In addition of HX to unsymmetrical alkene, H goes to carbon with more H atoms; X to carbon with fewer H atoms. Driven by carbocation stability. Anti-Markovnikov in presence of peroxide (Kharasch effect): radical mechanism.
-- NCERT Class 11 Chemistry, Ch. 9, p. 311Electrophilic Addition
8 MCQs2 revision cards9-step worked example
Source: NCERT HydrocarbonsOfficial key: NTA-verifiedLast updated: 27 Sep 2026
Electrophilic Addition, explained for NEET
The trap first: NEET questions on electrophilic addition to alkenes frequently test whether you remember the peroxide exception. The same reagent — HBr — gives opposite products depending on whether peroxide is present. Aspirants who apply Markovnikov's rule mechanically without checking the reaction conditions lose marks on what should be a straightforward question.
The mechanism. Electrophilic addition is the characteristic reaction of alkenes. The electron-rich π-bond of the C=C double bond acts as a nucleophile, attacking an incoming electrophile. The general sequence:
- Electrophile attacks the π-bond — one carbon bonds to the electrophile, generating a carbocation intermediate on the adjacent carbon.
- Nucleophile attacks the carbocation — the remaining fragment (anion or nucleophilic species) bonds to the positively charged carbon.
For HX addition (X = Cl, Br, I) to an unsymmetrical alkene, Markovnikov's rule governs regioselectivity: the hydrogen adds to the carbon already bearing more hydrogens, placing X on the more substituted carbon. The driving force is carbocation stability — the more substituted carbocation is more stable (NCERT Class 11 Chemistry, Chapter 9, page 311).
The peroxide exception (Kharasch effect). When HBr reacts with an alkene in the presence of peroxide (ROOR), the mechanism switches from ionic to radical. The radical intermediate is stabilised at the more substituted carbon, so Br ends up on the less substituted carbon — anti-Markovnikov addition. This reversal is specific to HBr only. HCl's C–Cl bond is too strong for radical initiation; HI's radical chain terminates too readily (NCERT Class 11 Chemistry, Chapter 9, page 312).
Watch-out for NEET: always scan the question stem for "in the presence of peroxide," "ROOR," or "Kharasch conditions." If absent, default to Markovnikov. If present and the acid is HBr, switch to anti-Markovnikov.
Can you answer these Electrophilic Addition MCQs?
Select an option to see the explanation. Wrong answers show why your choice was tempting — and name the exact trap it exploits.
MCQ 1Easy RecallPractice
In electrophilic addition to an alkene, the first step involves attack of the electrophile on the:
Show answer and why every option is right or wrong
Answer: B. B is correct. The electron-rich π-bond of the double bond acts as the nucleophilic site that attacks the incoming electrophile, initiating electrophilic addition (NCERT Class 11 Chemistry, Chapter 9, page 311).
Why A is wrong: A is wrong because the σ-bond of C–C is localised and lower in energy than the π-bond; it is not the site of electrophilic attack in addition reactions.
Why C is wrong: C is wrong because the lone pair on the halogen belongs to the electrophilic reagent (HX), not the alkene substrate undergoing attack.
Why D is wrong: D is wrong because σ* antibonding orbitals are relevant in elimination and SN2 back-side attack contexts, not in the initial step of electrophilic addition to alkenes.
MCQ 2Easy RecallPractice
Markovnikov's rule states that during addition of HX to an unsymmetrical alkene (without peroxide), hydrogen adds to the carbon that:
Show answer and why every option is right or wrong
Answer: D. D is correct. Markovnikov's rule: hydrogen goes to the carbon with more hydrogens, placing X on the more substituted carbon, because the resulting carbocation intermediate is more stable (NCERT Class 11 Chemistry, Chapter 9, page 311).
Why A is wrong: A is wrong because this describes anti-Markovnikov addition (radical mechanism with peroxide), not the standard ionic Markovnikov pathway.
