Organic Chemistry 1 • Radical Reactions

Radical Addition to Alkenes (HBr/Peroxides)

How peroxide-initiated radicals reverse the regiochemistry of HBr addition, yielding anti-Markovnikov products.

A molecular ring with laboratory flask and atom symbols for organic chemistry.
Inside this lesson 10 sections 5 questions 8 figures

SECTION 1

Historical Context & the Peroxide Effect

Throughout the 1920s and 1930s, organic chemists encountered a persistent and deeply frustrating puzzle: the addition of HBr to unsymmetrical alkenes sometimes yielded the product predicted by Markovnikov's rule, and other times yielded the opposite regiochemistry. Laboratories that used freshly purified reagents consistently obtained Markovnikov products, while those using older or impure reagents frequently reported anti-Markovnikov addition. The irreproducibility was so severe that it became known as the peroxide effect, once Morris Kharasch and Frank Mayo demonstrated that trace organic peroxides—formed by autoxidation of ethers used as solvents—were the culprit. Their work established that radical intermediates, rather than carbocations, governed the regiochemistry under these conditions.

1869 • Markovnikov's Rule Published

Vladimir Markovnikov formalized the observation that HX adds to alkenes such that hydrogen bonds to the less substituted carbon of the double bond, a rule grounded in carbocation stability.

1929–1933 • The Peroxide Effect Identified

Morris S. Kharasch and Frank R. Mayo at the University of Chicago showed that trace peroxides in reagents and solvents caused anti-Markovnikov addition of HBr to propylene and other alkenes.

1937 • Radical Chain Mechanism Proposed

Kharasch, Mayo, and Chao published the full radical chain mechanism involving initiation, propagation, and termination steps, explaining why only HBr—not HCl or HI—exhibits this effect.

1960s–present • Synthetic Applications

Radical additions became tools in polymer chemistry and total synthesis, with modern variants using tin hydrides (Bu₃SnH) and other radical mediators for selective C–C bond formation.

The central question that emerged from this historical episode remains essential today: why does the presence of peroxides reverse the regiochemistry of HBr addition, and why is this reversal unique to HBr among the hydrogen halides? Answering this requires a thorough understanding of radical chain mechanisms, the thermodynamics of individual propagation steps, and the kinetic factors that govern radical selectivity.

SECTION 2

Core Principles of Radical Chain Reactions

Radical addition to alkenes differs fundamentally from ionic addition in both the nature of the reactive intermediates and the regiochemical outcome. In ionic (electrophilic) addition, a proton adds first to generate the more stable carbocation, which then captures the nucleophilic halide—producing the Markovnikov product. In radical addition initiated by peroxides, a bromine radical adds first to generate the more stable carbon radical, and hydrogen is subsequently transferred from HBr—yielding the anti-Markovnikov product. The entire process operates through a radical chain mechanism consisting of three phases: initiation, propagation, and termination.

1

Initiation

Homolytic cleavage of the peroxide O–O bond generates alkoxy radicals (RO·), which abstract H from HBr to produce a bromine radical (Br·). This step requires energy input (heat or UV light) and produces the chain-carrying radical.
2

Propagation

The Br· adds to the alkene π bond, forming a new C–Br bond and a carbon radical at the more substituted position. This carbon radical then abstracts H from HBr, regenerating Br· and completing the cycle. Both steps must be exothermic for the chain to sustain itself.
3

Termination

Any combination of two radicals (Br· + Br·, R· + R·, or R· + Br·) destroys chain carriers. Because radical concentrations are low, termination is statistically infrequent, allowing each initiation event to produce hundreds of product molecules.
4

Regioselectivity

Br· adds to the less substituted carbon of the alkene, placing the resulting radical on the more substituted carbon (3° > 2° > 1°). This parallels carbocation stability: radical stability follows the same order due to hyperconjugation.
5

HBr Exclusivity

Only HBr undergoes this radical addition with alkenes. For HCl, the Cl· + alkene step is too exothermic and unselective, while for HI, the I· + alkene step is endothermic and thermodynamically unfavorable. HBr occupies the thermodynamic sweet spot.
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KEY TAKEAWAY

Think of the radical chain like a relay race. The peroxide fires the starting gun (initiation), producing the first runner—Br·. That runner passes the baton (an unpaired electron) to a carbon in the alkene during the first propagation step. The carbon radical then passes the baton back to a new Br· by abstracting H from HBr (second propagation step). The race keeps going until two runners collide and drop the baton (termination). The key insight is that Br· always chooses the end of the alkene that gives the carbon radical at the more substituted, more stable position—anti-Markovnikov regiochemistry is the direct consequence.

