Rules of the Day CH320M/328M

If you have not already, you must Watch the Terpene video. You are responsible for this information for the next midterm! Last time we will mention it!

Featured Golden Rules of Chemistry: Nothing today

1. For each reaction of an alkene we have covered you need to remember to call a NIRRS:

Nature of the reaction; what is the starting material/product? (i.e. alkene converted to an alcohol)

Intermediate (or "Important transition state" if applicable) of the reaction, the key to the mechanism (carbocation, halonium ion, etc.)

Reagents Learn the exact way to designate the reagents for each reaction

Regiochemistry What is the regiochemistry of addition? (Markovnikov, non-Markovinikov, etc.)

Stereochemistry of addition (anti, syn or mixed)

2. When studying for the exam, remember this material is very different then the first midterm material. You CAN do this. I suggest you follow these steps before looking through old exams: 1) Read each Rule of the Day so you can identify those things you are not familiar with. 2) Binge-watch the lectures recordings, skipping the things you undersrtand and spending time watching the things you do not yet understand (last year's class used this study method and they did great!) 3) Ask questions at office hours and the Review Session on Monday night 4) Start working through the old exams.

Study Tip for Today: Make sure you not only "know" the mechanisms, make sure you understand the role the different reagents play during each step (nucleophile, electrophile, strong acid, etc.)

3. An oxidation reaction involves a net loss of electrons and replaces C-H bonds with C-O bonds or pi bonds. A reduction reaction involves a net gain of electrons and replaces C-O or pi bonds with C-H bonds.

4. Alkenes react with 1. Hg(OAc)2 followed by 2) NaBH4 to give alchohols with Markovnikov regiochemistry and mixed stereochemistry. This cannot rearrange because there is no carbocation intermediate. You are not responsible for this reaction, but we wanted you to see it.

5. Alkynes with a hydrogen atom attached to one of the sp hybridized atoms are relatively acidic. This is because the anion produced upon deprotonation is on the sp atom, and more s character is stabilizing. Use the wicked strong base NaNH2 to deprotonate a terminal alkyne.

6. Deprotonated alkyne anions react as nucleophiles with primary haloalkanes in a single step in which the carbon-halogen bond is replaced with a carbon-carbon bond (Make a Bond between a nucleophile/electrophile and simultaneously Break a Bond to create stable molecules or ions). Big, mega deal here; We just made our first C-C bond!!! Time capsule: this is an SN2 reaction.

7. The chemistry of alkynes is predictable based on where the electrons are located: In two orthogonal pi bonds. Like alkenes, alkynes have the chemical personality of being a weak nucleophile, weak base.

8. Alkynes react with 2 equivalents of X2 and HX to give vicinal tetrahaloalkanes and dihaloalkanes, respectively. The HX reaction follows Markovnikov regiochemistry with terminal alkynes.

9. Alkynes are prepared from the base (NaNH2) promoted double elimination of a vicinal dihalide. For making an internal alkynes you need two equivalents of NaNH2, but for making terminal alkynes you need to 1. Add 3 equivalents of NaNH2 followed by 2. HCl/H2O to protonate the alkyne product. This provides a way to produce an alkyne from an alkene via the vicinal dihalide produced from an alkene and X2.

10.The enol form of a compound rapidly tautomerizes to the more stable keto form. The keto form is more stable because the C=O pi bond is stronger than the C=C pi bond.

HOMEWORK:

Read: Study for the exam this week.

There are no Daily Quizzes during the exam week.

Continue working on Homework Problem Set 7, due at 10 PM on Wednesday, October 21.

Click here to go to the Aktive Homework Problem Set 7

Click here to get a copy of the Gradescope Homework Problem Set 7