Alkyne Questions in English

(C) The hydration of alkynes in the presence of $Hg^{2+}$ and $H^+/H_2O$ (Kuccherov reaction) follows Markovnikov's addition of water to the triple bond.
$1$. For terminal alkynes like $CH_3-C \equiv CH$,the reaction yields a ketone $(CH_3-CO-CH_3)$.
$2$. For internal alkynes like $CH_3-C \equiv C-CH_3$ or $CH_3-C \equiv C-CH_2CH_3$,the reaction yields ketones.
$3$. For ethyne $(HC \equiv CH)$,the hydration product is ethenol $(CH_2=CH-OH)$,which tautomerizes to form ethanal $(CH_3-CHO)$,an aldehyde,not a ketone.
Therefore,$HC \equiv CH$ cannot give a ketone.

(C) Propene $(CH_3-CH=CH_2)$ is an alkene and does not react with ammoniacal silver nitrate $(AgNO_3 + NH_4OH)$.
Propyne $(CH_3-C \equiv CH)$ is a terminal alkyne with an acidic hydrogen atom.
It reacts with ammoniacal silver nitrate to form a white precipitate of silver propynide $(CH_3-C \equiv C^- Ag^+)$.
Therefore,$AgNO_3 + NH_4OH$ (Tollens' reagent) is used to distinguish between terminal alkynes and alkenes.

(A) The reaction of an alkyne with excess $HOCl$ proceeds through electrophilic addition.
First,$HOCl$ adds across the triple bond to form a chlorohydrin intermediate.
Since $HOCl$ is in excess,the addition occurs twice at the same carbon atom (following Markovnikov's rule,where the $OH$ group attaches to the more substituted carbon).
This results in a geminal diol intermediate,which is unstable and undergoes dehydration to form a ketone.
The final product is $Cyclohexyl-C(=O)-CHCl_2$.

(A) Lindlar's catalyst is a poisoned palladium catalyst ($Pd/CaCO_3$ or $Pd/BaSO_4$ poisoned with quinoline or sulfur).
It performs the partial hydrogenation of alkynes to alkenes.
The reaction is stereospecific and yields the $cis-$alkene.
The reaction is:
$CH_3-C \equiv C-CH_3 + H_2 \xrightarrow{Pd/CaCO_3} cis-CH_3-CH=CH-CH_3$
Thus,the hydrogenation of $2-$Butyne with Lindlar's catalyst produces $cis-2-$Butene.

(C) Let us analyze each reaction:
$(A)$ Dehydrohalogenation of vicinal dihalides with $Alc. KOH$ followed by $NaNH_2$ gives an alkyne.
$(B)$ Reaction of $CH_3-CCl_3$ with $Ag$ powder is not a standard method to produce $CH_3-C \equiv C-CH_3$. The reaction of $CH_3-CCl_3$ with $Ag$ typically leads to coupling products or other derivatives,not an alkyne.
$(C)$ Dehalogenation of vicinal dihalides with $Zn$ dust in alcohol gives an alkene $(CH_3-CH_2-CH=CH_2)$,not an alkyne.
$(D)$ Kolbe's electrolysis of sodium maleate or fumarate gives ethyne $(HC \equiv CH)$.
Note: Both $(B)$ and $(C)$ do not produce an alkyne. However,in the context of standard textbook problems,$(C)$ is the most direct example of a reaction producing an alkene instead of an alkyne from a vicinal dihalide.

(B) The reaction sequence is as follows:
$1.$ Chlorination of stilbene: $Ph-CH=CH-Ph + Cl_2 \xrightarrow{CCl_4} Ph-CHCl-CHCl-Ph$ $(X)$
$2.$ Double dehydrohalogenation: $Ph-CHCl-CHCl-Ph + 2NaNH_2 \rightarrow Ph-C \equiv C-Ph$ $(Y)$
$3.$ Partial hydrogenation: $Ph-C \equiv C-Ph + H_2 \xrightarrow{Pd/BaSO_4} \text{cis}-Ph-CH=CH-Ph$ $(Z)$.
Lindlar's catalyst $(Pd/BaSO_4)$ performs syn-addition of hydrogen to alkynes,yielding cis-alkenes. The structure of cis-stilbene corresponds to option $B$.