SECTION 3

Mechanism Diagram: The Radical Chain

Radical Addition of HBr to Propene (with ROOR)INITIATIONRO–OR → 2 RO·(homolysis, Δ or hν)RO· + H–Br → ROH + Br·(H-abstraction generates chain carrier)PROPAGATIONStep 1: Br· adds to the less substituted carbonBr· + CH₂=CH–CH₃ → BrCH₂–ĊH–CH₃2° radical (more stable)ΔH ≈ −20 kJ/mol (exothermic — π bond broken, C–Br σ bond formed)Step 2: Carbon radical abstracts H from HBrBrCH₂–ĊH–CH₃ + H–Br → BrCH₂–CH₂–CH₃ + Br·ΔH ≈ −38 kJ/mol (exothermic — Br· regenerated, chain continues)1-bromopropane(anti-Markovnikov)TERMINATIONBr· + Br· → Br₂R· + R· → R–RR· + Br· → R–BrOVERALL REACTIONCH₂=CH–CH₃ + HBr → CH₂Br–CH₂–CH₃(1-bromopropane)cycleFig 1. Complete radical chain mechanism for anti-Markovnikov addition of HBr to propene
The diagram traces all three phases of the radical chain mechanism. Initiation (top) generates Br· from peroxide homolysis followed by H-abstraction. Propagation (middle) consists of two exothermic steps that form the anti-Markovnikov product and regenerate Br·. Termination (bottom left) destroys radicals by coupling, ending the chain.

Examining the diagram closely, note that in propagation step 1, Br· adds to the terminal (less substituted) carbon of propene, generating a secondary carbon radical on C-2. This regioselectivity is thermodynamically driven: a secondary radical is approximately 12 kJ/mol more stable than the corresponding primary radical, and the transition state for radical addition reflects this stability difference. In propagation step 2, the carbon radical abstracts hydrogen from HBr (the weakest H–X bond among the hydrogen halides, at ~366 kJ/mol), regenerating the Br· chain carrier. The self-sustaining nature of this cycle means that a single initiation event can produce thousands of product molecules before termination intervenes, reflecting the high kinetic chain length of this process.

SECTION 4

Thermodynamic Analysis of Each Step

Understanding why the radical chain is self-sustaining—and why only HBr participates—requires a quantitative analysis of the enthalpy changes in each propagation step. We use bond dissociation energies (BDEs) to compute ΔH° for each elementary step. Recall that ΔH°rxn = Σ(BDEs of bonds broken) − Σ(BDEs of bonds formed). For a radical chain to be viable, both propagation steps must be exothermic or at least thermoneutral.

PROPAGATION STEP 1 (Br· + ALKENE)

ΔH°=BDE(πbond)−BDE(C–Br)≈268−285=−17kJ/molΔH° = BDE(π bond) − BDE(C–Br) ≈ 268 − 285 = −17 kJ/mol

The π component of the C=C bond (~268 kJ/mol) is broken, and a C–Br bond (~285 kJ/mol) is formed. The step is mildly exothermic, making it both thermodynamically and kinetically feasible.

PROPAGATION STEP 2 (R· + HBr)

ΔH°=BDE(H–Br)−BDE(C–H)≈366−410=−44kJ/molΔH° = BDE(H–Br) − BDE(C–H) ≈ 366 − 410 = −44 kJ/mol

The H–Br bond (366 kJ/mol) is broken, and a C–H bond (~410 kJ/mol) is formed. This step is strongly exothermic, providing the major thermodynamic driving force for the chain.

OVERALL ENTHALPY CHANGE

ΔH°(overall)=ΔH°1+ΔH°2≈(−17)+(−44)=−61kJ/molΔH°(overall) = ΔH°₁ + ΔH°₂ ≈ (−17) + (−44) = −61 kJ/mol

The overall addition of HBr across the double bond is exothermic by about 61 kJ/mol. Since both individual propagation steps are exothermic, the chain is self-sustaining.

Why Not HCl or HI?

