Alkyne Questions in English

(D) Partially deactivated palladised charcoal $(H_2 / Pd / CaCO_3)$ is known as Lindlar catalyst.
It is used for the partial hydrogenation of alkynes to cis-alkenes.

(C) The reaction of an unsaturated hydrocarbon $X$ with ozone (ozonolysis) produces compound $A$.
Compound $A$ gives a positive Tollen's test (forms a silver mirror with ammoniacal silver nitrate),which indicates that $A$ is an aldehyde or formic acid $(HCOOH)$.
Let us analyze the options:
$(A)$ $CH_3-C(CH_3)=C(CH_3)-CH_3$ on ozonolysis gives $CH_3-CO-CH_3$ (acetone),which does not give Tollen's test.
$(B)$ $CH_3-C(CH_3)=C(CH_2)_2$ on ozonolysis gives $CH_3-CO-CH_3$ and cyclobutanone,neither of which gives Tollen's test.
$(C)$ $CH_3-CH_2-C \equiv CH$ on ozonolysis gives $CH_3-CH_2-COOH$ and $HCOOH$. Formic acid $(HCOOH)$ gives a positive Tollen's test.
$(D)$ $CH_3-C \equiv C-CH_3$ on ozonolysis gives $2CH_3-COOH$ (acetic acid),which does not give Tollen's test.
Therefore,the correct hydrocarbon is $CH_3-CH_2-C \equiv CH$.

(C) The hydration of alkynes in the presence of $HgSO_4$ and $H_2SO_4$ follows Markovnikov's rule.
$1$. $HC \equiv CH + H_2O \xrightarrow{HgSO_4, H_2SO_4} CH_2=CH(OH) \rightleftharpoons CH_3-CHO$ (Ethanal).
$2$. Cyclohexylacetylene + $H_2O \xrightarrow{HgSO_4, H_2SO_4} \text{Cyclohexyl methyl ketone}$.
$3$. $CH_3-C \equiv CH + H_2O \xrightarrow{HgSO_4, H_2SO_4} CH_3-C(OH)=CH_2 \rightleftharpoons CH_3-CO-CH_3$ (Propanone).
$4$. $CH_3-C \equiv C-CH_3 + H_2O \xrightarrow{HgSO_4, H_2SO_4} CH_3-C(OH)=CH-CH_3 \rightleftharpoons CH_3-CO-CH_2-CH_3$ (Butan$-2-$one).
$CH_3-CH_2-CHO$ (Propanal) is an aldehyde that cannot be formed by the hydration of any alkyne because the hydration of terminal alkynes (other than ethyne) always yields a ketone due to the formation of a more stable secondary enol intermediate.

(B) Metallic sodium reacts with compounds containing acidic hydrogen atoms.
$tert-$butyl alcohol ($CH_3)_3COH$ has an acidic hydroxyl hydrogen.
Ethyne $(HC \equiv CH)$ has acidic terminal hydrogen atoms.
Gaseous ammonia $(NH_3)$ reacts with sodium to form sodamide $(NaNH_2)$.
$But-2-yne$ $(CH_3-C \equiv C-CH_3)$ does not have any acidic hydrogen atoms attached to the triply bonded carbons,therefore it does not react with metallic sodium.

(C) The structure of the molecule is $CH_3-CH_2-CH_2-C \equiv CH$.
To find the number of $\sigma$ bonds,we count all single bonds and one bond from each multiple bond.
- $C-H$ bonds: $3$ (in $CH_3$) + $2$ (in $CH_2$) + $2$ (in $CH_2$) + $1$ (in $CH$) = $8$ $\sigma$ bonds.
- $C-C$ bonds: $1$ $(C-C)$ + $1$ $(C-C)$ + $1$ $(C-C)$ = $3$ $\sigma$ bonds.
- $C \equiv C$ bond: $1$ $\sigma$ bond.
Total $\sigma$ bonds = $8 + 3 + 1 = 12$.

(A) The reaction of vinyl bromide $(CH_2=CHBr)$ with a strong base like $NaNH_2$ is a dehydrohalogenation reaction.
$NaNH_2$ is a very strong base that abstracts a proton from the vinyl bromide,leading to the elimination of $HBr$ and the formation of the triple bond in acetylene $(HC \equiv CH)$.
The reaction is:
$CH_2=CHBr + NaNH_2 \rightarrow HC \equiv CH + NaBr + NH_3$
Thus,the major product is acetylene $(HC \equiv CH)$.

