Alkyne Questions in English

(B) $1$. The reaction of $CH_3-CH_2-CH_2-C\equiv CH$ with $LiNH_2$ is an acid-base reaction where the terminal alkyne proton is removed to form the lithium acetylide intermediate $(A)$,which is $CH_3-CH_2-CH_2-C\equiv C^-Li^+$.
$2$. The intermediate $(A)$ then reacts with dimethyl sulfate $(CH_3)_2SO_4$ in a nucleophilic substitution reaction $(S_N2)$.
$3$. The acetylide ion acts as a nucleophile and attacks the methyl group of $(CH_3)_2SO_4$,resulting in the methylation of the alkyne.
$4$. The final product $(B)$ is $CH_3-CH_2-CH_2-C\equiv C-CH_3$ (pent$-2-$yne).

(B) The conversion involves the alkylation of a terminal alkyne.
First,$NaNH_2$ (sodium amide) acts as a strong base to deprotonate the terminal alkyne,forming a nucleophilic sodium alkynide salt.
Second,this alkynide ion undergoes an $S_N2$ reaction with a primary alkyl halide,such as $CH_3-CH_2-CH_2-Br$ (propyl bromide),to extend the carbon chain and form the desired internal alkyne product.

(D) Catalytic hydrogenation of an alkyne involves the addition of hydrogen across the triple bond to form the corresponding alkane.
$3$-ethylhexane has the molecular formula $C_8H_{18}$.
All the given alkynes ($3$-ethylhex-$1$-yne,$4$-ethylhex-$1$-yne,and $3$-ethylhex-$3$-yne) have the same carbon skeleton as $3$-ethylhexane.
Therefore,upon complete catalytic hydrogenation,all of them will yield $3$-ethylhexane as the final product.
Thus,the correct option is $D$.

(A, D) Reagent $(I)$ $(2Na/liq. NH_3)$ is a dissolving metal reduction that converts an alkyne into a $trans$-alkene $(Q)$.
Reagent $(II)$ ($H_2/Pd/CaCO_3$ with quinoline,also known as Lindlar's catalyst) is a partial hydrogenation that converts an alkyne into a $cis$-alkene $(R)$.
Reagent $(III)$ $(2H_2/Pd/C)$ is a complete hydrogenation that converts an alkyne into an alkane.
Therefore,statement $(a)$ is correct as $Q$ is the $trans$-product,and statement $(d)$ is correct as $R$ is the $cis$-product.

(B) $1$. The terminal alkyne $Br-(CH_2)_{12}-C\equiv CH$ reacts with $NaNH_2$ to form an acetylide ion,which then undergoes intramolecular nucleophilic substitution to form a cyclic alkyne $(A)$.
$2$. The partial hydrogenation of the cyclic alkyne $(A)$ using $Lindlar$ catalyst ($Pd/CaCO_3$ poisoned with $PbO$ or quinoline) specifically yields the $cis$-alkene $(B)$.

(B) The reaction involves the nucleophilic addition of the methoxide ion $(MeO^{-})$ to the alkyne.
Since the terminal alkyne $(Ph-C \equiv CH)$ is not acidic enough to be deprotonated by $MeO^{-}$ to a significant extent,the methoxide ion acts as a nucleophile.
The $MeO^{-}$ attacks the electrophilic carbon of the triple bond,followed by protonation by the solvent $(MeOH)$ to yield the enol ether.
The reaction proceeds as follows:
$Ph-C \equiv CH + MeO^{-} \rightarrow Ph-C^{-}=CH-OMe$
$Ph-C^{-}=CH-OMe + MeOH \rightarrow Ph-CH=CH-OMe + MeO^{-}$
Thus,the major product is $Ph-CH=CH-OMe$.

(D) The reaction of $Ph-CCl_2-CH_3$ with $2$ equivalents of $NaNH_2$ undergoes dehydrohalogenation to form the alkyne $Ph-C \equiv CH$.
Since $3$ equivalents of $NaNH_2$ are provided,the third equivalent acts as a strong base to abstract the acidic acetylenic proton from $Ph-C \equiv CH$.
This results in the formation of the sodium acetylide salt: $Ph-C \equiv C^{\Theta} Na^{\oplus} (A)$.

