Alkyne Questions in English

(C) The general formula for alkynes is $C_nH_{2n-2}$.
For the simplest alkyne,we set $n = 2$ because a triple bond requires at least two carbon atoms.
Substituting $n = 2$ into the formula: $C_2H_{2(2)-2} = C_2H_2$.
This compound is known as ethyne or acetylene.

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

(D) The reaction of calcium carbide $(CaC_2)$ with water $(H_2O)$ is a standard laboratory method for the preparation of acetylene ($C_2H_2$ or $CH \equiv CH$).
The balanced chemical equation is:
$CaC_2 + 2H_2O \to CH \equiv CH + Ca(OH)_2$
Therefore,the correct option is $(D)$.

(C) The addition of $HCN$ to ethyne $(CH \equiv CH)$ in the presence of $Ba(CN)_2$ as a catalyst is an electrophilic addition reaction.
The reaction proceeds as follows:
$CH \equiv CH + HCN \xrightarrow{Ba(CN)_2} CH_2 = CH - CN$
The product formed is $CH_2 = CH - CN$,which is commonly known as vinyl cyanide (or acrylonitrile).

(C) The reagent $Ag(NH_3)_2^+ OH^-$ is known as Tollens' reagent.
It reacts with terminal alkynes (alkynes having an acidic hydrogen atom attached to an $sp$ hybridized carbon) to form a white precipitate of silver acetylide.
Among the given options,$CH_3CH_2C \equiv CH$ is a terminal alkyne with an acidic hydrogen atom at the end of the chain.
Therefore,it will react with the reagent to form a white precipitate,while the others will not.

(D) Terminal alkynes contain acidic hydrogen atoms attached to $sp$-hybridized carbon atoms.
These acidic hydrogens react with strong bases like $Na$ or $NaNH_2$ to release $H_2$ gas.
Among the given options,$C_2H_2$ (acetylene) is a terminal alkyne.
The reaction is: $HC \equiv CH + 2Na \rightarrow NaC \equiv CNa + H_2 \uparrow$.

(C) Ozonolysis of alkynes using $O_3$ followed by oxidative workup with $H_2O_2$ leads to the cleavage of the triple bond to form carboxylic acids.
The reaction is: $CH_3-C \equiv CH + O_3/H_2O_2 \rightarrow CH_3COOH + HCOOH$.
Therefore, both $CH_3COOH$ and $HCOOH$ are formed as products.

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

(C) The acidity of hydrocarbons depends on the $s$-character of the carbon atom involved in the $C-H$ bond.
Terminal alkynes like $CH_3 - C \equiv CH$ are weakly acidic because the $sp$-hybridized carbon atom is more electronegative,allowing the proton to be removed by a strong base.
$CH_2 = CH_2$ $(sp^2)$ and $C_6H_6$ $(sp^2)$ are much less acidic.
$CH_3 - C \equiv C - CH_3$ has no acidic hydrogen atom attached to the $sp$-hybridized carbon,so it is not acidic.
Therefore,$CH_3 - C \equiv CH$ is the correct answer.

(D) Alkynes contain a triple bond $(C \equiv C)$,which makes them highly reactive and capable of undergoing various types of reactions:
$1$. Addition: Alkynes undergo electrophilic addition reactions,such as hydrogenation: $CH \equiv CH + 2H_2 \xrightarrow{Ni} CH_3 - CH_3$.
$2$. Substitution: The terminal hydrogen atoms in alkynes are acidic and can be replaced by metals: $CH \equiv CH + Na \xrightarrow{liq. NH_3} CH \equiv C^{-} Na^{+} + \frac{1}{2} H_2$.
$3$. Polymerization: Alkynes can undergo cyclic polymerization to form aromatic compounds: $3CH \equiv CH \xrightarrow{\text{hot } Cu \text{ tube}} C_6H_6$ (Benzene).
Since alkynes exhibit all these reactions,the correct option is $(d)$.

