Alkyne Questions in English

(A) Acetylene $(HC \equiv CH)$ is a terminal alkyne with acidic hydrogen atoms attached to $sp$-hybridized carbon atoms.
$1$. It reacts with $Na$ to form sodium acetylide $(NaC \equiv CNa)$ and releases $H_2$ gas.
$2$. It reacts with ammoniacal $AgNO_3$ (Tollens' reagent) to form a white precipitate of silver acetylide $(AgC \equiv CAg)$.
$3$. It reacts with $HCl$ via electrophilic addition to form vinyl chloride and eventually ethylidene chloride.
$4$. $NaOH$ is a base,but it is not strong enough to deprotonate the terminal alkyne (which has a $pK_a$ of approximately $25$). Therefore,acetylene does not react with $NaOH$.

(D) $1$-butyne $(CH_3-CH_2-C \equiv CH)$ is a terminal alkyne containing an acidic hydrogen atom attached to the $sp$-hybridized carbon.
Terminal alkynes react with ammoniacal cuprous chloride $(Cu_2Cl_2)$ to form a red precipitate of copper$(I)$ acetylide.
$2$-butyne $(CH_3-C \equiv C-CH_3)$ is a non-terminal alkyne and does not have an acidic hydrogen atom.
Therefore,it does not react with ammoniacal $Cu_2Cl_2$ solution.
Thus,ammoniacal $Cu_2Cl_2$ is used to distinguish between them.

(B) The reaction of propyne $(CH_3-C \equiv CH)$ with water in the presence of $HgSO_4$ and $H_2SO_4$ is an example of hydration of alkynes.
This reaction follows Markovnikov's rule.
Initially,an enol $(CH_3-C(OH)=CH_2)$ is formed.
This enol undergoes tautomerization to form a more stable ketone,which is acetone $(CH_3-CO-CH_3)$.

(A) The acidity of hydrocarbons depends on the hybridization of the carbon atom attached to the hydrogen.
Terminal alkynes like $CH_3 - C\equiv CH$ have a terminal hydrogen atom attached to an $sp$-hybridized carbon.
Due to the high $s$-character $(50\%)$ in $sp$-hybridized orbitals,the electrons are held more tightly by the nucleus,making the $C-H$ bond polar and the hydrogen atom acidic.
Other options like alkanes $(sp^3)$ and alkenes $(sp^2)$ have lower $s$-character and are not acidic enough to react with bases like $NaNH_2$ or $Cu_2Cl_2$.

(C) The reaction between calcium carbide $(CaC_2)$ and water $(H_2O)$ is a standard method for the preparation of ethyne (acetylene).
The chemical equation is:
$CaC_2 + 2H_2O \rightarrow Ca(OH)_2 + C_2H_2$
Thus,the product formed is ethyne $(C_2H_2)$.

(D) Ammoniacal $AgNO_3$ (Tollens' reagent) reacts with terminal alkynes $(R-C \equiv CH)$ to form white precipitates of silver acetylides.
This reaction occurs because the terminal hydrogen atom in terminal alkynes is acidic in nature.
$CH_3-CH(CH_3)-C \equiv CH$ is a terminal alkyne.
$HC \equiv CH$ (Acetylene) is a terminal alkyne.
$1$-Butyne $(CH_3-CH_2-C \equiv CH)$ is a terminal alkyne.
Since all the given compounds are terminal alkynes,they all react with ammoniacal $AgNO_3$.

(A) The reaction of propyne $(CH_3-C\equiv CH)$ with two moles of $HBr$ follows Markovnikov's rule.
In the first step,$HBr$ adds to the triple bond to form $2$-bromopropene $(CH_3-C(Br)=CH_2)$.
In the second step,another molecule of $HBr$ adds to the double bond of $2$-bromopropene,again following Markovnikov's rule,to form $2, 2$-dibromopropane $(CH_3-C(Br)_2-CH_3)$.