(C) The target compound $(A)$ is a chiral alkyne. To synthesize it,we need to perform an $S_N2$ reaction between a chiral alkyl halide and a nucleophile. The reaction involves the nucleophilic attack of the acetylide ion on the primary carbon of the alkyl bromide. The stereocenter remains unaffected because the reaction occurs at the terminal primary carbon. The correct approach is to use the chiral alkyl bromide $(S)-2\text{-bromobutane}$ derivative to maintain the configuration at the chiral center. Option $(C)$ correctly represents the sequence where the chiral center is preserved and the alkyne chain is extended.

(C) The reaction involves the alkylation of a terminal alkyne.
Step $A$: The terminal alkyne is treated with a strong base like $NaNH_2$ to form an acetylide ion.
Step $B$: The acetylide ion acts as a nucleophile and undergoes an $S_N2$ reaction with an alkyl halide.
To ensure the reaction occurs selectively at the $Br$ end,we use an alkyl halide with a better leaving group on the other side,such as $I-CH_2-CH_2-CH_2-Br$. Since $I^-$ is a better leaving group than $Br^-$,the nucleophilic attack occurs at the carbon attached to the iodine atom.
Therefore,$(A) = NaNH_2$ and $(B) = I-CH_2-CH_2-CH_2-Br$.

(A) The reaction involves the terminal alkyne $Ph-C \equiv C-H$ reacting with $I_2$ in the presence of a base (morpholine).
$1$. The base (morpholine) abstracts the acidic terminal proton from the alkyne to form an acetylide anion: $Ph-C \equiv C^-$.
$2$. The acetylide anion then acts as a nucleophile and attacks the iodine molecule $(I-I)$,displacing an iodide ion $(I^-)$.
$3$. This results in the formation of the iodoalkyne product: $Ph-C \equiv C-I$.

(C) Step $1$: The terminal alkyne reacts with $NaNH_2$ in $NH_3$ to form a sodium acetylide intermediate,which then undergoes an $S_N2$ reaction with $CH_3Br$ to form the methylated alkyne $(A)$,which is $1-(cyclohex-1-en-1-yl)prop-1-yne$.
Step $2$: The hydrogenation of the alkyne $(A)$ using Lindlar catalyst $(H_2/Pd/BaSO_4)$ is a stereoselective reaction that adds hydrogen across the triple bond in a $syn$-fashion,resulting in the formation of the $cis$-alkene $(B)$,which is $1-(cyclohex-1-en-1-yl)prop-1-ene$ (cis isomer).

(D) The hydrogenation of an alkyne with $2 \ mole$ of $H_2$ in the presence of $Pt$ catalyst results in the corresponding alkane.
For a compound to be optically active,it must contain a chiral center (a carbon atom bonded to four different groups).
Let's analyze the products:
$A$) $3-$Methyl-$1-$pentyne $(CH_3CH_2CH(CH_3)C \equiv CH)$ on hydrogenation gives $3-$methylpentane $(CH_3CH_2CH(CH_3)CH_2CH_3)$. This molecule has a chiral center at $C-3$ $(CH_3CH_2-CH(CH_3)-CH_2CH_3)$,so it is optically active.
$B$) $4-$Methyl-$1-$hexyne $(CH_3CH_2CH(CH_3)CH_2C \equiv CH)$ on hydrogenation gives $3-$methylhexane $(CH_3CH_2CH(CH_3)CH_2CH_2CH_3)$. This molecule has a chiral center at $C-3$,so it is optically active.
$C$) $3-$Methyl-$1-$heptyne $(CH_3CH_2CH_2CH_2CH(CH_3)C \equiv CH)$ on hydrogenation gives $3-$methylheptane $(CH_3CH_2CH_2CH_2CH(CH_3)CH_2CH_3)$. This molecule has a chiral center at $C-3$,so it is optically active.
Since all the given options result in optically active alkanes,the correct answer is $D$.

(C) $1$. Calcium carbide $(CaC_2)$ reacts with water $(H_2O)$ to produce acetylene $(C_2H_2)$,which is compound $(A)$.
$CaC_2 + 2H_2O \rightarrow Ca(OH)_2 + C_2H_2 (A)$
$2$. When acetylene $(C_2H_2)$ is passed through a red-hot copper tube at high temperature,it undergoes cyclic polymerization to form benzene $(C_6H_6)$,which is compound $(B)$.
$3C_2H_2 \xrightarrow{\text{Red hot Cu tube}} C_6H_6 (B)$
$3$. Therefore,the product $(B)$ is Benzene.