(A) The reaction sequence is as follows:
$1$. The terminal alkyne $Ph-C \equiv CH$ reacts with $NaNH_2$ in $NH_3$ to form the acetylide ion $Ph-C \equiv C^- Na^+$.
$2$. This nucleophilic acetylide ion undergoes an $S_N2$ reaction with $CH_3I$ to form the substituted alkyne $Ph-C \equiv C-CH_3$.
$3$. The catalytic hydrogenation of the internal alkyne $Ph-C \equiv C-CH_3$ using $H_2$ and $Pd/C$ (Lindlar's catalyst is not specified,but $Pd/C$ typically reduces alkynes to alkenes) results in the formation of the cis-alkene,$Ph-CH=CH-CH_3$.

(D) The hydrocarbon $(X)$ is but$-1-$yne $(CH_3-CH_2-C \equiv CH)$.
$1$. It decolourises bromine water due to the presence of a triple bond.
$2$. It forms a white precipitate with ethanolic $AgNO_3$ (Tollens' reagent) because it is a terminal alkyne,which has an acidic hydrogen.
$3$. Treatment of but$-1-$yne with $HgCl_2$ in aqueous $H_2SO_4$ (hydration) produces butan$-2-$one $(CH_3-CH_2-CO-CH_3)$.
$4$. Butan$-2-$one contains a methyl ketone group $(CH_3-CO-)$,which gives a positive iodoform test (yellow precipitate of $CHI_3$) when treated with $I_2$ and $NaOH$.

(A) The reaction proceeds in three steps:
$1$. Treatment of $1,2-dibromo-1-phenylethane$ with excess alcoholic $KOH$ leads to dehydrohalogenation,forming $bromostyrene$ $(Ph-CH=CHBr)$.
$2$. Further treatment with $NaNH_2$ (a strong base) causes another dehydrohalogenation to form the acetylide intermediate $(Ph-C \equiv C^- Na^+)$.
$3$. Finally,protonation with $H_3O^+$ yields the terminal alkyne,$phenylacetylene$ $(Ph-C \equiv CH)$.

(A) The molecular formula $C_6H_{10}$ corresponds to the general formula $C_nH_{2n-2}$,indicating an alkyne.
Treatment with $HBr$ to form a $gem$-dibromide indicates that the alkyne is a terminal alkyne (Markownikoff addition).
Acid-catalyzed hydration of a terminal alkyne using $HgSO_4$ and dil. $H_2SO_4$ produces a methyl ketone.
$A$ methyl ketone $(CH_3CO-R)$ gives a positive iodoform test.
Among the options,hex$-1-$yne $(CH_3CH_2CH_2CH_2C \equiv CH)$ is a terminal alkyne.
Upon hydration,it forms hexan$-2-$one $(CH_3CH_2CH_2CH_2COCH_3)$,which is a methyl ketone and gives a positive iodoform test.
Therefore,the compound $X$ is hex$-1-$yne.

(C) The reaction sequence is as follows:
$1$. Propyne $(CH_3C\equiv CH)$ reacts with $NaNH_2$ to form the sodium salt of propyne $(CH_3C\equiv C^-Na^+)$.
$2$. This salt reacts with methyl iodide $(CH_3I)$ as reagent $x$ to perform an $S_N2$ reaction,yielding but$-2-$yne $(CH_3C\equiv CCH_3)$.
$3$. But$-2-$yne is then reduced to trans-but$-2-$ene using sodium in liquid ammonia $(Na/liq. NH_3)$ as reagent $y$ (Birch reduction).
Thus,$x= CH_3I$ and $y= Na/liq. NH_3$.

(D) The oxidation of acetylene $(HC \equiv CH)$ with alkaline $KMnO_4$ proceeds through the formation of glyoxal $(CHO-CHO)$ as an intermediate.
Further oxidation of glyoxal leads to the formation of oxalic acid $(HOOC-COOH)$ as the major product.
The reaction is:
$HC \equiv CH + 2[O]$ $\xrightarrow{Alk. KMnO_4} CHO-CHO$ $\xrightarrow{2[O]} HOOC-COOH$
Therefore,the correct option is $(D)$.

