)$: The conjugate base is a neutral amine $(Ph-CH_2-NH_2)$. The positive charge on the nitrogen makes the proton highly acidic.
$4$. $Ph-CH_2-OH$ $(B)$: The conjugate base is an alkoxide ion $(Ph-CH_2-O^-)$,which is destabilized by the electron-donating inductive effect of the benzyl group.
Comparing these,the order of acidity is $C > A > D > B$.

(C) Dimethyl phthalate contains two ester groups $(-COOCH_3)$ attached to a benzene ring.
Each ester group reacts with $2$ moles of Grignard reagent $(CH_3MgI)$.
Step $1$: The first mole of $CH_3MgI$ attacks the carbonyl carbon of the ester to form a ketone intermediate,releasing $CH_3OMgI$.
Step $2$: The second mole of $CH_3MgI$ attacks the ketone intermediate to form a tertiary alcohol after acidic workup.
Since there are two ester groups,the total number of moles of $CH_3MgI$ required is $2 \times 2 = 4$ moles.
Therefore,$x = 4$.

(C) Diethyl carbonate reacts with excess Grignard reagent $(CH_3MgBr)$.
The first two equivalents of $CH_3MgBr$ replace the two ethoxy groups to form acetone $(CH_3-CO-CH_3)$.
The third equivalent then reacts with acetone to form tert-butyl alcohol $(CH_3-C(OH)(CH_3)_2)$ after hydrolysis.

(D) Grignard reagents $(RMgX)$ react with carbonyl groups in aldehydes,ketones,and esters to form alcohols.
However,they react with compounds containing acidic hydrogens (such as $-OH$ in carboxylic acids or phenols) to produce alkanes.
In option $(d)$,$Ph-O-CO-OH$ (phenyl hydrogen carbonate) contains an acidic $-OH$ group,which reacts with ethylmagnesium bromide $(CH_3CH_2MgBr)$ to form ethane $(C_2H_6)$ gas instead of an alcohol.

(B) The reaction of a carboxylic acid with organolithium reagents involves multiple steps due to the acidity of the protons present.
$1$. The first equivalent of $CH_3Li$ reacts with the carboxylic acid proton $(-COOH)$ to form a carboxylate salt $(R-COO^-Li^+)$ and methane $(CH_4)$.
$2$. The second equivalent of $CH_3Li$ reacts with the hydroxyl proton $(-OH)$ of the $HOCH_2-$ group to form an alkoxide salt $(Li^+O^-CH_2-)$.
$3$. The third equivalent of $CH_3Li$ attacks the carbonyl carbon of the carboxylate group to form a gem-diol dianion intermediate.
$4$. Upon workup with $NH_4Cl/H_2O$,the intermediate is protonated to give the final ketone product.
Since the substrate contains both a carboxylic acid group and a primary alcohol group,it requires $3$ equivalents of $CH_3Li$ to fully deprotonate both acidic sites and then attack the carbonyl group. Thus,$(x) = 3$.

(B) The reaction involves a nucleophilic substitution where the organolithium compound (bicyclo[$2.2$.$1$]hept$-2-$en$-2-$yllithium) acts as a nucleophile.
It attacks the electrophilic carbonyl carbon of methyl chloroformate $(Cl-CO-OMe)$.
The chloride ion $(Cl^-)$ acts as a leaving group.
This results in the formation of methyl bicyclo[$2.2$.$1$]hept$-2-$ene$-2-$carboxylate.

(C) Ethyl acetoacetate $(CH_3-CO-CH_2-CO_2Et)$ contains an active methylene group $(-CH_2-)$ situated between two electron-withdrawing carbonyl groups.
The protons on this carbon are relatively acidic $(pK_a \approx 11)$.
Grignard reagents like methyl magnesium iodide $(CH_3MgI)$ act as very strong bases.
In the presence of acidic protons,the acid-base reaction (deprotonation) occurs much faster than nucleophilic addition to the carbonyl group.
Therefore,one mole of $CH_3MgI$ will abstract a proton from the active methylene group to form a magnesium enolate salt and methane gas.
$CH_3-CO-CH_2-CO_2Et + CH_3MgI \rightarrow CH_3-CO-CH(MgI)-CO_2Et + CH_4 \uparrow$

(B) The reaction of diethyl carbonate with excess Grignard reagent $(CH_3MgBr)$ proceeds as follows:
$EtO-C(=O)-OEt$ $\xrightarrow{CH_3MgBr}$ $CH_3-C(=O)-OEt$ $\xrightarrow{CH_3MgBr}$ $CH_3-C(=O)-CH_3$ $\xrightarrow{CH_3MgBr}$ $(CH_3)_3C-OMgBr$ $\xrightarrow{H^{+}}$ $(CH_3)_3C-OH$.
The final product is $tert$-butyl alcohol ($2$-methylpropan-$2$-ol).