(A) Step $1$: The terminal alkyne $CH_3CH_2-C \equiv CH$ reacts with $NaNH_2$ to form an acetylide ion,which then undergoes an $S_N2$ reaction with $CH_3CH_2Br$ to yield $A$ ($hex-3-yne$,$CH_3CH_2-C \equiv C-CH_2CH_3$).
Step $2$: The reduction of $hex-3-yne$ $(A)$ with $Na$ in liquid $NH_3$ (Birch reduction) is a stereoselective reaction that produces $trans-hex-3-ene$ $(B)$.
Step $3$: The addition of $Br_2$ in $CH_2Cl_2$ to $trans-hex-3-ene$ $(B)$ is an anti-addition reaction,which results in the formation of $3,4-dibromohexane$ $(C)$.
Thus,the final product $C$ is $CH_3CH_2CH(Br)CH(Br)CH_2CH_3$.

(C) The synthesis of but$-1$-ene from ethanol $(CH_3CH_2OH)$ involves the following steps:
$1$. Conversion of ethanol to bromoethane: $CH_3CH_2OH + HBr \rightarrow CH_3CH_2Br + H_2O$.
$2$. Nucleophilic substitution with sodium acetylide: $CH_3CH_2Br + NaC \equiv CH \rightarrow CH_3CH_2C \equiv CH$ (but$-1$-yne).
$3$. Partial hydrogenation of the alkyne to an alkene using Lindlar catalyst: $CH_3CH_2C \equiv CH + H_2 \xrightarrow{Lindlar \ Pd} CH_3CH_2CH=CH_2$ (but$-1$-ene).
Therefore,the correct sequence is $1. HBr$,heat; $2. NaC \equiv CH; 3. H_2$,Lindlar $Pd$.

(B) Ozonolysis of an internal alkyne $(R-C \equiv C-R')$ followed by hydrolysis yields carboxylic acids $(RCOOH + R'COOH)$.
For the product to be only butanoic acid $(CH_3CH_2CH_2COOH)$,the alkyne must be symmetrical with a four-carbon chain on each side of the triple bond.
Thus,the alkyne is $CH_3CH_2CH_2-C \equiv C-CH_2CH_2CH_3$,which is $4-$octyne.
Reaction: $CH_3CH_2CH_2-C \equiv C-CH_2CH_2CH_3 \xrightarrow{O_3, H_2O} 2CH_3CH_2CH_2COOH$.

(C) $1$. Hydrolysis of maleic anhydride with $H_3O^+$ yields maleic acid $(A)$,which is $cis$-but$-2-$enedioic acid.
$2$. Reaction of maleic acid with $NaOH$ gives the sodium salt of maleic acid $(B)$,which is sodium maleate.
$3$. Kolbe electrolysis of sodium maleate involves the decarboxylation of the carboxylate ions at the anode,leading to the formation of a triple bond between the carbons that were originally attached to the carboxylate groups.
$4$. The reaction is: $cis-NaOOC-CH=CH-COONa \xrightarrow{\text{electrolysis}} HC \equiv CH + 2CO_2 + H_2 + 2NaOH$.
$5$. Thus,the final product $(C)$ is ethyne $(HC \equiv CH)$.

(C) The conversion of $1$-butyne $(CH_3CH_2C \equiv CH)$ to $1-D$-butanal $(CH_3CH_2CH_2CDO)$ requires the introduction of a deuterium atom at the $C-1$ position and the conversion of the alkyne to an aldehyde.
Step $I$ (Sodium amide,then $D_2O$) would produce $1-D$-butyne $(CH_3CH_2C \equiv CD)$.
Step $II$ (Hydroboration-oxidation using disiamylborane followed by $H_2O_2/NaOH$) converts $1$-butyne to butanal $(CH_3CH_2CH_2CHO)$.
To obtain $1-D$-butanal,one would need to perform hydroboration-oxidation on $1-D$-butyne using $D_2O_2/NaOD$ instead of $H_2O_2/NaOH$.
Since the provided reagents in $I$ and $II$ do not lead to the desired product $1-D$-butanal,option $III$ is correct.

$(A)$ The molecular formula $C_4H_6$ corresponds to the general formula $C_nH_{2n-2}$, which indicates an alkyne or a diene.
Since $(A)$ reacts with excess $Br_2$ to form $C_4H_6Br_4$, it indicates the presence of two double bonds or one triple bond.
Terminal alkynes react with ammoniacal silver nitrate $(\text{Tollens' reagent})$ to form a white precipitate of silver acetylide.
But-$1$-yne $(CH_3-CH_2-C \equiv CH)$ is a terminal alkyne, whereas but-$2$-yne $(CH_3-C \equiv C-CH_3)$ is a non-terminal alkyne.
Therefore, $(A)$ is but-$1$-yne.

(C) The reaction proceeds in two main stages:
$1$. Dehydrohalogenation: $1,2$-dibromopropane $(CH_3-CH(Br)-CH_2Br)$ reacts with $2$ moles of $NaNH_2$ to undergo double dehydrohalogenation,forming propyne $(CH_3-C \equiv CH)$.
$2$. Alkylation: The terminal alkyne (propyne) reacts with another $1$ mole of $NaNH_2$ to form the sodium acetylide intermediate $(CH_3-C \equiv C^-Na^+)$,which then reacts with ethyl bromide $(CH_3-CH_2Br)$ via an $S_N2$ mechanism to form pentyne $(CH_3-C \equiv C-CH_2-CH_3)$.
Thus,the total number of moles of $NaNH_2$ required is $2 + 1 = 3$ moles. Therefore,$X = 3$.

(C) The addition of $HBr$ to alkynes follows Markovnikov's rule.
In the presence of excess $HBr$,two molecules of $HBr$ add to the triple bond to form a geminal dihalide.
Step $1$: $CH_3-CH(CH_3)-C \equiv CH + HBr \rightarrow CH_3-CH(CH_3)-C(Br)=CH_2$
Step $2$: $CH_3-CH(CH_3)-C(Br)=CH_2 + HBr \rightarrow CH_3-CH(CH_3)-C(Br)_2-CH_3$ ($2,2$-dibromo-$3$-methylbutane).

(B) Cold $KMnO_4$ (Baeyer's reagent) acts as an oxidizing agent for alkynes.
It oxidizes internal alkynes to form vicinal diketones ($1,2$-diketones).
The reaction is: $CH_3-C \equiv C-CH_3 \xrightarrow{\text{Cold } KMnO_4} CH_3-C(=O)-C(=O)-CH_3$ (Butane-$2,3$-dione).

(D) All the given reactions result in the formation of phenylacetylene $(Ph-C \equiv CH)$.
$(a)$ Geminal dihalide $Ph-C(Br)_2-CH_3$ undergoes double dehydrohalogenation with $NaNH_2$ to form the alkyne.
$(b)$ Styrene $(C_6H_5-CH=CH_2)$ reacts with $Br_2$ to form a vicinal dihalide,which then undergoes double dehydrohalogenation.
$(c)$ Acetophenone $(C_6H_5-C(=O)-CH_3)$ reacts with $PCl_5$ to form a geminal dichloride,which then undergoes double dehydrohalogenation.
In all cases,$NH_4Cl$ is used for the protonation of the acetylide ion formed.

(C) Step $1$: Ethyne $(HC \equiv CH)$ reacts with excess $NaNH_2$ (a strong base) to remove both acidic protons,forming the disodium acetylide intermediate,$Na^{\oplus} [C \equiv C]^{2-} Na^{\oplus}$.
Step $2$: The disodium acetylide acts as a nucleophile and undergoes an $S_N2$ reaction with excess $I-CH_2-CH_2-CH_2-CH_3$ (butyl iodide). Each carbon of the acetylide attacks a butyl group,displacing the iodide ion.
Step $3$: The final product formed is $CH_3-CH_2-CH_2-CH_2-C \equiv C-CH_2-CH_2-CH_2-CH_3$,which is $5$-decyne.