HXBDE (H–X) kJ/molProp. Step 1 ΔH°Prop. Step 2 ΔH°Chain Viable?
HCl431−21 (exo)+21 (endo)No — Step 2 endothermic
HBr366−17 (exo)−44 (exo)Yes — both exothermic
HI297+13 (endo)−113 (exo)No — Step 1 endothermic

Thermodynamic analysis of radical chain viability for HCl, HBr, and HI addition to alkenes

The table reveals that HBr occupies a unique thermodynamic window. The H–Cl bond is too strong: although Cl· adds exothermically to the alkene, the subsequent H-abstraction from HCl by the carbon radical is endothermic because the H–Cl BDE (431 kJ/mol) exceeds the C–H BDE being formed (~410 kJ/mol). Conversely, the H–I bond is too weak: I· lacks the energetic driving force to add to the alkene in the first place because the C–I bond formed (~222 kJ/mol) does not compensate for breaking the π bond (~268 kJ/mol). Only for HBr are both propagation steps thermodynamically downhill, allowing the chain to proceed efficiently.

SECTION 5

Regiochemistry Compared: Ionic vs. Radical Addition

The most consequential feature of radical addition is its reversed regiochemistry relative to ionic electrophilic addition. In the absence of peroxides, HBr undergoes electrophilic addition: the proton adds to the less substituted carbon (following Markovnikov's rule) to generate the more stable carbocation at the more substituted position. In the presence of peroxides, Br· adds to the less substituted carbon for a different but analogous reason—the resulting carbon radical at the more substituted position is more stable. The outcome is that Br ends up on opposite carbons in the two pathways.

Markovnikov vs. Anti-Markovnikov Addition of HBr to PropeneIONIC (No Peroxides)RADICAL (With ROOR)Markovnikov productAnti-Markovnikov productStarting MaterialCH₂=CH–CH₃ (propene)Starting MaterialCH₂=CH–CH₃ (propene)Step 1: H⁺ adds to less substituted CCH₃–C⁺H–CH₃2° carbocation (more stable)Step 1: Br· adds to less substituted CBrCH₂–ĊH–CH₃2° radical (more stable)Step 2: Br⁻ captures carbocationCH₃–CHBr–CH₃2-bromopropaneStep 2: R· abstracts H from HBrBrCH₂–CH₂–CH₃1-bromopropaneBr on MORE substituted CBr on LESS substituted CFig 2. Side-by-side comparison of ionic (Markovnikov) vs. radical (anti-Markovnikov) HBr addition to propene
In ionic addition (left), H+ adds first to the terminal carbon, placing Br on C-2 (2-bromopropane, Markovnikov). In radical addition (right), Br· adds first to the terminal carbon, placing Br on C-1 (1-bromopropane, anti-Markovnikov). In both pathways, the intermediate (carbocation or radical) is formed at the more substituted, more stable position.

A subtle but critical point emerges from this comparison: the guiding principle in both pathways is identical—the more stable intermediate is preferred. In ionic addition, H+ adds first (forming the more substituted carbocation), so Br ends up on the more substituted carbon. In radical addition, Br· adds first (forming the more substituted radical), so Br ends up on the less substituted carbon. The reversal in regiochemistry arises not from a reversal in selectivity principles but from a reversal in which atom adds first.

⚠

Common Misconception

Students sometimes believe that Markovnikov's rule is "violated" in radical addition. More precisely, Markovnikov's rule as originally stated applies only to ionic additions. In radical addition, the regiochemistry is governed by radical stability, not carbocation stability. Both ionic and radical pathways favor the more stable intermediate—the difference lies in the order of addition.

SECTION 6

Worked Example: Radical Addition to 1-Methylcyclohexene

Consider the reaction of 1-methylcyclohexene with HBr in the presence of di-tert-butyl peroxide (DTBP). Predict the major product and draw the complete radical mechanism.

