MS 903 - Carbocation and Alkene Review Drills
Submitted by
Matt on August 6, 2011.
Textbook and Chapters: Carey and Giuliano 8th Ed. (2010) , Chapters 4, 5, 6
Keywords: alkene addition, carbocation
Description: Identify the intermediates (carbocation, radical, borane intermediate, etc.) and products for important reactions dealing with alkenes. Good review for an orgo1 midterm.
Problem # 342
For the reaction below, draw the structures of the carbocation intermediate and the final product.
Solution This is a acid-catalyzed hydration of an alkene, which is a Markovnikov addition. The product is an alcohol.
Problem Details MendelSet practice problem #
342 submitted by
Matt on June 7, 2011.
Problem # 343
For the reaction below, draw the structures of the chloronium ion intermediate and the final product.
Solution Addition of a chlorine, bromine, or iodine to an alkene goes through a halonium ion intermediate, which results in an anti-addition (trans product).
Problem Details MendelSet practice problem #
343 submitted by
Matt on June 7, 2011.
Problem # 344
For the reaction below, draw the structures of the borane intermediate and the final product.
Solution This is a hydroboration reaction (also called hydroboration-oxidation in some textbooks). The product is an alcohol.
Boron bonds to the less substituted carbon, and is replaced with an oxygen after the borane intermediate is treated with peroxide, resulting in an anti-Markovnikov addition.
Note that one equivalent of BH3 reacts with three equivalents of alkene; the borane intermediate is not BH2 (alkyl) but rather B(alkyl)3 .
Problem Details MendelSet practice problem #
344 submitted by
Matt on June 7, 2011.
Problem # 345
For the reaction below, draw the structures of the radical intermediate and the final product.
Solution When HBr reacts with in an alkene in the presence of peroxides (which are free radical initiators), the reaction goes through a free radical mechanism and adds in an anti-Markovnikov manner. The product is an alkyl bromide.
The reaction goes anti-Markovnikov because the more substituted radical intermediate is only formed if the bromine adds to the less substituted carbon.
Problem Details MendelSet practice problem #
345 submitted by
Matt on June 7, 2011.
Problem # 346
For the reaction below, draw the structures of the radical intermediate and the final product.
Solution This is a free radical halogenation. The hint is that you see light (hν) in the reaction arrow, indicating radicals are involved.
The more substituted the carbon atom, the more stable the corresponding radical, so the major product of this reaction is the 3º alkyl halide.
Problem Details MendelSet practice problem #
346 submitted by
Matt on June 7, 2011.
Problem # 347
For the reaction below, draw the structures of the carbocation intermediate and the final product.
Solution This is a substitution reaction. Because the reaction takes place in acid and the leaving group is on a 2º carbon, it will probably form a carbocation (SN 1 mechanism).
- OH is a poor leaving group so the alcohol will protonate first, so it can leave as H2 O, which is neutral.
Problem Details MendelSet practice problem #
347 submitted by
Matt on June 7, 2011.
Problem # 348
For the reaction below, draw the structures of the carbocation intermediate and the final product.
Solution The delta (Δ) in the reaciton arrows means that heat is being added to this reaction, which tends to favor elimination over substitution. Also, the reaction is using a non-nucleophilic acid (H2 SO4 ), which tends to favor elimination reactions (H3 PO4 is another common reagent for E1 reactions, while HCl or HBr tend to go SN 1).
Because this reaction is taking place in acid, a carbocation is likely to form, so this is an E1 reaction. Since water is lost over the course of the reaction, this is a dehydration, which is a type of elimination reaction.
Problem Details MendelSet practice problem #
348 submitted by
Matt on June 7, 2011.
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