(D) The bond length decreases as the bond order increases.
$A$. $CH_3-CH_2-CH_3$ (alkane) has $C-C$ single bonds with a bond length of approximately $1.54 \ \mathring{A}$.
$B$. $CH_3-CH_2-CH_2-CH_3$ (alkane) has $C-C$ single bonds with a bond length of approximately $1.54 \ \mathring{A}$.
$C$. $CH_2=CH-CH=CH_2$ (butadiene) has $C-C$ single bonds and $C=C$ double bonds. The $C-C$ single bond length is shortened due to conjugation (approx $1.46 \ \mathring{A}$).
$D$. $CH \equiv C-C \equiv CH$ (buta$-1,3-$diyne) contains a $C-C$ single bond between two $sp$ hybridized carbons. Due to the high $s$-character $(50\%)$ of $sp$ hybridized carbons,the $C-C$ bond length is the shortest (approx $1.37 \ \mathring{A}$).
Therefore,the shortest $C-C$ bond length is in $CH \equiv C-C \equiv CH$.

(B) The correct option is $(B)$.
Acetylene $(C_2H_2)$ is obtained by heating iodoform $(CHI_3)$ with silver powder $(Ag)$.
The chemical reaction is:
$2CHI_3 + 6Ag \xrightarrow{\Delta} C_2H_2 + 6AgI$
In this reaction,silver acts as a dehalogenating agent,removing iodine from iodoform to form silver iodide and coupling the remaining $CH$ fragments to form acetylene.

(D) The conversion of an internal alkyne like $2$-hexyne into a $trans$-alkene is achieved by Birch reduction using an alkali metal (like $Li$ or $Na$) in liquid ammonia $(NH_3)$.
This reaction proceeds via a $trans$-vinylic radical anion intermediate,which leads to the formation of the thermodynamically more stable $trans$-alkene product.

(A) The bond length between carbon atoms depends on the hybridization state of the carbon atoms involved.
$C-C$ single bond (alkane) length is $1.54 \ \mathring{A}$.
$C=C$ double bond (alkene) length is $1.34 \ \mathring{A}$.
$C \equiv C$ triple bond (alkyne) length is $1.20 \ \mathring{A}$.
Propyne $(CH_3-C \equiv CH)$ contains a carbon-carbon triple bond,which has the shortest bond length among the given options.

(B) The reaction of acetylene $(CH \equiv CH)$ with hydrogen $(H_2)$ in the presence of Lindlar's catalyst $(Pd/BaSO_4)$ is a partial hydrogenation reaction.
$CH \equiv CH + H_2 \xrightarrow[Lindlar \ Catalyst]{Pd/BaSO_4} CH_2 = CH_2$
Thus,acetylene is reduced to ethylene $(CH_2 = CH_2)$.

(A) The structure of acetylene $(C_2H_2)$ is $H-C \equiv C-H$.
In the triple bond between the two carbon atoms,there is one sigma $(\sigma)$ bond and two pi $(\pi)$ bonds.

(A) Propyne $(CH_3C \equiv CH)$ is a terminal alkyne containing an acidic hydrogen atom attached to an $sp$ hybridized carbon.
Terminal alkynes react with ammoniacal silver nitrate ($AgNO_3 + NH_4OH$,also known as Tollen's reagent) to form a white precipitate of silver acetylide.
$CH_3C \equiv CH + [Ag(NH_3)_2]^+ + OH^- \to CH_3C \equiv CAg \downarrow + 2NH_3 + H_2O$
Non-terminal alkynes like $2-$butyne do not have acidic hydrogen atoms and therefore do not react with ammoniacal $AgNO_3$.

(C) The reaction of magnesium carbide with water is as follows:
$Mg_2C_3 + 4H_2O \to CH_3C \equiv CH + 2Mg(OH)_2$
Propyne $(CH_3C \equiv CH)$ is obtained from $Mg_2C_3$.

(B) The hydrocarbon $C_6H_6$ reacts with $Br_2$ water,indicating the presence of unsaturation (double or triple bonds).
Ammoniacal $AgNO_3$ (Tollens' reagent) reacts specifically with terminal alkynes $(R-C\equiv CH)$ to form a white precipitate of silver acetylide.
$1,3,5-$Cyclohexatriene (Benzene) does not react with $Br_2$ water or ammoniacal $AgNO_3$.
$1,5-$Hexadiyne $(HC\equiv C-CH_2-CH_2-C\equiv CH)$ is a terminal alkyne and would give a precipitate,but its formula is $C_6H_6$.
However,$1,5-$Hexadiyne is a valid structure for $C_6H_6$ that contains terminal alkyne groups.
Given the options,$1,5-$Hexadiyne is the correct structure that satisfies both conditions.

