Predict the product of adding two equivalents of $ ext{HBr}$ to propyne, $ ext{CH}_3 ext{C}{\equiv} ext{CH}$.

Question

Predict the product of adding two equivalents of $	ext{HBr}$ to propyne, $	ext{CH}_3	ext{C}{\equiv}	ext{CH}$.

Accepted Answer

Concept understanding — Electrophilic Addition Reactions

Electrophilic Aromatic Substitution – The First Meeting

Imagine you have a benzene ring — that perfect, flat hexagon of six carbons with alternating double bonds. It's stable, almost stubbornly so. You want to attach something new to it, say a bromine atom or a nitro group. But benzene doesn't react like an alkene. It doesn't just add across a double bond. Instead, it does something more elegant: it kicks out a hydrogen and keeps its aromatic ring intact.

That's the heart of Electrophilic Aromatic Substitution (EAS).

The Intuition: Why "Substitution" and Not "Addition"?

Benzene's stability comes from its delocalised $\pi$ electrons — a cloud above and below the ring. This cloud is electron-rich, so it attracts electrophiles (electron-loving species). But if an electrophile simply added to a double bond, the ring would break its aromaticity, losing that huge stabilisation. That would be energetically costly.

So benzene does something smarter: it lets the electrophile attack, temporarily breaks aromaticity to form a high-energy intermediate (the arenium ion), and then loses a proton to restore the aromatic ring. The net result? A hydrogen is replaced by the electrophile. The ring is back to its stable, aromatic self.

Note

The key trade-off: temporary loss of aromaticity is acceptable because the final product regains it. Addition reactions would permanently destroy aromaticity — benzene avoids that.

The Precise Statement

Electrophilic Aromatic Substitution is a reaction in which an electrophile ( $E^+$ ) replaces a hydrogen atom on an aromatic ring, proceeding through a sigma complex (arenium ion) intermediate, and restoring aromaticity after deprotonation.

The general equation:

\ce{Ar-H + E+ -> Ar-E + H+}

where Ar represents an aromatic ring.

The Mechanism in Three Steps

Generation of the electrophile – Many EAS reactions need a catalyst to create a strong enough $E^+$ . For example, bromination uses $\ce{FeBr3}$ to polarise $\ce{Br2}$ into $\ce{Br^+}$ .
Attack by the aromatic ring – The $\pi$ electrons of benzene attack the electrophile, forming a sigma complex (also called the arenium ion or Wheland intermediate). This intermediate is non-aromatic — it has four $\pi$ electrons delocalised over five carbons, and one $sp^3$ carbon bearing the electrophile and a hydrogen.

\text{Benzene} + E^+ \longrightarrow \text{Sigma complex (non-aromatic)}

Deprotonation – A base (often the counterion of the catalyst, like $\ce{FeBr4-}$ ) removes the proton from the $sp^3$ carbon. The pair of electrons from the C–H bond flows back into the ring, restoring the aromatic sextet.

\text{Sigma complex} + \text{Base} \longrightarrow \text{Product} + \ce{H+}

Watch out

The sigma complex is not aromatic. It's a high-energy intermediate. Students often mistakenly think it's still aromatic — it isn't. That's why the step is fast and the complex is short-lived.

Why This Matters for Exams

EAS is the gateway to understanding how to put groups onto benzene rings. The rate-determining step is usually the formation of the sigma complex (step 2). The regiochemistry (where the electrophile goes) depends on whether the ring already has a substituent — that's the topic of activating/deactivating groups and ortho/para vs. meta directors. …

Why this formula?

Electrophilic Addition Reactions: Why the Mechanism Works

Electrophilic addition is a cornerstone of alkene and alkyne chemistry. Instead of memorising the "arrow pushing," let's understand why the reaction proceeds the way it does — driven by electron density, stability, and charge.

1. The Core Idea: Why Alkenes React This Way

Alkenes have a π-bond — a cloud of electrons above and below the plane of the σ-bond. This π-electron cloud is:

Electron-rich (nucleophilic)
Exposed (not shielded by σ-bonds like in alkanes)

An electrophile (electron-lover) is attracted to this high electron density. The reaction is electrophilic addition because the electrophile attacks first.

Key principle: The π-bond acts as a Lewis base (electron donor). The electrophile is a Lewis acid (electron acceptor).

2. The General Mechanism (Two-Step)

Step 1: Formation of a Carbocation (or Bridged Intermediate)

The electrophile (E⁺) attacks the π-bond. The π-electrons form a new σ-bond to E⁺, leaving the other carbon with a positive charge — a carbocation.