(C) The reaction of a lactone (cyclic ester) with an excess of Grignard reagent $(CH_3MgBr)$ followed by acid hydrolysis results in a diol.
The Grignard reagent attacks the carbonyl carbon twice: the first attack opens the ring to form a keto-alkoxide,and the second attack on the ketone group forms a di-alkoxide.
Upon hydrolysis,the product is a diol where one end is a primary alcohol and the other is a tertiary alcohol.
Reaction: $\delta\text{-valerolactone} + 2CH_3MgBr \xrightarrow{H_3O^{+}} HO-CH_2-CH_2-CH_2-CH_2-C(OH)(CH_3)_2$.

(B) The reaction of $4$-pentenoic acid $(CH_2=CH-CH_2-CH_2-COOH)$ with $I_2$ and $NaHCO_3$ is an example of iodo-lactonization.
In this reaction,the electrophilic iodine $(I^+)$ attacks the double bond to form a cyclic iodonium ion intermediate.
The carboxylate group $(-COO^-)$ formed by the reaction of the acid with $NaHCO_3$ then performs an intramolecular nucleophilic attack on the more substituted carbon of the iodonium ion.
This leads to the formation of a five-membered lactone ring,specifically $5$-iodomethyl-dihydrofuran-$2$-one.

(D) The starting material is $1$-chloroethen$-1-$ol,which is the enol form of acetyl chloride $(CH_3COCl)$.
Reaction with $Br_2$ leads to $\alpha$-bromination of the carbonyl compound (or electrophilic addition followed by elimination),yielding bromoacetyl chloride,$BrCH_2COCl$,as intermediate $(P)$.
Reaction of $BrCH_2COCl$ with methanol $(CH_3OH)$ is a nucleophilic acyl substitution reaction.
The methoxy group $(-OCH_3)$ replaces the chloride atom $(-Cl)$ to form methyl bromoacetate,$BrCH_2COOCH_3$,which is product $(Q)$.

(A) $LiAlH_4$ is a strong reducing agent that reduces esters to primary alcohols. In the given molecule,the ester group $(-CO_2CH_3)$ is reduced to a primary alcohol $(-CH_2OH)$,while the acetal group $(-C(OCH_3)_2-)$ remains unaffected because acetals are stable towards basic and nucleophilic reagents like $LiAlH_4$. Therefore,the product is $4,4-$dimethoxycyclohexylmethanol.

(C) The reaction of a carboxylic acid with organolithium reagents $(EtLi)$ proceeds through the formation of a dilithio species.
First,$1 \ equivalent$ of $EtLi$ deprotonates the carboxylic acid to form a lithium carboxylate.
Then,$2 \ equivalents$ of $EtLi$ attack the carbonyl carbon to form a stable tetrahedral intermediate (a gem-diol dianion derivative).
Upon acidic workup $(H_3O^+)$,this intermediate collapses to form a ketone.
Since the starting material is a hydroxy acid,the reaction specifically yields an $\alpha$-hydroxy ketone as shown in the provided mechanism.

(B) The reduction of an ester $R-CO-O-R'$ with $LiAlH_4$ followed by aqueous workup yields two alcohols: $R-CH_2OH$ and $R'-OH$.
For the reaction to yield two molecules of only a single alcohol,$R-CH_2OH$ must be identical to $R'-OH$.
In option $(B)$,$C_6H_5-CO_2-CH_2-C_6H_5$ (Benzyl benzoate):
$R = C_6H_5$ and $R' = -CH_2-C_6H_5$.
Reduction products: $C_6H_5-CH_2OH$ (Benzyl alcohol) and $C_6H_5-CH_2OH$ (Benzyl alcohol).
Thus,it yields two molecules of the same alcohol.

(C) The reaction of an ester with a base (saponification) is: $CH_3COOCH_2CH_3 + Na^{18}OH \rightarrow CH_3CO^{18}ONa + CH_3CH_2OH$.
Here,$(A)$ is $CH_3CO^{18}ONa$ (sodium acetate containing $^{18}O$) and $(B)$ is $CH_3CH_2OH$ (ethanol).
When $(A)$ is treated further,the carboxylate ion $(C)$ is $CH_3-C(=O)-^{18}O^-$.

