Practice Midterm Questions (Predict products, nomenclature, and synthesis)
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This exam is overall more difficult than the typical MT1 exam, but covers many of the scenarios you are likely to see.
Chapter 12 Summary from review session:
Reaction #7: Addition of carbene (:CR2) to form a cyclopropane (Could not find a suitable video for cyclopropane synthesis)
Reaction #8: Halogenation-- addition of Br2, Cl2, with or without a protic solvent
-Forms two adjacent anti-halogens in aprotic solvent.
-Forms two adjacent anti- halogen + solvent in protic solvent (ROH or H2O)
Note: the opening of the halonium ion (halogen ring) can occur not just with H2O, but with other alcohols (ROH)
Reaction 9: Epoxidation by peroxycarboxylic acid
-Forms an epoxide
This reaction forms an epoxide. This is a good time to review epoxide ring opening from 118A: https://ericpdotutoring.weebly.com/week-six
Note for synthesis: Opening epoxides with NaOH or H+, H2O will form two alcohols anti to one another. This is in contrast to the next reaction, reaction 10- KMnO4 or OsO4, which forms syn diols (2 alcohols).
Reaction 10: Dihydroxylation by KMnO4 or OsO4
-Forms two adjacent, syn- alcohols
Note: Nasiri teaches different reagents than what is shown in the video. Conceptually and mechanistically, the video is still helpful.
Also: we did not cover the last reaction shown: cleavage of alkenes (beginning 9:45~ onwards)
Instead: We learned Ozonolysis for cleavage of alkenes
*Feel free to use reactions that Nasiri did not teach on synthesis questions on exams*
Reaction 11: Ozonolysis
-Cleaves alkenes and replaces them with carbonyls
Note: Khan Academy uses DMS as their second step. DMS = dimethyl Sulfide = CH3SCH3
Note: The first step of Ozonolysis's mechanism is very similar to reaction 10 with KMnO4/ OsO4
Reaction #1: Synthesizing alkynes by double dehydrohalogenation
-Turns Alkane with two halogen leaving groups into a alkyne using strong base (NH2)
Note: Nasiri does not use "mineral oil" in her dehalogenations.
For the second reaction (geminal dihalide), rather than using excess NaNH2 and heat, Nasiri writes two steps: 1.) 3 equivalents of NaNH2 2.) H2O
Reaction #2: Reduction of alkynes by catalytic hydrogenation
-Turns Alkyne into either an alkane, cis alkene, or trans alkene (depending on reagent)
Note: we can form a Grignard's reagent or Cuprate reagent from a haloalkene
Reaction #5: Addition of OH to alkyne
First Video = Markovnikov addition of OH and Acid-catalyzed tautomerization
Second Video = Anti-Markovnikov addition of OH and base-catalyzed tautomerization
Recommended use of website:
Watch the videos, review the respective reaction sheet summaries, and then do the practice questions attached. Or try the practice questions first and review videos of topics that you struggled with.
Additional practice problems: