Alkyne Questions in English

(C) The structure of ethyne is $H-C \equiv C-H$.
In a triple bond,there is one strong sigma $(\sigma)$ bond formed by the head-on overlap of orbitals and two weak pi $(\pi)$ bonds formed by the lateral overlap of $p$-orbitals.
Therefore,a triple bond consists of one sigma bond and two pi bonds.

(C) The bond length depends on the bond order and the hybridization of the carbon atoms.
$C_2H_6$ $(ethane)$ has $sp^3$ hybridization with a $C-C$ single bond (bond length $\approx 154 \text{ pm}$).
$C_2H_4$ $(ethene)$ has $sp^2$ hybridization with a $C=C$ double bond (bond length $\approx 134 \text{ pm}$).
$C_2H_2$ $(ethyne)$ has $sp$ hybridization with a $C \equiv C$ triple bond (bond length $\approx 120 \text{ pm}$).
As the bond order increases, the bond length decreases.
Therefore, $C_2H_2$ has the shortest $C-C$ bond length.

(C) Acetylene $(C_2H_2)$ contains a carbon-carbon triple bond.
In the structure $H-C \equiv C-H$,each carbon atom is bonded to one hydrogen atom and one carbon atom via a triple bond.
According to the valence bond theory,each carbon atom undergoes $sp$ hybridization by mixing one $2s$ and one $2p$ orbital.
These two $sp$ hybrid orbitals are oriented at an angle of $180^{\circ}$ to each other,resulting in a linear geometry.

(B) The correct answer is $(B)$.
In acetylene $(C_2H_2)$,each carbon atom is bonded to one hydrogen atom and one other carbon atom via a triple bond.
The steric number for each carbon is $2$ (one sigma bond with $H$ and one sigma bond with $C$),which corresponds to $sp$ hybridisation.
Structure: $H - \mathop{C}\limits_{sp} \equiv \mathop{C}\limits_{sp} - H$.

(D) The bond length is inversely proportional to the bond order.
In $CH_3NH_2$,the bond is $C-N$ (single bond,bond order = $1$).
In $C_6H_5CH=NOH$,the bond is $C=N$ (double bond,bond order = $2$).
In $CH_3CONH_2$,the bond has partial double bond character due to resonance (bond order between $1$ and $2$).
In $CH_3CN$,the bond is $C \equiv N$ (triple bond,bond order = $3$).
Since $CH_3CN$ has the highest bond order $(3)$,it has the minimum bond length between carbon and nitrogen. Therefore,the correct option is $(D)$.

(A) The reaction between calcium carbide $(CaC_2)$ and heavy water $(D_2O)$ is analogous to the reaction with ordinary water $(H_2O)$.
$CaC_2 + 2D_2O \to C_2D_2 + Ca(OD)_2$
Thus,the product formed is dideuteroacetylene $(C_2D_2)$.

(B) The hydrolysis of magnesium carbide $(Mg_2C_3)$ yields propyne $(CH_3C \equiv CH)$.
The chemical reaction is as follows:
$Mg_2C_3 + 4H_2O \to CH_3C \equiv CH + 2Mg(OH)_2$
Therefore,the correct option is $(B)$.

(C) Alkynes are unsaturated hydrocarbons containing at least one carbon-carbon triple bond.
The general formula for an acyclic alkyne is given by $C_nH_{2n - 2}$,where $n$ is the number of carbon atoms $(n \ge 2)$.

(C) The structure of $3$-hexyne-$1$-ene is $CH_2=CH-CH_2-C \equiv C-CH_3$.
$A$ double bond contains $1$ $\pi$ bond and a triple bond contains $2$ $\pi$ bonds.
Therefore,the total number of $\pi$ bonds is $1 + 2 = 3$.

(C) The structure of acetylene $(HC \equiv CH)$ is $H - C \equiv C - H$.
In this molecule, there are two $C-H$ sigma bonds and one $C-C$ sigma bond, totaling $3\sigma$ bonds.
Additionally, there are two $\pi$ bonds between the two carbon atoms.
Thus, the molecule contains $3\sigma$ and $2\pi$ bonds.