(C) $1$. Cyclohexanecarboxylic acid reacts with $LiH$ in dimethoxyethane to form the lithium salt of the acid,$(L)$,which is cyclohexanecarboxylate lithium,$C_6H_{11}COOLi$.
$2$. The lithium salt $(L)$ then reacts with methyllithium $(CH_3Li)$. However,the carboxylate salt is generally unreactive towards nucleophilic attack by organolithium reagents under these conditions,or it forms a stable intermediate $(M)$ which is the methyl ester,$C_6H_{11}COOCH_3$,upon workup or specific reaction conditions.
$3$. The final step is the hydrolysis of $(M)$ with $H_3O^+$. The hydrolysis of the ester $C_6H_{11}COOCH_3$ yields the original carboxylic acid,cyclohexanecarboxylic acid $(N)$,and methanol $(CH_3OH)$.
$4$. Therefore,the product $(N)$ is cyclohexanecarboxylic acid.

(C) The reaction proceeds in two steps:
$1$. Esterification: The reaction of $C_6H_5-CH(OH)-COOH$ with ethanol $(CH_3CH_2OH)$ in the presence of dry $HCl(g)$ gas forms the ester $C_6H_5-CH(OH)-COOEt$.
$2$. Reaction with $SOCl_2$: Thionyl chloride $(SOCl_2)$ reacts with the hydroxyl group $(-OH)$ to replace it with a chlorine atom $(-Cl)$. Thus,the final product $(X)$ is $C_6H_5-CH(Cl)-COOEt$.

(A) The reaction is an acid-catalyzed transesterification.
Methyl acrylate reacts with $n$-butyl alcohol in the presence of $TsOH$ ($p$-toluenesulfonic acid) to form $n$-butyl acrylate and methanol.
Methanol,having the lowest boiling point $(65^\circ C)$,is removed by distillation to shift the equilibrium toward the formation of the product $(A)$ ($n$-butyl acrylate).
$CH_2=CH-CO-OCH_3 + CH_3-CH_2-CH_2-CH_2-OH \xrightarrow{TsOH, \Delta} CH_2=CH-CO-OCH_2-CH_2-CH_2-CH_3 + CH_3OH$

(B) $LiAlH_4$ is a strong reducing agent that reduces esters into two alcohols: $RCOOR' \xrightarrow{LiAlH_4} RCH_2OH + R'OH$.
For the product to be $2B$ (where $B$ is a single chiral alcohol),the ester must be symmetric such that both fragments produce the same chiral alcohol.
In option $B$,$CH_3-CH_2-CH(CH_3)-COOCH(CH_3)-CH_2-CH_3$ is a sec-butyl $2-$methylbutanoate ester.
Upon reduction,it yields $CH_3-CH_2-CH(CH_3)-CH_2OH$ ($2$-methylbutan$-1-$ol) and $CH_3-CH_2-CH(CH_3)OH$ (butan$-2-$ol).
However,if we consider the structure $CH_3-CH(CH_3)-COOCH(CH_3)-CH_2-CH_3$ (sec-butyl $2-$methylpropanoate),it yields different alcohols.
Given the requirement for $2B$ (identical chiral alcohols),the ester must be $sec-butyl$ $2-methylbutanoate$ is not correct,but rather an ester like $sec-butyl$ $2-methylbutanoate$ is not symmetric.
The correct structure for $2B$ where $B$ is chiral is $sec-butyl$ $2-methylbutanoate$ is not the case,but rather $sec-butyl$ $sec-butyl$ ester derivative. The correct answer is $B$.

(C) The reaction shown is a Fischer esterification,where a carboxylic acid reacts with an alcohol in the presence of an acid catalyst (like $HCl$) to form an ester and water.
In this reaction,Mandelic acid $(Ph-CH(OH)-COOH)$ reacts with ethanol $(EtOH)$ in the presence of $HCl$.
The carboxylic acid group $(-COOH)$ is converted into an ester group $(-COOEt)$,while the hydroxyl group $(-OH)$ attached to the alpha-carbon remains unaffected under these specific conditions.
The reaction is:
$Ph-CH(OH)-COOH + C_2H_5OH \xrightarrow{HCl} Ph-CH(OH)-COOC_2H_5 + H_2O$
Thus,the product is ethyl mandelate,which corresponds to option $C$.