(C) The given reaction is the reductive ozonolysis of $2-butyne$.
In the first step,$2-butyne$ reacts with ozone $(O_3)$ to form an ozonide intermediate.
In the second step,the ozonide is reduced by $Zn/H_2O$ to form $butane-2,3-dione$ $(CH_3-CO-CO-CH_3)$.
Therefore,$X$ is $O_3$.

(B) The reaction of ethyne $(HC \equiv CH)$ with ammoniacal silver nitrate $(AgNO_3 + NH_4OH)$ is a characteristic test for terminal alkynes.
Terminal alkynes have acidic hydrogen atoms.
These hydrogen atoms are replaced by silver ions to form silver acetylide $(Ag_2C_2)$,which appears as a white precipitate.
The balanced chemical equation is:
$HC \equiv CH + 2AgNO_3 + 2NH_4OH \rightarrow AgC \equiv CAg + 2NH_4NO_3 + 2H_2O$
Thus,'$X$' is $Ag_2C_2$.

(C) The reaction of acetylene $(CH \equiv CH)$ with hydrogen cyanide $(HCN)$ in the presence of a catalyst like $Ba(CN)_2$ is an addition reaction.
$CH \equiv CH + HCN \xrightarrow{Ba(CN)_2} CH_2 = CHCN$
The product formed is $CH_2 = CHCN$,which is commonly known as vinyl cyanide or acrylonitrile.

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

(A) $X$ is $C_2H_2$ (acetylene).
$HC \equiv CH + H_2O$ $\xrightarrow{HgSO_4 / H_2SO_4} [CH_2 = CHOH]$ $\rightarrow CH_3CHO$ ($Y$ is acetaldehyde).
$CH_3CHO \xrightarrow{[O]} CH_3COOH$ (acetic acid).
Therefore,the compound $X$ is $C_2H_2$.

(A) Calcium carbide $(CaC_2)$ reacts with heavy water $(D_2O)$ to produce deuteroacetylene $(C_2D_2)$ and calcium deuteroxide $(Ca(OD)_2)$.
The chemical reaction is:
$CaC_2 + 2D_2O \xrightarrow{} C_2D_2 + Ca(OD)_2$

(D) The reaction of acetylene $(CH \equiv CH)$ with $HBr$ proceeds via electrophilic addition.
First,one molecule of $HBr$ adds to acetylene to form vinyl bromide $(CH_2 = CHBr)$.
Then,a second molecule of $HBr$ adds to the vinyl bromide following Markovnikov's rule to form ethylidene dibromide $(CH_3CHBr_2)$.
The reaction is: $CH \equiv CH$ $\xrightarrow{HBr} CH_2 = CHBr$ $\xrightarrow{HBr} CH_3CHBr_2$.

(B) The reaction of ethyne $(CH \equiv CH)$ with one equivalent of hydrogen bromide $(HBr)$ follows electrophilic addition.
$CH \equiv CH + HBr \to CH_2 = CHBr$
The product formed is $CH_2 = CHBr$,which is known as vinyl bromide.
Therefore,the correct option is $B$.

(B) Acetylene $(CH \equiv CH)$ reacts with $HCl$ in a two-step electrophilic addition reaction.
Step $1$: $CH \equiv CH + HCl \to CH_2=CHCl$ (Vinyl chloride).
Step $2$: $CH_2=CHCl + HCl \to CH_3CHCl_2$ (Ethylidene chloride).
Since the reaction typically proceeds to completion in the presence of excess $HCl$,the final product is $CH_3CHCl_2$.

(A) When iodoform $(CHI_3)$ is heated with silver $(Ag)$ powder,it undergoes a dehalogenation reaction to form acetylene $(HC \equiv CH)$.
The balanced chemical equation is:
$2CHI_3 + 6Ag \to HC \equiv CH + 6AgI$

(C) The reaction $CH_2Br-C \equiv C-CH_2Br \xrightarrow[\Delta]{Zn} CH_2=C=C=CH_2$ is a debromination reaction.
When $1,4-dibromobut-2-yne$ is treated with zinc dust in the presence of heat,it undergoes elimination of two bromine atoms to form $buta-1,2,3-triene$ (also known as $allene$ derivative or $cumulene$).