(B)
For the conversion of $2,3$-dibromobutane to $2$-butyne,the following steps are used:
Step $1$: Dehydrohalogenation of $2,3$-dibromobutane using alcoholic $KOH$ yields $2$-bromobut-$2$-ene.
$CH_3-CH(Br)-CH(Br)-CH_3 \xrightarrow{\text{alc. } KOH} CH_3-C(Br)=CH-CH_3 + HBr$
Step $2$: Further dehydrohalogenation of $2$-bromobut-$2$-ene using a strong base like $NaNH_2$ yields $2$-butyne.
$CH_3-C(Br)=CH-CH_3 \xrightarrow{NaNH_2} CH_3-C \equiv C-CH_3 + NaBr + NH_3$

(B) The reaction is an electrophilic addition of $HBr$ to propyne $(H_3C-C\equiv CH)$.
Step $1$: The first molecule of $HBr$ adds to the triple bond following Markownikoff's rule,where the electrophile $H^+$ attaches to the terminal carbon to form a more stable secondary carbocation $(H_3C-C^+=CH_2)$. The bromide ion $(Br^-)$ then attacks the carbocation to form $2-$bromopropene $(H_3C-C(Br)=CH_2)$.
Step $2$: The second molecule of $HBr$ adds to the double bond of $2-$bromopropene,again following Markownikoff's rule. The $H^+$ adds to the terminal carbon $(CH_2)$ to form a more stable carbocation $(H_3C-C^+(Br)-CH_3)$,which is stabilized by the resonance effect of the bromine atom. The $Br^-$ then attacks this carbocation to yield $2,2-$dibromopropane $(H_3C-C(Br)_2-CH_3)$ as the major product.

(C) .
When calcium carbide is added to water,acetylene gas is evolved.
The reaction is as follows:
$\underset{\text{(Calcium Carbide)}}{CaC_2} + 2H_2O$ $\longrightarrow Ca(OH)_2 + \underset{\text{(Acetylene)}}{CH \equiv CH}$

(B) $1$. First,the $trans$-alkene reacts with $Br_2$ to form a vicinal dibromide: $Ph-CH(Br)-CH(Br)-CH_3$.
$2$. Next,dehydrohalogenation occurs using $alc. KOH$ followed by $NaNH_2$ to remove two moles of $HBr$,resulting in the formation of an alkyne: $Ph-C \equiv C-CH_3$.
$3$. Finally,the partial hydrogenation of the alkyne to a $cis$-alkene is achieved using $H_2$ with $Lindlar \text{ } Catalyst$ (a poisoned $Pd$ catalyst).
$4$. Therefore,the correct sequence is $Br_2, \text{alc. } KOH, NaNH_2, H_2, \text{Lindlar Catalyst}$.

(B) The reaction is: $CH_3-C \equiv CH + Na \rightarrow CH_3-C \equiv C^- Na^+ (A) + \frac{1}{2} H_2$.
Compound $A$ is sodium prop$-1-$ynide $(CH_3-C \equiv C^- Na^+)$.
Compound $C$ is $CH_3-C \equiv C-CH_2-CH(CH_3)_2$ (based on the structure provided in the image).
For this product to form,$B$ must be $2$-bromobutane $(CH_3-CH_2-CH(Br)-CH_3)$ or a similar alkyl halide that provides the branched chain.
However,looking at the provided image structure $CH_3-C \equiv C-CH_2-CH_2-CH(CH_3)$,the reaction is an $S_N2$ substitution.
Given the options,$A$ is $CH_3-C \equiv C^- Na^+$. The structure of $C$ in the image corresponds to $CH_3-C \equiv C-CH_2-CH_2-CH_3$ if we assume the branch is a typo in the image or $B$ is $1$-bromobutane. Based on standard textbook problems,$B$ is $CH_3-CH_2-CH_2-Br$.