(B) In the Fischer esterification reaction,the oxygen atom of the alcohol is incorporated into the ester product,while the hydroxyl group $(-OH)$ of the carboxylic acid is eliminated as water.
Specifically,the $O^{18}$ isotope from the methanol $(CH_3-O^{18}H)$ ends up in the ester linkage.
Therefore,the product $(X)$ is $Ph-CO-O^{18}-CH_3$.
This process is an esterification reaction.

(C) The given reaction involves the exchange of an alkoxy group $(OR')$ of an ester with the alkoxy group $(OR'')$ of an alcohol in the presence of an acid catalyst $(H^{\oplus})$.
This specific type of reaction,where one ester is converted into another ester,is known as $trans-esterification$.

(A) The reactivity of carbonyl compounds toward nucleophilic attack by $LiAlH_4$ depends on the electrophilicity of the carbonyl carbon.
$(i)$ Cyclopentanone is a ketone,which is highly reactive toward nucleophilic attack as there is no resonance stabilization of the carbonyl carbon.
(ii) $N$-Methylpyrrolidin$-2-$one is an amide. The lone pair on the nitrogen atom participates in resonance with the carbonyl group ($+M$ effect),significantly reducing the electrophilicity of the carbonyl carbon.
(iii) $\gamma$-Butyrolactone is an ester. The lone pair on the oxygen atom participates in resonance with the carbonyl group ($+M$ effect),reducing the electrophilicity of the carbonyl carbon,but to a lesser extent than the amide because oxygen is more electronegative than nitrogen.
Thus,the order of reactivity is: Amide $(ii)$ < Ester $(iii)$ < Ketone $(i)$.
Therefore,the correct order is $ii < iii < i$.

(C) $1$. The starting material is $2\text{-formylmethylbenzoic acid}$.
$2$. $NaBH_4$ is a selective reducing agent that reduces the aldehyde group $(-CHO)$ to a primary alcohol $(-CH_2OH)$ but does not reduce the carboxylic acid group $(-COOH)$. Thus,$(A)$ is $2\text{-(2-hydroxyethyl)benzoic acid}$.
$3$. Upon treatment with $H^{\oplus}$ and heating $(\Delta)$,the carboxylic acid and the primary alcohol undergo an intramolecular esterification (cyclization) reaction,eliminating a water molecule to form a cyclic ester known as a lactone.
$4$. The product $(B)$ is $3,4\text{-dihydroisocoumarin}$ (also known as dihydroisocoumarin).

(D) The hydrolysis of the given cyclic anhydride (Meldrum's acid derivative) initially leads to the formation of an unstable intermediate,which is a substituted malonic acid derivative.
This intermediate undergoes decarboxylation upon heating to form a $\gamma$-hydroxy acid.
The $\gamma$-hydroxy acid then undergoes intramolecular cyclization (lactonization) to form a five-membered lactone ring ($\gamma$-lactone) with a carboxylic acid substituent.
The final product is a five-membered lactone ring with a $-CO_2H$ group attached.

(A) The reaction involves the alkaline hydrolysis of an ester using $KOH$ in $H_2O/\Delta$.
The reactant is a tert-butyl ester. Alkaline hydrolysis of an ester (saponification) yields a carboxylate salt and an alcohol.
$1$. The ester group $(CH_3)_3CO-CO-CH_2-CH_2-CH(OH)-Ar$ undergoes hydrolysis to produce tert-butanol $(CH_3)_3COH$ and the corresponding carboxylic acid,which immediately reacts with $KOH$ to form the carboxylate salt $HOOC-CH_2-CH_2-CH(OH)-Ar$ (which becomes the carboxylate anion).
$2$. The ether groups ($-OCH_3$ and $-OCH_2CH_2CH_3$) attached to the aromatic ring are stable under these basic conditions and do not react.
$3$. The hydroxyl group $(-OH)$ on the aliphatic chain is also stable under these conditions.
Thus,the products are tert-butanol and the carboxylate salt of the carboxylic acid.
Total number of products = $2$.

(B) The reaction of $CH_3MgBr$ (a Grignard reagent) with different functional groups yields different alcohols:
$1$. $C_2H_5CHO$ (an aldehyde) reacts with $CH_3MgBr$ to form a secondary alcohol.
$2$. $C_2H_5CO_2CH_3$ (an ester) reacts with excess $CH_3MgBr$ to form a tertiary alcohol. The ester first reacts to form a ketone,which then reacts with another equivalent of $CH_3MgBr$ to form the tertiary alcohol after hydrolysis.
$3$. $C_2H_5COOH$ (a carboxylic acid) reacts with $CH_3MgBr$ to release methane gas due to the acidic hydrogen,not forming a tertiary alcohol.
$4$. $2,3-$epoxybutane reacts with $CH_3MgBr$ to form a secondary alcohol.
Therefore,the ester $C_2H_5CO_2CH_3$ is the correct answer.