(C) The bond energy of a $C-H$ bond depends on the hybridization of the carbon atom.
As the $s$-character in the hybrid orbital increases,the electronegativity of the carbon atom increases,making the $C-H$ bond shorter and stronger.
In $Ethane$ $(C_2H_6)$,carbon is $sp^3$ hybridized ($25\% \ s$-character).
In $Ethene$ $(C_2H_4)$,carbon is $sp^2$ hybridized ($33.3\% \ s$-character).
In $Ethyne$ $(C_2H_2)$,carbon is $sp$ hybridized ($50\% \ s$-character).
Since $Ethyne$ has the highest $s$-character,it has the strongest and shortest $C-H$ bond,resulting in the maximum bond energy.

(A) The structure of $but-1-ene-3-yne$ is $CH_2=CH-C\equiv CH$.
Counting the bonds:
There are $4$ $C-H$ sigma bonds.
There are $3$ $C-C$ sigma bonds.
Total sigma bonds = $4 + 3 = 7$.
There is $1$ $\pi$ bond in the double bond and $2$ $\pi$ bonds in the triple bond.
Total $\pi$ bonds = $1 + 2 = 3$.
Therefore,the molecule contains $7\sigma$ and $3\pi$ bonds.

(C) The acidity of a hydrocarbon depends on the $s$-character of the hybrid orbital of the carbon atom attached to the hydrogen atom.
Terminal alkynes,such as $CH_3C \equiv CH$ (propyne),have a hydrogen atom attached to an $sp$-hybridized carbon atom.
Due to the high $s$-character $(50\%)$ of the $sp$-hybrid orbital,the electron density is held more tightly by the carbon,making the $C-H$ bond polar and the hydrogen atom acidic.
Therefore,$CH_3C \equiv CH$ is an acidic hydrocarbon.

(A) The structure of $pent-4-en-1-yne$ is $HC \equiv C-CH_2-CH=CH_2$.
Number of $\sigma$ bonds = $4$ ($C-C$ bonds) + $6$ ($C-H$ bonds) = $10$.
Number of $\pi$ bonds = $2$ (from $C \equiv C$) + $1$ (from $C=C$) = $3$.

(D) The acidity of hydrocarbons depends on the $s$-character of the hybridised carbon atom to which the hydrogen is attached.
$1-$butyne $(CH_3-CH_2-C \equiv CH)$ contains a terminal hydrogen atom attached to an $sp$ hybridised carbon atom.
Since $sp$ hybridised carbon has $50\% \ s$-character,it is more electronegative,making the $C-H$ bond more polar and the hydrogen more acidic.
Butane $(sp^3)$,$1-$butene $(sp^2)$,and $2-$butyne (no acidic hydrogen) are significantly less acidic than $1-$butyne.

(C) The $-C \equiv C-$ bond represents a carbon-carbon triple bond.
Ethyne $(C_2H_2)$,also known as acetylene,has the structure $HC \equiv CH$,which contains a carbon-carbon triple bond.
Therefore,the correct option is $(C)$.

(B) The $s-$character of the $C-H$ bond in acetylene ($sp$ hybridized) is $50\%$,which is higher in comparison to $sp^2$ hybridized ethene $(33.3\%)$ and $sp^3$ hybridized ethane $(25\%)$.
Due to higher $s-$character,the electrons of the $C-H$ bond in acetylene are more strongly held by the carbon nucleus.
This increases the electronegativity of the carbon atom,which facilitates the removal of the hydrogen atom as a proton $(H^+)$,making acetylene the most acidic among the given options.

(C)
The reaction is: $CH \equiv CH + H_2 \xrightarrow{Ni/Pd/Pt} CH_2 = CH_2$
In this reaction,hydrogen atoms are added across the triple bond of acetylene to form ethylene. Therefore,it is an example of an addition reaction.