Why B is wrong: B is wrong because in the reactant alkene before addition, neither carbon bears a formal positive charge. Carbocation formation occurs AFTER the electrophile attacks, and the rule predicts which carbocation forms preferentially.
Why C is wrong: C is wrong because hydrogen goes to the LESS substituted carbon (more H atoms), not the more substituted carbon. The halide ends up on the more substituted carbon.
MCQ 3Easy RecallPractice
The anti-Markovnikov addition of HBr in the presence of peroxide is also known as:
Show answer and why every option is right or wrong
Answer: D. D is correct. The peroxide-mediated anti-Markovnikov addition of HBr to alkenes proceeds via a radical mechanism and is called the Kharasch effect (NCERT Class 11 Chemistry, Chapter 9, page 312).
Why A is wrong: A is wrong because the Wurtz reaction involves coupling of two alkyl halides with sodium metal to form higher alkanes — it is unrelated to electrophilic addition.
Why B is wrong: B is wrong because Saytzeff's rule predicts that elimination reactions preferentially form the more substituted (more stable) alkene — it governs elimination, not addition.
Why C is wrong: C is wrong because Hofmann elimination refers to the degradation of quaternary ammonium salts to give the less substituted alkene — a completely different reaction type.
MCQ 4Direct ApplicationPractice
What is the major product when HBr is added to propene (CH₃–CH=CH₂) in the absence of peroxide?
Show answer and why every option is right or wrong
Answer: C. C is correct. Without peroxide, ionic Markovnikov addition occurs: H adds to the terminal carbon (more H atoms), forming a secondary carbocation on C-2, and Br⁻ then attacks C-2 to give 2-bromopropane (NCERT Class 11 Chemistry, Chapter 9, page 311).
Why A is wrong: A is wrong because 1-bromopropane would require anti-Markovnikov addition (Br on the less substituted carbon), which occurs only with peroxide present and only for HBr (trap: Markovnikov reversal with peroxide).
Why B is wrong: B is wrong because 1,2-dibromopropane is the product of Br₂ addition to propene, not HBr addition. HBr adds one H and one Br across the double bond.
Why D is wrong: D is wrong because propane would require reduction (addition of H₂), not HBr addition. HBr adds H and Br, not two hydrogen atoms.
MCQ 5Direct ApplicationPractice
What is the major product when HBr is added to propene in the presence of benzoyl peroxide?
Show answer and why every option is right or wrong
Answer: A. A is correct. With peroxide, the radical (anti-Markovnikov) mechanism operates for HBr: Br radical adds to the less substituted terminal carbon (C-1), and H ends up on C-2, yielding 1-bromopropane (NCERT Class 11 Chemistry, Chapter 9, page 312).
Why B is wrong: B is wrong because 2-bromopropane is the Markovnikov product formed WITHOUT peroxide (ionic mechanism). The question specifies peroxide, so the radical anti-Markovnikov pathway operates (trap: Markovnikov reversal with peroxide).
Why C is wrong: C is wrong because allyl bromide (CH₂=CHCH₂Br) is formed by allylic substitution (e.g., NBS bromination), not by addition across the double bond. Electrophilic addition saturates the C=C bond.
Why D is wrong: D is wrong because isopropyl bromide is (CH₃)₂CHBr — this is 2-bromopropane by another name, which is the Markovnikov product formed without peroxide, not the anti-Markovnikov product.
MCQ 6Direct ApplicationPractice
HCl is added to 2-methylpropene (isobutylene) in the presence of peroxide. The major product is:
Show answer and why every option is right or wrong
Answer: B. B is correct. The peroxide-mediated anti-Markovnikov pathway is specific to HBr only. HCl does not undergo anti-Markovnikov addition because the H–Cl bond is too strong for effective radical chain initiation. Therefore, standard Markovnikov (ionic) addition occurs regardless of peroxide, giving the tertiary chloride (NCERT Class 11 Chemistry, Chapter 9, page 312).