(D) The acidity of methylenic hydrogens $(-CH_2-)$ increases when they are attached to electron-withdrawing groups (EWGs).
In $CH_3-CO-CH_2-CN$,the methylene group is located between two strong EWGs: an acetyl group $(-COCH_3)$ and a cyano group $(-CN)$.
This makes these hydrogens the most acidic due to the strong $-I$ and $-M$ effects of both groups,which stabilize the resulting carbanion through resonance.

(A) Decarboxylation of $\beta$-keto acids typically proceeds through a cyclic six-membered transition state,which requires the formation of an enol intermediate. This mechanism is governed by Bredt's rule in bicyclic systems.
According to Bredt's rule,a double bond cannot be placed at the bridgehead of a bridged bicyclic system unless the rings are large enough.
In option $A$,the carboxyl group is at the bridgehead position. Decarboxylation would require the formation of an enol with a double bond at the bridgehead. Since this violates Bredt's rule,the transition state is highly unstable or impossible to form,preventing decarboxylation.
Therefore,the compound in option $A$ does not undergo decarboxylation.

(A) $1.$ $PCC$ oxidizes the primary alcohol to an aldehyde: $HOCH_2CH_2CH_2COOCH_2CH_3 \xrightarrow{PCC} O=CHCH_2CH_2COOCH_2CH_3$ $(A)$.
$2.$ One molar equivalent of vinyl magnesium bromide reacts with the more reactive aldehyde group: $O=CHCH_2CH_2COOCH_2CH_3$ $\xrightarrow{H_2C=CHMgBr}$ $\xrightarrow{NH_4Cl/H_2O} H_2C=CHCH(OH)CH_2CH_2COOCH_2CH_3$ $(C)$.
$3.$ Saponification of the ester with $KOH/H_2O$ followed by acidification with $H_3O^{+}$ gives the carboxylic acid: $H_2C=CHCH(OH)CH_2CH_2COOH$.
$4.$ Acetylation of the secondary alcohol with acetic anhydride in pyridine yields the final product $(D)$: $H_2C=CHCH(OCOCH_3)CH_2CH_2COOH$.

(B) The reaction involves the ozonolysis of an unsaturated lactone.
$1$. Ozonolysis of the given cyclic compound breaks the double bond to form formaldehyde $(HCHO)$ and a cyclic anhydride/lactone intermediate $(A)$,which is malonic anhydride (or a related cyclic structure).
$2$. The nucleophilic attack of aniline $(Ph-NH_2)$ on the carbonyl group of the intermediate $(A)$ leads to ring opening.
$3$. The reaction follows the mechanism of nucleophilic acyl substitution,resulting in the formation of $N$-phenylmalonamic acid,which is $Ph-NH-CO-CH_2-CO_2H$.

(C) In reaction $(c)$,the carboxylic acid reacts with $NaOH$ to form a sodium salt,which then reacts with methyl iodide to form an ester,not a ketone.
$R-C(=O)-OH$ $\xrightarrow{NaOH} R-C(=O)-ONa$ $\xrightarrow{CH_3-I} R-C(=O)-OCH_3$ (Ester).
In option $(a)$,the hydration of an alkyne yields a ketone.
In option $(b)$,the hydration of an alkyne yields a ketone.
In option $(d)$,the hydroboration-oxidation of an internal alkyne yields a ketone.

(D) The starting material is a dianion of ethyl acetoacetate,specifically the $O,O$-dianion formed by treating ethyl acetoacetate with two equivalents of a strong base like $LDA$.
$1$. The dianion is more nucleophilic at the $\alpha$-carbon (the carbon between the two carbonyl groups) than at the oxygen atoms.
$2$. When treated with an alkyl halide like $CH_3CH_2I$,the $\alpha$-carbon attacks the ethyl group in an $S_N2$ reaction.
$3$. Subsequent acidic workup $(H_3O^+)$ protonates the enolate oxygens to regenerate the keto-ester functionality.
$4$. The resulting product is ethyl $2$-ethylacetoacetate,which is ethyl $2$-ethyl-$3$-oxobutanoate. Looking at the options,option $D$ represents the structure of $2$-ethyl-$3$-oxobutanoate (ethyl $2$-ethylacetoacetate).

