Sapling Ch. 4 #1

For my question I chose a reaction between and alkene and H-Br. The double bond attacks the H and breaks the bond between H and Br. The hydrogen will bind with one of the carbons, forcing the other to become a carbocation, but the Br- will bind with the carbocation making it neutral. The problem is shown below.

Chapter 4 #47

This photo shows the solution for Chapter 4, #47.  The reactions start as a cyclohexane with an alkene substituent with 8 different reagents resulting in 8 different products. The reactions cover a dibromo substitution, a dehydration reaction with H2, a primary alcohol reaction via hydroboration, and a secondary alcohol formed via H2SO4 and H2O, and ether formation, a bromo alcohol formation, a hydrogen bromonation and epoxide formation.  The reaction is a great example of materials used in organic synthesis. 

Sapling Chapter 4 Question Number 1

This question shows a reaction between an alkene and HBr. The double bond will first attack the hydrogen which will then cause the bond between hydrogen and bromine to attack the bromine. The hydrogen will add to one side of the bond leaving a carbocation at the other side. The negative bromine ion will then attack the carbocation to create a neutral structure. The visual is seen below.

Ch.7 #39

Ch.7 #39 on page 321.
Explain why the following products are not optically active.
a. the product obtained from the reaction of 1,3-butadiene with cis-1,2 dichloroethene.
b. the product obtained from the reaction of 1,3-butadiene with trans-1,2-dichloroethene.

The product from a is not optically active because it is a meso compound.
The syn addition to trans-isomer gives threo enantiomeric mixture, therfore the product from b is also not optically active.

Ch 9: Question 56 (p. 409)

Given the conditions, these reactions will be both SN1 and E1 reactions, meaning they will happen in more than one step, and form carbocations. The first step is the same in each of the products, making a carbocation where the Br was and a Br ion. For the first set of products, we then add CH3OH to the carbocation, making the SN1 and E1 product as shown. In the second set of products, there is a 1,2 alkyl shift making the cyclohexane a cyclopentane and moving a carbon up. This shifts the carbocation. From there the CH3OH is added and makes the SN1 and E1 products shown. In the third set of products, there is a 1,2 methyl shift, where the cyclohexane remains a cyclohexane, but one of the methanes is shifted. This too moves the carbocation, making the SN1 and E1 products shown when CH3OH is added.

chapter 4 #60 p194

In this problem, you have to add an alkyl halide, specifically HBr, to an alkene. As we all know, when an alkylhalide adds to an alkene, a carbocation intermediate forms which means you have to look out for possible rearrangements such as a methyl shift or a hydride shift. The reaction is regioselective.

The top problem, however, is not regioselective; it will give both products, not one is formed over the other because both have the same stability.

In the second problem, the reaction is regioselective and we see a rearrangement. Without a rearrangement, the Br would have added to carbon 2, a secondary carbocation. A rearrangement is possible, a hydride shift. By moving a hydrogen atom from carbon 3 to carbon 2, you will get a tertiary carbocation on carbon 3, which is more stable than a secondary carbocation. As previously stated, the reaction is regioselective and wants to be as stable as possible. The hydride shift and addition of Br to carbon 3 and H to carbon 1 gives the most stable product, the major product, which is shown.

The third problem is similar to the second one, however instead of a hydride shift, a methyl shift is seen. A methyl group moves from carbon 3 to carbon 2, just as the hydride did in the previous problem. Once again, this occurs to give the most stable carbocation for the halide to add to and, in turn, the most stable product, or the major product. 

The final problem shows a hydride shift as well. This is to get the most stable carbocation for the halide to add to. So once H adds to the less substituted carbon of the double bond, a hydride ion moves from carbon 3 to carbon 2 to form a tertiary carbocation on carbon 3 for the Br to add on to. 

Chapter 4 Number 43

 

For the first part we react the alkene with water and Sulfuric acid. We add an alcohol to the second carbon of the double bond. 1,2-hydride shift is performed to get a tertiary alcohol which is more stable than the secondary. For the second part we go through oxymercuration-reduction so we all the alcohol to the second carbon of the double bond. There is no hydride shift because there is no carbocation rearrangement. The third part is hydroboration-oxidation. For that we add the alcohol to the least stable carbon, that's why its added to the first carbon, which is primary, instead of the second carbon.