Why A is wrong: A is wrong because anti-Markovnikov addition with peroxide works ONLY with HBr, not HCl. The H–Cl bond dissociation energy is too high for radical initiation, so the ionic Markovnikov pathway operates even when peroxide is present (trap: Markovnikov reversal with peroxide — HBr-specific).
Why C is wrong: C is wrong because the reaction follows one dominant pathway — ionic Markovnikov addition — giving predominantly one product, not a 50:50 mixture.
Why D is wrong: D is wrong because HCl does react with alkenes via ionic electrophilic addition (Markovnikov). Peroxide simply fails to switch it to a radical pathway; the ionic reaction still proceeds.
MCQ 7Concept TrapPractice
In the electrophilic addition of HBr to propene (without peroxide), the intermediate carbocation is:
Show answer and why every option is right or wrong
Answer: A. A is correct. H⁺ adds to C-1 (terminal carbon with more hydrogens, per Markovnikov's rule), generating a secondary carbocation on C-2 (CH₃CH⁺CH₃). This is more stable than the primary alternative and drives the regioselectivity of the addition (NCERT Class 11 Chemistry, Chapter 9, page 311).
Why B is wrong: B is wrong because a primary carbocation on C-1 would form only if H⁺ added to C-2 — this is the less stable intermediate and is NOT favoured in the ionic mechanism without peroxide.
Why C is wrong: C is wrong because a tertiary carbocation requires four carbons bonded to the cationic centre. Propene has only three carbons total; the most substituted carbocation possible is secondary.
Why D is wrong: D is wrong because an allylic carbocation requires the positive charge to be adjacent to a remaining C=C double bond. In electrophilic addition, the π-bond is consumed during the first step, so no allylic stabilisation is available.
MCQ 8CalculationPractice
When HBr is added to 3-methylbut-1-ene (CH₂=CH–CH(CH₃)₂) without peroxide, a secondary carbocation initially forms. However, the major product corresponds to a tertiary halide. The best explanation is:
Show answer and why every option is right or wrong
Answer: C. C is correct. H⁺ adds to C-1 per Markovnikov's rule, initially forming a secondary carbocation on C-2. A 1,2-hydride shift from C-3 converts this to a tertiary carbocation on C-3, which is more stable. Br⁻ then attacks C-3, yielding 2-bromo-2-methylbutane as the major product. This is a two-step reasoning problem: Markovnikov addition followed by carbocation rearrangement.
Why A is wrong: A is wrong because anti-Markovnikov addition requires peroxide and specifically HBr. The question states 'without peroxide,' so the ionic Markovnikov pathway operates. The tertiary product arises from carbocation rearrangement, not anti-Markovnikov regioselectivity.
Why B is wrong: B is wrong because cyclic bromonium ion intermediates form during Br₂ addition to alkenes, not HBr addition. In HBr addition, the electrophile is H⁺ (from polarised H–Br), generating a carbocation, not a bromonium ion.
Why D is wrong: D is wrong because elimination (dehydrohalogenation) is the reverse of HX addition and requires a base. Under acidic addition conditions (HBr with no base), addition is the dominant pathway, not elimination.
Free NEET study resources
Get a structured 30-day study plan and a complete formula booklet — delivered to your inbox instantly.
Electrophilic Addition: quick recall before you leave
How do you solve a Electrophilic Addition question? A worked example
- 1
Given
• Substrate: but-1-ene (CH₃CH₂CH=CH₂)• Reagent: HBr• Condition: in the presence of organic peroxide (ROOR)
- 2
Required
Identify the major product of the reaction.
- 3
Concept
HBr addition to an unsymmetrical alkene in the presence of peroxide follows the radical (anti-Markovnikov) mechanism. This is the Kharasch effect, specific to HBr only.
- 4
Formula / Rule
Anti-Markovnikov addition: Br radical adds to the less substituted carbon (terminal carbon); H ends up on the more substituted carbon. The radical intermediate on the more substituted carbon is more stable (analogous to carbocation stability reasoning, but for radicals).