(A) $1$. The first reaction is the catalytic hydrogenation of an alkyne using $Pd/C$ (Lindlar's catalyst is typically poisoned,but $Pd/C$ generally reduces alkynes to alkenes). The product shown is a cis-alkene,specifically $cis$-$3$-methylpent-$2$-ene (or similar structure based on the image). Looking at the reactant $A$ in the image,it is $CH_3-C\equiv C-C_2H_5$,which is $2$-pentyne.
$2$. The second reaction is the Birch reduction of an internal alkyne ($CH_3-C\equiv C-CH_3$,which is but-$2$-yne) using $Na/\text{liquid } NH_3$. This reaction is stereospecific and yields the trans-alkene,which is trans-$2$-butene.
$3$. Therefore,$A$ is $2$-pentyne and $B$ is trans-$2$-butene.

(B) The reaction proceeds in two steps:
$1$. The reduction of but-$2$-yne $(CH_3-C \equiv C-CH_3)$ with $Na/liq. NH_3$ (Birch reduction) gives trans-but-$2$-ene $(CH_3-CH=CH-CH_3)$.
$2$. The subsequent reaction with cold dilute alkaline $KMnO_4$ (Baeyer's reagent) performs syn-hydroxylation of the alkene to form butane-$2,3$-diol $(CH_3-CH(OH)-CH(OH)-CH_3)$.
$3$. The product $P$ is butane-$2,3$-diol,which contains $2$ oxygen atoms.

(D) In ethyne $(HC \equiv CH)$,the carbon-carbon bond is a triple bond,which is shorter and stronger than the double bond in ethene $(CH_2=CH_2)$.
Ethyne is a linear molecule where both carbon atoms are $sp$ hybridised.
Therefore,the statement that the carbon-carbon bond in ethyne is weaker than that in ethene is incorrect.

(B) The conversion of $1,2-dibromoethane$ to ethyne involves dehydrohalogenation.
First,treatment with alcoholic $KOH$ removes one molecule of $HBr$ to form vinyl bromide $(CH_2=CHBr)$.
Second,because the $C-Br$ bond in vinyl bromide has partial double bond character due to resonance,it is less reactive towards elimination. Therefore,a stronger base like $NaNH_2$ (sodamide) is required to remove the second molecule of $HBr$ to yield ethyne $(HC \equiv CH)$.
Thus,the correct sequence is alcoholic $KOH$ followed by $NaNH_2$.

(D) The synthesis of $3-$octyne $(CH_3CH_2C \equiv CCH_2CH_2CH_2CH_3)$ involves the alkylation of an acetylide ion.
First,$1-$butyne $(CH_3CH_2C \equiv CH)$ reacts with sodium amide $(NaNH_2)$ to form the sodium butynide nucleophile: $CH_3CH_2C \equiv CH + NaNH_2 \rightarrow CH_3CH_2C \equiv C^{-}Na^{+} + NH_3$.
Next,this nucleophile performs an $S_N2$ reaction with $1-$bromobutane $(CH_3CH_2CH_2CH_2Br)$: $CH_3CH_2C \equiv C^{-}Na^{+} + CH_3CH_2CH_2CH_2Br \rightarrow CH_3CH_2C \equiv CCH_2CH_2CH_2CH_3 + NaBr$.
Thus,the required reagents are $1-$bromobutane and $1-$butyne.

(B) Step $1$: Analyze the final product. The final product is $2$ moles of acetone $(CH_3COCH_3)$. This implies that the intermediate $Q$ must be $2,3$-dimethylbut-$2$-ene,which upon ozonolysis gives $2$ moles of acetone.
Step $2$: Analyze the conversion of $Q$ to acetone. $Q$ undergoes acid-catalyzed dehydration followed by ozonolysis. $2,3$-dimethylbut-$2$-ene is formed by the dehydration of $2,3$-dimethylbutan-$2$-ol. Thus,$Q$ is $(CH_3)_2CH-CH(OH)-CH_3$ (Option $B$).
Step $3$: Analyze the formation of $Q$ from $P$. $P$ $(C_6H_{10})$ reacts with $dil. H_2SO_4/HgSO_4$ to form a ketone,which is then reduced by $NaBH_4$ to an alcohol $Q$. For $Q$ to be $2,3$-dimethylbutan-$2$-ol,the ketone must be $3,3$-dimethylbutan-$2$-one. This ketone is formed from the hydration of $3,3$-dimethylbut-$1$-yne $(P)$. Thus,$P$ is $(CH_3)_2CH-C \equiv C-H$ (Option $D$).
Therefore,$P$ is $D$ and $Q$ is $B$.

