{"id":7801,"date":"2013-12-09T08:00:00","date_gmt":"2013-12-09T14:00:00","guid":{"rendered":"https:\/\/www.masterorganicchemistry.com\/?p=7801"},"modified":"2026-05-07T04:54:49","modified_gmt":"2026-05-07T09:54:49","slug":"in-summary-free-radicals","status":"publish","type":"post","link":"https:\/\/www.masterorganicchemistry.com\/2013\/12\/09\/in-summary-free-radicals\/","title":{"rendered":"In Summary: Free Radicals"},"content":{"rendered":"<p><strong>A Summary Of Topics\u00a0 For Free-Radical Reactions<\/strong><\/p>\n<p>So what have we learned in this series about free radicals? Lots of things. We&#8217;ve seen that\u00a0 they are neutral species with a half-filled valence shell that tend to be highly reactive with alkanes\u00a0 and double bonds.\u00a0 We&#8217;ve seen that they are stabilized by electron donors and delocalization.\u00a0 We&#8217;ve shown their mechanisms proceed through three distinct stages (initiation, propagation, termination) and that although reactions tend to be selective for formation of the more stable free radical, bromine radicals are far more selective than chlorine radicals in halogenation of alkanes. Finally we&#8217;ve also explored allylic (and benzylic) halogenation and seen examples of allylic rearrangement.<\/p>\n<p><strong>Table of Contents<\/strong><\/p>\n<ol>\n<li><a href=\"#one\">What Are Free Radicals?<\/a><\/li>\n<li><a href=\"#two\">Homolytic Cleavage And &#8220;Single-Barbed&#8221; Curved Arrows<\/a><\/li>\n<li><a href=\"#three\"><span class=\"s1\"><span class=\"s1\">Factors That Affect Free Radical Stability<\/span><\/span><\/a><\/li>\n<li><a href=\"#four\">Factors That Destabilize Free Radicals<\/a><\/li>\n<li><a href=\"#five\">A Handy Shortcut For Determining Radical Stability<\/a><\/li>\n<li><a href=\"#six\">How Do Free Radical Reactions Work? Initiation, Propagation, Termination<\/a><\/li>\n<li><a href=\"#seven\">Free-Radical Halogenation Leads To Mixtures Of Isomers &#8211; But There Is Some Selectivity<\/a><\/li>\n<li><a href=\"#eight\">Bromination Is More Selective Than Chlorination<\/a><\/li>\n<li><a href=\"#nine\">Allylic Bromination,\u00a0 And Allylic Bromination With Allylic Rearrangement<\/a><\/li>\n<li><a href=\"#ten\">Free-Radical Addition Of Radicals To Alkenes<\/a><\/li>\n<li><a href=\"#notes\">Notes<\/a><\/li>\n<li><a href=\"#quizzes\">Quiz Yourself!<\/a><\/li>\n<\/ol>\n<hr \/>\n<h2><strong><a id=\"one\"><\/a>1. What Are Free Radicals?\u00a0<\/strong><\/h2>\n<p><a href=\"https:\/\/www.masterorganicchemistry.com\/2013\/07\/30\/free-radical-reactions\/\">Free radicals<\/a> are chemical species that contain a singly occupied orbital. They are\u00a0<span style=\"text-decoration: underline;\">neutral<\/span> and tend to be highly reactive.<br \/>\nOne of the most common classes of free-radical reactions is\u00a0<span style=\"text-decoration: underline;\">free-radical substitution<\/span>. Here is an example of the reaction between methane and Cl<sub>2<\/sub>:<\/p>\n<p><img fetchpriority=\"high\" decoding=\"async\" class=\"aligncenter wp-image-14611\" src=\"https:\/\/www.masterorganicchemistry.com\/wp-content\/uploads\/2019\/12\/1-free-radical-substitution-reaction-of-methane-with-cl2-giving-methyl-chloride-initiated-with-light.gif\" alt=\"free-radical-substitution-reaction-of-methane-with-cl2-giving-methyl-chloride-initiated-with-light\" width=\"555\" height=\"160\" \/><\/p>\n<h2><strong><a id=\"two\"><\/a>2. Homolytic Cleavage And &#8220;Single-Barbed&#8221; Curved Arrows<\/strong><\/h2>\n<p>Unlike many of the reactions we have seen previously, free radical reactions do not involve the donation or acceptance of an electron\u00a0<span style=\"text-decoration: underline;\">lone pair<\/span> and they are not\u00a0<span style=\"text-decoration: underline;\">ionic<\/span>. Instead, free radical reactions operate through\u00a0<span style=\"text-decoration: underline;\">homolytic cleavage<\/span> &#8211; that is, bonds break such that equal numbers of electrons are distributed to each atom.