(C) To prepare $2$-heptanone $(CH_{3}COCH_{2}CH_{2}CH_{2}CH_{2}CH_{3})$,we need a terminal alkyne that,upon hydration,yields the ketone at the $C-2$ position.
$1$-hexyne $(CH_{3}CH_{2}CH_{2}CH_{2}C \equiv CH)$ reacts with $NaNH_{2}$ to form the acetylide ion $(CH_{3}CH_{2}CH_{2}CH_{2}C \equiv C^-)$.
This ion reacts with $CH_{3}Br$ via an $S_{N}2$ mechanism to form $2$-heptyne $(CH_{3}CH_{2}CH_{2}CH_{2}C \equiv C-CH_{3})$.
Finally,the acid-catalyzed hydration of $2$-heptyne using $H_{2}O, H_{2}SO_{4}, Hg^{2+}$ follows Markovnikov's rule to produce $2$-heptanone.
Thus,option $C$ is the correct sequence.

(C) The reaction is the hydration of an alkyne using $HgSO_4$ and $dil. H_2SO_4$ (Kucherov reaction).
Step $-I$: The $Hg^{2+}$ ion acts as an electrophile and forms a non-classical carbocation intermediate with the triple bond.
Step $-II$: Water (nucleophile) attacks the more substituted carbon atom to form a stable carbocation intermediate,which eventually leads to the formation of an enol.
Step $-III$: The unstable enol undergoes tautomerization to form the more stable ketone.
In the given substrate $C_6H_5-C \equiv C-(C_6H_4)-OCH_3$,the attack of water occurs such that the carbonyl group is formed on the carbon closer to the electron-donating $-OCH_3$ group,resulting in $C_6H_5-CH_2-CO-(C_6H_4)-OCH_3$ as the major product.

(B) In reaction $1$,the hydration of propyne with $HgSO_4/dil. H_2SO_4$ follows Markovnikov's rule to form acetone $(CH_3-CO-CH_3)$.
In reaction $2$,hydroboration-oxidation follows anti-Markovnikov's rule to form propanal $(CH_3-CH_2-CHO)$.
Acetone,being a methyl ketone,gives a positive iodoform test with $NaOI$ (forming a yellow precipitate of $CHI_3$),while propanal does not.
Therefore,they can be differentiated using $NaOI$.

(C) The reaction sequence is as follows:
$1$. The hydration of an alkyne $(A)$ in the presence of $HgSO_4$ and $dil. H_2SO_4$ yields a ketone $(B)$.
$2$. The reduction of ketone $(B)$ with $LiAlH_4$ yields a secondary alcohol $(C)$.
$3$. For the product $(C)$ to be a racemic mixture,it must contain a chiral center. This implies that the ketone $(B)$ must be an unsymmetrical ketone such that its reduction creates a chiral carbon.
$4$. Looking at the options,$CH_3-C \equiv C-CH_3$ (but$-2-$yne) is symmetrical and would yield butan$-2-$one,which upon reduction gives a racemic mixture of butan$-2-$ol.
$5$. However,the provided image shows the reaction of but$-2-$yne $(CH_3-C \equiv C-CH_3)$ to form butan$-2-$one,which is then reduced to a racemic mixture of butan$-2-$ol.
$6$. Therefore,reactant $(A)$ is $CH_3-C \equiv C-CH_3$.

(A) The reaction is an elimination reaction (dehydrohalogenation) using a strong base,potassium tert-butoxide $(K^+ \, ^-OC(CH_3)_3)$.
This reaction proceeds via an $E2$ mechanism.
In the $E2$ elimination of vinyl halides,the base removes the proton $(H^+)$ and the leaving group $(Br^-)$ must be in an anti-periplanar orientation.
The starting material is $Ph-C^*(=CHBr)-C_6H_4-Br$.
The proton $(H)$ on the terminal carbon and the bromine $(Br)$ on the internal carbon are eliminated to form an alkyne.
Since the $H$ is on the terminal carbon and the $Br$ is on the carbon attached to the phenyl ring,the triple bond forms between the two carbons.
The carbon marked with the asterisk $(C^*)$ is the one attached to the phenyl group $(Ph)$.
Following the anti-elimination,the product is $Ph-C^* \equiv C-C_6H_4-Br$.

(B) The cyclic polymerization of acetylene $(CH \equiv CH)$ at high temperature (red hot iron tube) yields benzene $(C_6H_6)$,which is a poly-ene (a compound with multiple double bonds in a ring structure).
$3CH \equiv CH \xrightarrow{\Delta, \text{Fe tube}} C_6H_6$ (Benzene).
Similarly,propyne undergoes cyclic polymerization to form $1,3,5$-trimethylbenzene.