(D) $1$. Compound $P$ $(C_9H_{12})$ on hydrogenation with $Pt/H_2$ gives propylcyclohexane,indicating $P$ has a cyclohexyl ring and a propyl chain with a triple bond.
$2$. The reaction with $NaNH_2$ followed by $C_6H_5COCH_3$ and $H_3O^+/\Delta$ confirms $P$ is a terminal alkyne,specifically $3-$cyclohexylprop$-1-$yne.
$3$. $X$ is a reagent that partially reduces the alkyne to an alkene,which is $Pd-C/\text{quinoline}/H_2$ (Lindlar's catalyst).
$4$. Oxidation of the resulting alkene with $KMnO_4/H_2SO_4/\Delta$ yields the optically active acid $Q$.
$5$. Thus,$P$ is $3-$cyclohexylprop$-1-$yne (Option $A$) and $X$ is $Pd-C/\text{quinoline}/H_2$ (Option $B$).
$6$. Therefore,the correct options are $A$ and $B$.

(B) The reaction involves the reduction of an internal alkyne to a trans-alkene using sodium in liquid ammonia $(Na/liq. NH_3)$.
This is a dissolving metal reduction,which is stereoselective for the formation of the trans-isomer.
The terminal alkyne group remains unaffected by this reagent under these conditions.
Therefore,the internal alkyne is reduced to a trans-alkene while the terminal alkyne remains unchanged.

(C) $1.$ In Scheme $1$,$HO-CH_2-CH_2-C \equiv CH$ reacts with $NaNH_2$ to form the alkoxide-acetylide dianion. Subsequent reaction with $CH_3CH_2I$ alkylates the terminal alkyne,and $CH_3I$ methylates the hydroxyl group. Finally,$H_2$ with Lindlar catalyst performs syn-hydrogenation of the alkyne to a cis-alkene,yielding $X$ as $(Z)-1-methoxyhex-3-ene$.
$2.$ In Scheme $2$,$CH_3CH_2-C \equiv CH$ reacts with $NaNH_2$ to form the acetylide,which reacts with $2-bromopropan-1-ol$. After acidic workup,hydrogenation,and oxidation with $CrO_3$,the product $Y$ is $heptan-2-one$ $(CH_3-CO-CH_2-CH_2-CH_2-CH_3)$.
$Y$ $(heptan-2-one)$ contains a methyl ketone group,so it gives a positive iodoform test. $X$ is an ether,while $Y$ is a ketone; they are functional isomers. Thus,the correct options are $1 \rightarrow B$ and $2 \rightarrow C$.

(A) Statement $I$: Propyne $(CH_3-C \equiv CH)$ has an acidic hydrogen atom attached to the $sp$ hybridized carbon. It reacts with sodium metal as follows:
$CH_3-C \equiv CH + Na \rightarrow CH_3-C \equiv C^- Na^+ + \frac{1}{2} H_2 \uparrow$
Thus,$1 \ mole$ of propyne liberates $0.5 \ mole$ of $H_2$ gas. Statement $I$ is correct.
Statement $II$: Molar mass of propyne $(C_3H_4)$ is $40 \ g/mol$.
Moles of propyne = $\frac{4 \ g}{40 \ g/mol} = 0.1 \ mole$.
The reaction with $NaNH_2$ is:
$CH_3-C \equiv CH + NaNH_2 \rightarrow CH_3-C \equiv C^- Na^+ + NH_3 \uparrow$
$0.1 \ mole$ of propyne will liberate $0.1 \ mole$ of $NH_3$ gas.
Volume of $NH_3$ at $STP$ = $0.1 \ mole \times 22400 \ mL/mol = 2240 \ mL$.
The statement says $224 \ mL$,which is incorrect. Therefore,Statement $II$ is incorrect.

(A) Step $1$: $Ph-CO-CH_3$ reacts with $PCl_5$ to form $Ph-CCl_2-CH_3$ $(A)$.
Step $2$: $A$ reacts with $3 \text{ eq. } NaNH_2/NH_3$ to undergo dehydrohalogenation to form the acetylide ion $Ph-C \equiv C^- Na^+$ $(B)$.
Step $3$: Acidification of $B$ gives phenylacetylene $Ph-C \equiv CH$ $(C)$.
Step $4$: Hydroboration-oxidation of $C$ $(1. B_2H_6, 2. H_2O_2/OH^-)$ gives an enol $Ph-CH=CH-OH$,which tautomerises to phenylacetaldehyde $Ph-CH_2-CHO$ $(D)$.
Product $D$ is $Ph-CH_2-CHO$.
The phenyl ring contains $6$ $sp^2$ carbon atoms,and the carbonyl carbon in the aldehyde group is also $sp^2$ hybridised.
Total $sp^2$ carbon atoms $= 6 + 1 = 7$.

(A) The reaction sequence is as follows:
$1$. $C_2H_2$ (acetylene) undergoes hydration in the presence of $\text{dil. } H_2SO_4$ and $HgSO_4$ to form $A$,which is acetaldehyde $(CH_3CHO)$.
$2$. Acetaldehyde $(A)$ on oxidation $([O])$ gives $B$,which is acetic acid $(CH_3COOH)$.
$3$. Acetic acid $(B)$ reacts with $NaOH$ to form $C$,which is sodium acetate $(CH_3COONa)$.
$4$. Sodium acetate $(C)$ undergoes decarboxylation with soda lime $(NaOH/CaO)$ to form $D$,which is methane $(CH_4)$.

(A) The conversion of an internal alkyne $(CH_3-C \equiv C-CH_3)$ to a $cis$-alkene is a stereoselective reduction.
Lindlar catalyst ($H_2$ / $Pd$ poisoned with $CaCO_3$ or $BaSO_4$ and quinoline) is used for the syn-addition of hydrogen to alkynes,which yields the $cis$-isomer.
$Na$ / liquid $NH_3$ (Birch reduction) is used for the anti-addition of hydrogen to alkynes,which yields the $trans$-isomer.
Therefore,the correct reagent is $H_2$ / Lindlar catalyst.

(D) The reaction of acetylene $(C_2H_2)$ with water in the presence of $HgSO_4$ and $H_2SO_4$ is an electrophilic addition reaction.
First,water adds across the triple bond to form an unstable intermediate called vinyl alcohol $(CH_2=CH-OH)$.
This intermediate $(x)$ undergoes rapid tautomerization to form the more stable aldehyde,acetaldehyde $(CH_3-CHO)$.
Therefore,$(x)$ is $CH_2=CH-OH$.

(C) Step $1$: Reaction of ethyne with excess $HCl$ gives $1,1-$dichloroethane $(A)$.
$CH \equiv CH + 2HCl \rightarrow CH_3-CHCl_2 (A)$
Step $2$: Hydrolysis of $1,1-$dichloroethane $(A)$ followed by heating gives ethanal $(B)$.
$CH_3-CHCl_2 + H_2O \xrightarrow{\Delta} CH_3CHO (B) + 2HCl$
Step $3$: Reduction of ethanal $(B)$ with $Na-Hg / H_2O$ gives ethanol $(C)$.
$CH_3CHO + 2[H] \xrightarrow{Na-Hg / H_2O} CH_3CH_2OH (C)$
Thus,compound $C$ is ethanol.

(C) $I$. $CH_{3}-CH_{3}$ $(C_{2}H_{6})$: Contains only $\sigma$ bonds. (No $\pi$ bond)
$II$. $CH_{2}=CH_{2}$ $(C_{2}H_{4})$: Contains $1$ $\sigma$ bond and $1$ $\pi$ bond between carbons.
$III$. $CH \equiv CH$ $(C_{2}H_{2})$: Contains $1$ $\sigma$ bond and $2$ $\pi$ bonds between carbons.
$IV$. $CH_{3}-CH=CH_{2}$ $(C_{3}H_{6})$: Contains $1$ $\pi$ bond in the $C=C$ double bond.

(B) The reaction of acetylene $(CH \equiv CH)$ with excess of $HCl$ proceeds in two steps following Markovnikov's rule:
$1$. $CH \equiv CH + HCl \rightarrow CH_2 = CHCl$ (Vinyl chloride)
$2$. $CH_2 = CHCl + HCl \rightarrow CH_3CHCl_2$ (Ethylidene dichloride)
Thus,ethylidene dichloride is obtained from acetylene.