\u00a0\u00a0The curved arrow formalism is modified here: we draw curved arrows with a single barb, which shows the movement of a single electron:<\/p>\n<p><img decoding=\"async\" class=\"aligncenter wp-image-14612\" src=\"https:\/\/www.masterorganicchemistry.com\/wp-content\/uploads\/2019\/12\/2-homolytic-cleavage-of-chlorine-single-barbed-arrows-giving-chlorine-radicals.gif\" alt=\"homolytic-cleavage-of-chlorine-single-barbed-arrows-giving-chlorine-radicals\" width=\"555\" height=\"247\" \/><\/p>\n<h2><strong><a id=\"three\"><\/a>3. Factors That Affect Free Radical Stability<\/strong><\/h2>\n<p>Like carbocations, \u00a0free radicals are <span style=\"text-decoration: underline;\">electron-deficient<\/span> species. [Helpful to know: \u00a0the factors which affect the stability of free radicals are the same which influence the stability of carbocations.]\u00a0They can be stabilized through donation of electron density by neighbours; for this reason, radical stability increases in the order methyl &lt; primary &lt; secondary &lt; tertiary . [Radicals are also stabilized by <a href=\"https:\/\/www.masterorganicchemistry.com\/2013\/08\/02\/3-factors-that-stabilize-free-radicals\/#lonepairs\">adjacent atoms with lone pairs<\/a>, such as oxygen and nitrogen].<span style=\"color: #993366;\"><em> [See post: <a style=\"color: #993366;\" href=\"https:\/\/www.masterorganicchemistry.com\/2013\/08\/02\/3-factors-that-stabilize-free-radicals\/\">3 Factors Which Stabilize Free Radicals<\/a>]<\/em><\/span><\/p>\n<p><img decoding=\"async\" class=\"aligncenter wp-image-14613\" src=\"https:\/\/www.masterorganicchemistry.com\/wp-content\/uploads\/2019\/12\/3-order-of-radical-stability-methyl-least-stable-tertiary-most-stable.gif\" alt=\"order-of-radical-stability-methyl-least-stable-tertiary-most-stable\" width=\"705\" height=\"238\" \/><br \/>\nA second important factor which stabilizes free radicals is &#8220;delocalization&#8221; &#8211; that is, if the radical can be spread out over two or more carbons. A more familiar way of saying this is that <span style=\"text-decoration: underline;\">free radicals are stabilized by resonance<\/span>.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter wp-image-14614\" src=\"https:\/\/www.masterorganicchemistry.com\/wp-content\/uploads\/2019\/12\/4-delocalization-of-electrons-stabilizes-radicals-through-resonance-allylic-and-benzylic.gif\" alt=\"delocalization-of-electrons-stabilizes-radicals-through-resonance-allylic-and-benzylic\" width=\"705\" height=\"266\" \/><\/p>\n<h2><a id=\"four\"><\/a>4. Factors That Destabilize Free Radicals<\/h2>\n<p>As electron-deficient species, it is also helpful to keep in mind some of the factors which\u00a0<span style=\"text-decoration: underline;\">destabilize<\/span> free radicals. There are several important trends to keep in mind <span style=\"color: #993366;\"><em>[See post: <a style=\"color: #993366;\" href=\"https:\/\/www.masterorganicchemistry.com\/2013\/08\/05\/what-factors-destabilize-free-radicals\/\">Which Factors Destabilize Free Radicals<\/a>?]<\/em><\/span><\/p>\n<ul>\n<li>Free radicals decrease in stability as the % of s-character in the orbital increases [i.e. as the half-empty orbital becomes closer to the nucleus]. For that reason, free radical stability decreases as the atom goes from alkyl to alkenyl to alkynyl.<img loading=\"lazy\" decoding=\"async\" class=\"aligncenter wp-image-14615\" src=\"https:\/\/www.masterorganicchemistry.com\/wp-content\/uploads\/2019\/12\/5-sp3-stabilized-radicals-are-the-most-stable-followed-by-sp2-radicals-and-then-sp-radicals.gif\" alt=\"sp3-stabilized-radicals-are-the-most-stable-followed-by-sp2-radicals-and-then-sp-radicals\" width=\"485\" height=\"237\" \/><\/li>\n<li>Across a row of the periodic table, free radicals decrease in stability as the electronegativity increases<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"aligncenter wp-image-14616\" src=\"https:\/\/www.masterorganicchemistry.com\/wp-content\/uploads\/2019\/12\/6-across-a-row-of-the-periodic-table-free-radical-stability-is-inversely-proportional-to-electronegativity.gif\" alt=\"across-a-row-of-the-periodic-table-free-radical-stability-is-inversely-proportional-to-electronegativity\" width=\"605\" height=\"110\" \/><\/li>\n<li>Free radicals increase in stability going down a column of the periodic table, F\u2022 &lt; Cl\u2022 &lt; Br\u2022 &lt; I\u2022 \u00a0since the electron-deficient orbital is spread out over a greater volume.