(C) The reduction of an internal alkyne to a trans-alkene is achieved using dissolving metal reduction,typically with sodium or lithium in liquid ammonia $(NH_3(l))$.
$Ph-C \equiv C-Ph \xrightarrow{Li / NH_3(l)} \text{trans-}Ph-CH=CH-Ph$
Catalytic hydrogenation with $H_2 / \text{Lindlar catalyst}$ yields the cis-alkene,while complete catalytic hydrogenation yields the alkane. $LiAlH_4$ does not reduce alkynes to alkenes.

(C) The reaction of an alkyne with $H^{+}/Hg^{2+}$ (mercuric sulfate in dilute sulfuric acid) is a hydration reaction that follows Markovnikov's rule.
$Ph-C \equiv C-CH_3 + H_2O \xrightarrow{H^{+}/Hg^{2+}} Ph-C(OH)=CH-CH_3$ (enol intermediate).
The enol intermediate undergoes tautomerization to form the more stable ketone.
$Ph-C(OH)=CH-CH_3 \rightleftharpoons Ph-CO-CH_2CH_3$ (propiophenone).
Thus,the final product $A$ is $Ph-CO-CH_2CH_3$.

(A) The hydration of propyne $(CH_3-C\equiv CH)$ in the presence of $H_2SO_4$ and $HgSO_4$ follows Markovnikov's addition.
First,an unstable enol intermediate,$CH_3-C(OH)=CH_2$ (prop-$1$-en-$2$-ol),is formed.
This enol undergoes tautomerization to form the stable ketone,$CH_3-CO-CH_3$ (propan-$2$-one or acetone).
Reaction: $CH_3-C\equiv CH + H_2O$ $\xrightarrow[HgSO_4]{H_2SO_4} [CH_3-C(OH)=CH_2]$ $\rightarrow CH_3-CO-CH_3$.

(D) Acidic strength is directly proportional to the stability of the conjugate base.
Stability of the conjugate base depends on the electronegativity of the carbon atom bearing the negative charge.
The electronegativity order is $sp$ carbon ($50\% \ s$-character) > $sp^2$ carbon ($33.3\% \ s$-character) > $sp^3$ carbon ($25\% \ s$-character).
$1$. $CH \equiv CH$ (ethyne) has $sp$ hybridized carbons.
$2$. $CH_3 - C \equiv CH$ (propyne) has one $sp$ hybridized carbon,but the methyl group $(CH_3)$ is electron-donating ($+I$ effect),which destabilizes the conjugate base compared to ethyne.
$3$. $CH_2 = CH_2$ (ethene) has $sp^2$ hybridized carbons,which are less electronegative than $sp$ carbons.
Therefore,the correct order of acidic strength is $CH \equiv CH > CH_3 - C \equiv CH > CH_2 = CH_2$.

(D) The reaction of a vicinal dihalide with alcoholic $KOH$ followed by $NaNH_2$ results in double dehydrohalogenation to form an alkyne.
Step $1$: $CH_3-CH_2-CH(Br)-CH_2-Br \xrightarrow{KOH(alc)} CH_3-CH_2-CH=CHBr$ (Vinyl halide).
Step $2$: $CH_3-CH_2-CH=CHBr \xrightarrow{NaNH_2/liq. NH_3} CH_3-CH_2-C \equiv CH$ (But$-1-$yne).
The major product is but$-1-$yne.

(B) The reaction proceeds via electrophilic addition to the alkyne.
Step $1$: The addition of $DCl$ $(1 \ equiv)$ to $CH_3-C\equiv CH$ follows Markovnikov's rule. The electrophile $D^+$ adds to the terminal carbon to form the more stable secondary carbocation, followed by the attack of $Cl^-$. This yields $CH_3-C(Cl)=CHD$.
Step $2$: The addition of $DI$ to the alkene $CH_3-C(Cl)=CHD$ follows Markovnikov's rule. The electrophile $D^+$ adds to the carbon already bearing the deuterium (to form a stable carbocation adjacent to the chlorine atom), followed by the attack of $I^-$. This results in the geminal dihalo-dideutero product $CH_3-C(I)(Cl)-CHD_2$.

