CHEM 3153 1st EditionFinal Exam Study GuideLecture 30 (April 8)1. There are 2 enolates possible for the above starting material. Draw them and show a mech for their formation, which enolate do you expect to be more stable and why?2. Draw the transition states leading to the two enolates. Which TS do you expect to be more stable and why?3. Draw an energy diagram describing this process (reaction coordinate).Integrated Question:1. Show the electron flow associated with the formation of the enolate on the right hand side of the molecule.Nitrogen’s electrons attack an α H from the right side of the molecule. 2. Show the electron flow associated with the formation of the enolate on the left hand side of the molecule.Nitrogen’s electrons attack an α H from the left side of the molecule. The molecule is asymmetrical, so we have two different enolates.3. Which enolate do you expect to be more stable?The double bond in A is more substituted so it is more stable.4. Which transition state do you expect to be more stable?Molecule B is more stable because there is less steric hindrance.Reaction Coordinate:- For kinetic enolate (B)o Low temperatureo Not enough energy to overcome barrier to form the thermodynamic enolate Ex: 1. LDA at -78°C 2. R-X The alkyl group ends up on the less substituted α C- For thermodynamic enolate (A)o Higher temperature and less hindered base Ex: 1. NaOH at room temp 2. R-X Alkyl group ends up on more substituted α CMalonic Ester SynthesisTwo key steps:1. Enolate alkylation2. Hydrolysis/ decarboxylationKey reagent is diethyl malonateMech:Lecture 31 (April 13)Question: Show how you would achieve the following transformation.Brainstorming- Lost an ester (hydrolysis/ decarboxylation)o Decarboxylation = H3O+ and Δ- Have carboxylic acid (hydrolysis of ester)- Added 2 alkyl groups (methyl and ethyl) o Have 2 α H’s, one alkyl group through each enolateAcetoacetic Ester Synthesis- This mechanism is very similar to Malonic ester synthesis- Uses ethyl acetoacetateEx: - Can also add 2 different alkyl groups this wayConjugate Addition Reactions (Michael Additions)- Involves α, β unsaturated carbonyls- These molecules act as electrophilesIntegrated Question:1. Show the electron flow to produce the second greatest resonance contributor.It’s favorable that positive and negative charges are close. Having the charges very separated is bad. Remember, when drawing resonance structures to only use one arrow at a time.2. Show how the electrons flow to produce the third greatest resonance contributor.3. What are the most electrophilic positions?These are where the partial positive charges are.4. Arrange the resonance structures in order of prevalence and draw the hybrid.Nucleophilic Attack on α, β-unsaturated C==O’s- Numbering starts at the oxygen- The attack can occur 1, 2 (at carbon 2) or 1, 4 (at carbon 4)- Depends on type of nuc usedo “Hard” or “Soft”- A nuc is “hard” when it reacts mostly under electrostatic controlo Partial negative and positive charges Usually nuc with a high charge density (from 2nd row of the periodic table and above)o Ex: OH–, OR–, H2O, ROH, NH3 o Carbon nucs: R-mgBr (Grignard), R-Li, o Hard nucs will react at carbon 2 (do 1, 2 addition) Ex: o Why? It prefers to attack the greatest partial positive charge. The highest partial positive charge is at carbon 2 The largest partial positive charge is based on resonance structure In the above example, the 2nd structure contributed the most to the hybrid, thus it is the largest partial positive charge- “Soft” nucs react mostly under orbital controlo Usually nucs have diffused (large, spread out) orbitalso Usually nuc is from 3rd row elements and belowo Ex: I–, RS–,RSH, (doubly stabilized enolates)o R2CuLi (Lithium Dialkyl Cyprates)o Soft nucs react at carbon 4 (do 1, 4 addition) Ex: Lecture 32 (April 15)Question: Propose a mechanism for the following reaction.Mechanism (1, 4 addition):Reactions involving doubly stabilized enolates as nucs can be useful.Ex: Robinson Annulation- 1, 4 conjugate addition followed by intramolecular additionEx:Notes:- 1st part of the mechanism is conjugate addition- Hydrogen between two carbonyls is most acidic- Enolate is doubly stabilized  soft nuc  attacks at carbon 7o Double bond can move between oxygens- Can’t do intramolecular reaction at step 3 since it makes a 4 membered ring- Hydrogens on carbon 10 are easy to take since their on the end- π bond on carbon 10 can make a 6 membered ringSynthesis Involving Enols and EnolatesHalogenation1. Acidic conditionsEx: 2. Haloform reaction (basic)Ex: X = halogensLecture 33 (April 17)Aldol Reactions- β-OH carbonyl compound- α, β-unsaturated carbonyl1. No dehydrationEx: 2.With dehydrationEx: Question: Provide the dialdehyde starting material required to synthesize the following product.Integrated Question:1. Which bond corresponds to where we will do our first retrosynthesis disconnection?2. What will the structure look like after we do the first disconnection?3. Which bond corresponds to where we will do our second retrosynthesis disconnection?4. What will the structure look like after we do the second disconnection?5. What is the next step in our retrosynthesis?Convert the enolate to an aldehyde.Claisen CondensationLook for an ester with β C==OEx: Enolate AlkylationA carbon chain is added to the α position1. The carbon chain is at the most substituted positionEx: X = Cl, Br, I2. The carbon chain is at the least substituted position. The disconnections are the same as aboveEx: Malonic Ester SynthesisAcetoacetic Ester SynthesisConjugate Addition Reactants1, 4 – alkylation: new alkyl group is at β-position to C==OEx: Robinson AnnulationStart with the dicarbonyl compoundEx:Lecture 34 (April 20)Chapter 24: Carbohydrates- Large biological molecules consisting of carbons, hydrogens, and oxygens- Divided into four groupso Monosaccharideso Disaccharideso Oligosaccharideso Polysaccharides- We will only cover monosaccharides- Carbohydrates can exist in both the cyclic and open-chain formsEx: Glucose- Here we see how to open a ring- The hemiacetal is the most reactive part of the sugar molecule- The squiggle marking on the hemiacetal alcohol group means that the group can be in the axial or equatorial position- Fischer projections are commonly used to show structures of sugarso Drawn for open-chain form of sugarMonosaccharides- Simple