\ce{C=C} + \ce{E+} \longrightarrow \ce{C+-C-E}

Why does this happen?

The π-bond is weaker than a σ-bond (~260 kJ/mol vs ~350 kJ/mol). Breaking the π-bond to form a σ-bond is energetically favourable because the new σ-bond is stronger. The carbocation is a high-energy intermediate, but it's stabilised by:

Hyperconjugation (alkyl groups donate electron density)
Inductive effect (alkyl groups push electrons toward the positive carbon)

Step 2: Nucleophilic Attack

A nucleophile (Nu⁻) attacks the carbocation, forming a second σ-bond.

\ce{C+-C-E} + \ce{Nu-} \longrightarrow \ce{C-Nu-C-E}

Why does this happen?

The carbocation is electron-deficient (positive charge). The nucleophile is electron-rich. Opposite charges attract — this is electrostatic and orbital overlap driven.

3. The Key "Formula" — Markovnikov's Rule

Statement: In the addition of HX to an unsymmetrical alkene, the hydrogen attaches to the carbon with more hydrogens already, and the X attaches to the carbon with fewer hydrogens.

Why does this rule hold? (The reasoning)

Consider propene: $\ce{CH3-CH=CH2}$ + HBr.

Possible carbocations:
1. Primary carbocation: $\ce{CH3-CH+-CH2Br}$ (less stable)
2. Secondary carbocation: $\ce{CH3-CHBr-CH2+}$ (more stable)

The more substituted carbocation (secondary > primary) is more stable due to:

Hyperconjugation: More alkyl groups = more C–H σ-bonds that can donate electron density into the empty p-orbital of the carbocation.
Inductive effect: Alkyl groups are electron-donating, stabilising the positive charge.

Result: The reaction proceeds via the more stable carbocation, leading to Markovnikov addition.

Markovnikov's rule is not a law — it's a consequence of carbocation stability.

4. The "Anti-Markovnikov" Exception (Why It Happens)

With HBr in the presence of peroxides (ROOR), the addition is anti-Markovnikov — Br goes to the less substituted carbon.

Why? The mechanism changes from ionic to free-radical.

Peroxide decomposes to radicals: $\ce{ROOR -> 2 RO•}$
RO• abstracts H from HBr: $\ce{RO• + HBr -> ROH + Br•}$
Br• adds to the alkene — at the less substituted carbon (because the radical formed is more stable — tertiary > secondary > primary).
The new radical abstracts H from another HBr, regenerating Br•. …

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Propyne's triple bond has two pi bonds and so can accept two successive equivalents of $\text{HBr}$ , each addition following the same electrophilic-addition mechanism (and, since no peroxide is mentioned, the ordinary ionic Markovnikov pathway) as for an alkene. In the FIRST addition, $\text{H}^+$ adds to the terminal ( $\equiv\text{CH}$ ) carbon, generating the more stable vinylic/secondary-type cationic intermediate on the internal carbon, and $\text{Br}^-$ then adds there, giving the intermediate bromoalkene $\text{CH}_3\text{CBr=CH}_2$ (2-bromopropene) -- consistent with Markovnikov's rule applied to the triple bond. In the SECOND addition, the same logic is applied to this new alkene's double bond: $\text{H}^+$ adds to the terminal $\text{=CH}_2$ carbon (giving the more stable carbocation, now further stabilised by the adjacent bromine-bearing carbon), and the second $\text{Br}^-$ adds to the SAME carbon that already carries the fir …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

CBSE 2026Set ANNUAL1 markMCQ
$Q. CH3CH=CH2 --H+/H2O--> A. Major product is$
(a) CH3-CH-CH2 with an O bridging the CH and CH2 (a three-membered cyclic ether / epoxide, i.e. 2-methyloxirane)
(b) CH3-CH(OH)-CH3 (propan-2-ol)
(c) CH3CH2CH2OH (propan-1-ol)
(d) CH3-CH(OH)-CH2-OH (propane-1,2-diol)
›Reveal solutionSolution

Acid-catalysed hydration of an alkene (H+/H2O) is a Markovnikov addition: the -OH group ends up on the carbon that can best stabilise the intermediate carbocation, i.e. the more substituted carbon.

Mechanism: H+ protonates the double bond of CH3-CH=CH2. Protonation occurs so as to generate the more stable carbocation - here, protonating the terminal CH2 gives a secondary carbocation on the middle carbon, CH3-CH+-CH3, which is more stable than the alternative primary carbocation.