<\/li>\n<li>Free radicals adjacent to an electron-withdrawing group are less stable, since in effect, electron-density is being taken away from what is already an electron deficient species.<span style=\"color: #993366;\"><em> [Watch out, however &#8211; this only applies to electron withdrawing groups that cannot donate a pair of electrons, like CF<sub>3<\/sub> or CN.]<\/em><\/span><\/li>\n<\/ul>\n<h2><strong><a id=\"five\"><\/a>5. A Handy Shortcut For Determining Radical Stability<\/strong><\/h2>\n<p>Bond dissociation energies [BDE&#8217;s] measure the energy required for <span style=\"text-decoration: underline;\">homolytic bond cleavage<\/span>, and by proxy we can interpret C\u2013H bond strengths as a good measure of carbon free-radical stability. By examining trends in X-H bond strengths (where X is various atoms) we can discern trends in radical stability.<span style=\"color: #993366;\"><em> [See post: <a style=\"color: #993366;\" href=\"https:\/\/www.masterorganicchemistry.com\/2013\/08\/14\/bond-strengths-radical-stability\/\">Bond Strengths and Radical Stability<\/a>]<\/em><\/span><\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter wp-image-14617\" src=\"https:\/\/www.masterorganicchemistry.com\/wp-content\/uploads\/2019\/12\/7-a-handy-shortcut-in-determining-radical-stability-is-the-strength-of-the-c-h-bonds-as-seen-for-methyl-secondary-and-allylic.gif\" alt=\"a-handy-shortcut-in-determining-radical-stability-is-the-strength-of-the-c-h-bonds-as-seen-for-methyl-secondary-and-allylic\" width=\"805\" height=\"242\" \/><\/p>\n<p>This is also reflected in the strengths of &#8220;allylic&#8221; C-H bonds, which are weaker than normal due to the fact that a resonance-stabilized free-radical is produced. For example, the C-H bond strength of the methyl group of propane is about 98 kcal\/mol, whereas that of the methyl group of 1-propene is 85 kcal\/mol. For a complete list of factors, refer to the original post.<\/p>\n<h2><strong><a id=\"six\"><\/a>6. How Do Free Radical Reactions Work? Initiation, Propagation, Termination<\/strong><\/h2>\n<p>One &#8220;sure fire&#8221; way students find themselves identifying free-radical reactions is to look for the presence of &#8220;heat&#8221; [ \u0394] or &#8220;light&#8221; [h\u03bd\u00a0 ]. What does heat or light have to do with free-radical reactions? Well, it facilitates the first important step in any free radical reaction: <strong>initiation<\/strong>. This is the homolytic fragmentation of a weak bond to form two new free radicals.<span style=\"color: #993366;\"><em> [See post: <a style=\"color: #993366;\" href=\"https:\/\/www.masterorganicchemistry.com\/2013\/08\/30\/radical-reactions-why-is-light-or-heat-required\/\">Free Radical Reactions &#8211; Why is &#8220;Heat&#8221; or &#8220;Light&#8221; Required?<\/a>]<\/em><\/span><\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter wp-image-14618\" src=\"https:\/\/www.masterorganicchemistry.com\/wp-content\/uploads\/2019\/12\/8-initiation-step-of-free-radical-reaction-shows-equilibrium-between-chlorine-and-chlorine-radical.gif\" alt=\"initiation-step-of-free-radical-reaction-shows-equilibrium-between-chlorine-and-chlorine-radical\" width=\"555\" height=\"203\" \/><\/p>\n<p>It&#8217;s important to note here that &#8220;initiation&#8221; is not a reaction that goes to completion. At any given point, only a small concentration of the starting material is converted to free radicals. However, since the reaction is a \u00a0chain-reaction, this is generally sufficient.