Radical Addition of HBr to 1-Methylcyclohexene

  1. 1

    Step 1 — Identify the Alkene and Conditions

    The substrate is 1-methylcyclohexene, an endocyclic trisubstituted alkene. The double bond lies between C-1 (bearing the methyl group, trisubstituted) and C-2 (disubstituted). The presence of DTBP (a peroxide) signals that the reaction proceeds via a radical chain mechanism, so we expect anti-Markovnikov addition.
    Conditions: radical mechanism → anti-Markovnikov product expected
  2. 2

    Step 2 — Initiation

    The O–O bond in (CH₃)₃CO–OC(CH₃)₃ undergoes homolysis upon heating (BDE ≈ 155 kJ/mol), generating two tert-butoxy radicals. Each (CH₃)₃CO· then abstracts hydrogen from HBr (H–Br BDE = 366 kJ/mol, O–H BDE ≈ 440 kJ/mol; ΔH° ≈ −74 kJ/mol, highly exothermic) to produce Br· and tert-butanol.
    (CH₃)₃CO· + HBr → (CH₃)₃COH + Br·
  3. 3

    Step 3 — Propagation Step 1 (Br· Adds to Alkene)

    Br· adds to C-2 (the less substituted end of the double bond). This places the resulting radical on C-1—the more substituted carbon that already bears the methyl group. The radical at C-1 is tertiary, benefiting from hyperconjugation with three adjacent C–H/C–C bonds. Had Br· added to C-1 instead, the radical would reside on C-2 (secondary, less stable).
    Br adds to C-2; radical on C-1 (3° radical, more stable)
  4. 4

    Step 4 — Propagation Step 2 (H-Abstraction from HBr)

    The tertiary radical at C-1 abstracts H from another molecule of HBr. The C-1–H bond forms (~400 kJ/mol) while the H–Br bond breaks (366 kJ/mol), making this step exothermic by approximately −34 kJ/mol. A new Br· is generated to continue the chain.
    Product: 1-methyl-2-bromocyclohexane (Br on C-2, H on C-1)
  5. 5

    Step 5 — Identify the Major Product

    The major product is 1-methyl-2-bromocyclohexane, with bromine on C-2 (the less substituted carbon of the original alkene). This is the anti-Markovnikov product. In contrast, ionic addition of HBr (without peroxides) would place Br on C-1 (the more substituted carbon), yielding 1-bromo-1-methylcyclohexane via a 3° carbocation.
    Major product: 1-methyl-2-bromocyclohexane (anti-Markovnikov)
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Stereochemistry Note

Radical additions are generally not stereospecific. Because the carbon radical intermediate is planar (sp² hybridized), HBr can be delivered from either face, typically resulting in a mixture of cis and trans isomers for cyclic substrates. This is in contrast to the stereospecific anti addition observed in bromonium ion pathways for Br₂ addition.

SECTION 7

Ionic vs. Radical Addition: Strengths and Limitations

A practicing organic chemist must choose between ionic and radical conditions depending on the desired regiochemical outcome. Each pathway has distinct advantages and constraints that influence synthetic strategy.

FeatureIonic Addition (No Peroxides)Radical Addition (With ROOR)
RegiochemistryMarkovnikov — Br on more substituted CAnti-Markovnikov — Br on less substituted C
Key IntermediateCarbocation (sp² cation)Carbon radical (sp² radical)
Which Atom Adds First?H⁺ (electrophile → π bond)Br· (radical → π bond)
Applicable HXHCl, HBr, HI (all work)Only HBr (HCl and HI fail thermodynamically)
StereochemistryOften non-stereospecific (open cation); anti addition with Br₂Non-stereospecific (planar radical)
RearrangementsPossible (hydride/alkyl shifts in carbocations)Not observed (radicals do not rearrange)
Inhibited byNucleophilic solvents can competeRadical inhibitors (BHT, O₂, hydroquinone)

Comparison of ionic and radical HBr addition to alkenes

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KEY TAKEAWAY

An important practical advantage of radical addition is the absence of rearrangements. In ionic chemistry, carbocations are notorious for hydride and methyl shifts that scramble the carbon skeleton. Carbon radicals, however, do not undergo 1,2-shifts because the transition state for such a migration would require the migrating group to bridge a single electron—an energetically prohibitive geometry. This makes radical addition particularly valuable when working with substrates that would otherwise rearrange under ionic conditions.

SECTION 8

Connections to Advanced Radical Chemistry

The anti-Markovnikov addition of HBr to alkenes is the gateway reaction in a much broader landscape of radical-mediated transformations. In Organic Chemistry 2 and beyond, you will encounter increasingly powerful radical methods—from atom transfer radical polymerization (ATRP) to radical cyclizations (e.g., Baldwin's rules applied to radical ring closure), and modern photoredox catalysis that generates radicals under mild, visible-light conditions. The mechanistic reasoning you develop here—evaluating BDEs, assessing radical stability, and thinking about chain kinetics—transfers directly to these advanced contexts.

