Masterclass: 25 Ultra-Challenging JEE Advanced Problems on Alkynes
From the acidity of terminal sp-carbons and the stereocontrol of Birch reductions, to the nuances of hydration and alkyne isomerizations. Conquer the triple bond.
Alkynes present a unique paradox in organic chemistry. Despite possessing two pi-bonds, they are generally less reactive towards electrophiles than alkenes, yet they boast uniquely acidic terminal protons and serve as the premier starting materials for synthesizing complex carbon skeletons. Mastery of alkyne reduction stereochemistry (Syn vs Anti) is a mandatory requirement for JEE Advanced.
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Step 1: Analyze the $NaNH_2$ Reaction
Reaction: ${CH_3-C \equiv C-H} + {NH_2^-} \rightleftharpoons {CH_3-C \equiv C^-} + {NH_3}$
The $pK_a$ of a terminal alkyne is $\approx 25$. The conjugate acid formed is Ammonia (${NH_3}$), which has a $pK_a \approx 38$. Because the reaction proceeds from a stronger acid ($pK_a 25$) to a weaker acid ($pK_a 38$), the equilibrium lies massively to the right ($K_{eq} \approx 10^{13}$).
Step 2: Analyze the $NaOH$ Reaction
Reaction: ${CH_3-C \equiv C-H} + {OH^-} \rightleftharpoons {CH_3-C \equiv C^-} + {H_2O}$
The conjugate acid formed here is Water (${H_2O}$), which has a $pK_a \approx 15.7$.
Step 3: Thermodynamic Verdict
Water ($pK_a 15.7$) is a significantly stronger acid than the terminal alkyne ($pK_a 25$). The reaction attempts to proceed from a weak acid to a strong acid, which is thermodynamically forbidden. The equilibrium lies entirely to the left ($K_{eq} \approx 10^{-9}$).
View Solution
Step 1: First Electron Transfer
Sodium metal dissolves in liquid ammonia to provide solvated electrons. A single electron adds to the pi-antibonding orbital of the alkyne, forming a highly reactive Radical Anion ($[CH_3-C^{\bullet}=C^--CH_3]$). One carbon bears a radical (single electron), while the other bears a lone pair (negative charge).
Step 2: Stereochemical Lock (The Crucial Step)
The $sp^2$ hybridized radical anion can adopt two geometries. To minimize the extreme electrostatic repulsion between the single electron cloud and the lone pair electron cloud, the molecule rapidly adopts a trans-like (anti) geometry, placing the bulky methyl groups and the electron clouds as far apart as physically possible.
Step 3: Protonation and Second Electron
The carbanion site abstracts a proton from the ammonia solvent. A second solvated electron adds to the remaining radical to form a second carbanion (with the trans geometry now rigidly set), which abstracts a final proton to yield the alkene.
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Step 1: Why it stops at the Alkene (Thermodynamics of Adsorption)
For a molecule to be hydrogenated, it must chemically adsorb onto the Palladium metal surface. Alkynes, with their two pi-bonds, coordinate to the metal surface much more strongly than alkenes. The "poisons" (Lead acetate/Quinoline) partially deactivate the metal surface. They leave just enough active sites to bind the strongly adsorbing alkyne, but the resulting alkene is too weakly coordinating to compete for the remaining deactivated sites. The alkene detaches and escapes before it can be reduced further.
Step 2: Why it is exclusively cis (Syn-Addition)
Because the alkyne is physically plastered flat against the solid Palladium surface, the Hydrogen atoms (which are dissociated on the same metal surface) must be delivered from the exact same face of the molecule. This physical constraint mandates a syn-addition.
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Step 1: Bypassing the Vinylic Carbocation
Protonating an alkyne directly would form an $sp$-hybridized vinylic carbocation (${H_2C=C^+-H}$), which is highly unstable. The large, highly polarizable $Hg^{2+}$ ion acts as a potent electrophile, coordinating across the triple bond to form a stable, bridged mercurinium-like ion, avoiding the free carbocation entirely.