(B) The reduction of an internal alkyne to a $trans$-alkene is achieved using dissolving metal reduction.
Specifically,$2$-hexyne $(CH_3-C \equiv C-CH_2-CH_2-CH_3)$ reacts with lithium $(Li)$ in liquid ammonia $(NH_3)$ to produce $trans$-$2$-hexene.
This reaction proceeds via a radical anion intermediate,which favors the formation of the more stable $trans$-isomer.
Option $A$ $(H_2 / Pd / BaSO_4)$ is Lindlar's catalyst,which produces a $cis$-alkene.
Option $C$ $(H_2 / PtO_2)$ leads to complete hydrogenation to an alkane.

(B) Terminal alkynes,which contain an acidic hydrogen atom attached to an $sp$-hybridized carbon atom $(-C \equiv CH)$,can react with strong bases like sodium amide $(NaNH_2)$ or sodium metal in liquid ammonia to form acetylides.
In the given options,$CH_3CH_2C \equiv CH$ is a terminal alkyne.
The reaction is: $CH_3CH_2C \equiv CH + Na \rightarrow CH_3CH_2C \equiv C^-Na^+ + \frac{1}{2}H_2$.
Internal alkynes (options $A$ and $D$) and alkenes (option $C$) do not have acidic terminal hydrogen atoms and therefore do not react with sodium in liquid ammonia.

(C) The reaction of acetylene $(CH \equiv CH)$ with $HCl$ in the presence of $HgCl_2$ is an electrophilic addition reaction.
The triple bond undergoes addition to form vinyl chloride:
$CH \equiv CH + HCl \xrightarrow{HgCl_2} CH_2 = CH - Cl$

(C) The reactivity of carbon-carbon bonds depends on the presence of $\pi$-bonds.
$C - C$ is a single bond ($\sigma$-bond), which is stable.
$C = C$ contains one $\pi$-bond, and $C \equiv C$ contains two $\pi$-bonds.
$\pi$-bonds are electron-rich and more accessible for electrophilic attack compared to $\sigma$-bonds.
Among the given options, the triple bond $(C \equiv C)$ has the highest electron density and is generally the most reactive towards electrophilic addition reactions.

(C) Acetylene $(HC \equiv CH)$ is a terminal alkyne with acidic hydrogen atoms attached to $sp$-hybridized carbon atoms.
$1$. It reacts with $Na$ to form sodium acetylide $(NaC \equiv CNa)$ and releases $H_2$ gas.
$2$. It reacts with ammoniacal $AgNO_3$ (Tollens' reagent) to form a white precipitate of silver acetylide $(AgC \equiv CAg)$.
$3$. It reacts with $HCl$ via electrophilic addition to form vinyl chloride and eventually ethylidene chloride.
$4$. Acetylene is a very weak acid $(pK_a \approx 25)$ and does not react with a strong base like $NaOH$ ($pK_a$ of water is $15.7$) to form a salt,as the equilibrium lies far to the left.

(B) The reaction of acetylene $(HC \equiv CH)$ with dilute $H_2SO_4$ in the presence of $HgSO_4$ is a hydration reaction.
First,it forms vinyl alcohol $(CH_2=CH-OH)$,which is unstable and undergoes tautomerization to form acetaldehyde $(CH_3CHO)$.
The structure of acetaldehyde is $CH_3-C(=O)H$.
In the $C=O$ bond,there is $1$ $\sigma$ bond and $1$ $\pi$ bond.
Therefore,the final product acetaldehyde contains $1$ $\pi$ bond.

(B) The reaction of an alkyne with $H_2$ in the presence of $Pd/CaCO_3$ (Lindlar's catalyst) is a controlled hydrogenation reaction.
This catalyst partially reduces the alkyne to an alkene.
Specifically,this reaction follows syn-addition,which results in the formation of the cis-isomer.
Therefore,$4$-octyne reacts with $H_2$ over $Pd/CaCO_3$ to yield cis-$4$-octene.

(C) The reaction of ethyne $(CH \equiv CH)$ with water in the presence of $20\% \ H_2SO_4$ and $HgSO_4$ is a hydration reaction.
This process involves the formation of an unstable vinyl alcohol intermediate,which undergoes tautomerization to form acetaldehyde $(CH_3CHO)$.
Therefore,the reagent $Z$ is $20\% \ H_2SO_4 + HgSO_4$.

