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Determine a synthesis to prepare 2-chloro-4-methylpentane from 1-iodo-4-methylpentane.
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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.
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.
3° carbocations are more stable than 2º carbocations, so only the 3º carbocation is formed.
Carbocations aren't very stable and so don't last very long after they are formed.
Use curved arrows to show:
a) how a carbocation reacts with a halide ions to form an alkyl halide.
b) how a carbocation reacts with water to form an alcohol.
c) how a carbocation reacts with a base to form an alkene.
For a), the product is neutral and so you are done after the nucleophile (Cl-) attacks the carbocation.
For b), the the intermediate is a protonated alcohol, and so you must do a proton exchange step (also called a hydrogen exchange or deprotonation) to get the final alcohol product, which is neutral.
For c), there are two different types of beta hydrogens, and so two different alkenes are possible.
The more stable alkene is the one that will form, and this will always be the most highly substituted alkene. This is Zaitsev's rule. The rationale for this is hyperconjugation: neighboring carbon atoms stabilize an alkene.
Predict the product(s) of the reaction below, and used curved arrows to show a mechanism.
Due to Markovnikov's rule, only the more substituted alkyl halide is formed.
For the reaction below, draw the structures of the carbocation intermediate and the final product.
This is a acid-catalyzed hydration of an alkene, which is a Markovnikov addition. The product is an alcohol.
For the reaction below, draw the structures of the chloronium ion intermediate and the final product.
Addition of a chlorine, bromine, or iodine to an alkene goes through a halonium ion intermediate, which results in an anti-addition (trans product).
For the reaction below, draw the structures of the borane intermediate and the final product.
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.
For the reaction below, draw the structures of the radical intermediate and the final product.
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.
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.
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 (SN1) 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 RSCH2CH3 acts as a leaving group, or the sulfur deprotonates. Since the RSCH2CH3 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.
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.
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 (HSCH2CH3) 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 HSCH2CH3 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).
Indicate the major organic product of the reaction below. Include stereochemistry.
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).
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:
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.
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.
Total Problems: 7
Keywords: carbocation, carbocation formation, carbocation rearrangement
Description: This mendel set guides you through everything you have to know about carbocations:
Also includes some practice problems: addition to an alkene, dehydration (E1), and substitution (SN1).
Total Problems: 8
Textbook and Chapter: Carey and Giuliano 8th Ed. (2010), Chapter 6
Keywords: alkene addition, carbocation, dehydration, E1 mechanism
Description: This mendel set is a complement to Carbocation Drills. It is meant to prepare students on how to approach longer and more complicated mechanisms. Reaction mechanisms covered:
Total Problems: 4
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