Step 2: Nucleophilic Attack and Demercuration
Water attacks the bridged intermediate. Following the loss of a proton, an organomercury enol is formed. The acidic environment then replaces the Mercury group with a proton via protonolysis, yielding a pure Enol intermediate (${CH_2=CH-OH}$).
Step 3: Keto-Enol Tautomerization
The enol is thermodynamically unstable relative to its carbonyl counterpart. The pi-electrons shift to form a $C=O$ double bond, and the proton migrates to the adjacent carbon, rapidly tautomerizing into Acetaldehyde (${CH_3-CHO}$).
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Step 1: The Failure of Standard $BH_3$
Because ${BH_3}$ is small and has three active hydrogens, it can add across the triple bond to form a vinylborane, and then immediately add across the resulting double bond a second time, undergoing double hydroboration. Oxidation of this mess yields complex mixtures, not a clean aldehyde.
Step 2: The Solution (Bulky Boranes)
Disiamylborane ($Sia_2BH$) has massive, sterically hindering alkyl groups and only one active $B-H$ bond. When it adds across the terminal alkyne, the massive steric bulk physically prevents a second borane molecule from approaching the resulting alkene. The reaction stops cleanly at the vinylborane stage.
Step 3: Anti-Markovnikov Regioselectivity
The sheer size of the $Sia_2B$ group forces it to attach exclusively to the least hindered terminal carbon. Subsequent oxidation (${H_2O_2 / OH^-}$) replaces the boron with an $-OH$ group, yielding an enol at the terminal carbon, which tautomerizes cleanly to Hexanal.
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Step 1: Base-Catalyzed Internalization (Thermodynamic Control)
Heating with a strong base like ${KOH}$ causes a reversible cascade of deprotonations and protonations, migrating the triple bond up and down the chain via allene intermediates. Internal alkynes are thermodynamically more stable than terminal alkynes due to hyperconjugation. The equilibrium funnels into the internal alkyne.
Step 2: The Zipper Reaction (Kinetic Trapping)
$KAPA$ is an exceptionally strong, specialized base. It also causes the triple bond to rapidly migrate ("zip") along the chain. However, when the triple bond reaches the terminal position to become 1-pentyne, something unique happens.
Step 3: The Irreversible Trap
Terminal alkynes have a highly acidic $sp$-hybridized proton ($pK_a \approx 25$). The hyper-strong $KAPA$ base instantly and irreversibly deprotonates this terminal alkyne to form a stable alkynide anion (${C_3H_7-C \equiv C^- K^+}$). Because the anion cannot easily migrate further, the molecule is kinetically trapped at the terminus until aqueous workup protonates it back to 1-pentyne.
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Step 1: The Successful $S_N2$ Reaction
1-bromopropane is a primary ($1^{\circ}$) alkyl halide. It has minimal steric hindrance. The acetylide ion acts as a nucleophile, easily performing a backside $S_N2$ attack to displace the bromide ion and form the $C-C$ bond.
Step 2: The Steric Blockade
2-bromo-2-methylpropane is a tertiary ($3^{\circ}$) alkyl halide. The electrophilic carbon is completely surrounded by three massive methyl groups. Backside $S_N2$ attack is physically impossible due to extreme steric hindrance.
Step 3: The E2 Elimination Pathway
Denied the opportunity to act as a nucleophile, the acetylide ion exerts its other property: it is a very strong base (conjugate base of a weak acid, $pK_a 25$). It abstracts a beta-proton from one of the exposed methyl groups. This triggers a rapid E2 elimination, expelling the bromide ion to form an alkene.
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Step 1: Nucleophilic Attack
The carbonyl carbon of acetone is highly electrophilic due to the polar $C=O$ bond. The negatively charged carbon of the propynide ion attacks the carbonyl carbon, pushing the pi-electrons onto the oxygen atom to form an alkoxide intermediate: ${CH_3-C \equiv C-C(CH_3)_2-O^-}$.