(C) The given reaction is a dehydrohalogenation reaction where a vicinal or geminal dihalide is converted into an alkyne.
To convert a geminal dihalide $(R-CH_2-CCl_2-R)$ into an alkyne $(R-C \equiv C-R)$,a strong base is required to remove two molecules of $HCl$.
$NaNH_2$ (sodamide) in liquid ammonia $(liq. NH_3)$ is a very strong base commonly used for this purpose.
Therefore,the correct reagent is $NaNH_2$ in $liq. NH_3$.

(B) The reaction of $1,2$-dibromoethane $(BrCH_2-CH_2Br)$ with alcoholic potassium hydroxide $(KOH)$ is a dehydrohalogenation reaction.
First,it undergoes elimination to form vinyl bromide $(CH_2=CHBr)$.
Further treatment with alcoholic $KOH$ leads to the removal of another molecule of $HBr$ to form acetylene $(HC \equiv CH)$.

(A) The reaction of propyne $(CH_3-C \equiv CH)$ with two equivalents of hydrogen bromide $(HBr)$ follows Markovnikov's rule.
In the first step,$HBr$ adds to the triple bond to form $2$-bromopropene $(CH_3-C(Br)=CH_2)$.
In the second step,another molecule of $HBr$ adds to the double bond,again following Markovnikov's rule,to form $2,2$-dibromopropane $(CH_3-C(Br)_2-CH_3)$.

(D) Acetylene $(HC \equiv CH)$ is a terminal alkyne with acidic hydrogen atoms attached to the $sp$-hybridized carbon atoms.
$1$. It reacts with $HCl$ to form vinyl chloride and eventually ethylidene chloride.
$2$. It reacts with ammoniacal $AgNO_3$ (Tollens' reagent) to form a white precipitate of silver acetylide $(AgC \equiv CAg)$.
$3$. It reacts with $Na$ in liquid ammonia to form sodium acetylide $(NaC \equiv CNa)$.
$4$. It does not react with $NaOH$ because $NaOH$ is a base and acetylene is a very weak acid (weaker than water),so the acid-base reaction does not proceed to form a salt.

(A) Acetylene $(CH \equiv CH)$ reacts with arsenic trichloride $(AsCl_3)$ in the presence of a catalyst (like anhydrous $AlCl_3$) to form chlorovinyldichloroarsine,commonly known as Lewisite.
The reaction is: $CH \equiv CH + AsCl_3 \rightarrow ClCH=CHAsCl_2$ (Lewisite).

(D) Propyne $(CH_3-C \equiv CH)$ is a terminal alkyne,which contains an acidic hydrogen atom.
Terminal alkynes react with ammoniacal silver nitrate ($AgNO_3$ in $NH_4OH$) to form a white precipitate of silver acetylide $(CH_3-C \equiv C-Ag)$.
Propene $(CH_3-CH=CH_2)$ is an alkene and does not contain acidic hydrogen,so it does not react with ammoniacal $AgNO_3$.
Thus,ammoniacal $AgNO_3$ is used to distinguish between them.

(D) 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 is: $CaC_2 + 2H_2O \rightarrow Ca(OH)_2 + C_2H_2$ (Ethyne).

(C) The hydration of propyne $(CH_3-C \equiv CH)$ occurs in the presence of $Hg^{2+}$ and dilute $H_2SO_4$.
First,water adds to the triple bond to form an enol intermediate,$CH_3-C(OH)=CH_2$.
This enol undergoes tautomerization to form a more stable ketone,which is acetone $(CH_3-CO-CH_3)$.

(C) The hydration of alkynes in the presence of $HgSO_4$ and $H_2SO_4$ follows Markovnikov's rule.
Propyne $(CH_3-C\equiv CH)$ reacts with water to form an enol intermediate $(CH_3-C(OH)=CH_2)$.
This enol undergoes tautomerization to form the more stable ketone,which is acetone $(CH_3-CO-CH_3)$.