<br \/>\nThe second step in a free-radical process is at least one (but often two) &#8220;propagation&#8221; steps. For the free-radical halogenation of alkanes, the first propagation step is removal of hydrogen by a halogen radical, giving a new carbon radical. Note here that the number of free radicals remains constant (one on both sides of the equation).<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter wp-image-14619\" src=\"https:\/\/www.masterorganicchemistry.com\/wp-content\/uploads\/2019\/12\/9-first-propagation-step-of-free-radical-chlorination-is-chlorine-radical-plus-methane-giving-hcl-and-methyl-radical.gif\" alt=\"first-propagation-step-of-free-radical-chlorination-is-chlorine-radical-plus-methane-giving-hcl-and-methyl-radical\" width=\"605\" height=\"265\" \/><\/p>\n<p>This is followed by a second propagation step. While it may be tempting for many new students to form CH<sub>3<\/sub>-Cl through combination of CH<sub>3<\/sub>\u2022 with a chlorine radical, that&#8217;s not what happens next [<span style=\"color: #993366;\"><em>that would technically be a &#8220;termination&#8221; step, see below<\/em><\/span>]. What happens next is the reaction of alkyl radical with one of the starting molecules of Cl<sub>2<\/sub>, forming a new C\u2013Cl bond and a new Cl\u2022 radical which continues the chain reaction. Again, note: no net increase in the number of free radicals here:<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter wp-image-14620\" src=\"https:\/\/www.masterorganicchemistry.com\/wp-content\/uploads\/2019\/12\/10-second-propagation-step-of-free-radical-halogenation-is-reaction-of-radical-with-cl2.gif\" alt=\"second-propagation-step-of-free-radical-halogenation-is-reaction-of-radical-with-cl2\" width=\"555\" height=\"225\" \/><\/p>\n<p>At any given point the concentration of free radicals in solution is quite low. However, after most of the Cl<sub>2<\/sub> is consumed, free-radical concentrations will start to increase until the rate of radical-radical combination reactions is significant. This is called &#8220;termination&#8221; since it results in a net destruction of free radicals. There are many potential ways this can occur &#8211; here is one example:<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter wp-image-14621\" src=\"https:\/\/www.masterorganicchemistry.com\/wp-content\/uploads\/2019\/12\/11-termination-step-is-net-reduction-in-number-of-electrons.gif\" alt=\"termination-step-is-net-reduction-in-number-of-electrons.\" width=\"555\" height=\"164\" \/><\/p>\n<p>The bottom line is that by counting the number of radicals created or destroyed in each step you can determine if the step is initiation, propagation, or termination.<br \/>\n<span style=\"color: #993366;\"><em>[See post: <a style=\"color: #993366;\" href=\"https:\/\/www.masterorganicchemistry.com\/2013\/09\/06\/initiation-propagation-termination\/\">Initiation, Propagation, Termination<\/a>]<\/em><\/span><\/p>\n<h2><strong><a id=\"seven\"><\/a>7. Free-Radical Halogenation Leads To Mixtures Of Isomers &#8211; But There Is <em>Some<\/em> Selectivity<\/strong><\/h2>\n<p>Methane and ethanes are simple examples. Replacement of any one of the 4 hydrogens of methane by chlorine will result in the same product. However when we are dealing with a more complex alkane, like propane, butane, or 2,3-dimethyl hexane, we must be alert for the possibility that a variety of compounds will form. It&#8217;s an important skill to be able to recognize all the possible compounds that could result from, say, the mono-chlorination of an alkane like 2-methylpentane [there are 5]. These types of questions are best approached systematically: start with replacement of hydrogen by chlorine on one side of the molecule, and gradually work to the other side &#8211; it can help to double check that there are no duplicates by trying to name them (IUPAC). <span style=\"color: #993366;\"><em>\u00a0[See post: <a style=\"color: #993366;\" href=\"https:\/\/www.masterorganicchemistry.com\/2013\/09\/17\/isomers-from-free-radical-reactions\/\">Isomers