Q.How will you bring about the following conversions?
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🔒 Start your 14-day free trial to unlock the full solution →Concept understanding — Markovnikov Addition
The Intuition First
Imagine you have an alkene — a carbon-carbon double bond. That double bond is like a crowded room with two doors. When a molecule like HBr comes along, it wants to break that double bond and add across it. The question is: which carbon gets the hydrogen, and which gets the bromine?
You might think it doesn't matter — after all, the two carbons look similar. But they aren't. One carbon usually has more alkyl groups (methyl, ethyl, etc.) attached to it than the other. That carbon is more "electron-rich" — it has more friends pushing electrons toward it.
The hydrogen, being small and positively charged, is picky. It goes to the carbon that already has more hydrogens. Why? Because that carbon is less crowded and can stabilise the positive charge that forms temporarily during the reaction. The bromine, being large and negatively charged, goes to the other carbon — the one with more alkyl groups.
That's the intuition: the rich get richer. The carbon with more hydrogens gets another hydrogen. The carbon with more alkyl groups gets the halogen.
The Precise Statement
Markovnikov's Rule: When an unsymmetrical reagent (like HX, H₂O, etc.) adds to an unsymmetrical alkene, the hydrogen atom attaches to the carbon of the double bond that already has the greater number of hydrogen atoms.
In other words, for an alkene like CH3CH=CH2 (propene) reacting with HBr:
- Carbon 1 (the CH₂ end) has 2 hydrogens.
- Carbon 2 (the CH end) has 1 hydrogen.
- The H goes to carbon 1 (more hydrogens).
- The Br goes to carbon 2 (fewer hydrogens).
So the product is CH3CHBrCH3 (2-bromopropane), not CH3CH2CH2Br (1-bromopropane).
Why Does This Happen? The Real Chemistry
The reaction proceeds through a carbocation intermediate. When the H⁺ attacks the double bond, it can form one of two possible carbocations:
- A primary carbocation (if H⁺ goes to the more substituted carbon) — unstable.
- A secondary carbocation (if H⁺ goes to the less substituted carbon) — more stable.
The reaction chooses the path that gives the more stable carbocation. Alkyl groups stabilise carbocations through hyperconjugation and inductive effect — they donate electron density to the positively charged carbon.
The stability order of carbocations is: tertiary > secondary > primary > methyl. Markovnikov addition always proceeds through the most stable carbocation possible.
A Common Misconception
Many students think Markovnikov's rule means "hydrogen goes to the carbon with more hydrogens" because that carbon already has more hydrogens. That's backwards. The hydrogen goes there because that path leads to a more stable carbocation — the number of hydrogens is just a convenient way to predict the outcome, not the cause.
The One Big Exception …
Why this formula?
Markovnikov Addition: Why the Rule Holds
Markovnikov's rule is not a formula in the algebraic sense — it's a predictive principle for electrophilic addition to unsymmetrical alkenes. The "why" comes from carbocation stability and reaction mechanism.
The Rule in Words
When H–X adds to an unsymmetrical alkene, the hydrogen attaches to the carbon with more hydrogen atoms already attached, and the halogen (or X group) attaches to the carbon with fewer hydrogen atoms.
Example:
Propene (CHX3−CH=CHX2) + HBr → 2-bromopropane (major product), not 1-bromopropane.
Why This Happens: The Step-by-Step Reasoning
1. The Mechanism (Electrophilic Addition)
The reaction proceeds in two steps:
- Slow step (rate-determining): The alkene's π bond attacks the electrophilic HX+ from H–X, forming a carbocation intermediate.
- Fast step: The carbocation is attacked by the nucleophilic XX−.