(A) When $CHCl_3$ (chloroform) is heated with silver powder $(Ag)$,it undergoes a dehalogenation reaction to form $C_2H_2$ (acetylene or ethyne).
The chemical equation is: $2CHCl_3 + 6Ag \rightarrow C_2H_2 + 6AgCl$.

(D) The reaction of chloroform $(CHCl_3)$ with silver powder $(Ag)$ is a dehalogenation reaction that yields acetylene $(CH \equiv CH)$.
The balanced chemical equation is:
$2CHCl_3 + 6Ag \to CH \equiv CH + 6AgCl$
Therefore,the correct option is $(D)$.

(D) When chloroform $(CHCl_3)$ is heated with silver $(Ag)$ powder,it undergoes a dehalogenation reaction.
The reaction is as follows:
$2CHCl_3 + 6Ag \xrightarrow{\Delta} HC \equiv CH + 6AgCl$
In this reaction,two molecules of chloroform react with six atoms of silver to produce acetylene $(C_2H_2)$ and silver chloride $(AgCl)$.

(D) Step $1$: Dehydration of propan$-1-$ol with concentrated $H_2SO_4$ at $160-180\,^{\circ}C$ yields propene $(X = CH_3CH=CH_2)$.
Step $2$: Addition of $Br_2$ to propene yields $1,2$-dibromopropane $(Y = CH_3CH(Br)CH_2Br)$.
Step $3$: Dehydrohalogenation of $1,2$-dibromopropane with excess alcoholic $KOH$ yields propyne $(Z = CH_3-C \equiv CH)$.

(A) The reaction is the hydration of an alkyne in the presence of an acid and a mercury catalyst.
$CH_3-CH_2-C \equiv CH + H_2O \xrightarrow{Hg^{2+}, H^+} CH_3-CH_2-C(=O)-CH_3$ (Butanone).
$Hg^{2+}$ acts as a catalyst for the hydration of terminal alkynes to form ketones (via an enol intermediate).

(D) The reaction of acetylene $(C_2H_2)$ with water in the presence of $HgSO_4$ and dilute $H_2SO_4$ is known as Kucherov's reaction.
First,acetylene undergoes hydration to form an unstable intermediate called vinyl alcohol $(CH_2=CH-OH)$.
This vinyl alcohol then undergoes tautomerization to form a more stable isomer,acetaldehyde $(CH_3CHO)$.
The overall reaction is: $HC \equiv CH + H_2O \xrightarrow{HgSO_4 / H_2SO_4} [CH_2=CH-OH] \rightleftharpoons CH_3CHO$.

(B) The hydration of alkynes in the presence of $Hg^{2+}$ salts (like $HgSO_4$) and dilute acid (like $H_2SO_4$) follows Markovnikov's rule.
For propyne $(CH_3-C \equiv CH)$,the addition of water occurs across the triple bond.
The initial product is an enol $(CH_3-C(OH)=CH_2)$,which is unstable and undergoes tautomerization to form a stable ketone.
Specifically,propyne yields acetone $(CH_3-CO-CH_3)$.

(B) The hydration of ethyne $(HC \equiv CH)$ in the presence of $HgSO_4$ and $H_2SO_4$ is a classic reaction known as Kucherov's reaction.
This reaction proceeds via the formation of an unstable vinyl alcohol (enol) intermediate,which undergoes tautomerization to form a stable carbonyl compound.
$HC \equiv CH + H_2O$ $\xrightarrow{HgSO_4 / H_2SO_4} [CH_2=CH-OH]$ $\rightarrow CH_3CHO$ (Acetaldehyde).

(A) The reaction of an alkyne with water in the presence of $HgSO_4$ and $H_2SO_4$ is an electrophilic addition reaction that follows Markovnikov's rule.
First,the alkyne $CH_3 - CH_2 - C \equiv CH$ undergoes hydration to form an unstable enol intermediate: $CH_3 - CH_2 - C(OH) = CH_2$.
This enol intermediate then undergoes tautomerization to form a more stable keto form.
The final product $A$ is butan$-2-$one,which is represented as $CH_3 - CH_2 - C(=O) - CH_3$.