ConceptThis Lesson (Introductory)Advanced Treatment
Radical generationPeroxide homolysis (thermal)Photoredox catalysis, SET from metal complexes, electrochemistry
Radical acceptorSimple alkene (intermolecular addition)Intramolecular cyclization, radical cascades (5-exo-trig, etc.)
Chain carrierBr· / carbon radicalSn-centered radicals (Bu₃Sn·), Si radicals, thiol–ene chemistry
Selectivity controlThermodynamic (radical stability)Polarity matching (SOMO–LUMO interactions), persistent radical effect
StereochemistryNon-stereospecificChiral auxiliaries, enantioselective radical reactions via chiral Lewis acids

Connections between introductory radical addition and advanced radical chemistry

One area of particular modern significance is thiol–ene chemistry, which is mechanistically almost identical to HBr radical addition: a thiyl radical (RS·) adds to an alkene to form a carbon radical, which then abstracts hydrogen from RSH to regenerate the chain carrier. This reaction, sometimes called a "radical click" reaction, is widely used in materials science, bioconjugation, and polymer cross-linking. The principles of regioselectivity and chain kinetics that you learn in the context of HBr/peroxides apply directly.

SECTION 9

Practice Problems

Question 1: CONCEPTUAL

Explain why radical addition of HBr to alkenes is described as "anti-Markovnikov" even though both ionic and radical pathways favor formation of the more stable intermediate. What is the fundamental difference that reverses the regiochemistry?
Show answer
In both pathways, the more stable intermediate is preferred: a more substituted carbocation in ionic addition and a more substituted carbon radical in radical addition. The regiochemical difference arises from which atom adds first. In ionic addition, H⁺ adds first to form the carbocation, so Br⁻ ends up on the more substituted carbon (Markovnikov). In radical addition, Br· adds first to generate the radical, so Br ends up on the less substituted carbon (anti-Markovnikov). The same stability principle—more substituted = more stable—operates in both cases; the reversal is due to the reversed order of bond formation, not a different selectivity rule.

Question 2: BASIC CALCULATION

Using the following BDEs, calculate ΔH° for each propagation step in the radical addition of HBr to ethylene (CH₂=CH₂): π bond of C=C = 264 kJ/mol, C–Br = 285 kJ/mol, H–Br = 366 kJ/mol, C–H = 410 kJ/mol. Is the overall reaction exothermic?
Show answer
Propagation Step 1 (Br· + CH₂=CH₂ → BrCH₂–ĊH₂): ΔH° = BDE(π) − BDE(C–Br) = 264 − 285 = −21 kJ/mol (exothermic). Propagation Step 2 (BrCH₂–ĊH₂ + HBr → BrCH₂–CH₃ + Br·): ΔH° = BDE(H–Br) − BDE(C–H) = 366 − 410 = −44 kJ/mol (exothermic). Overall: ΔH° = (−21) + (−44) = −65 kJ/mol. Yes, the overall reaction is exothermic, and both individual propagation steps are exothermic, confirming that the radical chain is thermodynamically viable.

Question 3: INTERMEDIATE

Predict the major product of the reaction of 2-methylpropene (isobutylene, (CH₃)₂C=CH₂) with HBr under each of the following conditions: (a) no peroxides present; (b) di-tert-butyl peroxide and heat. Draw the key intermediate in each case and justify the regiochemistry.
Show answer
(a) Without peroxides, ionic electrophilic addition occurs. H⁺ adds to the terminal CH₂ (less substituted carbon), forming a tertiary carbocation at the central carbon: (CH₃)₃C⁺. Br⁻ then captures this cation to give 2-bromo-2-methylpropane ((CH₃)₃CBr)—the Markovnikov product. (b) With peroxides, radical addition occurs. Br· adds to the terminal CH₂ (less substituted carbon), generating a tertiary radical: (CH₃)₂Ċ–CH₂Br. The radical abstracts H from HBr to form 1-bromo-2-methylpropane ((CH₃)₂CHCH₂Br)—the anti-Markovnikov product. In both cases, the intermediate at the more substituted carbon (3° carbocation or 3° radical) is more stable, but the atom that adds first differs, reversing the position of Br.