(B) The molecular formula $C_5H_8$ corresponds to the general formula $C_nH_{2n-2}$,which indicates an alkyne.
The possible structural isomers for $C_5H_8$ are:
$1$. $Pent-1-yne$ $(CH_3-CH_2-CH_2-C \equiv CH)$
$2$. $Pent-2-yne$ $(CH_3-CH_2-C \equiv C-CH_3)$
$3$. $3-Methylbut-1-yne$ $(CH_3-CH(CH_3)-C \equiv CH)$
Thus,there are $3$ possible alkynes.

(B) The given compound is $CH_3-C \equiv C-CH(CH_3)-CH=CH_2$.
Hydrogenation with a poisoned palladium catalyst (Lindlar's catalyst) results in $syn$-addition of hydrogen to the triple bond,converting it into a $cis$-double bond.
The resulting product is $CH_3-CH=CH-CH(CH_3)-CH=CH_2$.
Due to the specific geometry and the presence of internal symmetry elements or the formation of a meso-like structure depending on the configuration,the product formed is optically inactive.

(C) Acetylene $(CH \equiv CH)$ reacts with ammoniacal cuprous chloride to form a red precipitate of copper acetylide $(Cu-C \equiv C-Cu)$.
$CH \equiv CH + 2[Cu(NH_3)_2]OH \to Cu-C \equiv C-Cu + 4NH_3 + 2H_2O$
Methane $(CH_4)$ and ethylene $(C_2H_4)$ do not possess acidic hydrogen atoms and therefore do not react with ammoniacal cuprous chloride.
$CH_4 + \text{Ammoniacal } Cu_2Cl_2 \to \text{No reaction}$
$C_2H_4 + \text{Ammoniacal } Cu_2Cl_2 \to \text{No reaction}$
Thus,the gases that pass through the bottle without reacting are methane and ethylene.

(A) Silver acetylide $(Ag-C \equiv C-Ag)$ reacts with hydrochloric acid $(HCl)$ to undergo a double displacement reaction.
The reaction is: $Ag-C \equiv C-Ag + 2HCl \to HC \equiv CH + 2AgCl$.
Thus,the product formed is acetylene $(C_2H_2)$.

(D) The reaction between acetylene $(C_2H_2)$ and arsenic trichloride $(AsCl_3)$ in the presence of anhydrous aluminium chloride $(AlCl_3)$ is an addition reaction.
$C_2H_2 + AsCl_3 \xrightarrow{AlCl_3} ClCH=CHAsCl_2$
This product is known as Lewisite or $ \beta- $chlorovinyl dichloroarsine.
Therefore,both $(a)$ and $(b)$ are correct.

(A) The reaction of molecular bromine $(Br_2)$ with an enyne $(CH_2=CH-CH_2-C \equiv CH)$ is regioselective.
At low temperatures,the electrophilic addition of $Br_2$ occurs preferentially at the more nucleophilic site.
In this molecule,the terminal alkyne group is less reactive towards electrophilic addition compared to the alkene group,but the addition of $Br_2$ to the triple bond is a characteristic reaction.
However,under controlled conditions,the addition occurs across the triple bond to form a dibromoalkene:
$CH_2=CH-CH_2-C \equiv CH + Br_2 \to CH_2=CH-CH_2-CBr=CHBr$.

(D) Acetylene $(HC \equiv CH)$ undergoes nucleophilic addition reactions due to the presence of $\pi$-electrons.
$1$. Addition of water: In the presence of $Hg^{2+}$ and $H_2SO_4$,water adds to acetylene to form vinyl alcohol,which tautomerizes to acetaldehyde. This is a nucleophilic addition process.
$2$. Addition of $HCN$: The cyanide ion $(CN^-)$ acts as a nucleophile and attacks the triple bond of acetylene.
$3$. Addition of $AsCl_3$: The reaction of acetylene with $AsCl_3$ in the presence of $AlCl_3$ (Lewis acid) is known as the Lewis-Arsenic reaction,which is a type of nucleophilic addition resulting in the formation of $\beta$-chlorovinylarsine derivatives.
Since all the given reactions involve nucleophilic attack on the alkyne,the correct option is $D$.

