Alkyne Questions in English

(D) Terminal alkynes,such as ethyne $(C_2H_2)$,contain acidic hydrogen atoms attached to $sp$-hybridized carbon atoms.
These acidic hydrogens can be replaced by active metals like sodium $(Na)$ to form sodium acetylides with the evolution of hydrogen gas $(H_2)$.
The reaction is: $2CH \equiv CH + 2Na \to 2CH \equiv C^{-}Na^{+} + H_2$.

(A) Acetylene $(CH \equiv CH)$ is a terminal alkyne with acidic hydrogen atoms.
When treated with ammoniacal silver nitrate $(AgNO_3)$,it forms silver acetylide,which is a white precipitate: $CH \equiv CH + 2[Ag(NH_3)_2]^+ + 2OH^- \to Ag-C \equiv C-Ag \downarrow + 4NH_3 + 2H_2O$.
When treated with ammoniacal cuprous chloride $(Cu_2Cl_2)$,it forms copper$(I)$ acetylide,which is a red precipitate: $CH \equiv CH + 2[Cu(NH_3)_2]^+ + 2OH^- \to Cu-C \equiv C-Cu \downarrow + 4NH_3 + 2H_2O$.

(A) The bond length decreases as the $s$-character of the hybrid orbital increases.
An $s$-orbital is closer to the nucleus than a $p$-orbital,so higher $s$-character results in a shorter bond length.
$sp$ hybridization has $50\% \ s$-character,$sp^2$ has $33.3\% \ s$-character,and $sp^3$ has $25\% \ s$-character.
Therefore,the $C - H$ bond length is minimum in $sp - s$ overlapping (alkynes).

Compound	Hybridization	$C - H$ bond length $(\mathring{A})$
Alkane	$sp^3$	$1.093$
Alkene	$sp^2$	$1.087$
Alkyne	$sp$	$1.057$

(A) The hydration of $1$-butyne $(CH_3-CH_2-C \equiv CH)$ in the presence of $HgSO_4$ and $H_2SO_4$ follows Markovnikov's rule.
Water adds across the triple bond to form an unstable enol intermediate $(CH_3-CH_2-C(OH)=CH_2)$.
This enol then tautomerizes to form the more stable ketone,$2$-butanone $(CH_3-CH_2-C(=O)-CH_3)$.

(A) $NaNH_2$ is a strong base that can deprotonate terminal alkynes because the acidic hydrogen attached to the $sp$-hybridized carbon is sufficiently acidic.
$CH \equiv CH + NaNH_2 \to CH \equiv C^{-} Na^{+} + NH_3$
Among the given options,$C_2H_2$ (acetylene) is a terminal alkyne,while $C_6H_6$ (benzene),$C_2H_6$ (ethane),and $C_2H_4$ (ethene) do not have sufficiently acidic protons to react with $NaNH_2$ to form a sodium salt.

(A) The gas that decolourises bromine in $CCl_4$ must be an unsaturated hydrocarbon (alkene or alkyne).
Formation of a white precipitate with ammoniacal silver nitrate (Tollens' reagent) is a characteristic test for terminal alkynes $(R-C\equiv CH)$.
$C_2H_2$ (acetylene) is a terminal alkyne that reacts as follows:
$HC\equiv CH + 2Br_2 \xrightarrow{CCl_4} CHBr_2-CHBr_2$ (decolourisation of bromine).
$HC\equiv CH + 2[Ag(NH_3)_2]^+ + 2OH^- \to AgC\equiv CAg \downarrow + 4NH_3 + 2H_2O$ (white precipitate of silver acetylide).

(D) . Ethyne $(C_2H_2)$ belongs to the alkyne series with the general formula $C_nH_{2n-2}$.
Homologues differ by a $-CH_2-$ group.
For $n=2$,the formula is $C_2H_2$ (ethyne).
For $n=3$,the formula is $C_3H_{2(3)-2} = C_3H_4$ (propyne).
Thus,$C_3H_4$ is the next homologue of ethyne.