(B) The reduction of an alkyne to a $Cis$-alkene is achieved by catalytic hydrogenation using Lindlar's catalyst.
Lindlar's catalyst consists of palladium deposited on calcium carbonate $(Pd-CaCO_3)$ or barium sulfate $(Pd-BaSO_4)$,which is partially poisoned with quinoline or lead acetate.
This specific reagent prevents the complete hydrogenation of the alkyne to an alkane and stereoselectively yields the $Cis$ isomer.
Therefore,the correct reagent is $Pd-C / \text{quinoline}$ (often referred to as Lindlar's catalyst).

(B) The reduction of an alkyne to a $cis$-alkene is achieved using Lindlar's catalyst,which consists of $Pd$ supported on $CaCO_3$ or $BaSO_4$ and poisoned with quinoline or lead acetate. This process is known as partial hydrogenation. Therefore,the correct reagent is $Pd-C, \text{ quinoline}$.

(D) Step $1$: $1,2-\text{Dibromoethane}$ $(BrCH_2-CH_2Br)$ reacts with alcoholic $KOH$ (dehydrohalogenation) to form $A$,which is $Bromoethene$ $(CH_2=CHBr)$.
Step $2$: $Bromoethene$ $(CH_2=CHBr)$ reacts with $NaNH_2$ (sodamide),a strong base,to undergo further dehydrohalogenation to form $B$,which is $Ethyne$ $(HC \equiv CH)$.

(B) The reaction sequence is as follows:
$1$. $A$ reacts with lithium amide $(LiNH_2)$ to form ethynyl lithium $(HC \equiv CLi)$. This indicates that $A$ must be ethyne $(HC \equiv CH)$,as terminal alkynes react with strong bases like lithium amide to form acetylides.
$2$. Ethynyl lithium $(HC \equiv CLi)$ then reacts with bromoethane $(CH_3CH_2Br)$ via an $S_N2$ reaction to form but-$1$-yne $(HC \equiv C-CH_2CH_3)$.
Therefore,$A$ is ethyne.

(D) Grignard reagents like methyl magnesium iodide $(CH_3MgI)$ act as strong bases and react with compounds containing acidic hydrogen atoms.
Terminal alkynes,which have a hydrogen atom attached to a triply bonded carbon atom $(R-C \equiv C-H)$,possess acidic hydrogen.
Among the given options,$CH_3CH_2CH_2C \equiv CH$ is a terminal alkyne.
The reaction is:
$CH_3CH_2CH_2C \equiv CH + CH_3MgI \rightarrow CH_3CH_2CH_2C \equiv CMgI + CH_4 \uparrow$
Therefore,option $(D)$ is correct.

(A) The reaction of calcium carbide $(CaC_2)$ with water $(H_2O)$ is a standard laboratory method for the preparation of ethyne $(C_2H_2)$.
The chemical equation for this reaction is:
$CaC_2 + 2H_2O \rightarrow Ca(OH)_2 + C_2H_2$
Thus,the alkyne formed is ethyne.

(C) The reaction sequence is as follows:
$1$. The reaction of a ketone with $PCl_5$ replaces the carbonyl oxygen with two chlorine atoms,forming a gem-dichloride: $\text{Cyclohexyl-CO-CH}_3 + PCl_5 \rightarrow \text{Cyclohexyl-CCl}_2\text{-CH}_3 (P) + POCl_3$.
$2$. The treatment of the gem-dichloride $(P)$ with excess $\text{NaNH}_2$ (a strong base) leads to double dehydrohalogenation,resulting in the formation of an alkyne: $\text{Cyclohexyl-CCl}_2\text{-CH}_3 + \text{excess NaNH}_2$ $\rightarrow \text{Cyclohexyl-C}\equiv\text{CH} (Q) + 2\text{NaCl} + 2\text{NH}_3$.
Therefore,the final product $Q$ is cyclohexylacetylene,which corresponds to option $C$.

(B) The general formula for alkynes is $C_{n}H_{2n-2}$. For $n=5$,the formula is $C_{5}H_{8}$.
There are three possible isomeric alkynes for this formula:
$(I)$ $CH \equiv C-CH_{2}-CH_{2}-CH_{3}$ (Pent-$1$-yne)
$(II)$ $CH_{3}-C \equiv C-CH_{2}-CH_{3}$ (Pent-$2$-yne)
$(III)$ $CH \equiv C-CH(CH_{3})_{2}$ ($3$-Methylbut-$1$-yne)
Thus,the total number of possible alkynes is $3$.