(B) The reaction sequence is as follows:
$1$. $2\text{-methylcyclohexanecarboxylic acid}$ reacts with $SOCl_2$ to form the acid chloride,$A$ $(2\text{-methylcyclohexanecarbonyl chloride})$.
$2$. The acid chloride $A$ reacts with $NH_3$ to form the amide,$B$ $(2\text{-methylcyclohexanecarboxamide})$.
$3$. The amide $B$ undergoes dehydration with $P_4O_{10}$ upon heating to form the nitrile,$C$ $(2\text{-methylcyclohexanecarbonitrile})$.
The stereochemistry is retained throughout the reaction sequence. The starting material has the $-COOH$ group and $-CH_3$ group in a specific relative configuration (trans). Therefore,the final product $C$ will also have the $-CN$ group and $-CH_3$ group in the same relative configuration.

(C) $1$. The molecular formula $C_4H_8O_3$ and the evolution of $CO_2$ with $NaHCO_3$ indicate that $X$ is a carboxylic acid.
$2$. $X$ is optically active,meaning it must contain a chiral center.
$3$. Reduction of $X$ with $LiAlH_4$ yields an achiral compound. Let us analyze option $C$: $CH_3CH(CH_2OH)COOH$.
$4$. $CH_3CH(CH_2OH)COOH$ has a chiral center at the $C-2$ position.
$5$. Upon reduction with $LiAlH_4$,the carboxylic acid group $-COOH$ is reduced to $-CH_2OH$,resulting in $CH_3CH(CH_2OH)CH_2OH$.
$6$. In $CH_3CH(CH_2OH)CH_2OH$,the central carbon is attached to two identical $-CH_2OH$ groups,making the molecule achiral.
$7$. Thus,$X$ is $CH_3CH(CH_2OH)COOH$.

(C) The reaction is a nucleophilic acyl substitution where the nucleophile $H-\bullet^-$ (where $\bullet = ^{18}O$) attacks the carbonyl carbon of the ester $CH_3-C(=O)-O-CH_2-CH_3$.
$1$. The nucleophile $H-\bullet^-$ attacks the electrophilic carbonyl carbon,leading to the formation of a tetrahedral intermediate.
$2$. The tetrahedral intermediate collapses,expelling the ethoxide ion $(EtO^-)$ as a leaving group.
$3$. This results in the formation of $CH_3-C(=O)-\bullet-H$ (an ester with $^{18}O$ in the carbonyl oxygen position).
$4$. The ethoxide ion $(EtO^-)$ then acts as a base and abstracts the proton from the newly formed ester,resulting in the formation of $CH_3-C(=O)-\bullet^-$ and ethanol $(EtOH)$.
Thus,one of the products is $CH_3-C(=O)-\bullet^-$.

(A) The reaction sequence is an example of acetoacetic ester synthesis.
$1$. Acetoacetic ester $(CH_3COCH_2COOC_2H_5)$ reacts with $NaOEt$ to form an enolate ion at the $\alpha$-carbon.
$2$. This enolate reacts with $CH_3I$ via an $S_N2$ mechanism to produce ethyl $2$-methylacetoacetate $(CH_3COCH(CH_3)COOC_2H_5)$.
$3$. Treatment with another equivalent of $NaOEt$ generates a new enolate at the same $\alpha$-carbon.
$4$. This enolate reacts with $CH_3CH_2CH_2Br$ ($n$-propyl bromide) via $S_N2$ to yield the final product,ethyl $2$-methyl-$2$-propylacetoacetate,which is $CH_3-C(=O)-C(CH_3)(CH_2CH_2CH_3)-COOC_2H_5$.

(C) The reaction involves the formation of a carbanion from dimethyl malonate $(CH_2(CO_2Me)_2)$ using sodium $(Na)$.
The carbanion $Na^+[CH(CO_2Me)_2]^-$ acts as a nucleophile.
This nucleophile attacks the electrophilic carbon of methyl chloroformate $(ClCO_2Me)$.
The reaction proceeds as follows:
$CH_2(CO_2Me)_2 + Na \rightarrow Na^+[CH(CO_2Me)_2]^- + \frac{1}{2}H_2$
$Na^+[CH(CO_2Me)_2]^- + ClCO_2Me \rightarrow CH(CO_2Me)_3 + NaCl$
Thus,the missing reactant is $ClCO_2Me$.