(C) The reaction of acetylene $(CH \equiv CH)$ with $HCl$ in the presence of $HgCl_2$ as a catalyst is an electrophilic addition reaction.
The reaction proceeds as follows:
$CH \equiv CH + HCl \xrightarrow{HgCl_2} CH_2 = CH - Cl$
The product formed is vinyl chloride.

(C) The reaction of propyne with water in the presence of $40\% \ H_2SO_4$ and $1\% \ HgSO_4$ is an example of hydration of alkynes.
The reaction proceeds via the formation of an enol intermediate:
$CH_3-C \equiv CH + H_2O \xrightarrow{H_2SO_4, HgSO_4} CH_3-C(OH)=CH_2$.
This enol intermediate undergoes tautomerization to form the more stable ketone,which is acetone:
$CH_3-C(OH)=CH_2 \rightarrow CH_3-C(=O)-CH_3$ (Acetone).

(A) When $3$ molecules of propyne $(CH_3-C \equiv CH)$ undergo cyclic polymerization in the presence of red hot iron tube at $873 \ K$,they form $1,3,5$-trimethylbenzene,which is commonly known as Mesitylene.
The reaction is as follows:
$3 CH_3-C \equiv CH \xrightarrow{\Delta, \text{Fe tube}} \text{Mesitylene}$

(C) Ammoniacal cuprous chloride (Tollens' reagent for alkynes) reacts with terminal alkynes to form a red precipitate of copper acetylide.
$2NH_4OH + Cu_2Cl_2 \to 2CuOH + 2NH_4Cl$
$CuOH + 2NH_4OH \to [Cu(NH_3)_2]OH + 2H_2O$
$HC \equiv CH + 2[Cu(NH_3)_2]OH \to CuC \equiv CCu + 4NH_3 + 2H_2O$
Since $C_2H_2$ (acetylene) is a terminal alkyne,it forms the copper derivative (copper acetylide).

(C) The cyclic polymerization (cyclotrimerization) of ethyne $(CH \equiv CH)$ to benzene $(C_6H_6)$ is typically carried out by passing the gas through a red-hot iron tube at $873 \ K$. However,transition metal halides like $NbCl_3$ (Niobium$(III)$ chloride) are also known to catalyze this specific reaction.
$3 CH \equiv CH \xrightarrow{NbCl_3} C_6H_6$

(C) Acetylene $(HC \equiv CH)$ is oxidized by $KMnO_4$ to form oxalic acid $(HOOC-COOH)$.
The reaction is: $HC \equiv CH + 4[O] \xrightarrow{KMnO_4} HOOC-COOH$.

(B) The reaction of ethyne with dilute $H_2SO_4$ in the presence of $Hg(II)$ salts (like $HgSO_4$) is a hydration reaction.
The reaction proceeds as follows:
$CH \equiv CH + H_2O$ $\xrightarrow{40\% \ H_2SO_4, \ 1\% \ HgSO_4} [CH_2 = CH - OH] \text{ (vinyl alcohol)}
[CH_2 = CH - OH]$ $\xrightarrow{\text{tautomerization}} CH_3 - CHO \text{ (ethanal)}$
Thus,the final product is ethanal.

(C) Acetylene $(HC \equiv CH)$ is a terminal alkyne with acidic hydrogen atoms.
It reacts with ammoniacal cuprous chloride $(Cu_2Cl_2 + NH_4OH)$ to form a red-brown precipitate of copper$(I)$ acetylide $(CuC \equiv CCu)$.
Ethylene $(CH_2 = CH_2)$ does not contain acidic hydrogen and does not react with ammoniacal cuprous chloride.
Therefore,this reagent is used to distinguish between them.

(A) Terminal alkynes contain acidic hydrogen atoms attached to the $sp$-hybridized carbon atom.
These terminal alkynes react with Tollens' reagent,$Ag(NH_3)_2^+$,to form a white precipitate of silver acetylide.
This reaction is specific to terminal alkynes and is used to distinguish them from internal alkynes and other hydrocarbons.
Therefore,the correct option is $A$.