2. The Key: Carbocation Stability
The more stable carbocation intermediate forms faster and determines the major product.
| Carbocation Type | Stability Order | Reason |
|---|---|---|
| Tertiary (3∘) | Most stable | +3 alkyl groups donate electron density via hyperconjugation and inductive effect |
| Secondary (2∘) | Intermediate | +2 alkyl groups |
| Primary (1∘) | Least stable | +1 alkyl group |
| Methyl (CHX3X+) | Unstable | No alkyl stabilization |
3. Applying to Propene + HBr
Propene: CHX3−CH=CHX2
Two possible protonation sites:
- Path A (Markovnikov): HX+ adds to CHX2 (terminal carbon) → forms secondary carbocation:
CHX3−CHX+−CHX3(2∘)
- Path B (Anti-Markovnikov): HX+ adds to CH (middle carbon) → forms primary carbocation:
CHX3−CHX2−CHX2X+(1∘)
Result: The secondary carbocation is more stable (by ~25–30 kJ/mol), so Path A is faster. The BrX− then attacks the positively charged carbon, giving 2-bromopropane.
The "Formula" — A Stability-Based Prediction
There is no algebraic formula, but a decision rule:
Major product=Product from the more stable carbocation
For alkenes with alkyl substituents, the stability order is:
Tertiary>Secondary>Primary>Methyl …
Concept: Each conversion uses a specific organic reaction — elimination, substitution, addition, or coupling. The key is to identify the functional group change and choose the correct reagent.
- Ethanol → but-1-yne Step 1: Dehydrate ethanol to ethene (conc. H2SO4, 170°C). Step 2: Brominate ethene to 1,2-dibromoethane (Br2). Step 3: Double dehydrohalogenation with NaNH2 in liquid NH3 gives ethyne. Step 4: Alkylate ethyne with CH3CH2Br using NaNH2 to get but-1-yne.
- Ethane → bromoethene Step 1: Free-radical bromination of ethane gives bromoethane (Br2, hv). Step 2: Dehydrohalogenation with alcoholic KOH yields ethene. Step 3: Brominate ethene to 1,2-dibromoethane, then dehydrobrominate with alcoholic KOH to bromoethene.
- Propene → 1-nitropropane Step 1: Add HBr to propene (peroxide effect, anti-Markovnikov) to get 1-bromopropane. Step 2: Treat with aqueous AgNO2 (nitrite) — SN2 gives 1-nitropropane.
- Toluene → benzyl alcohol Step 1: Free-radical chlorination of toluene (Cl2, hv) gives benzyl chloride. Step 2: Hydrolysis with aqueous NaOH yields benzyl alcohol.
- Propene → propyne Step 1: Brominate propene to 1,2-dibromopropane (Br2). Step 2: Double dehydrohalogenation with NaNH2 in liquid NH3 gives propyne.
- Ethanol → ethyl fluoride Step 1: Convert ethanol to ethyl chloride using PCl5 or SOCI2. Step 2: Swarts reaction — treat ethyl chloride with AgF (or Hg₂F₂, CoF₂, SbF₃) to get ethyl fluoride. (Simple NaF does not drive this exchange the way NaI does in a true Finkelstein reaction — fluorination needs one of the Swarts-reaction metal fluorides.)
- Bromomethane → propanone Step 1: Convert CH3Br to CH3MgBr (Mg, dry ether — Grignard reagent). Step 2: React CH3MgBr with CH3CN (acetonitrile), then hydrolyse the resulting imine-magnesium complex with dilute acid to get propanone. (A nitrile is used rather than an acid chloride because the initial addition product is stable to further Grignard attack until hydrolysis — an acid chloride would over-react with excess Grignard to give a tertiary alcohol instead.)
- But-1-ene → but-2-ene Step 1: Add HBr to but-1-ene (no peroxide, Markovnikov addition) to get 2-bromobutane. …
Each conversion is achieved by a specific sequence of organic reactions — the key is to identify the functional-group transformation needed and then apply the correct reagents stepwise. The final products are obtained via standard named reactions like dehydrohalogenation, Wurtz reaction, Sandmeyer reaction, etc.
Let's work through each conversion one by one, focusing on the why behind each step.
(i) Ethanol to but-1-yne
Concept: We need to increase the carbon chain from 2 to 4 carbons and introduce a terminal triple bond. The strategy: convert ethanol to ethene (dehydration), then to 1,2-dibromoethane (bromination), then to ethyne (double dehydrohalogenation), and finally alkylate the terminal alkyne.
-
Ethanol → Ethene: Dehydrate ethanol using concentrated H2SO4 at 170°C.