(C) $1$. The formation of a white precipitate with ammoniacal $AgNO_3$ indicates that $X$ is a terminal alkyne $(R-C \equiv CH)$.
$2$. The reaction with excess $KMnO_4$ (oxidative cleavage) yields $(CH_3)_2CHCOOH$ (isobutyric acid).
$3$. The structure of the acid $(CH_3)_2CHCOOH$ contains $4$ carbon atoms. Since the original compound $X$ has $5$ carbon atoms,the terminal alkyne must be $(CH_3)_2CH-C \equiv CH$.
$4$. Oxidative cleavage of $(CH_3)_2CH-C \equiv CH$ breaks the triple bond to form $(CH_3)_2CHCOOH$ and $CO_2$.

(B) The starting material is $1,1$-dichlorobutane,$CH_3-CH_2-CH(Cl)_2$.
Treatment with excess $NaNH_2$ (a strong base) causes dehydrohalogenation to form an alkyne.
$CH_3-CH_2-CH(Cl)_2 \xrightarrow{NaNH_2} CH_3-CH_2-C \equiv CH$ (Compound $P$,$1$-butyne).
$1$-butyne has an acidic terminal hydrogen.
$CH_3-CH_2-C \equiv CH \xrightarrow{NaNH_2} CH_3-CH_2-C \equiv C^- Na^+$.
Reaction with $CH_3Cl$ (an alkyl halide) leads to nucleophilic substitution (alkylation).
$CH_3-CH_2-C \equiv C^- + CH_3Cl \rightarrow CH_3-CH_2-C \equiv C-CH_3 + Cl^-$.
Thus,$Q$ is $CH_3-CH_2-C \equiv C-CH_3$ ($2$-pentyne).

(A) $1.$ Hydration of ethyne: $HC \equiv CH + H_2O \xrightarrow{dil. H_2SO_4, HgSO_4} CH_3CHO$ (Ethanal,$A$)
$2.$ Oxidation: $CH_3CHO \xrightarrow{[O]} CH_3COOH$ (Ethanoic acid,$B$)
$3.$ Neutralization: $CH_3COOH + NaOH \rightarrow CH_3COONa$ (Sodium ethanoate,$C$)
$4.$ Decarboxylation: $CH_3COONa + NaOH \xrightarrow{CaO, \Delta} CH_4$ (Methane,$D$) + $Na_2CO_3$.
Therefore,$[D]$ is $CH_4$.

(D) The reaction of an internal alkyne with $BH_3$ in $THF$ followed by treatment with acetic acid $(CH_3COOH)$ is a stereoselective reduction process.
$1$. Hydroboration of $2$-butyne $(CH_3 - C \equiv C - CH_3)$ with $BH_3$ involves $syn$-addition to form a vinyl borane.
$2$. Subsequent protonolysis with $CH_3COOH$ replaces the boron group with a hydrogen atom, retaining the $syn$ stereochemistry.
This sequence converts the alkyne into a $cis$-alkene. Therefore, the major product is $\text{cis}-2-\text{butene}$.

(D) Propene and propyne can be distinguished by using the Ammoniacal silver nitrate test (Tollens' reagent).
Propyne $(CH_3-C \equiv CH)$ contains a terminal acidic hydrogen atom.
This acidic hydrogen reacts with Ammoniacal silver nitrate ($AgNO_3$ in $NH_4OH$) to form a white precipitate of silver propynide $(CH_3-C \equiv C-Ag)$.
Propene $(CH_3-CH=CH_2)$ does not contain an acidic hydrogen and therefore does not react with Ammoniacal silver nitrate to form a precipitate.

(C) Step $1$: The ozonolysis of $CH_3-C \equiv C-CH_3$ (but$-2-$yne) followed by hydrolysis yields acetic acid $(CH_3COOH)$ as the product $X$.
$CH_3-C \equiv C-CH_3 \xrightarrow{O_3, H_2O} 2CH_3COOH$.
Step $2$: The reaction of $CH_3COOH$ with soda lime $(NaOH + CaO)$ is a decarboxylation reaction.
$CH_3COOH + NaOH \rightarrow CH_3COONa + H_2O$.
$CH_3COONa + NaOH \xrightarrow{CaO, \Delta} CH_4 + Na_2CO_3$.
Thus,the final product $Y$ is methane $(CH_4)$.