(A) When acetylene $(C_2H_2)$ is passed through an electric arc in the presence of nitrogen $(N_2)$,the reaction leads to the formation of hydrogen cyanide $(HCN)$.
The chemical equation for this reaction is:
$C_2H_2 + N_2 \xrightarrow{\text{electric arc}} 2HCN$

(B) The ozonolysis of acetylene $(HC \equiv CH)$ proceeds via the formation of an ozonide intermediate.
Upon hydrolysis,the ozonide breaks down to form glyoxal $(CHO-CHO)$.
If the reaction conditions are oxidative or if further oxidation occurs,it can lead to the formation of formic acid $(HCOOH)$.
Therefore,the primary product is glyoxal,and under certain conditions,formic acid is also obtained.

(B) The bond length between carbon atoms depends on the hybridization state of the carbon atoms involved.
$sp^3-sp^3$ single bond (as in $C_2H_6$) has a bond length of $154 \ pm$.
$sp^2-sp^2$ double bond (as in $C_2H_4$) has a bond length of $134 \ pm$.
$sp-sp$ triple bond (as in $C_2H_2$) has a bond length of $120 \ pm$.
In $Propyne$ $(CH_3-C \equiv CH)$, there is a $C \equiv C$ triple bond, which has the shortest bond length among the given options.

(A) Acetylene $(HC \equiv CH)$ reacts with ammoniacal silver nitrate $(AgNO_{3} + NH_{4}OH)$ to form a white precipitate of silver acetylide $(AgC \equiv CAg)$.
The chemical reaction is:
$HC \equiv CH + 2AgNO_{3} + 2NH_{4}OH \rightarrow AgC \equiv CAg + 2NH_{4}NO_{3} + 2H_{2}O$
This reaction is specific to terminal alkynes,which have acidic hydrogen atoms,and results in the formation of a white precipitate of silver acetylide.

(B) Ethylidene chloride is $1,1-$dichloroethane $(CH_3CHCl_2)$.
It is prepared by the addition of $HCl$ to acetylene $(C_2H_2)$ in two steps:
Step $1$: $C_2H_2 + HCl \rightarrow CH_2=CHCl$ (Vinyl chloride)
Step $2$: $CH_2=CHCl + HCl \rightarrow CH_3CHCl_2$ (Ethylidene chloride)
Thus,the reaction of $C_2H_2$ with $HCl$ yields ethylidene dichloride.

(A) The synthesis of $1, 3-$butadiene from acetylene $(C_2H_2)$ involves two main steps:
$1$. Dimerization of acetylene: Acetylene undergoes linear dimerization in the presence of $CuCl$ and $NH_4Cl$ to form vinylacetylene $(CH_2=CH-C \equiv CH)$.
$2$. Partial hydrogenation: Vinylacetylene is then selectively hydrogenated using $H_2/Pd(BaSO_4)$ (Lindlar's catalyst) to reduce the triple bond to a double bond,yielding $1, 3-$butadiene $(CH_2=CH-CH=CH_2)$.

(D) Benzene $(C_6H_6)$ is formed by the cyclic trimerization of ethyne $(C_2H_2)$.
When ethyne gas is passed through a red-hot iron tube at $873 \ K$,it undergoes cyclic polymerization to form benzene.
The reaction is: $3C_2H_2 \xrightarrow{873 \ K, \text{Red hot Fe tube}} C_6H_6$.
Therefore,the correct option is $(d)$.

(A) The reaction of acetylene $(CH \equiv CH)$ with acetic acid $(CH_3COOH)$ in the presence of $Hg^{2+}$ catalyst proceeds via the formation of vinyl acetate as an intermediate.
$HC \equiv CH + CH_3COOH \xrightarrow{Hg^{2+}} CH_2=CH-OOCCH_3$ (vinyl acetate)
Further addition of another molecule of acetic acid to the vinyl acetate leads to the formation of ethylidene diacetate.
$CH_2=CH-OOCCH_3 + CH_3COOH \xrightarrow{Hg^{2+}} CH_3-CH(OOCCH_3)_2$ (ethylidene diacetate)
Thus,the final product is $CH_3-CH(CH_3COO)_2$.