Step 2: Acidic Workup
The addition of dilute acid (${H_3O^+}$) protonates the alkoxide oxygen, converting it into a hydroxyl group ($-OH$).
Step 3: Structural Identification
The final molecule is ${CH_3-C \equiv C-C(CH_3)_2-OH}$. The IUPAC name is 2-methyl-3-pentyn-2-ol. Because the hydroxyl group is situated on a carbon directly adjacent to a carbon-carbon triple bond, this specific motif is classified as a propargylic alcohol.
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Step 1: Electrophilic Attack
The pi-electrons of the alkyne attack the polarizable ${Br_2}$ molecule. The Bromine atom simultaneously uses a lone pair to bond back to the other carbon. However, unlike alkenes, an alkyne has a remaining pi-bond. This forms a highly strained, bridged bromonium-like ion that still possesses a double bond in the ring: a cyclic bromirenium ion.
Step 2: The Stereochemical Lock
Because the massive Bromine atom bridges one entire face of the molecule, it acts as a physical shield. The positive charge is distributed across the three-membered ring.
Step 3: Anti-Attack
The expelled bromide ion (${Br^-}$) must now act as a nucleophile to open the ring. To avoid the massive steric bulk of the bridging Bromine, it must attack from the strictly opposite face (backside attack). This forces the two Bromine atoms onto opposite sides of the newly formed double bond.
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Step 1: The First Addition
Protonation of propyne yields the more stable secondary vinylic carbocation (${CH_3-C^+=CH_2}$) over the primary one. Bromide attacks to yield 2-bromopropene.
Step 2: The Second Protonation (The Paradox)
Now we add $H^+$ to 2-bromopropene. Bromine is highly electronegative ($-I$ effect), which should destabilize a positive charge. If it adds to the terminal carbon, the carbocation forms on C2: ${CH_3-C^+(Br)-CH_3}$. If it adds to C2, the carbocation forms on C1: ${CH_3-CH(Br)-C^+H_2}$.
Step 3: Resonance Stabilization (The Deciding Factor)
While Bromine exerts a strong $-I$ effect, it also possesses lone pairs. When the positive charge forms directly on the carbon bearing the Bromine atom, the Bromine can donate a lone pair to form a pi-bond via resonance: $[{CH_3-C^+(Br)-CH_3} \leftrightarrow {CH_3-C(=Br^+)-CH_3}]$. This resonance stabilization (+$M$ effect) overwhelmingly dominates the inductive destabilization. Therefore, the second proton strictly adds to the terminal carbon, forming a geminal dihalide.
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Step 1: Initiation
Peroxides homolytically cleave to form alkoxy radicals, which abstract hydrogen from ${HBr}$ to generate the highly reactive Bromine radical (${Br^{\bullet}}$).
Step 2: Regioselective Radical Addition
The ${Br^{\bullet}}$ attacks the pi-bond of 1-butyne. It has two choices:
- Attack C2 to form a primary vinylic radical at C1 (highly unstable).
- Attack C1 to form a secondary vinylic radical at C2: ${CH_3-CH_2-C^{\bullet}=CH-Br}$.
The secondary vinylic radical is more stable due to hyperconjugation from the adjacent ethyl group. Thus, Bromine bonds to the terminal carbon (Anti-Markovnikov).
Step 3: Hydrogen Abstraction and Stereochemistry
The vinylic radical abstracts a Hydrogen atom from a fresh molecule of ${HBr}$. Because the vinylic radical is $sp^2$ hybridized and undergoes rapid inversion (or is a shallow pyramid), the hydrogen can approach from either face. This scrambling results in a thermodynamic mixture of both (E) and (Z) isomers of 1-bromo-1-butene.
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Step 1: Cleavage of an Internal Alkyne (2-pentyne)
Structure: ${CH_3-C \equiv C-CH_2CH_3}$.
The ozone molecule adds across the triple bond to form an unstable intermediate, which breaks down in the presence of water. The triple bond is entirely severed, and each involved carbon is fully oxidized to a carboxylic acid group.