(A) The hydration of alkynes in the presence of $Hg^{2+}$ and $H_2SO_4$ follows Markovnikov's rule. The reaction proceeds as follows:
$CH_3-CH_2-C \equiv CH + H_2O \xrightarrow{Hg^{2+}, H_2SO_4, 333 \ K} [CH_3-CH_2-C(OH)=CH_2]$
This enol intermediate undergoes tautomerization to form a stable ketone:
$[CH_3-CH_2-C(OH)=CH_2] \rightleftharpoons CH_3-CH_2-CO-CH_3$ (Butan$-2-$one).

(D)
The reduction of an alkyne to a $trans-alkene$ is achieved using dissolving metal reduction,such as $Na$ in liquid $NH_3$ (Birch reduction).
The reaction is:
$CH_3-C \equiv C-CH_3 \xrightarrow{Na \mid liq. NH_3} trans-CH_3-CH=CH-CH_3$

(B) The given reaction shows the partial hydrogenation of an alkyne to a $cis$-alkene.
Lindlar's catalyst,which is partially deactivated palladised charcoal $(Pd/C)$ poisoned with quinoline or sulfur,is used for this purpose.
Therefore,the reagent '$X$' is $Pd/C$ (Lindlar's catalyst).

(D) Alkynes are converted into cis-alkenes by partial hydrogenation using Lindlar's catalyst,which is $H_{2}$ gas in the presence of $Pd$ supported on $CaCO_{3}$ and poisoned with lead or quinoline. The reaction is represented as:
$R-C \equiv C-R \xrightarrow{H_{2} / Pd-CaCO_{3}} R-CH=CH-R$ (cis-alkene).

(A) The reaction sequence is: $\text{Alkyne}$ $\xrightarrow{H_2, \text{Lindlar's catalyst}} A$ $\xrightarrow{\text{Ozonolysis}} B \text{ (only)}$.
In the Wacker process,ethene $(CH_2=CH_2)$ is oxidized to acetaldehyde $(CH_3CHO)$,which is $B$.
Since ozonolysis of alkene $A$ yields $B$ $(CH_3CHO)$ only,$A$ must be $CH_3CH=CHCH_3$ (but$-2-$ene).
Lindlar's catalyst reduces an alkyne to a $cis$-alkene.
Therefore,the starting alkyne must be $CH_3-C \equiv C-CH_3$ (but$-2-$yne).

(C) The acidic character of hydrocarbons depends on the percentage of $s$-character in the $C-H$ bond. Higher $s$-character leads to higher electronegativity of the carbon atom,making the $C-H$ bond more polar and the hydrogen more acidic.
$1$. $a: HC \equiv CH$ ($sp$ hybridized,$50\% \ s$-character)
$2$. $d: CH_3 - C \equiv CH$ ($sp$ hybridized terminal alkyne,$50\% \ s$-character,but the $+I$ effect of the methyl group decreases acidity compared to $a$)
$3$. $b: CH_2 = CH_2$ ($sp^2$ hybridized,$33.3\% \ s$-character)
$4$. $c: CH_3 - CH_3$ ($sp^3$ hybridized,$25\% \ s$-character)
Therefore,the order of acidity is $a > d > b > c$.

(C) Let us analyze each reaction:
$(A)$ $CH_2Br-CH_2Br + Zn \rightarrow CH_2=CH_2 + ZnBr_2$. This is a correct dehalogenation reaction.
$(B)$ $CH_3-C \equiv C-CH_3 + H_2 \xrightarrow{Pd/BaSO_4} cis-CH_3-CH=CH-CH_3$. This is a correct Lindlar's catalyst reduction.
$(C)$ $CH_3-C \equiv CH + H_2O \xrightarrow{Hg^{2+}, H^+} CH_3-CO-CH_3$ (acetone). The given product $CH_3-CH_2-CHO$ (propanal) is incorrect. Hydration of propyne yields acetone.
$(D)$ $2Ph-CH_2-Br + 2Na \xrightarrow{\text{dry ether}} Ph-CH_2-CH_2-Ph + 2NaBr$. This is a correct Wurtz reaction.
Therefore,option $(C)$ is the incorrect reaction.