Water then attacks this secondary carbocation, and loss of a proton gives the final alcohol:

…

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CBSE 2026Set ANNUAL1 markMCQ
$Q. The general molecular formula of an alkene is$
(a) CnH2n+2
(b) CnH2n
(c) CnH2n+1
(d) CnH2n-2
›Reveal solutionSolution

General formula of an alkene = CnH2n.

Alkenes contain one C=C double bond and have the general formula CnH2n (e.g. ethene C2H4, propene C3H6). Al …

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CBSE 2025Set X11 mark
$\text{H}_3\text{O}^{\oplus}$
›Reveal solutionSolution
The electrophilic attack of H3O+ on an alkene protonates the double bond to form a carbocation (the alcohol is only formed later, after water addition and deprotonation). Answer: carbocation.

Acid-catalysed hydration of an alkene proceeds in steps:

Electrophilic attack of H3O+ (protonation) on the alkene forms a carbocation.

Water then attacks the carbocation.

Loss of a proton gives the alcohol. …
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CBSE 2025Set ANNUAL1 markMCQ
$Q. CH2=CH2 + Br2 --(CCl4)--> X. Here 'X' is:$
(a) CH2Br-CH2Br
(b) CH2=CHBr
(c) CH≡CH
(d) CHBr=CHBr
›Reveal solutionSolution

Br $_2$ adds across the C=C double bond of ethene (electrophilic addition), giving the vicinal dibromide.

Alkenes are electron-rich due to the $\pi$ bond, so they readily undergo electrophilic addition with bromine. In CCl $_4$ (an inert non-aqueous solvent, used so no other nucleophile interferes), $Br_2$ adds directly across the double bond:

$CH_2=CH_2 + Br_2 \xrightarrow{CCl_4} CH_2Br-CH_2Br$

…

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CBSE 2023Set ANNUAL1 markMCQ
$Q. Hydration of propene in the presence of dil. H2SO4 gives$
(a) CH3-CH2-CH2-OH
(b) CH3-CH(OH)-CH3
(c) CH3-CH2-OH
(d) CH3-OH
›Reveal solutionSolution

Markovnikov addition of water (via a more stable secondary carbocation intermediate) places -OH on the middle carbon of propene, giving 2-propanol.

CH3-CH=CH2 + H2O --(dil. H2SO4)--> CH3-CH(OH)-CH3 …

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CBSE 2023Set annual31 mark
$Q. Why are alkenes more reactive in nature?$
›Reveal solutionSolution
The pi bond in the C=C double bond of alkenes is weak and electron-rich, so it is easily attacked by electrophiles — this makes alkenes far more reactive than the saturated alkanes.

In an alkene, the doubly-bonded carbons are sp2-hybridised. Each carbon forms three sigma bonds in a plane (120° apart) using sp2 orbitals, and the double bond consists of one sigma bond (head-on sp2-sp2 overlap) plus one pi bond, formed by sideways overlap of the unhybridised p-orbitals on the two carbons.

The pi bond has two key features that make alkenes reactive:

It is weaker than a sigma bond (sideways p-orbital overlap is less effective than head-on overlap), so it breaks more easily than a C-C sigma bond.

Its electron cloud is spread above and below the molecular plane, away from the nuclei, making these electrons loosely held and easily accessible/polarisable. …
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CBSE 2022Set ANNUAL1 mark
$Q. How will you carry out the following conversion? Propene to propan-1-ol$
›Reveal solutionSolution

Direct acid-catalysed hydration of propene follows Markovnikov's rule and gives propan-2-ol; to reach the anti-Markovnikov propan-1-ol instead, propene is converted via hydroboration–oxidation.

Why simple hydration doesn't work

Direct acid-catalysed addition of water to propene ( $CH_3-CH=CH_2$ ) follows Markovnikov's rule — $H^+$ adds to the terminal (less substituted) carbon and $OH$ ends up on the more substituted (secondary) carbon — giving propan-2-ol, not the target propan-1-ol.

Hydroboration–oxidation route to propan-1-ol

Step 1 — Hydroboration: diborane ( $B_2H_6$ , or $BH_3$ ·THF) adds across the double bond with boron attaching to the less substituted (terminal) carbon (anti-Markovnikov, because it is a concerted, steric/electronic-controlled syn addition where boron preferentially bonds to the less hindered carbon):

$3\,CH_3-CH=CH_2 + B_2H_6 \longrightarrow (CH_3CH_2CH_2)_3B$

…

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CBSE 2022Set ANNUAL1 markMCQ
$Q. Butene-1 is changed into Butane$
(a) H2/Pd
(b) Zn/HCl
(c) Sn/HCl
(d) Zn-Hg
›Reveal solutionSolution

But-1-ene → butane by catalytic hydrogenation, H₂/Pd — option (a).