Products: Acetic acid (${CH_3COOH}$) and Propanoic acid (${CH_3CH_2COOH}$).
Step 2: Cleavage of a Terminal Alkyne (1-pentyne)
Structure: ${HC \equiv C-CH_2CH_2CH_3}$.
The internal carbon (C2) oxidizes normally to a carboxylic acid, yielding Butanoic acid (${CH_3CH_2CH_2COOH}$).
The terminal carbon (C1), however, is bonded only to a hydrogen. It initially oxidizes to Formic acid (${HCOOH}$). However, under oxidative cleavage conditions, formic acid is violently unstable and immediately oxidizes further into Carbon dioxide (${CO_2}$) and water.
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Step 1: The Acidity Factor
1-butyne is a terminal alkyne (${CH_3CH_2-C \equiv C-H}$). The hydrogen is attached to an $sp$-hybridized carbon. Because $sp$-orbitals have $50\%$ s-character, they hold electrons very tightly to the nucleus, rendering that terminal hydrogen unusually acidic ($pK_a \approx 25$). 2-butyne (${CH_3-C \equiv C-CH_3}$) is internal and lacks this acidic proton.
Step 2: The Reaction
When 1-butyne is treated with Tollens' reagent ($[Ag(NH_3)_2]^+ OH^-$), the basic hydroxide ion removes the acidic proton, and the Silver cation coordinates with the resulting alkynide anion. This forms Silver butynide (${CH_3CH_2-C \equiv C^- Ag^+}$), which is highly insoluble and instantly drops out of solution as a heavy white precipitate.
Step 3: Conclusion
Because 2-butyne has no acidic proton, it cannot form this salt and remains unreactive.
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Step 1: Ideal Alkyne Geometry
The two carbons of a triple bond are $sp$ hybridized. The ideal bond angle for an $sp$ hybridized carbon is exactly $180^{\circ}$. This means an alkyne and its two directly attached atoms strongly prefer to be in a perfectly straight, linear line ($-C \equiv C-$).
Step 2: The Strain of Small Rings
If you force a triple bond into a small 6- or 7-membered ring, the geometry of the ring forcibly bends the $180^{\circ}$ bond angles down to accommodate the ring closure. This severe bending causes massive angle strain. In cyclohexyne, the strain is so catastrophic that the molecule instantly reacts with itself or anything nearby to break the triple bond.
Step 3: The 8-Carbon Threshold
A cyclooctane ring (8 carbons) is physically large enough and possesses enough conformational flexibility (puckering) to bridge the linear alkyne segment without bending the $sp$ bond angles beyond their breaking point. While cyclooctyne is still highly strained and very reactive (frequently used in "click chemistry"), it crosses the threshold of isolation stability.
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Step 1: Visualize the Alignment
Three molecules of propyne align in a hexagonal arrangement inside the red-hot tube. For the most sterically favorable and electronically coordinated cyclization, the molecules arrange head-to-tail: The $CH_3$ group of molecule 1 is adjacent to the $CH$ group of molecule 2, and so on.
Step 2: The Cyclization Mechanism
Under intense thermal energy, one pi-bond from each of the three triple bonds breaks and reaches out to form a new sigma bond with the adjacent propyne molecule. This concerted electron shift closes the ring, leaving three alternating double bonds inside a six-membered ring (an aromatic benzene core).
Step 3: Locating the Substituents
Because the propyne molecules aligned head-to-tail, the three methyl groups end up symmetrically distributed around the newly formed benzene ring on alternating carbons (positions 1, 3, and 5).
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Step 1: Ground State Stability ($sp$ vs $sp^2$)
The carbons of an alkyne are $sp$ hybridized ($50\%$ s-character). Because s-orbitals are closer to the nucleus, the pi-electrons in an alkyne are held much more tightly by the carbon nuclei than the pi-electrons in an $sp^2$ hybridized alkene ($33\%$ s-character). Therefore, the alkyne's pi-electrons are less "available" or nucleophilic to attack an incoming electrophile.