CH3CH2OHH2SO4,170∘CCH2=CH2+H2O
-
Ethene → 1,2-Dibromoethane: Add bromine across the double bond.
CH2=CH2+Br2→CH2Br−CH2Br
-
1,2-Dibromoethane → Ethyne: Double dehydrohalogenation with alcoholic KOH (or NaNH2).
CH2Br−CH2Bralc.KOH,heatHC≡CH
-
Ethyne → But-1-yne: Alkylate the terminal alkyne. First, treat ethyne with NaNH2 in liquid NH3 to form sodium acetylide. Then react with ethyl bromide (CH3CH2Br).
HC≡CHNaNH2HC≡C−Na+CH3CH2BrHC≡C−CH2CH3 …
Let's tackle these conversions one by one. The key is to think backwards from the product to the reactant, identifying the functional group transformations needed.
Here is one clear, named method for each, with step-by-step reasoning.
(i) Ethanol to but-1-yne
Method: Build ethyne first, then alkylate its acetylide with an ethyl halide
Why this works: But-1-yne is a terminal alkyne — the cleanest NCERT route is to make ethyne from ethanol, deprotonate it to sodium acetylide, and add the remaining two carbons as bromoethane.
Steps:
- Ethanol → Ethene: Acid-catalysed dehydration.
CH3CH2OHconc.H2SO4,443KCH2=CH2
- Ethene → 1,2-Dibromoethane: Electrophilic addition of bromine.
CH2=CH2Br2/CCl4BrCH2CH2Br
- 1,2-Dibromoethane → Ethyne: Double dehydrohalogenation.
BrCH2CH2Br2NaNH2HC≡CH
- Ethyne → Sodium acetylide: Deprotonation of the terminal C–H.
HC≡CHNaNH2HC≡C−Na+
- Acetylide + Bromoethane → But-1-yne: SN2 alkylation adds the two-carbon chain (bromoethane itself is made from ethanol + PBr3).
HC≡C−Na++CH3CH2Br→HC≡C−CH2CH3
Final Product: But-1-yne
(ii) Ethane to bromoethene
Method: Four Steps: Halogenation → Elimination → Addition → Selective Elimination
Why this works: Dehydrohalogenation of bromoethane would remove its ONLY bromine and give plain ethene — a two-step route can never give bromoethene. The vinyl bromide must come from a 1,2-dibromide that loses just ONE HBr.
Steps:
- Free Radical Halogenation: React ethane with Br2 in the presence of UV light (hv). This gives bromoethane.
CH3CH3+Br2hvCH3CH2Br+HBr
- Dehydrohalogenation: Treat bromoethane with alcoholic KOH. This eliminates HBr to form ethene.
CH3CH2Bralc.KOH,ΔCH2=CH2
- Addition of Bromine: Add Br2 (in CCl4) across the double bond to get 1,2-dibromoethane.
CH2=CH2+Br2→CH2Br−CH2Br
- Selective Dehydrohalogenation: Treat with ONE equivalent of alcoholic KOH to eliminate a single HBr.
CH2Br−CH2Bralc.KOH(1eq.),ΔCH2=CHBr
Final Product: Bromoethene (Vinyl bromide)
(iii) Propene to 1-nitropropane
Method: Anti-Markovnikov Addition (via Free Radical Mechanism)
Why this works: Direct nitration of an alkene with HNO3 is messy. We need to add the nitro group (−NO2) to the terminal carbon (C-1), which is the anti-Markovnikov position.
Steps:
- Anti-Markovnikov Addition of HBr: React propene with HBr in the presence of a peroxide (e.g., benzoyl peroxide). This gives 1-bromopropane.
CH3CH=CH2+HBrPeroxideCH3CH2CH2Br
- Nucleophilic Substitution: React 1-bromopropane with alcoholic AgNO2 (silver nitrite). The nitrite ion (NO2−) acts as an ambident nucleophile, but with AgNO2, the major product is the nitroalkane.
CH3CH2CH2Br+AgNO2ΔCH3CH2CH2NO2+AgBr
Final Product: 1-Nitropropane
(iv) Toluene to benzyl alcohol
Method: Free Radical Halogenation followed by Nucleophilic Substitution
Why this works: We need to selectively oxidize the methyl group to a −CH2OH group without touching the ring. Free radical bromination targets the benzylic position.