(C) The reaction of a terminal alkyne with $BH_3.THF$ followed by oxidation with $H_2O_2/OH^-$ is a hydroboration-oxidation reaction.
This reaction follows anti-Markovnikov addition of water across the triple bond.
For a terminal alkyne like $Ph-C \equiv CH$,the initial product is an enol,$Ph-CH=CH-OH$.
This enol tautomerizes to form an aldehyde.
Therefore,the final product is $Ph-CH_2-CHO$ (phenylacetaldehyde).

(D) The given reaction is the Reppe carbonylation of acetylene: $CH \equiv CH + CO + H_2O \xrightarrow{Ni(CO)_4} CH_2 = CH - COOH$.
The product formed is acrylic acid $(CH_2 = CH - COOH)$.
Statement $A$ is true: Acrylic acid is an $\alpha, \beta$-unsaturated acid.
Statement $B$ is true: The reaction involves the addition of $H$ and $COOH$ across the triple bond.
Statement $C$ is true: The product is indeed acrylic acid.
Statement $D$ is false: Acrylic acid $(CH_2 = CH - COOH)$ reacts with ethyl alcohol $(C_2H_5OH)$ to form ethyl acrylate $(CH_2 = CH - COOC_2H_5)$,not ethyl butanoate.

(A) The addition of $HBr$ to alkynes in the presence of peroxide follows the anti-Markovnikov rule.
First,$CH_3-C \equiv CH + HBr \xrightarrow{\text{peroxide}} CH_3-CH=CHBr$ ($1$-bromopropene).
Second,the addition of another mole of $HBr$ to $CH_3-CH=CHBr$ in the presence of peroxide follows the anti-Markovnikov rule again,leading to the terminal dibromide.
$CH_3-CH=CHBr + HBr \xrightarrow{\text{peroxide}} CH_3-CH_2-CHBr_2$ ($1$,$1$-dibromopropane).
Therefore,the final product is $CH_3-CH_2-CHBr_2$.

(B) The reduction of an alkyne with $Na$ in liquid $NH_3$ (Birch reduction) proceeds via anti-addition of hydrogen,resulting in the formation of a $trans$-alkene $(A)$.
The reduction of an alkyne with $H_2$ in the presence of Lindlar's catalyst ($Pd + CaCO_3$ or $Pd + BaSO_4$) proceeds via syn-addition of hydrogen,resulting in the formation of a $cis$-alkene $(B)$.

(B) For reaction $(1)$: The starting material is a gem-dibromide $Ph-C(Br)_2-CH_3$. Treatment with $NaNH_2$ (a strong base) causes dehydrohalogenation to form an alkyne. Two equivalents of $NaNH_2$ are required to remove two molecules of $HBr$ to form the terminal alkyne $Ph-C \equiv CH$ (which is $(A)$). Thus,$x = 2$.
For reaction $(2)$: The starting material is $CH_3-CH_2-C(Br)_2-CH_3$. Treatment with $NaNH_2$ causes dehydrohalogenation to form the internal alkyne $CH_3-C \equiv C-CH_3$. Two equivalents of $NaNH_2$ are required to remove two molecules of $HBr$. Thus,$y = 2$.
The sum $(x + y) = 2 + 2 = 4$.

(D) $1$. The first reaction is hydroboration-oxidation of propyne: $CH_3-C \equiv CH + BH_3-THF \rightarrow CH_3-CH=CH-BH_2$. Subsequent oxidation with $H_2O_2/OH^{\Theta}$ yields the enol $CH_3-CH=CH-OH$,which tautomerizes to propanal,$CH_3-CH_2-CHO$ $(P)$.
$2$. The second reaction is oxymercuration-demercuration of propyne: $CH_3-C \equiv CH + Hg(OAc)_2/H_2O \rightarrow CH_3-C(OH)=CH-HgOAc$. Reduction with $NaBH_4$ yields the enol $CH_3-C(OH)=CH_2$,which tautomerizes to acetone (propanone),$CH_3-CO-CH_3$ $(Q)$.
$3$. $(P)$ is an aldehyde and $(Q)$ is a ketone. Both have the molecular formula $C_3H_6O$.
$4$. Since they have the same molecular formula but different functional groups,$(P)$ and $(Q)$ are functional isomers.