Step 2: Transition State/Intermediate Instability
Electrophilic addition is a two-step process. The rate-determining step is the formation of a carbocation intermediate.
- An alkene forms a standard, $sp^2$ hybridized alkyl carbocation.
- An alkyne forms an $sp$ hybridized vinylic carbocation (${R-C^+=CH_2}$).
An $sp$ hybridized carbon is highly electronegative and strongly resists bearing a positive charge. The vinylic carbocation is vastly higher in energy (more unstable) than a standard carbocation, creating a massive activation energy barrier that severely slows down the reaction.
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Step 1: Attack on 1-Pentyne (Terminal)
1-pentyne is strongly unsymmetrical. Attack yields a vinylic carbocation. The internal carbocation is secondary, while the terminal is primary. The massive energy difference dictates perfect Markovnikov regiocontrol, yielding exclusively 2-pentanone.
Step 2: Attack on 2-Pentyne (Internal)
2-pentyne is an internal, unsymmetrical alkyne. The triple bond is between C2 and C3.
- If hydration occurs at C2, the intermediate carbocation is at C3. This is a secondary vinylic carbocation stabilized by an ethyl group.
- If hydration occurs at C3, the intermediate carbocation is at C2. This is a secondary vinylic carbocation stabilized by a methyl group.
Step 3: Thermodynamic Similarity
The stabilities of these two possible secondary vinylic carbocations are nearly identical (the difference between methyl and ethyl stabilization is negligible here). Therefore, the water molecule attacks both C2 and C3 with almost equal probability.
Step 4: The Products
Attack at C2 yields an enol that tautomerizes to 2-pentanone. Attack at C3 yields an enol that tautomerizes to 3-pentanone.
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Step 1: Evaluate the Proposed Route
Propyne + $NaNH_2 \rightarrow$ Propynide ion (${CH_3-C \equiv C^-}$).
Reactant: tert-butyl bromide (a tertiary alkyl halide).
As discussed in Problem 7, $S_N2$ attack on a tertiary halide is sterically impossible. The strongly basic propynide ion will simply cause E2 elimination of the tert-butyl bromide, yielding isobutylene and recovering propyne. The synthesis fails completely.
Step 2: Design the Correct Route
To successfully form the $C-C$ bond, the alkyl halide MUST be primary (or methyl) to ensure a clean $S_N2$ attack without competitive elimination. Therefore, the required steric bulk must already be present on the alkyne starting material.
Step 3: The Successful Pairing
Target: ${CH_3-C \equiv C-C(CH_3)_3}$.
Disconnect the bond next to the triple bond to leave the bulk on the alkyne.
Starting Alkyne: 3,3-dimethyl-1-butyne (${HC \equiv C-C(CH_3)_3}$). Deprotonate with $NaNH_2$.
Alkyl Halide: Methyl Bromide (${CH_3Br}$).
The bulky alkynide effortlessly performs an $S_N2$ attack on the unhindered methyl bromide to yield the target molecule.
View Solution
Step 1: The Binding Affinity
Catalytic hydrogenation occurs on the solid surface of the metal catalyst ($Pt$, $Pd$, or $Ni$). The alkyne must physically adsorb onto the metal surface. Because it has a flat, linear geometry and two pi-bonds, the alkyne binds to the metal surface incredibly tightly.
Step 2: The First Reduction
Hydrogen atoms transfer to the alkyne, reducing it to an alkene.
Step 3: The Kinetic Trap
Here is the problem: an alkene is actually more reactive towards hydrogenation than an alkyne! Furthermore, once the alkene is formed on the metal surface, it does not immediately desorb (detach) into the solution. Since it is already sitting on the active catalyst surrounded by hydrogen atoms, and it is highly reactive, it instantly undergoes the second reduction to become an alkane.