From Free Radical Reactions<\/a>]<\/em><\/span><\/p>\n<p>That being said, isomers will not be formed in equal quantities. There are two factors to consider. The first is the fact that <span style=\"text-decoration: underline;\">not all C-H bonds are of equal strength<\/span>; as we mentioned above, tertiary C-H bonds are easier to break than secondary C-H bonds, which are easier to break than primary C-H bonds. So there will be an inherent difference in reactivity between these bonds simply by the fact that it requires less activation energy to break a weaker bond. \u00a0A second factor is\u00a0<span style=\"text-decoration: underline;\">statistics<\/span>. In propane, for example, there are six primary C-H bonds and two secondary C-H bonds. Based on statistics alone, we would expect mono-chlorination propane to yield a 3:1 ratio of 1-chloropropane to 2-chloropropane.<br \/>\nWhat&#8217;s interesting is that the actual ratio of 2-chloropropane to 1-chloropropane is 45:55 !<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter wp-image-14622\" src=\"https:\/\/www.masterorganicchemistry.com\/wp-content\/uploads\/2019\/12\/12-selectivity-in-the-free-radical-chlorination-of-propane-gives-45-to-55-radio-of-1-chloropropane-to-2-chloropropane.gif\" alt=\"selectivity-in-the-free-radical-chlorination-of-propane-gives-45-to-55-radio-of-1-chloropropane-to-2-chloropropane\" width=\"555\" height=\"206\" \/><\/p>\n<p>It&#8217;s possible to use this information to quantify the <b>reactivity\u00a0<\/b>of each type of hydrogen in a molecule. Using the method from<a href=\"https:\/\/www.masterorganicchemistry.com\/2013\/09\/23\/selectivity-in-free-radical-reactions\/\"> this post<\/a>, we can determine that the secondary C-H is 3.66 times more reactive than the primary C\u2013H under these conditions.<span style=\"color: #993366;\"><em> [See Post: <a style=\"color: #993366;\" href=\"https:\/\/www.masterorganicchemistry.com\/2013\/09\/23\/selectivity-in-free-radical-reactions\/\">Selectivity In Free Radical Reactions<\/a>]<\/em><\/span><\/p>\n<h2><b><a id=\"eight\"><\/a>8. Bromination Is More Selective Than Chlorination<br \/>\n<\/b><\/h2>\n<p>Interestingly, bromination of alkanes is even more selective than chlorination for tertiary and secondary C\u2013H bonds. For example in the bromination of propane at 25\u00b0C, the reaction is selective for secondary C\u2013H over primary C\u2013H by a factor of 97 to 1 .<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter wp-image-14623\" src=\"https:\/\/www.masterorganicchemistry.com\/wp-content\/uploads\/2019\/12\/13-bromine-has-much-higher-selectivity-for-secondary-position-of-propane-than-primary.gif\" alt=\"bromine-has-much-higher-selectivity-for-secondary-position-of-propane-than-primary\" width=\"555\" height=\"257\" \/><\/p>\n<p>Why the huge difference? The most rigorous way to look at it is to apply Hammond&#8217;s Postulate: chlorination is exothermic, bromination is endothermic; therefore, chlorination will have a &#8220;reactant-like&#8221; transition state and bromination will have a &#8220;product-like&#8221; transition state. In this case the difference in energies between <span style=\"text-decoration: underline;\">products<\/span> is greater than the difference in energies between\u00a0<span style=\"text-decoration: underline;\">reactants<\/span> ; and since selectivity is determined by differences in energies between transition states, therefore, bromination will be more selective than chlorination. If this sounds complex, <a href=\"https:\/\/www.masterorganicchemistry.com\/2013\/10\/31\/selectivity-in-free-radical-reactions-bromine-vs-chlorine\/\">the full post is here<\/a>. Alternatively, a <a href=\"https:\/\/www.masterorganicchemistry.com\/2013\/11\/13\/halogenation-at-tiffanys\/\">simple analogy<\/a> I use to explain this concept is here. [See Post: <a href=\"https:\/\/www.masterorganicchemistry.com\/2013\/10\/31\/selectivity-in-free-radical-reactions-bromine-vs-chlorine\/\">Selectivity In Free Radical Reactions &#8211; Chlorine vs. Bromine<\/a>]<\/p>\n<h2><strong><a id=\"nine\"><\/a>9. Allylic Bromination,\u00a0 And Allylic Bromination With Allylic Rearrangement<\/strong><\/h2>\n<p>Weak C\u2013H bonds (like tertiary C\u2013H bonds, for instance) are very easily halogenated. Allylic and benzylic C\u2013H bonds are particularly weak (about 86 kcal\/mol) which makes them excellent substrates for bromination reactions. Molecular bromine (Br<sub>2<\/sub>) can be useful for benzylic bromination, but attempts to use this reagent with alkenes like, say, propene, would result in <a href=\"https:\/\/www.masterorganicchemistry.com\/2013\/03\/15\/bromination-of-alkenes-the-mechanism\/\">formation of a dibromide<\/a>.<br \/>\nThe way around this is to use the useful source of bromonium ion, N-bromo succinimide (NBS), which serves as a dilute source of Br<sub>2<\/sub> . This allows replacement of the allylic C\u2013H bond with Br. This process is called &#8220;<a href=\"https:\/\/www.masterorganicchemistry.com\/2013\/11\/25\/allylic-bromination\/\">allylic bromination<\/a>&#8220;.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter wp-image-14624\" src=\"https:\/\/www.masterorganicchemistry.com\/wp-content\/uploads\/2019\/12\/14-allylic-bromination-of-cyclopentene-gives-substitution-adjacent-to-c-c-pi-bond.gif\" alt=\"allylic-bromination-of-cyclopentene-gives-substitution-adjacent-to-c-c-pi-bond\" width=\"505\" height=\"202\" \/><\/p>\n<p>Interestingly, changing the substrate just a little bit can result in some surprising results. The example below differs from the example above by only a single methyl group. However, this can lead to two potential products &#8211; the &#8220;normal&#8221; product of allylic bromination (A) and a closely related product, allylic bromination with allylic rearrangement (B). One factor which can dictate which final product is formed is determining which alkene is more substituted.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter wp-image-14625\" src=\"https:\/\/www.masterorganicchemistry.com\/wp-content\/uploads\/2019\/12\/15-free-radical-substitution-of-3-methylcyclopentene-gives-two-products-one-with-allylic-rearrangement-giving-more-substituted-double-bond.gif\" alt=\"free-radical-substitution-of-3-methylcyclopentene-gives-two-products-one-with-allylic-rearrangement-giving-more-substituted-double-bond\" width=\"705\" height=\"318\" \/><\/p>\n<h2><a id=\"ten\"><\/a>10. Free-Radical Addition Of Radicals To Alkenes<\/h2>\n<p>Finally, it&#8217;s important to note that H\u2013Br can add to alkenes in a completely different fashion depending whether or not &#8220;peroxides&#8221; are present. In the absence of peroxides, H-Br adds through a &#8220;normal&#8221; electrophilic addition [<a href=\"https:\/\/www.masterorganicchemistry.com\/2013\/02\/22\/addition-pattern-1-the-carbocation-pathway\/\">post here<\/a> &#8211; not radical related]. However, if peroxides are added and the mixture is heated, it leads to homolytic cleavage of the peroxide, forming new free radicals (initiation). Peroxide can then remove a hydrogen from H-Br forming the Br\u2022 radical, and then Br\u2022 can add to alkenes. What&#8217;s important to note is that Br\u2022 adds to alkenes in a such a way that we end up with the most stable radical intermediate, which then removes hydrogen from H\u2013Br, regenerating bromine radical. The net result is &#8220;anti-Markovnikov&#8221; addition of H-Br across a double bond, the opposite of what is observed in the absence of peroxides. [<a href=\"https:\/\/www.masterorganicchemistry.com\/2013\/04\/12\/a-fourth-alkene-addition-pattern-free-radical-addition\/\">full mechanism in this post<\/a>]<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter wp-image-14626\" src=\"https:\/\/www.masterorganicchemistry.com\/wp-content\/uploads\/2019\/12\/16-free-radical-addition-of-hbr-to-alkenes-gives-anti-markovnikov-addition-of-hbr-across-the-alkene.gif\" alt=\"free-radical-addition-of-hbr-to-alkenes-gives-anti-markovnikov-addition-of-hbr-across-the-alkene\" width=\"655\" height=\"281\" \/><\/p>\n<p>That does it for free radical reactions (for now).