Exercises — Markovnikov vs anti-Markovnikov (peroxide effect, Kharasch)

$H$ adds to the carbon already carrying more hydrogen atoms; $Br$ adds to the carbon carrying fewer hydrogens (the more substituted carbon). Why: the ionic mechanism builds the more stable carbocation on the more-substituted carbon (left branch of s01), and $B{r}^{-}$ then attacks that carbon.

Only $HBr$. $HF$ and $HCl$ have $H-X$ bonds too strong (see the BDE table: $H-F=567$, $H-Cl=431$ kJ/mol); $HI$ has addition of $I^{\bullet }$ too endothermic. Only $HBr$ makes both propagation steps exothermic.

(a) A carbocation (positively charged carbon). (b) A carbon free radical (carbon with one unpaired electron, written with a dot $\stackrel{\bullet }{C}$).

Ionic path → most stable carbocation. $H^{+}$ can land on C1 ($=C{H}_2$ end) → $+$ on C2 → tertiary ($3°$) cation $(C{H}_3{)}_2\stackrel{+}{C}-C{H}_3$. Why is this tertiary cation so stable? Two effects, both from the three attached carbons (see Carbocation Stability and Hyperconjugation):

1. Hyperconjugation: each neighbouring $C-H$ bond can overlap sideways with the empty orbital on the $+$ carbon, letting its electrons "leak" toward the positive centre. A tertiary cation has many such neighbouring $C-H$ bonds → lots of leaking → charge is spread out → lower energy.
2. +I (inductive) donation: alkyl groups gently push electron density toward the $+$ carbon through the sigma bonds, again cushioning the charge. More alkyl groups = more of both = more stable. That is why $3°>2°>1°$. The other option (H on C2) gives a primary ($1°$) cation with far fewer of these effects — rejected. $B{r}^{-}$ attacks C2: $(C{H}_3{)}_{2}CBr-C{H}_3=\text{2-bromo-2-methylpropane}$

Radical path → $B{r}^{\bullet }$ adds first to the less substituted carbon (C1), leaving the radical on the more substituted carbon (C2, now tertiary radical — most stable, same hyperconjugation/+I reasons as the cation). $(C{H}_3{)}_{2}C=C{H}_2+B{r}^{\bullet }\to (C{H}_3{)}_2\stackrel{\bullet }{C}-C{H}_{2}Br$ Then it grabs $H$ from $HBr$ (propagation step 2 of the loop in s03): $(C{H}_3{)}_{2}CH-C{H}_{2}Br=\text{1-bromo-2-methylpropane (isobutyl bromide)}$

Trick question! $HCl$ shows no peroxide effect, so the answer is the Markovnikov product regardless of peroxide: $C{H}_3-CHCl-C{H}_3=\text{2-chloropropane}$

Step 2 breaks an $H-X$ bond and forms a $C-H$ bond. Approximate enthalpy: $\Delta H_2\approx (\text{BDE of }H-X)-(\text{BDE of new }C-H)$

• $HCl$: $\Delta H_2\approx 431-413=+18$ kJ/mol → endothermic → slow → chain dies.
• $HBr$: $\Delta H_2\approx 366-413=-47$ kJ/mol → exothermic → fast → chain survives. So $H-Cl$ is too strong to give up its $H$ cheaply; $H-Br$ is weak enough.

A radical chain only runs fast if every propagation step is downhill (exothermic). Step 1 for $HI$ is uphill by $\approx +54$ kJ/mol: $I^{\bullet }$ barely adds to the double bond (the weak $C-I$ bond formed does not "pay" for breaking the $\pi$ bond). Instead, $I^{\bullet }$ radicals just recombine to $I_2$ (a termination event). With step 1 stalled, no chain → no anti-Markovnikov → $HI$ stays Markovnikov.

But-2-ene is symmetrical about the double bond: both alkene carbons carry exactly one $H$ and one $C{H}_3$. So whichever carbon $Br$ lands on, you get the same molecule: $C{H}_3-C{H}_2-CHBr-C{H}_3=\text{2-bromobutane}$ "More-H carbon" vs "fewer-H carbon" does not exist — the two carbons are equivalent. Peroxide changes nothing about which product, only the mechanism. Regiochemistry rules only matter for unsymmetrical alkenes.

Reagents: $HBr$ + organic peroxide (e.g. benzoyl peroxide), heat/light. Initiation (startup, happens once — top of s03): $RO-OR\stackrel{\Delta }{\to }2R{O}^{\bullet }R{O}^{\bullet }+HBr\to ROH+B{r}^{\bullet }$ Propagation (the repeating loop of s03): $C{H}_3-CH=C{H}_2+B{r}^{\bullet }\to C{H}_3-\stackrel{\bullet }{C}H-C{H}_{2}Br(2°\text{ radical, step 1})$ $C{H}_3-\stackrel{\bullet }{C}H-C{H}_{2}Br+HBr\to C{H}_3-C{H}_2-C{H}_{2}Br+B{r}^{\bullet }(\text{step 2})$ Termination (chain-ending, two radicals pair up), e.g.: $B{r}^{\bullet }+B{r}^{\bullet }\to B{r}_2$ Net product: 1-bromopropane (anti-Markovnikov), with $B{r}^{\bullet }$ regenerated each propagation turn until a termination stops that chain.