(A) The hydration of terminal alkynes like $but-1-yne$ in the presence of $HgSO_4$ and $H_2SO_4$ follows Markovnikov's rule.
Step $1$: The alkyne reacts with water to form an unstable enol intermediate: $CH_3-CH_2-C \equiv CH + H_2O \xrightarrow{HgSO_4/H_2SO_4} [CH_3-CH_2-C(OH)=CH_2]$.
Step $2$: The enol undergoes tautomerization to form a stable ketone: $[CH_3-CH_2-C(OH)=CH_2] ightarrow CH_3-CH_2-COCH_3$ $(Butan-2-one)$.
Therefore,the correct product is $Butan-2-one$.

(C) The reaction of an 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: $RCOOH$ and $R'COOH$.
In the given reaction,$CH_3-C\equiv C-CH_2-CH_3$ undergoes oxidative cleavage at the triple bond.
The $CH_3-C$ part is oxidized to acetic acid $(CH_3COOH)$.
The $-C-CH_2-CH_3$ part is oxidized to propanoic acid $(CH_3CH_2COOH)$.
Thus,the products are $CH_3COOH + CH_3CH_2COOH$.

(A) The reaction sequence is as follows:
$1$. $C_2H_2$ (acetylene) undergoes hydration in the presence of $HgSO_4$ and $H_2SO_4$ to form acetaldehyde $(CH_3CHO)$,which is $A$.
$2$. Acetaldehyde $(CH_3CHO)$ on oxidation $([O])$ yields acetic acid $(CH_3COOH)$,which is $B$.
$3$. Therefore,$B$ is an acid.

(C) When acetylene $(CH \equiv CH)$ reacts with acetic acid $(CH_3COOH)$ in the presence of $Hg^{2+}$ ions,it undergoes an addition reaction to form ethylidene diacetate $(CH_3-CH(OOCCH_3)_2)$.
$CH \equiv CH + 2CH_3COOH \xrightarrow{Hg^{2+}} CH_3-CH(OOCCH_3)_2$

(D) Calcium carbide is represented as $CaC_2$,which consists of $Ca^{2+}$ and $[C \equiv C]^{2-}$ ions.
In the acetylide ion $[C \equiv C]^{2-}$,there is a triple bond between the two carbon atoms.
$A$ triple bond consists of $1$ sigma $(\sigma)$ bond and $2$ pi $(\pi)$ bonds.

(A) In acetylene $(HC \equiv CH)$,the two carbon atoms are connected by a triple bond. $A$ triple bond consists of one $\sigma$ (sigma) bond and two $\pi$ (pi) bonds.

(B) The transformation of $1,2-dibromoethane$ to $ethyne$ involves dehydrohalogenation.
$1$. First,$1,2-dibromoethane$ reacts with alcoholic $KOH$ to undergo dehydrohalogenation,forming $vinyl$ $bromide$ $(CH_2=CHBr)$.
$2$. Then,the stronger base $NaNH_2$ is used to remove the second molecule of $HBr$ from $vinyl$ $bromide$ to form $ethyne$ $(HC \equiv CH)$.

(C) Step $1$: Reaction of $1$-butyne with $HCl$. According to Markovnikov's rule,the electrophile $H^+$ adds to the terminal carbon,and the nucleophile $Cl^-$ adds to the more substituted carbon. The product $B$ is $CH_3-CH_2-CCl=CH_2$ ($2$-chloro$-1-$butene).
Step $2$: Reaction of $B$ with $HI$. Again,following Markovnikov's rule,the $H^+$ adds to the $CH_2$ group and the $I^-$ adds to the $CCl$ carbon. The final product $C$ is $CH_3-CH_2-CCl(I)-CH_3$ ($2$-chloro$-2-$iodobutane).

(A) When $CHCl_3$ (chloroform) is heated with silver powder $(Ag)$,it undergoes a dehalogenation reaction to form acetylene $(C_2H_2)$.
The chemical reaction is: $2CHCl_3 + 6Ag \rightarrow C_2H_2 + 6AgCl$.

(A) When chloroform $(CHCl_3)$ is heated with silver $(Ag)$ powder,it undergoes dehalogenation to form acetylene $(C_2H_2)$.
The chemical reaction is: $2CHCl_3 + 6Ag \rightarrow C_2H_2 + 6AgCl$.