Question 4: APPLIED

A synthetic chemist wants to convert 1-hexene into 1-bromohexane selectively. They initially try adding HBr in acetic acid solvent without any additives, but find that the major product is 2-bromohexane. (a) Explain why this happens. (b) Propose modified reaction conditions that would give 1-bromohexane as the major product. (c) A colleague suggests using HI with peroxides instead of HBr. Would this work? Justify using BDEs.
Show answer
(a) Without peroxides, HBr undergoes ionic electrophilic addition. H⁺ adds to C-1 of 1-hexene, forming a secondary carbocation on C-2. Br⁻ captures at C-2, yielding 2-bromohexane (Markovnikov product). (b) To obtain 1-bromohexane (anti-Markovnikov), the chemist should add a radical initiator such as di-tert-butyl peroxide (or benzoyl peroxide) and heat. Under these conditions, Br· adds to C-1, generating a secondary radical on C-2, which abstracts H from HBr to give 1-bromohexane. (c) HI with peroxides would not work. The I· + alkene propagation step is endothermic (BDE of π ≈ 264 kJ/mol vs. C–I BDE ≈ 222 kJ/mol; ΔH° ≈ +42 kJ/mol). Since this step is thermodynamically unfavorable, the radical chain cannot sustain itself. Only HBr has both propagation steps exothermic.

Question 5: CRITICAL THINKING

Consider the radical addition of HBr to 3,3-dimethyl-1-butene ((CH₃)₃C–CH=CH₂) in the presence of peroxides. (a) Predict the major product. (b) If this same substrate were treated with HBr under ionic conditions, a rearranged product is sometimes observed. Explain why rearrangement can occur under ionic but not radical conditions, and predict the rearranged ionic product. (c) How could you experimentally distinguish between the radical and ionic products?
Show answer
(a) Under radical conditions, Br· adds to C-1 (terminal, less substituted), generating a secondary radical on C-2. H-abstraction from HBr gives 1-bromo-3,3-dimethylbutane ((CH₃)₃C–CH₂–CH₂Br), the anti-Markovnikov product. No rearrangement occurs because carbon radicals do not undergo 1,2-shifts. (b) Under ionic conditions, H⁺ adds to C-1, initially forming a secondary carbocation on C-2. However, this 2° cation is adjacent to a quaternary carbon, and a 1,2-methyl shift can occur to generate a more stable tertiary carbocation on C-3: (CH₃)₂(CH₃)Ċ⁺–CH₂–CH₃. Br⁻ capture then gives a rearranged product, 2-bromo-2,3-dimethylbutane. Rearrangement occurs in carbocations because the migrating group stabilizes the cation by forming a more substituted center, but radicals lack the electrophilic driving force for such 1,2-shifts—the half-filled p orbital does not provide sufficient stabilization for the bridged transition state. (c) The products differ in molecular connectivity and can be distinguished by ¹H NMR spectroscopy: the radical product shows a –CH₂Br group (triplet near δ 3.4), while the rearranged ionic product shows a C–Br bearing no hydrogens (quaternary carbon) or a –CHBr group depending on the shift pathway. Mass spectrometry fragmentation patterns would also differ.

SUMMARY

Lesson Summary

Summary

In the presence of organic peroxides (ROOR), HBr adds to alkenes via a radical chain mechanism consisting of initiation (peroxide homolysis → RO· + HBr → Br·), propagation (Br· adds to the less substituted carbon, forming a more stable radical at the more substituted position; the carbon radical then abstracts H from HBr, regenerating Br·), and termination (radical–radical coupling). The product exhibits anti-Markovnikov regiochemistry—Br ends up on the less substituted carbon—because Br· adds first (unlike ionic addition, where H⁺ adds first).

This reaction is unique to HBr among the hydrogen halides: for HCl, the second propagation step (R· + HCl) is endothermic because H–Cl is too strong, while for HI, the first step (I· + alkene) is endothermic because C–I is too weak. The thermodynamic analysis using bond dissociation energies (BDEs) confirms that both propagation steps must be exothermic for a viable chain. Key advantages of radical addition include the absence of carbocation rearrangements and predictable anti-Markovnikov selectivity, making it a valuable tool in synthetic planning.

Varsity Tutors • Organic Chemistry 1 • Radical Addition to Alkenes (HBr/Peroxides)