(B) The reactivity of hydrocarbons towards electrophilic addition reactions follows the order: $Alkyne > Alkene > Alkane$.
Ethyne $(C_2H_2)$ is an alkyne,Ethene $(C_2H_4)$ is an alkene,and Ethane $(C_2H_6)$ and Methane $(CH_4)$ are alkanes.
Due to the presence of a triple bond,Ethyne is highly reactive compared to the others.

(C) The dimerization of acetylene $(HC \equiv CH)$ in the presence of a catalyst system consisting of cuprous chloride $(Cu_2Cl_2)$ and ammonium chloride $(NH_4Cl)$ yields vinyl acetylene $(HC \equiv C-CH=CH_2)$.
This vinyl acetylene then reacts with hydrogen chloride $(HCl)$ to form $4$-chloro-$1,2$-butadiene,which subsequently rearranges to form $2$-chloro-$1,3$-butadiene,commonly known as chloroprene.

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

(C) $2 \, CH \equiv CH + 5O_2 \to 4 \, CO_2 + 2H_2O$,$\Delta H = -1300 \, kJ$
The combustion of acetylene is highly exothermic,and the heat produced during the combustion is used for welding purposes in the form of an oxy-acetylene flame.

(C) The metallic carbide is calcium carbide $(CaC_2)$.
When $CaC_2$ reacts with water,it produces acetylene ($C_2H_2$ or $CH \equiv CH$):
$CaC_2 + 2\,H_2O \to Ca(OH)_2 + C_2H_2$
Acetylene is a colourless gas that burns readily in air.
It reacts with ammoniacal silver nitrate (Tollens' reagent) to form a white precipitate of silver acetylide:
$CH \equiv CH + 2\,[Ag(NH_3)_2]OH \to AgC \equiv CAg \downarrow (\text{white ppt.}) + 4\,NH_3 + 2\,H_2O$
Therefore,the gas evolved is acetylene.

(C) The reaction of $but-1-yne$ $(CH_3-CH_2-C \equiv CH)$ with cold alkaline $KMnO_4$ (Baeyer's reagent) leads to the oxidative cleavage of the triple bond.
Since the terminal carbon is involved,it is oxidized to $CO_2$ and $H_2O$,while the remaining part forms a carboxylic acid.
The reaction is: $CH_3-CH_2-C \equiv CH \xrightarrow{\text{Cold alk. } KMnO_4} CH_3CH_2COOH + CO_2$.

(C) $1$. $CH_3COOH \xrightarrow{LiAlH_4} CH_3CH_2OH$ $(A)$
$2$. $CH_3CH_2OH \xrightarrow{H^{+}, 443 \ K} CH_2=CH_2$ $(B)$
$3$. $CH_2=CH_2 \xrightarrow{Br_2} CH_2Br-CH_2Br$ $(C)$
$4$. $CH_2Br-CH_2Br \xrightarrow{alc. KOH} CH \equiv CH$ $(D)$
The product $D$ is $CH \equiv CH$,which is Acetylene.

(A) The formation of $C-H$ bond in acetylene involves $sp-$hybridised carbon atom.
Since $s-$electrons are closer to the nucleus than $p$ electrons,$sp$ hybrid orbitals are more electronegative due to their smaller (in length) and thicker lobes,which increases the time spent by electrons near the nucleus compared to $sp^2$ and $sp^3$ orbitals.
Thus,the electrons present in a bond having more $s-$character will be closer to the nucleus.
In alkynes $(C \equiv C-H)$ bond,the $s-$character is $50\%$,so the electrons constituting this bond are more strongly attracted by the carbon nucleus.
Thus,the acetylenic carbon atom becomes more electronegative in comparison to $sp^2$ and $sp^3$ hybridized carbons,and hence the hydrogen atom present on the carbon atom $(C \equiv C-H)$ can be easily removed as a proton $(H^+)$.