(A) The hydration of alkynes in the presence of $Hg^{+2}/H_3O^{\oplus}$ follows Markovnikov's addition of water to form an enol,which then tautomerizes to a carbonyl compound.
$1$. For $CH \equiv CH$ (ethyne),the product is $CH_3CHO$ (acetaldehyde),which is an aldehyde.
$2$. For $CH_3 - C \equiv CH$ (propyne),the product is $CH_3 - CO - CH_3$ (acetone),which is a ketone.
$3$. For $CH_3 - C \equiv C - CH_3$ (but$-2-$yne),the product is $CH_3 - CO - CH_2 - CH_3$ (butanone),which is a ketone.
Therefore,only reaction $A$ produces an aldehyde.

(A) $But-1-yne$ $(CH_3CH_2C \equiv CH)$ is a terminal alkyne with an acidic hydrogen atom attached to the $sp$ hybridized carbon.
Terminal alkynes react with ammoniacal silver nitrate ($AgNO_3 + NH_4OH$,also known as Tollen's reagent) to form a white precipitate of silver acetylide $(CH_3CH_2C \equiv CAg)$.
$But-2-yne$ $(CH_3C \equiv CCH_3)$ is a non-terminal alkyne and does not have an acidic hydrogen atom,so it does not react with ammoniacal silver nitrate.
Therefore,$(AgNO_3 + NH_4OH)$ is used to distinguish between them.

(A) The electrolysis of aqueous potassium maleate (cis-butenedioate) is an example of the Kolbe electrolysis reaction.
During electrolysis,the maleate ion undergoes decarboxylation at the anode to form an alkyne.
The reaction is as follows:
$cis-KOOC-CH=CH-COOK \xrightarrow{electrolysis} HC \equiv CH + 2CO_2 + H_2 + 2KOH$.
Thus,the product formed is acetylene $(CH \equiv CH)$.

(A) $1$. The electrolysis of an aqueous solution of potassium maleate (or fumarate) is a Kolbe electrolysis reaction.
$2$. The reaction is: $CH(COOK)=CH(COOK) + 2H_2O \xrightarrow{\text{electrolysis}} CH \equiv CH + 2CO_2 + H_2 + 2KOH$.
$3$. Thus,$X$ is acetylene $(CH \equiv CH)$.
$4$. The hydration of acetylene in the presence of $Hg^{2+}$ and $dil. H_2SO_4$ (Kucherov reaction) yields acetaldehyde $(CH_3CHO)$.
$5$. The reaction is: $CH \equiv CH + H_2O$ $\xrightarrow{Hg^{2+}/H_2SO_4} [CH_2=CH-OH]$ $\rightarrow CH_3-CH=O$.
$6$. Therefore,$Y$ is $CH_3-CH=O$.

(C) The reaction proceeds as follows:
$CH_3 - C \equiv CH$ $\xrightarrow{Excess\ HCl} CH_3 - C(Cl)_2 - CH_3 (A)$ $\xrightarrow{H_2O/OH^{-}} [CH_3 - C(OH)_2 - CH_3]$ $\xrightarrow{-H_2O} CH_3 - C(=O) - CH_3 (x)$
The addition of $HCl$ to propyne follows Markovnikov's rule to form $2,2-$dichloropropane $(A)$. Hydrolysis of this geminal dihalide produces an unstable geminal diol,which undergoes dehydration to form acetone $(x)$.

(C) The reaction of an internal alkyne like $2-$butyne with sodium metal $(Na)$ in liquid ammonia $(NH_3)$ is a dissolving metal reduction.
This reaction proceeds via a radical anion intermediate and results in the anti-addition of hydrogen atoms across the triple bond.
Consequently,the product formed is the $trans-$isomer of the corresponding alkene.
$CH_3-C \equiv C-CH_3 \xrightarrow{Na/Liq. NH_3} trans-CH_3-CH=CH-CH_3$

(A) The reaction of a gem-dichloride $(CH_3CH_2CH_2CHCl_2)$ with a strong base like $NaNH_2$ ($2 \ g$ equivalent) undergoes dehydrohalogenation.
First,one molecule of $HCl$ is eliminated to form a vinyl chloride intermediate $(CH_3CH_2CH=CHCl)$.
Then,the second molecule of $HCl$ is eliminated to form the terminal alkyne,$CH_3CH_2C \equiv CH$ (but$-1-$yne).

