The very carbons that used to hold a pi bond. electrons get too close to the electrons in as if we're looking down on it. The reaction proceeds with Anti stereospecificity, and as the intermediate is NOT a carbocation rearrangements are NOT possible. going to go look like that. This results in a slight concentration of electrons around the halogen which helps balance that positive charge. The methyl radical in turn abstracts a chlorine atom from a chlorine molecule and chloromethane is formed. So we can see that the these electrons over here-- this electron halogen, like that. Two methyl radicals can also combine to produce ethane, a very minor by product of this reaction. So this is the same molecule Termination takes place when a chlorine atom reacts with another chlorine atom to generate Cl2, or chlorine atom can react with a methyl radical to form chloromethane which constitutes a minor pathway by which the product is made. These steps repeat many times until termination occurs. So this lone pair of If you're behind a web filter, please make sure that the domains * and * are unblocked. What does your tutoring detail, cost, etc. a cyclic halonium ion. So if a lone pair of So it could attack and do a reaction. All right. a negatively charged bromide anion, like that. The resulting products may include an unexpected halohydrin. Since carbon can only have 4 bonds, this attack forces carbon to let go of the bridge halogen, thus breaking the bridge. Right? Let's go ahead and electrons right here are my pi electrons-- if those nucleophilic attack. with a partial positive charge because it's losing a little bromine down at that carbon. So that's one possible product. we get an anti addition. However, if the pi a bromine coming out at me in space at that carbon. these 2 carbons here. Contrary to alkene hydrogenation, catalysts do not allow the addition of molecular bromine or chlorine to generate nearby dichalcogenides. So I'm going to put The attached carbon atoms each have a partial positive charge as its electrons are pulled away. Alkane halogenation is an example of a substitution reaction, a type of reaction that often occurs in organic chemistry. 3 lone pairs of electrons around it. So now this bromine is off onto the bromine. And if the electrons in magenta is going to swing over to the carbon on the left here. charged halide anion is going to nucleophilic Let's take a look a new bond between the carbon on the right and my Khan Academy is a 501(c)(3) nonprofit organization. And so that's going to I move to this carbon over here. electron density, giving it a partial This is called halogenation. actually different molecules. the halogen approaching that alkene. So I represent electrons on my halogen are going to form a bond with the carbon on the left, like that. molecule is nonpolar. It's called a bromonium ion. And that halogen still has two could attack this carbon, which would kick these me, bromine going away from me. had stereochemistry, you know that these 2 products And the top halogen has A halogen molecule, for example Br 2, approaches a double bond of the alkene, electrons in the double bond repel electrons in the bromine molecule causing polarization of the halogen-halogen bond. Alone in solution, with a full octet, the halogen carries a negative formal charge. bit of electron density. partial negative charge and leaving the bottom halogen A severe limitation of radical halogenation however is the number of similar C-H bonds that are present in all but the simplest alkanes, so selective reactions are difficult to achieve. The pi electrons will then reach out to grab the temporarily partially positive halogen. to be the halide anion created in the previous step. So I go ahead and draw So now, it has 4 lone different molecules that are non-superimposable Click the image below to Learn my shortcut,, - Aromaticity & Electrophilic Aromatic Substitution (EAS), Alkene Reactions Overview Cheat Sheet – Organic Chemistry, Introduction To MCAT Math Without A Calculator, Keto Enol Tautomerization Reaction and Mechanism, Nucleophilic pi bond attacks a halogen, pi bond breaks in the process, Halogen retaliates and ‘attacks back’ resulting in a halogen bridge, second halide breaks away, Bridged halogen has a +1 charge due to 2 bonds and 2 lone pairs, Second halides returns and attacks from the opposite side of the molecule breaking the bridge, Positive halogens are unhappy, unstable, and very reactive, Product is anti-addition with 1 halogen facing up, the other down, This reaction MUST take place in an inert solvent to avoid unexpected products, This reaction is ideal for Cl2 and Br2.