(B) reacts with $Ag_2O$ to give a precipitate,indicating it is a terminal alkyne.
$A$ undergoes hydration $(Hg^{2+}/H^{+})$ to form $B$,which is reduced by $NaBH_4$ to $C$.
$C$ gives turbidity with Lucas reagent $(conc. HCl / ZnCl_2)$ within $5 \text{ minutes}$,indicating $C$ is a secondary alcohol.
$CH_3-C \equiv CH (A) \xrightarrow{Ag_2O} CH_3-C \equiv CAg \downarrow (\text{ppt})$
$CH_3-C \equiv CH (A)$ $\xrightarrow{Hg^{2+}/H^{+}} CH_3-COCH_3 (B)$ $\xrightarrow{NaBH_4} CH_3-CH(OH)-CH_3 (C)$
$CH_3-CH(OH)-CH_3 (C) \xrightarrow{conc. HCl / ZnCl_2} CH_3-CH(Cl)-CH_3 + H_2O$ (Turbidity within $5 \text{ minutes}$).

(A) $1$-Alkynes (terminal alkynes) contain an acidic hydrogen atom attached to the $sp$-hybridized carbon $(R-C \equiv C-H)$.
When treated with Tollen's reagent (ammoniacal silver nitrate,$AgNO_3 + NH_4OH$),terminal alkynes form a white precipitate of silver acetylide $(R-C \equiv C-Ag)$.
$2$-Alkynes (internal alkynes) do not have an acidic hydrogen atom and therefore do not react with Tollen's reagent to form a precipitate.
Thus,Tollen's reagent is used to distinguish between $1$-alkyne and $2$-alkyne.

(C) Acetylene $(HC \equiv CH)$ undergoes hydration in the presence of $HgSO_4$ and $H_2SO_4$ to form vinyl alcohol,which tautomerizes to acetaldehyde $(CH_3CHO)$.
$HC \equiv CH + H_2O$ $\xrightarrow{HgSO_4/H_2SO_4} [CH_2=CH-OH]$ $\rightarrow CH_3CHO$

(B) The reaction of $CH_3-C \equiv C-CH_3$ with acidic $KMnO_4$ under heating conditions causes oxidative cleavage of the alkyne to form two molecules of acetic acid $(A)$:
$CH_3-C \equiv C-CH_3 \xrightarrow{H^{+}/KMnO_4, \Delta} 2 CH_3-COOH$
Next,the reduction of acetic acid $(CH_3-COOH)$ using $LiAlH_4$ followed by acidic workup $(H^{+})$ yields ethanol $(B)$:
$CH_3-COOH \xrightarrow{(1) LiAlH_4, (2) H^{+}} CH_3-CH_2-OH$

(C) The hydration of alkynes in the presence of $H_2SO_4$ and $HgSO_4$ is known as Kucherov's reaction.
Ethyne $(CH \equiv CH)$ is the only alkyne that produces an aldehyde ($CH_3CHO$,acetaldehyde) upon hydration.
All other terminal alkynes produce ketones due to the formation of a more stable carbocation intermediate or the specific tautomerization of the enol form.
For $CH \equiv CH + H_2O$ $\xrightarrow{H_2SO_4/HgSO_4} [CH_2=CH-OH]$ $\rightarrow CH_3-CHO$.

(A) The reaction sequence is as follows:
$1$. The hydration of an alkyne $A$ in the presence of $HgSO_4$ and $H_2SO_4$ yields a ketone $B$. For phenylacetylene $(C_6H_5-C \equiv CH)$,hydration follows Markovnikov's rule to give acetophenone $(C_6H_5-CO-CH_3)$ as the product $B$.
$2$. Acetophenone $(C_6H_5-CO-CH_3)$ contains a methyl ketone group,which undergoes the iodoform reaction with $I_2$ and $NaOH$ to produce iodoform ($CHI_3$,product $C$) and sodium benzoate $(C_6H_5COONa)$.
$3$. Therefore,the reactant $A$ is phenylacetylene.

(A) The acidity of hydrocarbons depends on the hybridization of the carbon atom to which the hydrogen is attached. The acidity order is: terminal alkynes ($sp$ hybridized) > alkenes ($sp^2$ hybridized) > alkanes ($sp^3$ hybridized).
$1$. In ethyne $(HC\equiv CH)$,the carbon is $sp$ hybridized,which has $50\%$ $s$-character. This makes the $C-H$ bond more polar and the conjugate base (acetylide ion) more stable.
$2$. In ethene $(CH_2=CH_2)$,the carbon is $sp^2$ hybridized ($33.3\%$ $s$-character).
$3$. In ethane $(CH_3-CH_3)$,the carbon is $sp^3$ hybridized ($25\%$ $s$-character).
$4$. In benzene $(C_6H_6)$,the carbon is $sp^2$ hybridized.
Therefore,ethyne contains the most acidic hydrogen.