(D) Terminal alkynes,such as compound $(iii)$,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.
$CH_3-CH_2-C\equiv CH + [Ag(NH_3)_2]^+ + OH^-$ $\rightarrow CH_3-CH_2-C\equiv C^- Ag^+ \text{ (white precipitate)} + 2NH_3 + H_2O$
Compounds $(i)$,$(ii)$,and $(iv)$ do not possess acidic terminal hydrogen atoms and therefore do not form a precipitate with this reagent.

(A) The synthesis of the chemical warfare agent $Lewisite$ involves the reaction of acetylene $(CH \equiv CH)$ with arsenic trichloride $(AsCl_3)$ in the presence of a catalyst like anhydrous aluminum chloride $(AlCl_3)$.
The reaction is: $HC \equiv CH + AsCl_3 \rightarrow ClCH = CHAsCl_2$ (Lewisite or $2-$chlorovinyl dichloroarsine).

(C) Ozonolysis of an internal alkyne followed by hydrolysis yields diketones as intermediate products,which are further oxidized to carboxylic acids.
The reaction proceeds as follows:
$CH_3-C \equiv C-CH_2-CH_3$ $\xrightarrow{O_3} \text{ozonide intermediate}$ $\xrightarrow{H_2O} CH_3-CO-CO-CH_2-CH_3 + H_2O_2$
The diketone $CH_3-CO-CO-CH_2-CH_3$ is then cleaved to form carboxylic acids:
$CH_3-CO-CO-CH_2-CH_3 \rightarrow CH_3COOH + CH_3CH_2COOH$
Thus,the final products are acetic acid and propanoic acid.

(A) The correct answer is $A$. The $C \equiv C$ bond (alkyne) is generally considered more reactive towards electrophilic addition reactions compared to $C = C$ (alkene) and $C - C$ (alkane) due to the presence of two $\pi$-bonds and the nature of $sp$-hybridized carbon atoms.

(D) The synthesis of the unsymmetrical alkyne $CH_3-C\equiv C-CH_2-CH_3$ involves the stepwise alkylation of ethyne $(CH\equiv CH)$.
Step $1$: Ethyne reacts with sodamide $(NaNH_2)$ to form sodium acetylide $(CH\equiv C^{-}Na^{+})$.
Step $2$: Sodium acetylide reacts with iodomethane $(CH_3I)$ to form propyne $(CH\equiv C-CH_3)$.
Step $3$: Propyne is treated with another equivalent of $NaNH_2$ to form the sodium salt of propyne $(CH_3-C\equiv C^{-}Na^{+})$.
Step $4$: This salt reacts with iodoethane $(C_2H_5I)$ via an $S_N2$ reaction to yield the final product $CH_3-C\equiv C-CH_2-CH_3$.

(D) When propyne $(CH_3-C \equiv CH)$ is treated with dilute $H_2SO_4$ and $HgSO_4$ (Kucherov reaction),hydration occurs according to Markovnikov's rule to form an unstable enol,which then tautomerizes to form propanone (acetone).
$CH_3-C \equiv CH + H_2O$ $\xrightarrow{HgSO_4/H_2SO_4} [CH_3-C(OH)=CH_2]$ $\rightarrow CH_3-CO-CH_3$

(B) The reaction of ethyne $(C_2H_2)$ with $HCl$ occurs in two steps.
In the first step,vinyl chloride is formed.
In the second step,another molecule of $HCl$ adds to vinyl chloride according to Markovnikov's rule to give $1,1$-dichloroethane $(CH_3-CHCl_2)$ as the final product.
$CH \equiv CH + HCl$ $\rightarrow CH_2=CHCl$ $\xrightarrow{HCl} CH_3-CHCl_2$

(C) The reaction sequence is as follows:
$1$. $CaC_2 + 2H_2O \rightarrow HC \equiv CH (A) + Ca(OH)_2$
$2$. $HC \equiv CH + H_2O \xrightarrow{Hg^{2+}, dil. H_2SO_4} CH_3CHO (B)$ (Kucherov reaction)
$3$. $CH_3CHO + H_2 \xrightarrow{Ni} CH_3CH_2OH (C)$ (Reduction of acetaldehyde to ethanol).
Therefore,the end product $C$ is $C_2H_5OH$.