(C) Calcium carbide $(CaC_2)$ reacts with heavy water $(D_2O)$ to produce deuteroacetylene $(C_2D_2)$ and calcium deuteroxide $(Ca(OD)_2)$.
The balanced chemical equation is:
$CaC_2 + 2D_2O \longrightarrow C_2D_2 + Ca(OD)_2$

(B) The reaction of $pent-1-yne$ with excess $HCl$ follows Markovnikov's rule.
In the first step,$HCl$ adds to the triple bond to form a geminal-like intermediate,but specifically,the $Cl$ atom attaches to the more substituted carbon (the $C-2$ position).
$CH_3-CH_2-CH_2-C \equiv CH + HCl \rightarrow CH_3-CH_2-CH_2-CCl=CH_2$
In the second step,another molecule of $HCl$ adds to the double bond,again following Markovnikov's rule,resulting in the formation of a geminal dichloride at the $C-2$ position.
$CH_3-CH_2-CH_2-CCl=CH_2 + HCl \rightarrow CH_3-CH_2-CH_2-CCl_2-CH_3$.

(A) The reaction of an alkyne with $H_2$ in the presence of Lindlar's catalyst (poisoned $Pd/CaCO_3$) is a stereoselective reduction.
This reaction specifically produces a $cis$-alkene as the major product because the hydrogen atoms are added to the same side of the triple bond (syn-addition).

(C) Acetylene $(HC \equiv CH)$ is acidic in nature due to the $sp$ hybridized carbon atoms.
$1$. Treatment with $NaNH_2$ (a strong base) removes the acidic proton to form sodium acetylide $(HC \equiv C^- Na^+)$,which then reacts with an alkyl halide $(R - X)$ via an $S_N2$ mechanism to form a higher alkyne $(HC \equiv C - R)$.
$2$. Alternatively,treatment with a Grignard reagent $(R - Mg - X)$ also removes the acidic proton to form the acetylide,which subsequently reacts with an alkyl halide $(R - X)$ to yield a higher alkyne.
Therefore,both methods are effective for the synthesis of higher alkynes.

(A) The acidity of terminal alkynes like acetylene $(HC \equiv CH)$ is due to the $sp$ hybridization of the carbon atom.
In $sp$ hybridization,the carbon atom has $50\%$ $s-$character.
Due to higher $s-$character,the electrons in the $C-H$ sigma bond are held more strongly by the carbon atom,making the hydrogen atom more acidic and easily removable as a proton $(H^+)$.

(C) Step $1$: $CaO + 3C \xrightarrow{\text{Heat}} CaC_2 (A) + CO$
Step $2$: $CaC_2 + 2H_2O \rightarrow Ca(OH)_2 + C_2H_2 (B) \text{ (Acetylene)}$
Step $3$: $C_2H_2 + H_2O \xrightarrow{H_2SO_4, Hg^{2+}} CH_3CHO (C) \text{ (Acetaldehyde)}$
Thus,the final product $C$ is Acetaldehyde.

(B) When chloroform $(CHCl_3)$ is heated with silver powder $(Ag)$,it undergoes a dehalogenation reaction to form ethyne $(C_2H_2)$.
The balanced chemical equation is:
$2CHCl_3 + 6Ag \rightarrow C_2H_2 + 6AgCl$

(D) The reactant $CH_3-C \equiv C-CH_3$ is an internal alkyne (but$-2-$yne).
$NaNH_2$ is a strong base,but it requires an acidic terminal hydrogen (alkyne hydrogen) to perform deprotonation or isomerization.
Since but$-2-$yne does not have any terminal acidic hydrogen atoms,it does not react with $NaNH_2$ under standard conditions.
Therefore,no reaction occurs.

(C) The oxidation of alkynes with selenium dioxide $(SeO_2)$ typically occurs at the allylic or propargylic position.
For propyne $(CH_3-C \equiv CH)$,the oxidation occurs at the methyl group adjacent to the triple bond.
The reaction converts the terminal methyl group into a carbonyl group,resulting in the formation of methylglyoxal $(CH_3COCHO)$.

(A) Ozonolysis of an alkyne with $O_3$ followed by hydrolysis $(H_2O)$ results in the cleavage of the triple bond to form diketones or carboxylic acids depending on the conditions. For internal alkynes,the triple bond is cleaved to form two carboxylic acids.
The reaction is: $(CH_3)_2CH-C \equiv C-CH_3 + O_3 / H_2O \rightarrow (CH_3)_2CHCOOH + CH_3COOH$.
Thus,the products are isobutyric acid and acetic acid.