<br \/>\nIn the next series we&#8217;ll start putting a lot of our Org 1 reactions together\u2026 and finally delve into one of my favourite topics &#8211; \u00a0<strong>synthesis.<br \/>\n<\/strong><\/p>\n<hr \/>\n<h2><a id=\"notes\"><\/a>Notes<\/h2>\n<div class=\"related-articles\"><p><strong>Related Articles<\/strong><\/p><ul><li><a href=\"https:\/\/www.masterorganicchemistry.com\/2010\/06\/25\/bond-dissociation-energies-homolytic-cleavage\/\" class=\"\"><span>Bond Dissociation Energies = Homolytic Cleavage<\/span><\/a><\/li><li><a href=\"https:\/\/www.masterorganicchemistry.com\/2013\/07\/30\/free-radical-reactions\/\" class=\"\"><span>Free Radical Reactions<\/span><\/a><\/li><li><a href=\"https:\/\/www.masterorganicchemistry.com\/2013\/08\/02\/3-factors-that-stabilize-free-radicals\/\" class=\"\"><span>3 Factors That Stabilize Free Radicals<\/span><\/a><\/li><li><a href=\"https:\/\/www.masterorganicchemistry.com\/2013\/08\/14\/bond-strengths-radical-stability\/\" class=\"\"><span>Bond Strengths And Radical Stability<\/span><\/a><\/li><li><a href=\"https:\/\/www.masterorganicchemistry.com\/2013\/08\/30\/radical-initiation-why-is-light-or-heat-required\/\" class=\"\"><span>Free Radical Initiation: Why Is \u201cLight\u201d Or \u201cHeat\u201d Required?<\/span><\/a><\/li><li><a href=\"https:\/\/www.masterorganicchemistry.com\/2013\/09\/06\/initiation-propagation-termination\/\" class=\"\"><span>Initiation, Propagation, Termination<\/span><\/a><\/li><li><a href=\"https:\/\/www.masterorganicchemistry.com\/2013\/09\/17\/monochlorination-isomers-from-free-radical-reactions\/\" class=\"\"><span>Monochlorination Products Of Propane, Pentane, And Other Alkanes<\/span><\/a><\/li><li><a href=\"https:\/\/www.masterorganicchemistry.com\/2013\/10\/31\/selectivity-in-free-radical-reactions-bromine-vs-chlorine\/\" class=\"\"><span>Selectivity in Free Radical Reactions: Bromination vs. Chlorination<\/span><\/a><\/li><li><a href=\"https:\/\/www.masterorganicchemistry.com\/2013\/11\/13\/halogenation-at-tiffanys\/\" class=\"\"><span>Halogenation At Tiffany\u2019s<\/span><\/a><\/li><li><a href=\"https:\/\/www.masterorganicchemistry.com\/2013\/11\/25\/allylic-bromination\/\" class=\"\"><span>Allylic Bromination<\/span><\/a><\/li><li><a href=\"https:\/\/www.masterorganicchemistry.com\/2013\/12\/02\/bonus-topic-allylic-rearrangements\/\" class=\"\"><span>Bonus Topic: Allylic Rearrangements<\/span><\/a><\/li><li><a href=\"https:\/\/www.masterorganicchemistry.com\/organic-chemistry-practice-problems\/radicals-practice-quizzes\/\" class=\"\"><span>Free Radicals Practice Quizzes (MOC Membership required)<\/span><\/a><\/li><\/ul><\/div>\n<p><strong><a id=\"noteone\"><\/a>Note 1. <\/strong>Related to the addition of H-Br to alkenes, is an interesting process called &#8220;free radical polymerization&#8221;.<\/p>\n<p>If certain alkenes are heated in the presence of a radical initator (let&#8217;s say &#8220;peroxides&#8221; as a typical example), the peroxy radical can add to the alkene, forming a new free radical on the most substituted position\u2026 and the resulting free radical can add to another alkene\u2026. generating another free radical\u2026. which adds to yet another alkene! &#8211; and so on.<\/p>\n<p>The process can repeat thousands of times before termination occurs. Some of the most common plastics of daily life can be made this way &#8211; such as polyethylene, polypropylene, and polystyrene.<\/p>\n<hr \/>\n<h2><a id=\"quizzes\"><\/a>Quiz Yourself!<\/h2>\n<p><br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"wp-image-36214 aligncenter\" src=\"https:\/\/www.masterorganicchemistry.com\/wp-content\/uploads\/quiz-previews\/3003-Front-Image-Only.png\" alt=\"\" width=\"600\" height=\"450\" \/><\/a><\/p>\n<p style=\"text-align: center;\"><a href=\"https:\/\/www.masterorganicchemistry.com\/moc-membership\/\"><strong>Become a MOC member<\/strong><\/a> to see the clickable quiz with answers on the back.