(B) For the molecular formula $C_5H_8$,the degree of unsaturation is $2$. Since we are looking for alkynes,each structure must contain one triple bond.
The possible structural isomers are:
$1.$ $CH \equiv C-CH_2-CH_2-CH_3$ ($1$-pentyne)
$2.$ $CH_3-C \equiv C-CH_2-CH_3$ ($2$-pentyne)
$3.$ $CH \equiv C-CH(CH_3)_2$ ($3$-methyl-$1$-butyne)
Therefore,there are $3$ possible structural alkynes.

(A) The reaction of an alkyne with water in the presence of $dil. H_2SO_4$ and $HgSO_4$ is a hydration reaction.
For the given substrate $CH_3-C \equiv C-O-CH_3$,the hydration follows Markovnikov's rule.
The $OH^-$ group attacks the more substituted carbon (or the carbon attached to the electron-donating group) to form an enol intermediate.
However,in the presence of an alkoxy group $(-OCH_3)$,the reaction proceeds to form an ester.
The hydration of the alkyne $CH_3-C \equiv C-O-CH_3$ leads to the formation of methyl propionate $(CH_3CH_2COOCH_3)$ as the major product.

(B) The molecular formula $C_6H_{10}$ corresponds to an alkyne (general formula $C_nH_{2n-2}$). The structural isomers are as follows:
$1$. Hex$-1-$yne: $CH_3-CH_2-CH_2-CH_2-C \equiv CH$
$2$. Hex$-2-$yne: $CH_3-CH_2-CH_2-C \equiv C-CH_3$
$3$. Hex$-3-$yne: $CH_3-CH_2-C \equiv C-CH_2-CH_3$
$4$. $3-$Methylpent$-1-$yne: $CH_3-CH_2-CH(CH_3)-C \equiv CH$
$5$. $4-$Methylpent$-1-$yne: $CH_3-CH(CH_3)-CH_2-C \equiv CH$
$6$. $4-$Methylpent$-2-$yne: $CH_3-CH(CH_3)-C \equiv C-CH_3$
$7$. $3,3-$Dimethylbut$-1-$yne: $(CH_3)_3C-C \equiv CH$
There are a total of $7$ structural isomers.

(B) Step $1$: $CH_3-C \equiv CH$ reacts with $Na$ to form sodium propyne,$X$ $(CH_3-C \equiv C^-Na^+)$.
Step $2$: Sodium propyne $(X)$ reacts with methyl chloride $(CH_3-Cl)$ via an $S_N2$ mechanism to form but$-2-$yne $(Y)$,which is $CH_3-C \equiv C-CH_3$.

(C) The reaction between calcium carbide $(CaC_2)$ and nitrogen $(N_2)$ at high temperatures is given by:
$CaC_2 + N_2 \to CaCN_2 + C$
The product formed is calcium cyanamide $(CaCN_2)$.

(D) $1$. The reaction starts with propyne $(H_3C-C \equiv CH)$.
$2$. Treatment with $NaNH_2$ removes the acidic terminal proton to form the sodium acetylide intermediate $(A)$: $H_3C-C \equiv C^- Na^+$.
$3$. This intermediate undergoes an $S_N2$ reaction with bromoethane $(CH_3CH_2Br)$ to form pent$-2-$yne $(B)$: $H_3C-C \equiv C-CH_2CH_3$.
$4$. Finally,hydrogenation with Lindlar's catalyst $(H_2/Pd, BaSO_4)$ reduces the internal alkyne to a $cis$-alkene.
$5$. Therefore,the product $(C)$ is $cis$-pent$-2-$ene.

(D) The reaction of $CaC_2$ with $H_2O$ produces Acetylene $(X = HC \equiv CH)$.
When Acetylene is passed through a solution of $Cu_2Cl_2$ and $NH_4Cl$,it undergoes linear dimerization to form Vinylacetylene $(Y = CH_2 = CH - C \equiv CH)$.