SI week 2 answers; Activating alcohols and ethers and epoixides PDF

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The activation of alcohols ...

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Review: 1. Lets do this first for reviewz:

2. More review lol:

Spring 2020 pweaze

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SI

Spring 2020 pweaze

3. Last one I promise but they’re important:

4. Jk lmao:

5. Which of the following reactions is expected to produce 1-propanol (CH3CH2CH2OH) at a faster rate?

Explain: DMSO is a polar aprotic solvent. With this being the case and us reacting a strong base/ strong nucleophile with a 1 prime carbon, we can be sure this will act as SN1. SN1 reactions may be hindered by the possible protons donated by a protic solvent. An E2 reaction would also benefit from this. SN1 and E1 would work best with the stability coming from the polarity of a polar protic solvent, balancing the carbocation. 6. Which of the following reactions is expected to exhibit a faster rate?

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Iodine is much larger and polarizable. With this in mind it will be a better nucleophile, causing this SN1 reaction to occur quicker. It is located farther down on the periodic table, meaning it is larger. 7. Arrange each set of isomeric alkenes in order of stability.

The more substituted an alkene, so the more carbon molecules the double bond is directly attached to, the more stable. 8. Draw the expected products of this reaction. This is a tertiary benzylic stabilized carbon. Because we are dealing with a weak nucleophile/base we know it will go either SN1 or E1. With that in mind there is heat which favors elimination, making that the major product and Sn1 minor.

9. Complete the following reactions by reacting the indicated molecule with a nucleophile. Remember that S and R indicate the configuration of the stereochemistry. a.

S)-2-chloropentane and NaSH

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b.

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(R)-3-iodohexane and NaCl

R configuration shows clockwise while S is counter. We move from the largest to the smallest group around the carbon with the stereochemistry. When doing these SN2 reactions we will see inversion of configuration.

10. Major and minor products please.

First is major. It is more substituted. We gotta remember that when doing an E2 reaction, which this is, we have to have a hydrogen antiperiplanar to the leaving group. 11. Please sir may I have some major and minor product?

Major is correct because t-BuOK is a big bulky base, meaning we will get the Hoffman product in this reaction. Meaning zaitzev is the minor product.

Making alcohols good leaving groups 12. What are Sulfonate esters used for and what are their value when being used compared to other activating molecules.

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13. Provide the reagents:

We can either maintain configuration or we can flip it right away. When the OH maintained becomes the leaving group so we need the proper nucleophile to later come in. 14. What duh reagents? Also in this situation, why can we use the reagents we use without worry?

This is not a chiral center so we need not worry about stereochemistry. With that being said, this reaction would likely go through a sn1 reaction because we are at a tertiary carbon. 15. Reagents please:

Again, we gotta look at our configuration. We have to use these reactions because they are always going to go SN2 for SOCL2 and the TsCl group will maintain the configuration. 16. Predict the product here and explain why we use sulfuric and phosphoric acid for these reactions.

When reacting an alcohol with either concentrated sulfuric or phosphoric acid we will protonate the alcohol and the week conjugate base will act basically in an E1 reaction here. Again, this is different to

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halogen based acids because those conjugate bases will act nucleophilically while these will work just as bases.

Ethers and epoxides: 17. Perform the ether reaction. Remember the rule that determines what type of reaction is performed.

There is excess and in this instance we are unable to form a carbocation intermediate, eliminating the ability to go sn1 and making this an sn2 reaction at both sites. 18. Predict the product: decide whether this ring opening is acid catalyzed and then choose where to attack with the nucleophile.

Not acid catalyzed so when we attack its at the less substituted and then secondly the water protonates the later Oxygen. 19. Predict the product for this reaction:

We cant attack at an SP2 carbon, meaning there is no actual attack on the benzene. We then only attack with the bromine on the side at the right. 20. Give duh product plez:

Here we have a reaction where it would be possible to form a carbocation intermediate. This means the bromine would appear first on the ring and then since there is excess we would get a bromine on the other carbon group.

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21. Same as the last problem. I added this one because I worried you might not recognize the reagent added to the epoxide.

The LiAlH4 is basically like a hydride anion, which usually acts basically, but now its acting as a nucleophile and attacking the 1 prime carbon as this is not acid catalyzed. 22. Decide which way the reaction is going to move when doing this ring opening.

This is acid catalyzed, meaning we should attack with the nucleophile at the tertiary carbon. If it were between 1 and 2 we would attack at two. We have to first catalyze with the acid and then attack.

Organometallics: 23. Predict the product for this reaction:

Here we see the Mg get between the chloride and the carbon and then the new nucleophile created will be used to attack the ch2 and perform an sn2 reaction to knock off the iodine.

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24. Here are several predict the problems with organometallics. Rememeber, these boys can not be used with protic solvents because they will donate a proton and remove the metal, cancelling out the molecule....