From NCERT Class 11 Chemistry (Hydrocarbons): an alkene is reduced to an alkane by addition of hydrogen over a metal catalyst (Ni, Pd or Pt):

CH₂=CH–CH₂–CH₃ + H₂ →(Pd) CH₃–CH₂–CH₂–CH₃. …

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CBSE 2021Set OC1 mark
$CH_3-CH=CH_2 \xrightarrow{H^+/H_2O} ?$
›Reveal solutionSolution
Acid-catalysed hydration of propene proceeds through the more stable secondary carbocation, so OH ends up on the middle carbon, giving propan-2-ol (Markovnikov addition).

Mechanism

Protonation: $H^+$ (from $H_3O^+$ ) adds to one of the alkene carbons. Protonating the terminal $=CH_2$ carbon generates a secondary carbocation at C-2, $CH_3-\overset{+}{C}H-CH_3$ , which is more stable than the alternative primary carbocation that would form if $H^+$ added to the other carbon (Markovnikov's rule: the proton adds to the carbon that already bears more hydrogens, so as to generate the more stable, more substituted carbocation).

Nucleophilic attack: water attacks the electrophilic secondary carbocation: $CH_3-\overset{+}{C}H-CH_3 + H_2O \rightarrow CH_3-CH(OH_2^+)-CH_3$ . …
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CBSE 2021Set annual21 mark
$Q. Out of ethylene and acetylene which is more reactive towards nucleophilic addition reactions and why?$
›Reveal solutionSolution

Acetylene reacts faster with nucleophiles than ethylene because its sp carbons are more electronegative than ethylene's sp2 carbons.

In ethylene (CH2=CH2), each carbon of the double bond is sp2-hybridised, with 33% s-character. In acetylene (CH≡CH), each carbon of the triple bond is sp-hybridised, with 50% s-character.

Greater s-character means the hybrid orbital electrons (and hence the bonding electrons) are held closer to and more tightly by the nucleus, making sp carbon atoms more electronegative than sp2 carbon atoms. As a result, the carbon atoms of a triple bond are relatively more electron-deficient (carry a greater partial positive character) than those of a double bond, so they attract an electron-rich nucleophile more strongly.

…

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CBSE 2019Set ANNUAL1 mark
$Q. How would you convert propene to propan-1-ol?$
›Reveal solutionSolution

Direct acid-catalysed hydration of propene would give the Markovnikov (2°) alcohol; to get the anti-Markovnikov, terminal (1°) alcohol, hydroboration–oxidation is used instead.

Simple acid-catalysed addition of water to propene ( $CH_3\text{–}CH=CH_2$ ) follows Markovnikov's rule and would place $-OH$ on the more substituted carbon, giving propan-2-ol — not what is wanted here.

To obtain the terminal alcohol, propan-1-ol, the hydroboration–oxidation sequence is used, which adds $H$ and $OH$ with anti-Markovnikov regiochemistry (boron, and hence eventually $OH$ , ends up on the less substituted, terminal carbon):

Step 1 (hydroboration): propene reacts with diborane; boron adds to the less hindered (terminal) carbon: …

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CBSE 2018Set ANNUAL1 markMCQ
$Q. Rate of hydration in aqueous acid will be in the order - (I) cyclopropyl-CH=CH2 ; (II) cyclopropyl-CH=CH-CH3 ; (III) cyclopropyl-C(CH3)=CH2$
(a) I < II < III
(b) III < II < I
(c) I < III < II
(d) II < I < III
›Reveal solutionSolution
Rate ∝ stability of the intermediate cyclopropylcarbinyl cation → I < II < III.

Acid-catalysed (Markovnikov) hydration proceeds via protonation to the most stable carbocation, which here is always on the carbon next to the cyclopropyl ring (cyclopropyl strongly stabilises an adjacent + charge):

I (cyclopropyl-CH=CH₂): gives a 2° cyclopropylcarbinyl cation. …
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Q.Predict the product of adding two equivalents of HBr\text{HBr}HBr to propyne, CH3C≡CH\text{CH}_3\text{C}{\equiv}\text{CH}CH3​C≡CH.

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Q.Predict the product of adding two equivalents of $\text{HBr}$ to propyne, $\text{CH}_3\text{C}{\equiv}\text{CH}$ .

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