(B) Acetylene $(HC \equiv CH)$ is a terminal alkyne with acidic hydrogen atoms.
It reacts with ammoniacal silver nitrate $(AgNO_3)$ to form a white precipitate of silver acetylide $(AgC \equiv CAg)$.
It reacts with ammoniacal cuprous chloride $(Cu_2Cl_2)$ to form a red precipitate of copper$(I)$ acetylide $(CuC \equiv CCu)$.

(C) The given reaction is a dehydrohalogenation reaction where two molecules of $HCl$ are removed from a gem-dihalide to form an alkyne.
$R-CH_2-CCl_2-R$ undergoes elimination in the presence of a strong base like alcoholic $KOH$ or $NaNH_2$.
However,for the conversion of a gem-dihalide to an alkyne,alcoholic $KOH$ is the standard reagent used to facilitate double dehydrohalogenation.
Thus,the correct reagent is $KOH$ in $C_2H_5OH$.

(D) The reaction of $1,1,2,2$-tetrabromoethane $(CHBr_2-CHBr_2)$ with zinc dust $(Zn)$ is a dehalogenation reaction.
When $1,1,2,2$-tetrabromoethane is heated with zinc dust in an alcoholic solution,it undergoes debromination to form ethyne $(C_2H_2)$.
The reaction is: $CHBr_2-CHBr_2 + 2Zn \rightarrow CH \equiv CH + 2ZnBr_2$.

(C) $1$. The compound must react with $[Ag(NH_3)_2]^+$ (Tollens' reagent),which indicates it is a terminal alkyne $(R-C \equiv CH)$.
$2$. Upon oxidation with alkaline $KMnO_4$,a terminal alkyne undergoes oxidative cleavage of the triple bond to form a carboxylic acid $(R-COOH)$ and $CO_2$.
$3$. The product given is $(CH_3)_3C-COOH$ (pivalic acid). This implies the starting terminal alkyne must be $(CH_3)_3C-C \equiv CH$.
$4$. The reaction is: $(CH_3)_3C-C \equiv CH + [O] \xrightarrow{alkaline \ KMnO_4} (CH_3)_3C-COOH + CO_2$.

(C) The hydration of alkynes in the presence of dilute $H_2SO_4$ and $HgSO_4$ follows Markovnikov's addition.
For a symmetrical alkyne like $CH_3-CH_2-C \equiv C-CH_3$ ($2-$pentyne),the addition of water across the triple bond results in an enol intermediate.
$CH_3-CH_2-C \equiv C-CH_3 + H_2O \xrightarrow{Hg^{2+}, H^+} CH_3-CH_2-C(OH)=CH-CH_3$ (enol).
This enol undergoes tautomerization to form a mixture of ketones.
Specifically,the oxygen can attach to either $C-2$ or $C-3$ of the pentane chain.
Thus,the reaction yields a mixture of $2-$pentanone $(CH_3-CO-CH_2-CH_2-CH_3)$ and $3-$pentanone $(CH_3-CH_2-CO-CH_2-CH_3)$.

(C) Ozonolysis of an internal alkyne followed by hydrolysis results in the cleavage of the triple bond to form two carboxylic acid molecules.
For the given alkyne $CH_3 - C \equiv C - CH_2CH_3$ (pent$-2-$yne):
Step $1$: The triple bond is cleaved.
Step $2$: The carbon atoms of the triple bond are oxidized to carboxylic acid groups.
$CH_3 - C \equiv C - CH_2CH_3 \rightarrow CH_3COOH + CH_3CH_2COOH$.
Thus,the products are acetic acid $(CH_3COOH)$ and propanoic acid $(CH_3CH_2COOH)$.

(C) The synthesis of chloroprene $(2-chloro-1,3-butadiene)$ involves the dimerization of acetylene $(HC \equiv CH)$ to form vinyl acetylene $(CH_2=CH-C \equiv CH)$.
This reaction is catalyzed by a mixture of cuprous chloride $(Cu_2Cl_2)$ and ammonium chloride $(NH_4Cl)$.
Subsequently,vinyl acetylene reacts with hydrogen chloride $(HCl)$ in the presence of cuprous chloride to yield chloroprene.