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Step 1: The Susceptibility of Alkynes
Alkenes repel nucleophiles because their electron-rich pi-clouds are diffuse. In alkynes, the carbons are $sp$ hybridized. The high s-character ($50\%$) makes the carbon atoms significantly more electronegative, pulling the pi-electron cloud tighter and exposing the nuclei slightly more to nucleophilic attack.
Step 2: Nucleophilic Attack
The strong nucleophile, Cyanide ion (${CN^-}$), attacks the triple bond. A pair of pi-electrons is pushed onto the adjacent carbon, forming a vinylic carbanion: ${N \equiv C-CH=C^--H}$.
Step 3: Stability of the Intermediate
Normally, carbanions are highly unstable. However, a vinylic carbanion places the negative charge in an $sp^2$ hybridized orbital, which (with $33\%$ s-character) stabilizes the negative charge far better than an $sp^3$ orbital would.
Step 4: Protonation
The vinylic carbanion rapidly abstracts a proton from $HCN$ (regenerating the ${CN^-}$ catalyst) to form the final product, Acrylonitrile.
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Step 1: The Basic Environment
The Birch reduction uses Sodium metal dissolved in liquid ammonia, creating solvated electrons and amide ions (${NH_2^-}$). This is an incredibly powerful basic environment.
Step 2: The Acid-Base Priority
Acid-base reactions are kinetically vastly faster than redox or addition reactions. When a terminal alkyne (which possesses a highly acidic proton, $pK_a \approx 25$) is introduced to this basic mixture, it instantly undergoes an acid-base reaction. The amide ion (or solvated electron) strips the terminal proton away, forming a stable Sodium alkynide salt (${R-C \equiv C^- Na^+}$).
Step 3: Resistance to Reduction
The resulting alkynide anion is heavily negatively charged. It vehemently repels the approach of any solvated electrons from the Sodium metal. Consequently, the reduction mechanism is completely short-circuited.
Step 4: Recovery
The molecule sits harmlessly as a salt in the ammonia. When water is added at the end to quench the reaction, the water simply reprotonates the alkynide salt, yielding the exact starting terminal alkyne back.
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Step 1: The First Elimination
1,2-dibromopropane (${CH_3-CHBr-CH_2Br}$) is an ordinary alkyl halide. $KOH$ in ethanol easily abstracts a beta-proton to expel a bromide ion via standard E2, yielding a vinyl halide: 1-bromopropene (${CH_3-CH=CHBr}$).
Step 2: The Problem with Vinyl Halides
The second elimination requires removing a proton from an $sp^2$ hybridized carbon and expelling a halogen attached to an $sp^2$ carbon.
- $sp^2$ $C-H$ bonds are stronger and less acidic than $sp^3$ $C-H$ bonds.
- The lone pairs on the Bromine atom are in resonance with the adjacent pi-bond, giving the $C-Br$ bond partial double-bond character. This makes the leaving group extremely difficult to break off.
Step 3: The Need for an Extreme Base
Because the proton is hard to remove and the leaving group is tightly bound, the activation energy for this second E2 elimination is incredibly high. Standard $KOH$ simply lacks the thermodynamic power to force this reaction. Sodium amide ($NaNH_2$), being a significantly stronger base, provides the extreme driving force necessary to overcome this barrier and yield the alkyne.
View Solution
Step 1: Formation of the Copper Acetylide
The weak base removes the acidic terminal proton of phenylacetylene. The resulting anion coordinates with the $Cu(I)$ catalyst to form a Copper(I) phenylacetylide intermediate (${Ph-C \equiv C-Cu}$).
Step 2: Oxidative Coupling
The presence of atmospheric Oxygen acts as an oxidizing agent. It oxidizes the Copper from $Cu(I)$ to $Cu(II)$, triggering a single-electron transfer that generates an alkynyl radical (${Ph-C \equiv C^{\bullet}}$).
Step 3: Dimerization
Two highly reactive alkynyl radicals instantly couple together, forming a new Carbon-Carbon single bond between the two $sp$-hybridized carbons.