(A) The acidity of hydrogen atoms in hydrocarbons depends on the hybridization of the carbon atom to which it is attached.
In $Ethyne$ $(HC \equiv CH)$,the carbon atoms are $sp$ hybridized.
$sp$ hybridization has $50\%$ $s$-character,making the carbon atom more electronegative and thus pulling the electron density of the $C-H$ bond towards itself.
This makes the hydrogen atom acidic,allowing it to be removed by strong bases.
In $Ethene$ ($sp^2$ hybridized,$33.3\%$ $s$-character) and $Ethane$ ($sp^3$ hybridized,$25\%$ $s$-character),the $s$-character is lower,making the $C-H$ bond less polar and the hydrogen less acidic.

(A) The reaction of an internal alkyne with ozone followed by hydrolysis is an oxidative cleavage reaction.
For an internal alkyne $R-C \equiv C-R'$,the products are two carboxylic acids: $R-COOH$ and $R'-COOH$.
In the given reaction,$CH_3-C \equiv C-CH_2CH_3$ is cleaved at the triple bond.
The fragments are $CH_3-C \equiv$ and $\equiv C-CH_2CH_3$.
Upon oxidation,these yield $CH_3COOH$ (acetic acid) and $CH_3CH_2COOH$ (propanoic acid).
Therefore,the correct products are $CH_3COOH + CH_3CH_2COOH$.

(D) The synthesis of $3$-octyne $(CH_3CH_2C \equiv CCH_2CH_2CH_2CH_3)$ involves the nucleophilic substitution of a bromoalkane with an acetylide ion.
The reaction is: $CH_3CH_2C \equiv C^- Na^+ + BrCH_2CH_2CH_2CH_3 \rightarrow CH_3CH_2C \equiv CCH_2CH_2CH_2CH_3 + NaBr$.
Here,the alkyne used is $1$-butyne $(CH_3CH_2C \equiv CH)$ and the bromoalkane is $1$-bromobutane ($CH_3CH_2CH_2CH_2Br$ or $BrCH_2CH_2CH_2CH_3$).

(C) The reaction of an alkyne with $H^+ / Hg^{2+}$ is an acid-catalyzed hydration reaction,which follows Markovnikov's rule.
For the alkyne $C_6H_5-C \equiv C-CH_3$,the addition of water occurs across the triple bond.
The $OH^-$ group attaches to the carbon atom that is more substituted or stabilized by the phenyl group,leading to the formation of an enol intermediate.
Specifically,the intermediate is $C_6H_5-C(OH)=CH-CH_3$.
This enol then undergoes tautomerization to form the more stable ketone,which is $C_6H_5-CO-CH_2-CH_3$ (propiophenone derivative).
However,looking at the options provided,option $C$ represents the final ketone product.

(B) The reaction of a geminal dihalide like $CH_3CH_2CHCl_2$ with a strong base like $NaNH_2$ (sodamide) proceeds via dehydrohalogenation.
First,one molecule of $HCl$ is eliminated to form a vinyl halide $(CH_3CH=CHCl)$.
Then,the second molecule of $HCl$ is eliminated by the excess $NaNH_2$ to form an alkyne.
Thus,$CH_3CH_2CHCl_2 \xrightarrow{NaNH_2} CH_3-C\equiv CH$.

(B) The reaction of acetylene $(CH \equiv CH)$ with water in the presence of $dil. \ H_2SO_4$ and $HgSO_4$ is a hydration reaction.
$CH \equiv CH + H_2O \xrightarrow{HgSO_4 / H_2SO_4} [CH_2 = CH-OH] \to CH_3CHO$ (Acetaldehyde).
The final product is acetaldehyde $(CH_3-CHO)$.
In the structure of acetaldehyde $(CH_3-C(=O)-H)$,there is one $C=O$ double bond.
$A$ double bond consists of one $\sigma$-bond and one $\pi$-bond.
Therefore,the number of $\pi$-bonds in the product is $1$.

(A) The oxymercuration-demercuration (or hydration) of terminal alkynes like $1$-butyne $(CH_3-CH_2-C \equiv CH)$ in the presence of $HgSO_4$ and $H_2SO_4$ follows Markovnikov's rule.
$1$. The addition of water across the triple bond forms an enol intermediate: $CH_3-CH_2-C(OH)=CH_2$.
$2$. This enol undergoes tautomerization to form a stable ketone.
$3$. The final product is $2$-butanone $(CH_3-CH_2-CO-CH_3)$.