(C) The industrial preparation of acetylene $(C_2H_2)$ involves the reaction of carbon (graphite) with hydrogen gas in the presence of an electric arc.
The chemical equation for this process is: $2C + H_2 \xrightarrow{\text{Electric arc}} CH \equiv CH$.

(C) The hydration of acetylene in the presence of dilute $H_2SO_4$ and $Hg^{2+}$ ions proceeds as follows:
$CH \equiv CH + H_2O \xrightarrow{H_2SO_4, Hg^{2+}} [CH_2 = CH - OH]$
This intermediate,vinyl alcohol,is unstable and undergoes tautomerization (rearrangement) to form a more stable carbonyl compound:
$[CH_2 = CH - OH] \rightarrow CH_3 - CHO$ (Acetaldehyde).

(A) The acidic nature of hydrogen in hydrocarbons depends on the hybridization of the carbon atom to which it is attached.
In $CH \equiv CH$ (Ethyne),the carbon is $sp$ hybridized,which has $50\%$ $s$-character,making it more electronegative and thus the $C-H$ bond is polar,allowing the hydrogen to be acidic.
Reaction with sodium metal confirms this: $2CH \equiv CH + 2Na \to 2CH \equiv C^{-}Na^{+} + H_2 \uparrow$.
Therefore,ethyne has acidic hydrogen.

(C) $1.$ Acetylene reacts with $NaNH_2$ to form sodium acetylide: $HC \equiv CH + NaNH_2 \rightarrow HC \equiv C^{-}Na^{+} + NH_3$.
$2.$ Sodium acetylide reacts with acetone $(CH_3COCH_3)$ followed by acidification to form $X$: $HC \equiv C^{-}Na^{+} + CH_3COCH_3 \xrightarrow{H^{+}} HC \equiv C-C(CH_3)_2OH$ ($2$-methylbut-$3$-yn-$2$-ol).
$3.$ Hydration of the alkyne $X$ using $Hg^{2+}/H_2O/H^{+}$ follows Markovnikov's addition to form an enol,which tautomerizes to the ketone $Z$: $HC \equiv C-C(CH_3)_2OH \xrightarrow{Hg^{2+}, H_2O, H^{+}} CH_3-CO-C(OH)(CH_3)_2$ ($3$-hydroxy-$3$-methylbutan-$2$-one).
Therefore,the structure of $Z$ is $CH_3-CO-C(OH)(CH_3)_2$.

(C) The bond length between carbon atoms depends on the bond order. As the bond order increases, the bond length decreases.
The order of bond lengths is: $C-C > C=C > C \equiv C$.
$1.$ Ethane $(C_2H_6)$ has a $C-C$ single bond $(154 \ pm)$.
$2.$ Ethene $(C_2H_4)$ has a $C=C$ double bond $(134 \ pm)$.
$3.$ Ethyne $(C_2H_2)$ has a $C \equiv C$ triple bond $(120 \ pm)$.
Therefore, the carbon-carbon bond length is minimum in Ethyne.

(D) In ethyne $(CH \equiv CH)$,the carbon-carbon triple bond consists of $1$ $\sigma$ bond and $2$ $\pi$ bonds.
The cylindrical shape of the alkyne molecule arises due to the presence of these two $\pi$ bonds,which are formed by the sideways overlap of two sets of $p$-orbitals perpendicular to each other.

(D) Terminal alkynes (alkynes with a triple bond at the end of the chain) contain an acidic hydrogen atom attached to the $sp$-hybridized carbon.
These terminal alkynes react with ammoniacal silver nitrate (Tollens' reagent) to form a white precipitate of silver acetylide.
This reaction is a characteristic test for terminal alkynes,whereas alkenes and internal alkynes do not give this test.