The direct answer for 1-bromopropane itself is the peroxide/$HBr$ radical route of L4.1 — without peroxide, plain $HBr$ gives the Markovnikov product (2-bromopropane), so you do need the radical initiator. The conceptual comparison from the Connections list is Hydroboration-Oxidation, another anti-Markovnikov route — but note it installs an $-OH$, not $Br$, giving propan-1-ol. Different reason: hydroboration is anti-Markovnikov because boron (the bulky, electron-poor atom) adds to the less-hindered/less-substituted carbon in a single concerted step (steric + electronic control), not because of a radical chain. Same anti-Markovnikov outcome, entirely different mechanism.

$H^{+}$ adds to C1 (the $=C{H}_2$ end) → $+$ on C2 → this initial cation is secondary ($2°$): $(C{H}_3{)}_{2}CH-\stackrel{+}{C}H-C{H}_3$ This $2°$ cation sits right next to a tertiary $CH$. A 1,2-hydride shift ($H$ with its electron pair migrates from that adjacent tertiary carbon to the $+$ carbon) converts it into the more stable tertiary ($3°$) cation: $(C{H}_3{)}_2\stackrel{+}{C}-C{H}_2-C{H}_3$ $B{r}^{-}$ then attacks that tertiary carbon → 2-bromo-2-methylbutane (a rearranged Markovnikov product). Key idea: carbocations rearrange to greater stability; radicals essentially do not, so the peroxide route would not rearrange.

$\Delta H_2\approx 567-413=+154\text{ kJ/mol}$ Massively endothermic — even worse than $HCl$ ($+18$). Also $F^{\bullet }$ addition to alkenes is far too energetic/uncontrolled. So $HF$ never shows the peroxide effect. This confirms the Goldilocks picture: only $HBr$ sits in the narrow "both steps exothermic" window.

The slow (rate-determining) step forms the reactive intermediate. By the Hammond postulate (defined at the top of this page, visualised in s02), the transition state of this endothermic step resembles the intermediate, so the more stable intermediate has the lower-energy transition state → that path is faster → it decides the product. The final product's stability does not steer this. Why opposite placement (compare the two branches of s01):

• Ionic: the species that adds first is $H^{+}$. To leave a $3°$ cation, $H^{+}$ goes to the less-substituted carbon → $B{r}^{-}$ then lands on the more-substituted carbon.
• Radical: the species that adds first is $B{r}^{\bullet }$. To leave a $3°$ radical, $B{r}^{\bullet }$ goes to the less-substituted carbon → $Br$ ends up on the less-substituted carbon. Same stability order, but a different atom adds first, so the $Br$ ends up on opposite carbons. That single swap is the whole peroxide effect.

Initiation (top of s03) is a one-time, uphill cost paid to create the first radicals; it is not part of the repeating cycle. Heat/light supplies that startup energy. Once $B{r}^{\bullet }$ exists, the propagation cycle (the loop in s03, steps 1 and 2) is net exothermic and self-sustaining — each turn regenerates $B{r}^{\bullet }$ and releases energy, so no continuous external input is needed. A tiny amount of peroxide therefore triggers many product molecules. Termination (two radicals combining, e.g. $B{r}^{\bullet }+B{r}^{\bullet }\to B{r}_2$) is what eventually removes radicals and stops each chain — a few terminations per many propagations, so it barely dents the yield. The overall reaction ($alkene+HBr\to$ bromoalkane) is exothermic; the positive initiation term is negligible against that.

Product of propene + HBr, no peroxide? ::: 2-bromopropane Product of propene + HBr, with peroxide? ::: 1-bromopropane Does HCl show the peroxide effect? ::: No — H–Cl too strong, step 2 endothermic Symmetrical alkene + HBr + peroxide — does the product change vs no peroxide? ::: No — regiochemistry is meaningless for a symmetrical alkene What adds first in the radical path? ::: The bromine radical $B{r}^{\bullet }$ What does BDE stand for? ::: Bond Dissociation Energy — energy to break a bond into two radicals What does $3°$ mean for a carbon? ::: Tertiary — the carbon is directly attached to 3 other carbons What is a termination step? ::: Two radicals combine and pair their electrons, ending the chain (e.g. $B{r}^{\bullet }+B{r}^{\bullet }\to B{r}_2$)