(D) Compound $(3)$ is a terminal alkyne $(CH_3CH_2C \equiv CH)$.
Terminal alkynes contain an acidic hydrogen atom attached to the $sp$-hybridized carbon.
Ammoniacal silver nitrate $(Tollens' \ reagent)$ reacts with terminal alkynes to form a white precipitate of silver acetylide $(CH_3CH_2C \equiv CAg)$.
Compounds $(1)$,$(2)$,and $(4)$ do not have acidic hydrogen atoms and therefore do not react with ammoniacal silver nitrate.
Thus,ammoniacal silver nitrate is the most suitable reagent to distinguish compound $(3)$ from the others.

(D) The reaction of ethyne $({C_2}{H_2})$ with water in the presence of $HgSO_4$ and $H_2SO_4$ is a hydration reaction.
First,the addition of water to the triple bond occurs to form an unstable enol intermediate,which is $CH_2 = CH - OH$ (vinyl alcohol).
This enol intermediate undergoes tautomerization to form the more stable carbonyl compound,acetaldehyde $(CH_3CHO)$.
Therefore,$X$ is $CH_2 = CH - OH$.

(C) The reaction of ethyne $(C_2H_2)$ with ozone $(O_3)$ followed by reductive workup with $Zn/CH_3COOH$ is an ozonolysis reaction.
$1$. $C_2H_2 + O_3 \rightarrow$ Ozonide intermediate $(X)$.
$2$. The ozonide of ethyne on reductive cleavage with $Zn/CH_3COOH$ yields glyoxal $(CHO-CHO)$.
Therefore,the product $Y$ is $CHO-CHO$.

(D) Acetylene $(HC \equiv CH)$ contains terminal hydrogen atoms attached to $sp$-hybridized carbon atoms. Due to the high $s$-character $(50\%)$ in $sp$-hybridized carbon,the $C-H$ bond is polar,making the hydrogen atom acidic. When acetylene reacts with ammoniacal silver nitrate $(Tollens' \text{ reagent})$,it forms a white precipitate of silver acetylide $(Ag-C \equiv C-Ag)$. This reaction is a characteristic test for terminal alkynes and confirms their acidic nature.

(C) The reaction of acetylene $(HC \equiv CH)$ with acetic acid $(CH_3COOH)$ in the presence of $Hg^{2+}$ catalyst is an addition reaction.
This reaction proceeds as follows:
$HC \equiv CH + CH_3COOH \xrightarrow{Hg^{2+}} CH_2=CH-O-CO-CH_3$
The product formed is vinyl acetate $(CH_2=CH-O-CO-CH_3)$.

(B) The hydration of alkynes in the presence of $HgSO_4$ and $H_2SO_4$ follows Markovnikov's rule.
For a symmetrical alkyne like $CH_3-C \equiv C-CH_3$ ($2-$butyne),the addition of water results in the formation of an enol intermediate,which tautomerizes to a ketone.
$CH_3-C \equiv C-CH_3 + H_2O \xrightarrow{HgSO_4/H_2SO_4} CH_3-C(OH)=CH-CH_3 \rightleftharpoons CH_3-CO-CH_2-CH_3$.
The final product is $2-$butanone.

(C) The reaction of $1, 2-$dibromopropane $(CH_3-CHBr-CH_2Br)$ with $NaNH_2$ (a strong base) leads to dehydrohalogenation to form propyne $(CH_3-C \equiv CH)$.
This process requires $2$ moles of $NaNH_2$ to remove two molecules of $HBr$.
The resulting propyne is then treated with a base to form the sodium propynide $(CH_3-C \equiv C^-Na^+)$,which requires an additional $1$ mole of $NaNH_2$.
Finally,the sodium propynide reacts with ethyl bromide $(CH_3CH_2Br)$ via an $S_N2$ mechanism to form pentyne $(CH_3-C \equiv C-CH_2CH_3)$.
Total moles of $NaNH_2$ required: $2$ (for dehydrohalogenation) $+ 1$ (for deprotonation) $= 3$ moles.

(B) The reaction of $1,2-$dibromoethane $(BrCH_2-CH_2Br)$ with alcoholic $KOH$ is a dehydrohalogenation reaction.
This reaction proceeds via two successive elimination steps to form ethyne $(HC \equiv CH)$ as the final product $X$.
In ethyne $(HC \equiv CH)$,each carbon atom is bonded to one hydrogen atom and triple-bonded to the other carbon atom.
Since each carbon atom is involved in one triple bond and one single bond,the hybridization of each carbon atom is $sp$.