(C) $(C) \, R-CH_2-CCl_2-R \xrightarrow{KOH \text{ (alc.)}} R-C \equiv C-R + 2HCl$
This reaction is an example of dehydrohalogenation,where two molecules of $HCl$ are removed from a gem-dihalide to form an alkyne.
Alcoholic $KOH$ acts as a strong base to facilitate this elimination reaction.

(D) $1.$ Kolbe's electrolysis of potassium fumarate yields acetylene.
$2.$ Hydrolysis of calcium carbide $(CaC_2 + 2H_2O \rightarrow Ca(OH)_2 + CH \equiv CH)$ produces acetylene.
$3.$ Dehydrohalogenation of ethylene bromide $(Br-CH_2-CH_2-Br)$ using alcoholic $KOH$ also yields acetylene.
Therefore,all the given options are correct.

(B) The electrolysis of an aqueous solution of potassium fumarate or potassium maleate yields acetylene at the anode via the Kolbe electrolysis mechanism.
$KOOC-CH=CH-COOK + 2H_2O \xrightarrow{\text{Electrolysis}} CH \equiv CH + 2CO_2 + 2KOH + H_2$

(D) . The compound $C_3H_4$ is propyne $(CH_3-C\equiv CH)$.
Terminal alkynes contain an acidic hydrogen atom attached to the $sp$-hybridized carbon.
This acidic hydrogen reacts with ammoniacal silver nitrate (Tollens' reagent) to form a white precipitate of silver acetylide,which confirms the presence of a terminal triple bond.
$CH_3-C\equiv CH + [Ag(NH_3)_2]^+ OH^- \to CH_3-C\equiv C^-Ag^+ \downarrow + 2NH_3 + H_2O$.

(C) The reaction sequence is as follows:
$1$. $CH \equiv CH$ undergoes hydration in the presence of $Hg^{2+}/H_2SO_4$ (Kucherov reaction) to form acetaldehyde $(X)$: $CH \equiv CH + H_2O \xrightarrow{Hg^{2+}/H_2SO_4} CH_3-CHO$.
$2$. Acetaldehyde $(X)$ is reduced by $LiAlH_4$ to form ethanol $(Y)$: $CH_3-CHO \xrightarrow{LiAlH_4} CH_3-CH_2OH$.
$3$. Ethanol $(Y)$ reacts with $P_4/Br_2$ (which generates $PBr_3$ in situ) to form ethyl bromide $(Z)$: $CH_3-CH_2OH \xrightarrow{P_4/Br_2} CH_3-CH_2Br$.

(C) The reaction of acetylene $(CH \equiv CH)$ in the presence of a nickel catalyst like $Ni(CN)_2$ under high pressure leads to the cyclic tetramerization of acetylene.
$4 \ CH \equiv CH \xrightarrow{Ni(CN)_2, \text{Pressure}} C_8H_8$ (Cyclooctatetraene).
Therefore,the correct option is $C$.

(D) . Ethyne $(CH \equiv CH)$ contains acidic hydrogen atoms attached to $sp$-hybridized carbon atoms.
These hydrogen atoms can be easily removed by strong bases like $NaNH_2$ to form acetylide salts (e.g.,$NaC \equiv CNa$).

(A) The molecular formula $C_4H_6$ corresponds to the general formula $C_nH_{2n-2}$,which indicates an alkyne or a diene.
When $A$ reacts with excess $Br_2$ to form $C_4H_6Br_4$,it indicates the addition of $2$ moles of $Br_2$,confirming the presence of a triple bond (alkyne).
The formation of a white precipitate with ammoniacal silver nitrate (Tollen's reagent) indicates the presence of a terminal alkyne (acidic hydrogen).
Therefore,the compound $A$ is $But-1-yne$ $(CH_3-CH_2-C \equiv CH)$.