- 5
Substitution
• But-1-ene: CH₃CH₂CH=CH₂• Peroxide initiates radical chain → Br• attacks the terminal C-1 (less substituted)• H adds to C-2
- 6
Calculation
This is a product-prediction problem, not a numerical calculation.
Br• adds to C-1 → radical on C-2 (secondary, more stable than primary on C-1) → H abstracts to C-2 → product: 1-bromobutane (CH₃CH₂CH₂CH₂Br). - 7
Final answer
Major product: 1-bromobutane (CH₃CH₂CH₂CH₂Br)
This is the anti-Markovnikov product. Without peroxide, the product would be 2-bromobutane (Markovnikov product). - 8
Common trap
The high-frequency trap here: if you ignore the "presence of peroxide" condition, you'd predict 2-bromobutane (Markovnikov product) — and select the wrong option. Also, if the question changed the acid to HCl or HI while keeping peroxide, the answer switches back to Markovnikov because the peroxide effect is HBr-specific.
- 9
Similar NEET-style question
"When HBr reacts with 2-methylbut-2-ene in the presence of peroxide, the major product is ___." (Answer: the anti-Markovnikov product, 2-bromo-3-methylbutane. In CH₃–C(CH₃)=CH–CH₃ the C-2 carbon carries the extra methyl and is the MORE substituted one, so Br• adds to C-3 and the hydrogen ends up on C-2. That is the point of the peroxide effect: the radical adds so as to leave the more stable radical behind, and here the tertiary radical sits on C-2.)
---
What to remember before solving Electrophilic Addition questions
Addition of HBr, H₂O (acid catalyst), Br₂. Mechanism: (1) electrophile attacks C=C, forms cation; (2) nucleophile attacks cation. Regioselectivity by Markovnikov.
-- NCERT Class 11 Chemistry, Ch. 9, p. 311Which Electrophilic Addition formulas do you need for NEET?
Markovnikov's rule (and anti-Markovnikov)
Without peroxide: ionic mechanism — H goes to carbon with MORE hydrogens (carbocation stability rule). With peroxide (HBr only, Kharasch): radical mechanism — anti-Markovnikov.
| Symbol | Quantity | SI Unit |
|---|---|---|
| H,X | added atoms | - |
Valid when
- Asymmetric alkene
- H-X with X = Cl, Br, I
- Without peroxide for Markovnikov
Where do students lose marks on Electrophilic Addition?
These are the exact patterns that cause wrong answers in NEET. Each trap includes when it triggers and how to avoid it.
Category: Organic Reaction Conditions
HBr addition to alkene: WITHOUT peroxide → Markovnikov (H to C with more H). WITH peroxide (Kharasch effect) → anti-Markovnikov (radical mechanism). Specific to HBr only — not HCl, HI.
When it triggers
Question gives HX addition to alkene with explicit peroxide condition or hints (e.g. ROOR, light).
How to avoid
Without peroxide: ionic mechanism, carbocation stability → Markovnikov. With peroxide: radical mechanism, radical stability → anti-Markovnikov. Effect ONLY for HBr (HCl too strong, HI too weak).
More in Hydrocarbons: 3 exam traps and mistakes · 1 question pattern from its other lessons.
Electrophilic Addition questions from past NEET papers
No question in our NEET 2020–2025 set targets this topic directly.
How does NEET ask about Electrophilic Addition?
Recurring question shapes from past papers. Each pattern shows why wrong options look tempting.
Predict major product of HX addition to alkene. Markovnikov rule (without peroxide); anti-Markovnikov with peroxide.
InterpretationEasy
Common distractors
ignores peroxide effect
Same product regardless of conditions
Sources
NCERT refs: Class 11 Chemistry Chapter 9, p.311 | Class 11 Chemistry Chapter 9, p.312
Page numbers are the ones printed in the current NCERT textbook (2023 rationalised edition), unless marked pre-2023. The books are free at ncert.nic.in.
Test yourself on this topic with real past-paper questions:
Practice this topic →