Step 4: Final Structure
The resulting molecule is ${Ph-C \equiv C-C \equiv C-Ph}$. This is a conjugated diyne, specifically named 1,4-diphenyl-1,3-butadiyne. This classic reaction is known as the Glaser Coupling.
View Solution
Step 1: Initial Hydration (Markovnikov)
The solvent is heavy water ($D_2O$) and the acid source provides $D^+$. Following the Kucherov mechanism, $D^+$ adds to the terminal carbon, and the nucleophile $D_2O$ attacks the internal carbon (C2).
Intermediate enol formed: ${CH_3-C(OD)=CHD}$.
Step 2: Keto-Enol Tautomerization
The enol must tautomerize to the ketone. In this process, the pi-bond grabs a deuterium ion ($D^+$) from the acidic solvent to add to the terminal carbon, and the oxygen atom loses its deuterium ($D^+$) to restore the double bond.
Structure after one tautomerization: ${CH_3-C(=O)-CHD_2}$.
Step 3: Extended Enolization in Solvent
Ketones with alpha-hydrogens are in constant, dynamic equilibrium with their enol forms. In a massive excess of $D_2O/D^+$, every time the ketone forms an enol on the left side (${CH_2=C(OD)-CHD_2}$), it grabs a $D^+$ from the solvent and loses an $H^+$ back to the solvent. Over time, all enolizable alpha-hydrogens are completely swapped out for Deuterium.
Step 4: Final Structure
The original propyne had three alpha-hydrogens on the methyl group, and the terminal carbon eventually becomes a methyl group as well. All of these hydrogens are alpha to the new carbonyl and will be completely replaced.
View Solution
Step 1: Degree of Unsaturation (DU)
Formula is $C_6H_{10}$. Saturated hexane is $C_6H_{14}$. Missing 4 hydrogens means $DU = 2$. It absorbs 2 moles of $H_2$, confirming the absence of a ring. It is an acyclic molecule with either two double bonds or one triple bond.
Step 2: Silver Nitrate Test
It forms a white precipitate with Ammoniacal $AgNO_3$ (Tollens' reagent). This is a definitive test for a terminal alkyne. So, the molecule has the structure $R-C \equiv C-H$.
Step 3: Hydration and Iodoform Test
Hydration of a terminal alkyne ($R-C \equiv CH$) via $HgSO_4$ always follows Markovnikov's rule, yielding a methyl ketone ($R-CO-CH_3$).
A positive Iodoform test confirms the presence of this $CH_3-CO-$ group. So, Y is definitely a methyl ketone. This confirms our terminal alkyne deduction.
Step 4: Determine the 'R' group
The remaining 'R' group must contain 4 carbons ($C_6 - C_2$). It could be a butyl, isobutyl, sec-butyl, or tert-butyl group.
Wait, is there any more information? "yields a single major product Y". All terminal alkynes of $C_6H_{10}$ yield methyl ketones.
Let's re-read: "forms a white precipitate". So it must be a terminal alkyne.
If it's 1-hexyne: yields 2-hexanone.
If it's 3-methyl-1-pentyne: yields 3-methyl-2-pentanone.
If it's 3,3-dimethyl-1-butyne: yields 3,3-dimethyl-2-butanone.
Usually, without further degradation data, the straight-chain isomer is assumed as the primary answer in basic deductions, but technically multiple isomers fit. Let's provide the most common straight-chain isomer, 1-Hexyne.
Mastering the Triple Bond
Congratulations on conquering these 25 ultra-challenging problems on Alkyne chemistry! You have successfully navigated the stereochemical traps of the Birch reduction, the thermodynamic pitfalls of alkyne zipper reactions, and the deep mechanistic logic of Mercuric-catalyzed hydration. Remember, in JEE Advanced, the triple bond is a dynamic functional group—acting as a potent acid in the presence of strong bases, and a highly selective canvas for stereocontrolled reductions. Keep honing your mechanistic intuition, and visit Chemca.in for more elite masterclasses!
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