Steps:
- Benzylic Bromination: React toluene with N-bromosuccinimide (NBS) in the presence of light (hv) and a radical initiator (like benzoyl peroxide). This selectively brominates the methyl group.
C6H5CH3+NBShv,CCl4C6H5CH2Br
- Hydrolysis: Treat benzyl bromide with aqueous NaOH or Na2CO3. This substitutes the bromine with a hydroxyl group.
C6H5CH2Braq.NaOH,ΔC6H5CH2OH
Final Product: Benzyl alcohol
(v) Propene to propyne
Method: Two-Step: Halogenation then Double Dehydrohalogenation
Why this works: We need to introduce a triple bond. This is done by adding two halogen atoms to the double bond, then removing two molecules of HX.
Steps:
- Electrophilic Addition of Bromine: React propene with Br2 in CCl4. This gives 1,2-dibromopropane.
CH3CH=CH2+Br2CCl4CH3CHBrCH2Br
- Double Dehydrohalogenation: Treat the dibromide with a strong base like alcoholic KOH (first elimination) followed by sodamide (NaNH2) in liquid NH3 (second elimination, which is more difficult).
CH3CHBrCH2Br1.alc.KOH,Δ2.NaNH2,liq.NH3CH3C≡CH
Final Product: Propyne
(vi) Ethanol to ethyl fluoride
Method: Nucleophilic Substitution (via a good leaving group)
Why this works: Direct substitution of −OH by F− is poor because OH− is a bad leaving group. We must first convert the alcohol into a better leaving group.
Steps:
- Convert Alcohol to Alkyl Halide (Chloride): React ethanol with PCl5 or SOCl2 to form chloroethane.
CH3CH2OH+PCl5→CH3CH2Cl+POCl3+HCl
- Halogen Exchange (Swarts Reaction): React chloroethane with AgF (or Hg2F2, CoF2, SbF3). The driving force is the precipitation of AgCl. (The name Finkelstein reaction is reserved for the Cl/Br → iodide exchange with NaI in acetone.)
CH3CH2Cl+AgFDMFCH3CH2F+AgCl
Final Product: Ethyl fluoride
(vii) Bromomethane to propanone
Method: Grignard Reaction + Oxidation …
Common Mistakes in Markovnikov Addition & Organic Conversions
Students often lose marks in these conversions due to conceptual confusion and procedural errors. Let's break down the key mistakes and how to avoid them.
(i) Ethanol → But-1-yne
✗ Common Mistakes
- Trying to add 2 carbons directly to ethanol without first converting to a better leaving group.
- Using NaNH₂ directly on ethanol (alcohols are not acidic enough for this).
- Forgetting that but-1-yne has a terminal alkyne (triple bond at C1).
✓ Correct Approach
- Ethanol → Ethene (dehydration with conc. H₂SO₄, 170°C)
- Ethene → 1,2-dibromoethane (Br₂ addition)
- 1,2-dibromoethane → But-1-yne (2 moles NaNH₂ in liq. NH₃ to form sodium acetylide, then alkylate with CH₃CH₂Br — ethyl bromide, not methyl bromide, since ethyne (2 C) + ethyl (2 C) = the 4 carbons but-1-yne needs; using CH₃Br would only reach propyne)
Why this works: The double dehydrohalogenation creates the triple bond, and the alkylation adds the extra carbon.
(ii) Ethane → Bromoethene
✗ Common Mistakes
- Trying direct bromination of ethane (gives bromoethane, not bromoethene).
- Using Br₂/CCl₄ on ethane (no reaction — alkanes need light/heat for substitution).
- Confusing bromoethene (CH₂=CHBr) with bromoethane (CH₃CH₂Br).
✓ Correct Approach
- Ethane → Ethene (cracking or catalytic dehydrogenation, Cr₂O₃/Al₂O₃, 600°C)
- Ethene → 1,2-dibromoethane (Br₂/CCl₄ addition)
- 1,2-dibromoethane → Bromoethene (alc. KOH, heat — dehydrohalogenation)
Key insight: You need a double bond first, then add Br₂, then eliminate one HBr.
