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Chapter 8-3 - Addition of Hydrogen Halides to Alkenes
Organic Chemistry 307: Chapter 8 - Reactions of Alkenes; Section 3 - Addition of Hydrogen Halides to Alkenes
Organic Chemistry
Undergraduate 3

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The first step in the addition of HBr to an alkene such as 2-butene is __________ of the double bond.
When the proton adds to the __________ carbon, a tertiary carbocation results.
When the proton adds to the __________ carbon atom, a secondary carbocation results.
Overall, the __________ carbocation is more stable.
1) Protonation
2) Secondary
3) Tertiary
4) Tertiary
Note that protonation of one carbon atom of a double bond gives a carbocation on the carbon that was not __________.
Therefore, the proton adds to the end of the double bond that is (less/more) substituted to give the (less/more) substituted carbocation (the more stable carbocation).
1) Protonated
2) Less
3) More
*Mechanism 8-2: Ionic Addition of HX to an Alkene*
Step 1: Protonation of the __________ bond forms a carbocation
Step 2: Attack by the __________ ion gives the addition product.
1) Pi
2) Halide
The addition of HBr (and other hydrogen halides) to an alkene is said to be __________ because in each case, one of the two possible orientations of addition results preferentially over the other.
1) Regioselective
*__________ __________ states that the addition of a proton acid to the double bond of an alkene results in a product with the acid proton bonded to the carbon atom that already holds the greater number of hydrogen atoms.*
1) Markovnikov's Rule
Reactions that follow Markkovnikov's rule are said to follow __________ __________ and give the __________ __________.
1) Markovnikov orientation
2) Markovnikov product
Markovnikov's rule can be extended to include a wide variety of other additions; the extended version of Markovnikov's rule states that in an __________ addition to an alkene, the __________ adds in such a way as to generate the most stable intermediate.
1) Electrophilic
2) Electrophile
__________-__________ reactions are those that occur in the presence of peroxides (and free radicals).
1) Anti-Markovnikov
__________ gives rise to free radicals that initiate __________, causing it to occur by a radical mechanism.
The oxygen-oxygen bond in __________ is rather __________, so it can break to give two __________ radicals.
1) Peroxides
2) Addition
3) Peroxides
4) Weak
5) Alkoxy
__________ radicals (R-O) initiate the __________-__________ addition of HBr.
1) Alkoxy
2) Anti-Markovnikov
*Mechanism 8-3: Free-Radical Addition of HBr to Alkenes*
Initiation: Formation of _________
Propogation: A __________ reacts to generate another radical.
Step 1: A __________ radical adds to the double bond to generate an alkyl radical on the (less/more) substituted carbon atom.
Step 2: The alkyl radical abstracts a __________ atom from HBr to generate the product and a __________ radical.
1) Radicals
2) Bromine
3) More
4) Hydrogen
5) Bromine
Analyzing Free-Radical Addition of HBr More Closely:
In the __________ step, free radical generated form the __________ react with HBr to form bromine radicals.
The bromine radical lacks an __________ of electrons in its valence shell, making it electron-deficient and __________. It adds to a double bond, forming a new free radical with the odd electron on a __________ atom.
This free radical reacts with an __________ molecule to form a C-H bond and generate another bromine radical.
Both of these propagation steps are moderately __________, allowing them to proceed faster than the __________ steps.
The number of __________ __________ throughout the reaction is constant until the reactants are consumed.
1) Initiation
2) Peroxide
3) Octet
4) Electrophilic
5) Carbon
6) HBr
7) Exothermic
8) Termination
9) Free radicals
With an unsymmetrical alkene such as 2-methyl-2-butene, adding the __________ radical to the __________ end of the double bond forms a tertiary radical.
As in the protonation of an alkene, the electrophile (the __________ radical) adds to the (less/more) substituted end of the double bond, and the unpaired electron appears on the (less/more) substituted carbon to give the most stable free radical.
This intermediate reacts with HBr to give the __________-__________ product, in which hydrogen has added to the (less/more) substituted end of the double bond; the end that began with (fewer/greater) hydrogens.
1) Bromine
2) Secondary
3) Bromine
4) Less
5) More
6) Anti-Markovnikov
7) More
8) Fewer
Both mechanisms for the __________ of HBr to an alkene (with and without peroxides) follow the extended statement of the __________ __________: in both cases, an electrophile adds to the (less/more) substituted end of the double bond to give the (less/more) stable intermediate, either a carbocation or a free radical.
In the ionic reaction, the electrophile is __________; in the peroxide-catalyzed free-radical reaction, the electrophile is __________.
1) Addition
2) Markovnikov's Rule
3) Less
4) More
5) H+
6) Bromine radical
A __________ adds to the double bond to give the most stable radical in the intermediate; radicals, thus, follow this stability rule:
__________ > __________ > __________ > __________
1) Radical
2) 3° > 2° > 1° > °CH3
During free radical addition of HBr to alkenes, if a small amount of peroxide is present, a mixture of __________ and __________-__________ products result; if an appreciable amount is present, the reaction will occur at such a high rate that only the __________-__________ product will be observed.
1) Markovnikov
2) Anti-Markovnikov
3) Anti-Markovnikov
The reversal of orientation in the presence of peroxides is called the __________ __________ and only occurs with the addition of __________ to alkenes.
The effect is not seen with other hydrogen halides such as HCl because the reaction of an alkyl radical with HCl is strongly __________ (same with HI).
1) Peroxide effect
2) HBr
3) Endothermic
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