Term
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. |
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Definition
1) Protonation
2) Secondary
3) Tertiary
4) Tertiary |
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Term
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). |
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Definition
1) Protonated
2) Less
3) More |
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Term
*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. |
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Definition
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Term
| 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. |
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Definition
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Term
| *__________ __________ 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.* |
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Definition
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Term
| Reactions that follow Markkovnikov's rule are said to follow __________ __________ and give the __________ __________. |
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Definition
1) Markovnikov orientation
2) Markovnikov product |
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Term
| 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. |
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Definition
1) Electrophilic
2) Electrophile |
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Term
| __________-__________ reactions are those that occur in the presence of peroxides (and free radicals). |
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Definition
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Term
__________ 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. |
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Definition
1) Peroxides
2) Addition
3) Peroxides
4) Weak
5) Alkoxy |
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Term
| __________ radicals (R-O) initiate the __________-__________ addition of HBr. |
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Definition
1) Alkoxy
2) Anti-Markovnikov |
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Term
*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. |
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Definition
1) Radicals
2) Bromine
3) More
4) Hydrogen
5) Bromine |
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Term
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. |
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Definition
1) Initiation
2) Peroxide
3) Octet
4) Electrophilic
5) Carbon
6) HBr
7) Exothermic
8) Termination
9) Free radicals |
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Term
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. |
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Definition
1) Bromine
2) Secondary
3) Bromine
4) Less
5) More
6) Anti-Markovnikov
7) More
8) Fewer |
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Term
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 __________. |
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Definition
1) Addition
2) Markovnikov's Rule
3) Less
4) More
5) H+
6) Bromine radical |
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Term
A __________ adds to the double bond to give the most stable radical in the intermediate; radicals, thus, follow this stability rule:
__________ > __________ > __________ > __________ |
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Definition
1) Radical
2) 3° > 2° > 1° > °CH3 |
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Term
| 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. |
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Definition
1) Markovnikov
2) Anti-Markovnikov
3) Anti-Markovnikov |
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Term
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). |
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Definition
1) Peroxide effect
2) HBr
3) Endothermic |
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