MS 4638 - Sacramento City College Chem 420 Unit 9 Homework Alkenes (II)
Description: Please do the following Mendelset problems, and also all of problems #43-69 in Chapter 11 of Second Language.
Problem # 1336
Determine a synthesis to prepare 2-chloro-4-methylpentane from 1-iodo-4-methylpentane.
Solution 1-iodo-4-methylpentane is reacted with KOt-Bu(potassium tert-butoxide) to afford 4-methyl-1-pentene(E2 Reaction). Next, this product is reacted with HCl to perform a markovnikov addition addition across a double bond and afford the desired product, 2-chloro-4-methylpentane.
Problem Details MendelSet practice problem #
1336 submitted by
reifa on October 26, 2011.
Problem # 334
Let's go over how a carbocation can form from an alkene.
Use curved arrows to show the two carbocations that can from from 1-methylcyclohexene.
Solution 3° carbocations are more stable than 2º carbocations, so only the 3º carbocation is formed.
Problem Details MendelSet practice problem #
334 submitted by
Matt on June 7, 2011.
Problem # 336
Predict the product(s) of the reaction below, and used curved arrows to show a mechanism.
Solution Due to Markovnikov's rule, only the more substituted alkyl halide is formed.
Problem Details MendelSet practice problem #
336 submitted by
Matt on June 7, 2011.
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 # 520
Let's work through an alkene addition reaction. Draw the structures for each of the species in the three boxes below (3º carbocation, protonated thiol, and thiol). Also draw curved arrows to show electron movement. Note: thiol = RSH, like an alcohol, but with sulfur instead of oxygen.
Solution Note that this is the reverse of problem 519 . Instead of going from thiol (alcohol) to carbocation to alkene, we're going from alkene to carbocation to thiol. In each step ask yourself "what arrows can I draw?" and choose the step that doesn't go backwards.
The first step is the alkene picks up a proton to form the more stable carbocation. (See problem 334 if you don't understand why only the 3º carbocation is formed). To get rid of the carbocation, we can either do a beta-elimination (E1) to form an alkene, or an addition reaction (SN 1) to form a protonated thiol. Since forming the alkene would be going backwards, the only choice is addition.
Then you're left with a protonated thiol. How do we get rid of the positive charge on the sulfur? There are two legal moves- either the RSCH2 CH3 acts as a leaving group, or the sulfur deprotonates. Since the RSCH2 CH3 leaving would form a carbocation and take us backwards, the only option is for the sulfur to deprotonate.
Because the sulfur added to the more substituted carbon, this was a Markovnikov addition.
Problem Details MendelSet practice problem #
520 submitted by
Matt on June 30, 2011.
Problem # 521
Let's work through a halogenation reaction. Draw the structures for each of the species in the four boxes below (3º carbocation, halonium ion, protonated thiol, and thiol). Also draw curved arrows to show electron movement. Note: thiol = RSH, like an alcohol, but with sulfur instead of oxygen.
Solution Because the chlorine is more electronegative than iodine, the iodine will have a partial positive charge and will be attacked by the alkene. It forms the more stable carbocation as normal (as in problem 335 ). The nucleophile (HSCH2 CH3 ) will eventually attack the carbocation, but the iodine does something special first- it forms a cyclic intermediate.
The formation of the cyclic halonium ion is probably the most important concept you will learn in first semester organic chemistry. With iodine, it's called an iodonium ion. (with chlorine or bromine, it's called a chloronium or bromonium ion).
The HSCH2 CH3 will attack the carbon that was the carbocation, but it's forced to attack from the side opposite to that of the iodine. So the product will always have anti stereochemistry (you get the trans product).
Problem Details MendelSet practice problem #
521 submitted by
Matt on July 1, 2011.
Problem # 529
Indicate the major organic product of the reaction below. Include stereochemistry.
Solution This is a hydrogenation reaction. The product is an alkane. But this product has two stereocenters.
Normally, the product would be a racemic mixuture of enantiomers (equal amounts of each stereoisomer). But in this case, the starting material is optically active (chiral but not racemic), so the products won't be equal and opposite as usual.
Since there is a substitutent blocking the top face of the molecule (isopropyl is a wedge), the hydrogens will preferentially add to the opposite face of the molecule (dashes).
Problem Details MendelSet practice problem #
529 submitted by
Matt on July 2, 2011.
Problem # 530
Let's work through anti and syn additions to alkenes.
Show the product for each reaction below, and indicate whether the product will be a racemic mixture of enantiomers, or a meso compound (which is achiral).
Solution The reaciton of an alkene with Br2 is an anti -addition, and hydrogenation (H2 or D2 ) is a syn addition. From this we can figure our the relative stereochemistry of each product.
Each starting material is achiral, and therefore not optically active, so the products cannot be optically active.
There are three ways for products not to be optically active:
products can be achiral
products can be a racemic mixture
products can be meso
In both a) and b) each product has a sterocenter, so the products can't be achiral.
In a), one stereocenter is S and the other stereocenter is R, and both with the same substituents, so this is a mirror image relationship, and the products are meso.
In b), the products have no internal mirror planes, so the products are chiral, and must be a racemic mixture.
Problem Details MendelSet practice problem #
530 submitted by
Matt on July 2, 2011.
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