(A) Acetylene $(HC \equiv CH)$ is a terminal alkyne with acidic hydrogen atoms.
$1$. It reacts with strong bases like $NaNH_2$ or metals like $Na$ to form acetylides.
$2$. It reacts with ammoniacal $AgNO_3$ (Tollens' reagent) to form a white precipitate of silver acetylide.
$3$. It reacts with $HCl$ via electrophilic addition to form vinyl chloride.
$4$. $NaOH$ is a strong base but is not strong enough to deprotonate acetylene effectively under standard conditions,and it does not undergo addition reactions with alkynes. Therefore,it does not react with acetylene.

(C) Acetylene $(HC \equiv CH)$ reacts with hypochlorous acid $(HOCl)$ in a two-step addition process.
First,$HOCl$ adds across the triple bond to form an intermediate.
With excess $HOCl$,the reaction proceeds as follows:
$HC \equiv CH + 2 \, HOCl \to [CHCl_2 - CH(OH)_2]$.
This gem-diol intermediate is unstable and undergoes dehydration (loss of $H_2O$) to form $Cl_2CH - CHO$,which is Dichloroacetaldehyde.
Therefore,the correct option is $C$.

(A) The structure of $but-1-yne$ is $CH \equiv C-CH_2-CH_3$.
In terminal alkynes,the hydrogen atom attached to the $sp$-hybridized carbon atom is acidic in nature because of the high electronegativity of the $sp$ carbon.
In $but-1-yne$,there is only $1$ such acidic hydrogen atom.
Therefore,$1 \ mole$ of $but-1-yne$ can easily donate $1 \ mole$ of protons ($H^+$ ions).

(B) Ammoniacal $AgNO_3$ (Tollens' reagent) reacts with terminal alkynes $(R-C\equiv CH)$ due to the presence of acidic hydrogen atoms attached to the $sp$-hybridized carbon.
In the given options,$CH_3-CH=CH-C\equiv CH$ is a terminal alkyne.
Therefore,it contains an acidic hydrogen atom and will undergo a reaction with ammoniacal $AgNO_3$ to form a white precipitate of silver acetylide.

(B) The hydration of acetylene $(CH \equiv CH)$ to form acetaldehyde $(CH_3CHO)$ is a classic reaction known as Kucherov's reaction.
This reaction requires the presence of dilute $H_2SO_4$ and a mercuric salt catalyst,typically $HgSO_4$.
The reaction proceeds as follows:
$CH \equiv CH + H_2O$ $\xrightarrow{H_2SO_4, HgSO_4} [CH_2=CHOH]$ $\rightarrow CH_3CHO$
Therefore,the correct catalyst is $HgSO_4$.

(B) The reaction of propyne $(CH_3C \equiv CH)$ 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 $(CH_3C(Br)=CH_2)$.
In the second step,another molecule of $HBr$ adds to the double bond,again following Markovnikov's rule,to yield $2, 2-$dibromopropane $(CH_3C(Br)_2CH_3)$.

(C) Ammoniacal silver nitrate (Tollens' reagent) reacts with terminal alkynes,which possess an acidic hydrogen atom attached to the $sp$-hybridized carbon atom,to form white precipitates of silver acetylides.
$CH \equiv CH$,$CH_3 - C \equiv CH$,and $Cl - CH_2 - C \equiv CH$ are terminal alkynes and contain acidic hydrogen atoms.
$CH_3 - C \equiv C - CH_3$ (But$-2-$yne) is an internal alkyne and lacks an acidic hydrogen atom.
Therefore,it does not react with ammoniacal $AgNO_3$ to form a precipitate.

(A) The reaction of ethyne $(CH \equiv CH)$ with $O_3$ in the presence of $NaOH$ forms an ozonide intermediate $(X)$.
Upon treatment with $Zn/CH_3COOH$ (reductive workup),the ozonide is converted into glyoxal $(CHO-CHO)$.
Further reduction of glyoxal in the presence of $Zn/CH_3COOH$ leads to the formation of ethylene glycol $(HOCH_2-CH_2OH)$ as the final product '$Y$'.
The reaction sequence is:
$CH \equiv CH$ $\xrightarrow{O_3} \text{Ozonide}$ $\xrightarrow{Zn/CH_3COOH} CHO-CHO$ $\xrightarrow{[H]} HOCH_2-CH_2OH$.