(iii) Propene → 1-Nitropropane
✗ Common Mistakes
- Assuming direct nitration of propene (gives mixture of nitroalkenes, not 1-nitropropane).
- Forgetting that Markovnikov addition would give 2-nitropropane, not 1-nitropropane.
- Using HNO₃/H₂SO₄ (nitrating mixture) — this works for aromatics, not alkenes.
✓ Correct Approach (Anti-Markovnikov needed)
- Propene → 1-Bromopropane (HBr with peroxide — anti-Markovnikov addition)
- 1-Bromopropane → 1-Nitropropane (NaNO₂ in DMF or aqueous ethanol, SN2 reaction)
Critical point: The first step must be anti-Markovnikov to get the Br at C1. Without peroxide, HBr adds to C2 (Markovnikov product).
(iv) Toluene → Benzyl Alcohol
✗ Common Mistakes
- Trying direct oxidation of methyl group (gives benzoic acid, not benzyl alcohol).
- Using KMnO₄ or K₂Cr₂O₇ — these over-oxidize to benzoic acid.
- Forgetting that benzyl alcohol is C₆H₅CH₂OH (alcohol, not acid).
✓ Correct Approach
- Toluene → Benzyl chloride (Cl₂/hν or Cl₂/UV light — side-chain chlorination)
- Benzyl chloride → Benzyl alcohol (aq. NaOH or KOH, hydrolysis)
Why this works: Free radical chlorination attacks the benzylic position (CH₃ group), not the ring. Then SN2 hydrolysis gives the alcohol.
(v) Propene → Propyne
✗ Common Mistakes
- Trying one-step elimination (propene → propyne requires two eliminations).
- Using alc. KOH directly on propene (no reaction — no leaving group).
- Forgetting that propyne has a terminal triple bond.
✓ Correct Approach
- Propene → 1,2-Dibromopropane (Br₂/CCl₄ addition)
- 1,2-Dibromopropane → Propyne (2 moles NaNH₂ in liq. NH₃)
Mechanism: Two successive dehydrohalogenations — first gives bromopropene, second gives propyne.
(vi) Ethanol → Ethyl Fluoride
✗ Common Mistakes
- Trying direct reaction with HF (poor yield, dangerous, and HF is weak acid).
- Using F₂ gas (explosive, non-selective).
- Confusing with other halides (Cl, Br, I are easier).
✓ Correct Approach
- Ethanol → Ethene (conc. H₂SO₄, 170°C)
- Ethene → Ethyl fluoride (HF addition — Markovnikov gives ethyl fluoride)
Alternative: Ethanol → Ethyl chloride (SOCl₂) → Ethyl fluoride (metathesis with AgF or KF in polar aprotic solvent).
(vii) Bromomethane → Propanone
✗ Common Mistakes
- Trying direct alkylation of bromomethane (gives ethane, not propanone).
- Forgetting that propanone (acetone) has a carbonyl group (C=O).
- Using Grignard reagent incorrectly.
✓ Correct Approach
- Bromomethane → Methyl magnesium bromide (Mg/dry ether — Grignard reagent)
- CH₃MgBr + CH₃CN → Propanone (react with acetonitrile, then hydrolyse the resulting imine-magnesium complex with dilute acid)
Why a nitrile, not an acid chloride: an acid chloride (e.g. CH₃COCl) reacts with a Grignard reagent to first give a ketone, but that ketone is itself attacked by any remaining Grignard reagent, over-reacting to a tertiary alcohol. A nitrile's initial addition product is stable to further Grignard attack, so hydrolysis cleanly stops at the ketone.
(viii) But-1-ene → But-2-ene
✗ Common Mistakes
- Thinking this is a simple isomerization (it requires bond migration).
- Using H⁺ alone (gives carbocation rearrangement but also polymerization).
- Forgetting that double bond migration needs specific conditions.
✓ Correct Approach
- But-1-ene → But-2-ene (H₂SO₄ (dilute), heat — or Al₂O₃, 300°C)
Mechanism: Protonation gives secondary carbocation (more stable), then deprotonation gives the more substituted alkene (but-2-ene). …
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