(A) Among the given compounds,only ethyne $(CH \equiv CH)$ contains acidic hydrogen because the $sp$ hybridized carbon atom is more electronegative,allowing the hydrogen to be removed by strong bases or reactive metals like $Na$.
The reaction is: $2CH \equiv CH + 2Na \rightarrow 2CH \equiv C^{-}Na^{+} + H_2 \uparrow$.

(A) The addition of $HBr$ to propyne $(CH_3 - C \equiv CH)$ follows Markovnikov's rule.
Step $1$: $CH_3 - C \equiv CH + HBr \rightarrow CH_3 - C(Br) = CH_2$ ($2$-bromopropene).
Step $2$: $CH_3 - C(Br) = CH_2 + HBr \rightarrow CH_3 - C(Br)_2 - CH_3$ ($2$,$2$-dibromopropane).
Thus,the final product is $CH_3 - CBr_2 - CH_3$.

(D) Ethyne $(C_2H_2)$ belongs to the alkyne homologous series with the general formula $C_nH_{2n-2}$.
For $n=2$,the formula is $C_2H_{2(2)-2} = C_2H_2$.
For $n=3$,the formula is $C_3H_{2(3)-2} = C_3H_4$ (Propyne).
Therefore,$C_3H_4$ belongs to the same homologous series as ethyne.

(B) Lindlar's catalyst is a heterogeneous catalyst consisting of palladium deposited on calcium carbonate $(CaCO_3)$ and poisoned with lead or quinoline. It is used for the partial hydrogenation of alkynes to cis-alkenes.

(D) Terminal alkynes like ethyne $(HC \equiv CH)$ have acidic hydrogen atoms. When treated with a strong base like sodamide $(NaNH_2)$,they form a sodium salt known as sodium acetylide $(HC \equiv C^-Na^+)$.

(D) The synthesis of $3$-octyne $(CH_3CH_2C \equiv CCH_2CH_2CH_2CH_3)$ involves the alkylation of an acetylide ion.
First,the terminal alkyne reacts with sodium amide $(NaNH_2)$ to form a sodium acetylide:
$CH_3CH_2C \equiv CH + NaNH_2 \rightarrow CH_3CH_2C \equiv C^- Na^+ + NH_3$
Then,the acetylide ion acts as a nucleophile and attacks the bromoalkane $(CH_3CH_2CH_2CH_2Br)$ via an $S_N2$ mechanism:
$CH_3CH_2C \equiv C^- Na^+ + BrCH_2CH_2CH_2CH_3 \rightarrow CH_3CH_2C \equiv CCH_2CH_2CH_2CH_3 + NaBr$
Thus,the reactants are $1$-bromobutane $(CH_3CH_2CH_2CH_2Br)$ and $1$-butyne $(CH_3CH_2C \equiv CH)$.

(A) The reaction of propyne $(CH_3-C \equiv CH)$ with two moles of hydrogen bromide $(HBr)$ follows Markovnikov's rule.
In the first step,$HBr$ adds to the triple bond to form $2$-bromopropene $(CH_3-C(Br)=CH_2)$.
In the second step,another molecule of $HBr$ adds to the double bond,again following Markovnikov's rule,to form $2,2$-dibromopropane $(CH_3-C(Br)_2-CH_3)$.

(D) Terminal alkynes,such as ethyne $(C_2H_2)$,contain acidic hydrogen atoms attached to $sp$-hybridized carbon atoms.
These acidic hydrogen atoms can be replaced by active metals like sodium $(Na)$ to release hydrogen gas $(H_2)$.
The reaction is: $HC \equiv CH + 2Na \rightarrow NaC \equiv CNa + H_2 \uparrow$.
Alkanes $(CH_4, C_2H_6)$ and alkenes $(C_2H_4)$ do not have sufficiently acidic hydrogen atoms to react with sodium.