<\/p>\n<p><br \/>\n<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"wp-image-36214 aligncenter\" src=\"https:\/\/www.masterorganicchemistry.com\/wp-content\/uploads\/quiz-previews\/3495-Front-Image-Only.png\" alt=\"\" width=\"600\" height=\"450\" \/><\/a><\/p>\n<p style=\"text-align: center;\"><a href=\"https:\/\/www.masterorganicchemistry.com\/moc-membership\/\"><strong>Become a MOC member<\/strong><\/a> to see the clickable quiz with answers on the back.<\/p>\n<p><br \/>\n<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"wp-image-36214 aligncenter\" src=\"https:\/\/www.masterorganicchemistry.com\/wp-content\/uploads\/quiz-previews\/1166-Front-Image-Only.png\" alt=\"\" width=\"600\" height=\"450\" \/><\/a><\/p>\n<p style=\"text-align: center;\"><a href=\"https:\/\/www.masterorganicchemistry.com\/moc-membership\/\"><strong>Become a MOC member<\/strong><\/a> to see the clickable quiz with answers on the back.<\/p>\n<p><br \/>\n<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"wp-image-36214 aligncenter\" src=\"https:\/\/www.masterorganicchemistry.com\/wp-content\/uploads\/quiz-previews\/3258-Front-Image-Only.png\" alt=\"\" width=\"600\" height=\"450\" \/><\/a><\/p>\n<p style=\"text-align: center;\"><a href=\"https:\/\/www.masterorganicchemistry.com\/moc-membership\/\"><strong>Become a MOC member<\/strong><\/a> to see the clickable quiz with answers on the back.<\/p>\n<p><br \/>\n<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"wp-image-36214 aligncenter\" src=\"https:\/\/www.masterorganicchemistry.com\/wp-content\/uploads\/quiz-previews\/3496-Front-Image-Only.png\" alt=\"\" width=\"600\" height=\"450\" \/><\/a><\/p>\n<p style=\"text-align: center;\"><a href=\"https:\/\/www.masterorganicchemistry.com\/moc-membership\/\"><strong>Become a MOC member<\/strong><\/a> to see the clickable quiz with answers on the back.<\/p>\n<p><br \/>\n<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"wp-image-36214 aligncenter\" src=\"https:\/\/www.masterorganicchemistry.com\/wp-content\/uploads\/quiz-previews\/3439-Front-Image-Only.png\" alt=\"\" width=\"600\" height=\"450\" \/><\/a><\/p>\n<p style=\"text-align: center;\"><a href=\"https:\/\/www.masterorganicchemistry.com\/moc-membership\/\"><strong>Become a MOC member<\/strong><\/a> to see the clickable quiz with answers on the back.<\/p>\n<p><br \/>\n<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"wp-image-36214 aligncenter\" src=\"https:\/\/www.masterorganicchemistry.com\/wp-content\/uploads\/quiz-previews\/3644-Front-Image-Only.png\" alt=\"\" width=\"600\" height=\"450\" \/><\/a><\/p>\n<p style=\"text-align: center;\"><a href=\"https:\/\/www.masterorganicchemistry.com\/moc-membership\/\"><strong>Become a MOC member<\/strong><\/a> to see the clickable quiz with answers on the back.<\/p>\n<p><\/p>\n","protected":false},"excerpt":{"rendered":"<p>A Summary Of Topics\u00a0 For Free-Radical Reactions So what have we learned in this series about free radicals? Lots of things. We&#8217;ve seen that\u00a0 they <\/p>\n","protected":false},"author":1,"featured_media":14611,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"_acf_changed":false,"footnotes":""},"categories":[1411],"tags":[939,170,952,946,709,463,933,261,934],"post_folder":[],"class_list":["post-7801","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-free-radical-reactions","tag-allylic","tag-allylic-bromination","tag-bond-dissociation-energy","tag-free-radical","tag-homolytic","tag-initiation","tag-propagation","tag-radicals","tag-termination"],"acf":[],"yoast_head":"<!-- This site is optimized with the Yoast SEO plugin v27.7 - https:\/\/yoast.com\/product\/yoast-seo-wordpress\/ -->\n<title>In Summary: Free Radicals &#8211; Master Organic Chemistry<\/title>\n<meta name=\"description\" content=\"A summary of key free radical topics including examples of radical reactions, radical stability, mechanisms, selectivity, allylic halogenation &amp; 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