{"id":50,"date":"2018-11-19T20:16:42","date_gmt":"2018-11-19T20:16:42","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/suny-potsdam-organicchemistry2\/chapter\/10-4-simple-addition-to-alkenes\/"},"modified":"2018-11-19T20:16:42","modified_gmt":"2018-11-19T20:16:42","slug":"10-4-simple-addition-to-alkenes","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/suny-potsdam-organicchemistry2\/chapter\/10-4-simple-addition-to-alkenes\/","title":{"raw":"10.4. Simple addition to alkenes","rendered":"10.4. Simple addition to alkenes"},"content":{"raw":"\n
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Alkenes undergo addition reactions. Often, they add a proton to one end of the double bond and another group to the other end. These reactions happen in slightly different ways, however.<\/p>\n

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Alkenes are reactive because they have a high-lying pair of \u03c0-bonding electrons. These electrons are loosely held, being high in energy compared to \u03c3-bonds. The fact that they are not located between the carbon nuclei, but are found above and below the plane of the double bond, also makes these electrons more accessible.<\/p>\n

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Electrophilic addition of hydrogen halides (H-X) to alkenes<\/h1>\n

Hydrogen halides such as H-Br and H-Cl are suitable electrophiles for a simple addition to an alkene.  Since the halogen is much more electronegative than the hydrogen, the H-X bond is quite polarized, with the H carrying a partial positive charge (\u03b4+) and serving as the electrophilic atom.  In this example, HCl adds to but-1-ene to form racemic 2-chlorobutane:<\/p>\n

\"\"<\/p>\n

The mechanism begins with an electrophilic addition elementary step, where the alkene forms a new sigma bond to the electrophilic H, and breaks the H-Cl bond displacing Cl-.  The second elementary step is a coordination step, where the chloride ion attacks the carbocation to form a second sigma bond.  Note that overall we have lost one pi bond, but gained a sigma bond.  Notice that Markovnikov's Rule is followed here, since the secondary carbocation formed is more stable than the alternative, which would have been primary.<\/p>\n

\"\"<\/p>\n

Video on addition of HX to alkenes<\/h4>\n

\"\"<\/p>\n[embed]https:\/\/www.youtube.com\/watch?v=vH--iR5jwSk[\/embed]\n

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Problem EA1.1.<\/h4>\n

The reaction of 2-methylpropene (or isobutylene) with HBr, as depicted above, is really a 2-step process.  Draw this mechanism again and in each of the two steps label both the nucleophile and the electrophile (so, that's four labels).  Name the two elementary steps.<\/p>\n<\/div>\n

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Problem EA1.2.<\/h4>\n

Draw a reaction progress diagram for the reaction of 2-methylpropene with hydrogen bromide.<\/p>\n<\/div>\n

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Problem EA1.3.<\/h4>\n

Predict the rate law for the reaction of 2-methylpropene with hydrogen bromide.<\/p>\n<\/div>\n

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Hydration of alkenes<\/h1>\n

Water can be added to an alkene in the presence of an acid catalyst, to produce an alcohol.<\/p>\n

\"\"<\/p>\n

Water is not a strong enough electrophile to add an H+<\/sup> directly to an alkene, but H3<\/sub>O+<\/sup> is.  The H3<\/sub>O+<\/sup> is. formed when sulfuric acid dissociates in water:<\/p>\n

\"\"<\/p>\n

The alkene then reacts with the H3<\/sub>O+<\/sup> in the electrophilic addition elementary step.  The resultant carbocation is reactive, and water then easily reacts as a nucleophile to perform the coordination step.  A final acid-base step, regenerating the H3<\/sub>O+<\/sup> catalyst, produces the alcohol product.<\/p>\n

\"\"<\/p>\n

Khan Academy video on hydration of alkenes:<\/p>\n

\"\"<\/p>\n<\/div>\n<\/div>\n

\n[embed]https:\/\/www.youtube.com\/watch?v=dJhxphep_gY[\/embed]\n<\/div>\n<\/section>\n\n","rendered":"
\n
\n

Alkenes undergo addition reactions. Often, they add a proton to one end of the double bond and another group to the other end. These reactions happen in slightly different ways, however.<\/p>\n

\"\"<\/a>\n<\/p>\n

Alkenes are reactive because they have a high-lying pair of \u03c0-bonding electrons. These electrons are loosely held, being high in energy compared to \u03c3-bonds. The fact that they are not located between the carbon nuclei, but are found above and below the plane of the double bond, also makes these electrons more accessible.<\/p>\n

\"\"<\/a>\n<\/p>\n

Electrophilic addition of hydrogen halides (H-X) to alkenes<\/h1>\n

Hydrogen halides such as H-Br and H-Cl are suitable electrophiles for a simple addition to an alkene.  Since the halogen is much more electronegative than the hydrogen, the H-X bond is quite polarized, with the H carrying a partial positive charge (\u03b4+) and serving as the electrophilic atom.  In this example, HCl adds to but-1-ene to form racemic 2-chlorobutane:<\/p>\n

\"\"<\/p>\n

The mechanism begins with an electrophilic addition elementary step, where the alkene forms a new sigma bond to the electrophilic H, and breaks the H-Cl bond displacing Cl-.  The second elementary step is a coordination step, where the chloride ion attacks the carbocation to form a second sigma bond.  Note that overall we have lost one pi bond, but gained a sigma bond.  Notice that Markovnikov’s Rule is followed here, since the secondary carbocation formed is more stable than the alternative, which would have been primary.<\/p>\n

\"\"<\/p>\n

Video on addition of HX to alkenes<\/h4>\n

\"\"<\/p>\n