(B) $C_n H_{2n + 2}$ is the general formula for alkanes,which are saturated hydrocarbons.
$C_n H_{2n}$ is the general formula for alkenes.
$C_n H_{2n - 2}$ is the general formula for alkynes,which contain at least one carbon-carbon triple bond.

(A) The reaction of an alkyne with $1 \ mol$ of $Cl_2$ is an electrophilic addition reaction. The addition of halogens to alkynes typically proceeds via an anti-addition mechanism,resulting in the formation of a trans-dihaloalkene as the major product. The reaction is: $CH_3-C \equiv C-CH_2-CH_3 + Cl_2 \rightarrow CH_3-C(Cl)=C(Cl)-CH_2CH_3$ (trans-isomer).

(C) $1$. The formation of a white precipitate with ammoniacal $AgNO_3$ indicates the presence of a terminal alkyne $(-C\equiv CH)$.
$2$. Oxidation of a terminal alkyne with hot alkaline $KMnO_4$ results in the cleavage of the triple bond,where the terminal carbon is converted to $CO_2$ and the remaining part forms a carboxylic acid.
$3$. The product $(CH_3)_2CHCOOH$ (isobutyric acid) contains $4$ carbons. Since the original compound has $5$ carbons,the structure must be $(CH_3)_2CH-C\equiv CH$ ($3$-methyl$-1-$butyne).
$4$. Oxidation of $(CH_3)_2CH-C\equiv CH$ gives $(CH_3)_2CHCOOH + CO_2$.

(B) The reaction of $1, 2-$dibromoethane with alcoholic $KOH$ is a dehydrohalogenation reaction.
When $1, 2-$dibromoethane $(BrCH_2-CH_2Br)$ is heated with alcoholic $KOH$,it undergoes double dehydrohalogenation to form acetylene $(CH \equiv CH)$.
The reaction is: $BrCH_2-CH_2Br + 2KOH \text{ (alc.)} \xrightarrow{\Delta} CH \equiv CH + 2KBr + 2H_2O$.

(C) The reduction of internal alkynes like $2-$butyne with alkali metals (such as $Li$ or $Na$) in liquid $NH_3$ (Birch reduction) proceeds via a radical anion intermediate.
This mechanism favors the formation of the more stable $trans-$alkene isomer due to steric repulsion between the alkyl groups in the transition state.
Therefore,the reaction of $2-$butyne with $Li/NH_3$ yields $trans-2-$butene.

(A) $pent-1-yne$ is a terminal alkyne containing an acidic hydrogen atom.
It reacts with Tollen's reagent $([Ag(NH_3)_2]^+)$ to form a white precipitate of silver pent$-1-$ynide.
$pent-1-ene$ does not contain an acidic hydrogen and therefore does not react with Tollen's reagent.
Thus,Tollen's reagent is used to distinguish between them.

(A) Ethyne $(C_2H_2)$ contains acidic hydrogen atoms attached to $sp$-hybridized carbon atoms.
It reacts with ammoniacal $AgNO_3$ (Tollens' reagent) to form a white precipitate of silver acetylide $(Ag-C \equiv C-Ag)$.
Ethane $(C_2H_6)$ and ethene $(C_2H_4)$ do not contain acidic hydrogen atoms and therefore do not react with ammoniacal $AgNO_3$.
Thus,ethane and ethene remain unreacted.

(D) Acetylene $(HC \equiv CH)$ reacts with ammoniacal silver nitrate $(AgNO_3)$ to form silver acetylide $(AgC \equiv CAg)$.
This reaction occurs because the terminal hydrogen atoms in acetylene are acidic due to the $sp$ hybridization of carbon atoms,which makes them more electronegative and capable of releasing $H^+$ ions.
Therefore,acetylene shows acidic property.