Tuesday, December 10, 2013

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.

Ch. 9 #9 from the text

Sapling Chapter 9-Q29

Synthesis of pentanal from acetylene

      The first step in synthesising pentanal from acetylene is forming the acetylide ion by the addition of NaNH2. The hydrogen atom reacts with the amide ion in NaNHto form HCC- and ammonia.The carbon in the acetylide ion is a good nucleophile and reacts with 1-chloropropane in an Sn2 reaction. The chlorine atom in 1-chloropropane takes the electrons from the C-Cl bond forming a positively charged carbon atom on the hydrocarbon chain and a chloride ion. The carbanion attacks the electrophilic carbon forming 1-butyne. 1-butyne then reacts with BH3/THF. The reaction with BHis anti-Markovnikov and a syn hydration forming 1-penten-1-ol with mechanism shown below. This molecule then tautomerises to form pentanal.

R,S Configuration Chapter 5 Problem 26

Page 213:

Draw the stereoisomers of the following amino acids. Indicate pairs of enantiomers and pairs of diastereomers.

Pictured it has leucine and isoleucine as the structures to draw stereoisomers of. I am going to focus of the first part of the problem, labeling R and S of just the second molecule, isoleucine. In the solution there are four options of stereoisomers that have two chiral centers, such as in isoleucine. There are the R R, S S, R S, and S R configuation options. Leucine, with only one chiral center has only one sterioisomer, its enantiomer.

I mostly wanted to do this problem to show my "Idiot's" way to determine R and S. First thing is to be able to decipher which, R or S, is clockwise. R is easy to remember because when you turn a car Right, the steering wheel turns clockwise. If for some reason I forget that I remember S is counterclockwise because if you start an arror from the top tip of the line that makes the S and draw arrow barbs pointing downward midway down the S the arrow curves around in a counter clockwise fashion. That would be much easier to understand if drawn out...

Now, as far as determining configuration. Rather than spatially rearranging to get the lowest priority group in the back before finding the configuration of the first, second, and then third priority groups I use a rule a of enantiomers. An enantiomer is the opposite configuration of another molecule which occurs when two of its groups are switched. That noted, if the lowest priority group isn't in the back you can still determine the configuration, just make sure you switch it. If you have a Hyrodgen as a substituent off of a chiral center and is not in the back and you determine the first, second, and third priority groups turn clockwise, R fashion, then it has the S configuration.


Monday, December 9, 2013

Sapling Chapter 2 #40 Newman Projection

The Newman Projection on the left displayed the stereochemistry of a conformer. The Newman projection in this problem displayed a staggered formation, which is the most stable compared to the eclipsed form. There are two major carbons here. One that holds the halogen Br, I, and Cl and the other F and hydroxide and a hydrogen. If we look at the stereochemistry carefully, the wedge-and-dash drawing, we know that Br is trans to H, OH is trans to Cl, and F is trans to I. This labeled them as opposite to each other. The back carbon cannot be seen, but it has the Br, Cl, and I; therefore only the front-facing carbon can be seen through the Newman projection.   

Sapling chapter 13 #2

To determine the molecular ion in the mass spectrum we look at the highest m/z. In this case the highest is 100 m/z. This information is used to determine the molecular formula for the molecule. The other m/z are also used to help determine the formula however the total for the entire molecule is always the highest m/z.

Detailed solution for a ring expansion from Bruice (Problem 9 Chapter 9)

Detailed solution for a problem from Bruice (Problem 31 D)

Sapling Chapter 7 #22

Chapter 9 #27


The synthesis reaction I chose is pictured below. As well as following the picture I will explain what happens with each reagent.
1. The addition of H2SO4 and heat eliminates the alcohol group as well as a hydrogen from the hexane ring, forming a double bond between carbons 1 and 2.
2. The addition of Br2/H2O leads to the addition of a bromide and an alcohol on either side of the double bond.
3. The addition of NaOH pulls the H away from the alcohol (OH) group and breaks the bond with Br. This leads to intramolecular reactions between the now O- and carbon 2 which is now positive.  The O- and C+ form  a bond and this creates an epoxide.

Thank you!
Katilyn Wiggins

Chapter 3, problem 18

Which of the following are electrophiles and which nucleophiles

a) H-
b) CH3O-
c) CH3C(triple bond) CH
d) CH3C+HCH3
e) NH3

Electrophiles: d
Nucleophiles; a,b,c,e

Electrophiles are electron loving compounds, they are always looking for a pair of electrons. Electrophiles usually have a positive charge, which is why d is the correct answer

Nucleophile , Electron rich. It has electrons to spare. Most nucleophiles have lone pairs or a negative charge.

Sapling: Chapter 8, Problem 22

In this problem, 2-chloro-3-dimethyl-pentane is treated with H2O and -OH to show the results of both an SN1 and SN2 reactions. SN1 reactions react with weak bases, so the compound treated with H2O will cause SN1. SN2 happen in presence of strong bases so -OH will cause an SN1 reaction. First in the SN1 reaction, unlike an SN2 reaction, a carbocation can form. Because of this, when the OH goes to substitute for the Cl, the positive charge on the second carbon is moved to the third carbon with a methyl shift from 3 to 2 (moves from secondary to tertiary to form most stable carbocation). The OH is formed on the third carbon to form 2-methyl-3-methyl-3-pentanol. The SN2 reaction with -OH doesn not have a carbocation, so the OH will form on the secondary second carbon. However, SN2 reactions invert the stereochemistry on the carbon it attaches to (it has a back side attack and comes in on the side opposite the leaving group). because of this, the OH is shown attached at the bottom of the C below where the Cl left.

 I'm really sorry this isn't better drawn.

Chapter 3, problem 16

I hope no one has already done this one already -

Draw the isomers for the following compounds, and name each one:
A)  2-methyl-2,4-hexadiene
B)  2,4-heptadiene
C)  1,3-pentadiene

Ok, so before we start tackling this problems, I would like to say that it is important to take your time when figuring these out.  I always find that breaking down each part of the name and working backwards is the best way to go, as you'll see in my explanations...

A)  2-methyl-2,4-hexadiene

  Soooo, let's break it down -->   2-methyl-2,4-hexadiene ... ene means I have a double bond and di tells me I have tow of them, hexa tells me that I will have 6 of something (usually Carbon), 2,4- tells me where the 'enes' will be (ene means that I have an SP2 - a RC=CR), and 2-methyl, tells me I have a methyl on carbon two.  Sooo, we start on the left, and draw 6 C, then add our 'ene' (double bond) to Carbons 2 and 4.

B)  2,4-heptadiene

We continue with the same process --> 2,4-heptadiene...  ene means I have a double bond and di tells me I have tow of them, hepta tells me that I will have 7 of something (usually Carbon), and 2,4 tells me where I put my two 'enes'.

C)  1,3-pentadiene

Again, same thing... 1,3-pentadiene --> ene means double bond, di means I have two of them, penta means I have 5 of something (Usually Carbon), and 1,3 tells me where my two (di) 'enes' go.