(D) The bond length between two carbon atoms depends on the hybridization of the carbon atoms involved. The bond length decreases as the $s$-character in the hybrid orbitals increases.
$1$. In $CH_3-CH_2-CH_3$ (Propane),the bond is between $sp^3-sp^3$ carbons.
$2$. In $CH_3-CH=CH_2$ (Propene),the bond is between $sp^3-sp^2$ carbons.
$3$. In $CH_3-C\equiv CH$ (Propyne),the bond is between $sp^3-sp$ carbons.
Since the $sp$ hybridized carbon has $50\% \ s$-character,it is more electronegative and holds the bonding electrons closer to the nucleus,resulting in the shortest bond length compared to $sp^2$ or $sp^3$ hybridized carbons. Therefore,the $sp^3-sp$ bond in Propyne is the shortest.

(A) The reaction of calcium carbide with water produces acetylene $(A)$:
$CaC_2 + 2H_2O \to C_2H_2 + Ca(OH)_2$
Acetylene $(C_2H_2)$ undergoes hydration in the presence of $dil. H_2SO_4$ and $HgSO_4$ to form an unstable enol intermediate,which tautomerizes to form acetaldehyde $(B)$:
$C_2H_2 + H_2O \xrightarrow{H_2SO_4/HgSO_4} [CH_2=CHOH] \to CH_3CHO$

(D) The reaction of calcium carbide $(CaC_2)$ with water $(H_2O)$ produces acetylene $(C_2H_2)$ and calcium hydroxide $(Ca(OH)_2)$.
$CaC_2 + 2H_2O \rightarrow C_2H_2 + Ca(OH)_2$

(C) The structure of $but-1-yne$ is $CH_3-CH_2-C \equiv CH$.
In terminal alkynes,the hydrogen atom attached to the $sp$-hybridized carbon atom involved in the triple bond is acidic in nature.
In $but-1-yne$,there is only one such hydrogen atom attached to the terminal carbon of the triple bond.
Therefore,the number of acidic hydrogen atoms is $1$.

(C) Ethyne $(C_2H_2)$,also known as acetylene,is used in oxy-acetylene welding. When burnt in oxygen,it produces an oxy-acetylene flame which has a very high temperature (approx. $3000 \ ^\circ C$),sufficient for welding metals.

(D) An acyclic alkyne is an unsaturated hydrocarbon containing at least one carbon-carbon triple bond.
For an alkane,the general formula is $C_nH_{2n+2}$.
For an alkene,the general formula is $C_nH_{2n}$.
For an alkyne,the presence of a triple bond reduces the number of hydrogen atoms by four compared to an alkane,resulting in the general formula $C_nH_{2n-2}$.

(B) The partial reduction of alkynes to alkenes is achieved using Lindlar's catalyst,which is $Pd$ supported on $CaCO_3$ or $BaSO_4$ and poisoned with quinoline or sulfur.
This reaction produces a $cis$-alkene.
$R-C \equiv C-R' + H_2 \xrightarrow{Pd/BaSO_4} R-CH=CH-R'$.

(C) Alkynes are unsaturated hydrocarbons containing at least one carbon-carbon triple bond $(C \equiv C)$.
The general formula for alkynes is $C_nH_{2n-2}$.
For the simplest alkyne,$n = 2$,which gives $C_2H_{2(2)-2} = C_2H_2$.
This compound is known as ethyne or acetylene.

(C) The reaction of calcium carbide $(CaC_2)$ with water $(H_2O)$ produces calcium hydroxide $(Ca(OH)_2)$ and acetylene ($C_2H_2$ or $CH \equiv CH$).
$CaC_2 + 2H_2O \rightarrow Ca(OH)_2 + CH \equiv CH$
Thus,the colorless gas produced is acetylene.

(A) The reaction of an alkyne with $HgSO_4$ and $H_2SO_4$ is a hydration reaction (Kucherov reaction).
For an unsymmetrical alkyne like $C_6H_5 - C \equiv C - CH_3$,the addition of water follows Markovnikov's rule.
The $OH^-$ group attaches to the more substituted carbon atom.
Thus,the intermediate enol formed is $C_6H_5 - C(OH) = CHCH_3$.
This enol tautomerizes to form the ketone,$C_6H_5 - CO - CH_2CH_3$ ($1$-phenylpropan$-1-$one).