Hydrogen Halide Addition

Introduction

Hydrogen halides add across many alkenes to form alkyl halides. For an unsymmetrical double bond, the position of H and X is the key question. Under ordinary polar conditions, the product usually follows Markovnikov orientation because the first protonation step favours a more stable carbocation. HBr with a radical initiator such as a peroxide is an important exception: bromine adds first in a radical chain, often reversing the orientation.

Core explanation

In the ordinary polar pathway, the alkene pi electrons form a bond to H from HX while the H–X bond pair moves to X. Protonation can occur at either alkene carbon. The favoured route generally leaves the more stable carbocation on the other carbon. For propene, adding H to terminal C1 places positive charge at secondary C2, whereas adding H to C2 would create a primary cation at C1. Chloride, bromide or iodide then attacks the secondary cation, giving 2-halopropane. This is the common Markovnikov result: H goes to the carbon that initially bore more H, and X to the more substituted carbon.

The rule is a useful shortcut but the carbocation comparison is the explanation. If resonance can stabilise a cation, an allylic or benzylic route may dominate even if a simple substitution count is misleading. Once a carbocation forms, a 1,2-hydride or alkyl shift may produce a more stable cation before halide capture, giving a rearranged product. The halide anion can attack either face of a planar cation, so a new stereocentre may form as a mixture when no other chiral influence is present. A strongly asymmetric environment can bias that mixture.

With HBr and peroxides under radical-chain conditions, an initiator creates radicals that ultimately generate Br·. Bromine radical adds to the alkene first, preferably in the direction that leaves the more stable carbon radical. For propene, Br· bonds to terminal C1, leaving a secondary radical at C2. That radical abstracts H from HBr, giving 1-bromopropane and regenerating Br· for another propagation cycle. The overall product is anti-Markovnikov because Br is on the less substituted carbon. This route is not a proton-first carbocation mechanism.

The familiar peroxide effect is characteristic of HBr, not a general promise for HCl or HI. The propagation steps for HCl and HI do not have the same favourable energy balance under ordinary conditions, so adding peroxide should not be used as a universal way to reverse HX orientation. This statement concerns the common alkene-addition chain, not every conceivable photochemical or catalytic reaction involving those acids.

The reagent label alone is incomplete. HBr without radical initiation normally follows polar Markovnikov addition; HBr with peroxides, light or another suitable radical initiation condition may take the radical route. Radical inhibitors, oxygen, solvent and temperature can change the competition. A product's orientation is evidence about mechanism, but to make a reliable prediction, read the full condition line.

Step-by-step reasoning

For ordinary HX, mark both alkene carbons, protonate each one hypothetically, and compare the resulting carbocations. Draw X⁻ attack on the favoured cation and check for rearrangement. If HBr plus radical initiator is specified, reverse the order of addition: add Br· first to create the more stable radical, then transfer H from HBr. Balance atom positions and identify the final halide's carbon.

Visual explanation

Draw propene twice. In the upper panel, H⁺ adds to terminal C1, leaving C2⁺, then Br⁻ attacks C2 to make 2-bromopropane. In the lower panel, Br· adds to terminal C1, leaving a radical on C2, which abstracts H from HBr to make 1-bromopropane. Label the upper intermediate with a plus sign and the lower with a single radical dot.

Real-world analogy

Two guests enter a two-seat room in different orders. If the first guest is hydrogen, the room settles into a carbocation arrangement before bromide arrives; if bromine radical enters first, the room settles into a carbon-radical arrangement before hydrogen arrives. The different first step changes which seat each guest finally occupies. The analogy is about pathway order, not a literal preference of people.

Real-world example

To prepare 1-bromopropane from propene, a chemist can use HBr with appropriate radical initiation; ordinary polar HBr addition instead gives mainly 2-bromopropane. These two constitutional isomers have different subsequent substitution and elimination behaviour, so the peroxide condition is a synthetic design choice rather than a small decorative note over the reaction arrow.

Why?

Why do both pathways still prefer a relatively stable intermediate but give opposite orientation? In the ionic route, H adds first and the favoured carbocation is secondary, placing Br later at that carbon. In the radical route, Br adds first and the favoured radical is secondary, placing Br on the other, less substituted carbon. The same stability preference acts on different first-added fragments.

Common misconception

"Peroxide reverses addition of every hydrogen halide." The standard anti-Markovnikov radical-chain exception applies to HBr with suitable initiation. Do not predict the same straightforward effect for HCl or HI, and do not draw a carbocation for the radical path. Conditions determine which mechanistic sequence is justified.

Worked example

Question: Give the major constitutional product of propene with (a) HBr alone under polar conditions and (b) HBr plus peroxide under radical-chain conditions. Explain the carbon bearing Br in each.

Reasoning: (a) H protonates terminal carbon, giving a secondary cation at C2; bromide attacks C2. (b) Br radical adds to terminal carbon to leave a secondary radical at C2; that radical obtains H from HBr.

Answer: (a) 2-bromopropane, Markovnikov; (b) 1-bromopropane, anti-Markovnikov.

Quick check

1. In radical addition of HBr to propene, which fragment adds to the alkene first? Answer: A bromine radical adds first; the resulting carbon radical later abstracts H from HBr.

Exam focus

Read the conditions above the arrow. For polar HX addition, show protonation, a carbocation and halide attack, with rearrangement if plausible. For peroxide-initiated HBr, show radical initiation and propagation using single-headed arrows if required. State that the common peroxide exception is HBr-specific under ordinary conditions.

Advanced insight

The radical chain succeeds when both propagation steps are sufficiently favourable: bromine-radical addition creates a carbon radical, and H abstraction from HBr regenerates Br·. The corresponding chain energy balances for HCl and HI are unfavourable in different steps under standard teaching conditions. This energetic constraint explains why the peroxide effect is a specific mechanistic exception rather than a universal regiochemical switch.

Summary

Ordinary HX addition protonates an alkene to a relatively stable carbocation, then X⁻ attacks, usually giving Markovnikov orientation. Carbocation rearrangement and stereochemical mixing may occur. HBr with suitable radical initiation adds Br· first, then H, often giving anti-Markovnikov orientation. The standard peroxide effect does not extend automatically to HCl or HI. Always use the stated conditions to select the pathway.

Practice questions

1. What is the usual polar product of propene plus HCl? Answer: 2-Chloropropane, because the preferred protonation creates a secondary carbocation at C2. 2. What is the radical-chain product of propene plus HBr with peroxide initiation? Answer: 1-Bromopropane, because Br adds to terminal C1 while a secondary radical forms at C2. 3. Why can ordinary HX addition produce a rearranged alkyl halide? Answer: The carbocation formed after protonation can undergo a hydride or alkyl shift before halide capture. 4. Should HCl with peroxide be assumed to give anti-Markovnikov addition? Answer: No. The familiar favourable radical-chain peroxide effect is specific to HBr under ordinary conditions.