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Oxidation of Alcohols to Carbonyl Compounds:
1) Basic pattern (primary -> ?, secondary -> ?, tertiary --> ?)
2) 'Strong' reagent
5) Stop after aldehyde?
- 1) PAK or SAK the ox! Primary -> aldehyde -> ketone vs. secondary alcohol -> ketone.
- Basic pattern/similarity is removing an H from the alcohol/aldehyde carbon
- 2) Use a good oxidizing agent like chromic acid (H2CrO4). This will do what I discussed in 1). It goes all the way! This is found in Jones' reagent.
- 3) In solution, the chromic acid is orange-red. it turns greenish-blue after reacting with alcohols. This can thus test for alcohol.
- 4) Formation of chromate ester (O attacks the Cr to make good LG), elimination reaction with base taking proton and kickoff of the chromium (remember elimination creates double bond, C=O). This creates the aldehyde if we started with a primary alcohol.
- This can undergo another reaction by first hydrating with water (add OH and another H to the O on the double bond).
- - activates the C=O for another round of oxidation (same as the first round mech). This creates the final product, C.A.6) Use PCC (doesn't need water in the solvent) so we stop at aldehyde.
Formation of Organolithium compounds
4) reactions of organolithium compounds
- 1) lithium metal with organic halide in ether.
- 2) replace halide with lithium on the organic halide
- 3) reduction with lithium, fragmentation, recombination
- 4) so far I just know the reaction with water in the acid-base reaction and the reaction with carbonyl:
- - addition of polarized C-metal bond to carbonyl C
- - acid base reaction to protonate
- - final products alcohol and salt. we form a C-C bond!
reduction of amides to amines:
1) Reactants: more reactive stuff - the lithium aluminum hydride
added to any amide
- 2) the O in the C=O gets removed, replace those two bonds with hydrogen bonds. - these both come in as hydride (first step, then last step)
- 3) Mechanism: hydride addition, C=O 'revenge' and attacks the aluminum, lone pair movement and oxygen-metal leaving, hydride attack to complete the amine.
reduction of carbonyl compounds to alcohols:
1) Reactants --> products
3) Mechanism: carbonyl + sodium borohydride
4) Overall reaction with sodium borohydride?
5) Use of lithium aluminum hydride on C.A. and esters
6) Cyclic esters
- 1) Carbonyl compound + either sodium borohydride or lithium aluminum hydride
- - AK shooting out sodium pellets - only aldehyde and ketone reduction for the sodium - reduce down to the alcohol.
- 2) Hydride attack as a NP to the C=O to begin.
- Trick: C=O to single bond, add H to that O and to the carbon H-C-O-H
- 3) hydride attack, 'revenge' by the C=O, workup with H2O removes the boron and adds H to form alcohol.
4) 4 carbonyl compounds + 1 NaBH4 + 4 H2O ---> 4 alcohols + Na B(OH)4
6) Cyclic esters make di-ols
Silyl Ether Protecting Groups:
1) TMSCl (PMS...), react other part of molecule (add C-C), remove with TBAF
Creation of Epoxide Polymers
- 1) Strong base OR- and the epoxide.
- 2) First step is the base-catalyzed opening, but before we protonate to make di-ol, we use that O- to attack another epoxide
Base-Catalyzed Opening of Epoxides:
1) Reactants and products
- 1) Strong NP (O-) and epoxide to make di-ol.
- 2) Attack the least sterically hindered side of the carbon since it's SN2-like.
- 3) SN2 inversion.
Acid-Catalyzed Opening of Epoxides:
1) Reactants and Products
- 1) Acid protonates oxygen on the epoxide, Neutral O (H2O, ROH) attacks as NP to form di-ol.
- 2) We attack the more subbed carbon because the neutral NP is 'weaker' and needs more 'help' from the more electrophilic/charge concentrated carbon.
- 3) SN2 inversion!
Synthesis of Epoxides
1) reagents and products
1) Use mCPBA (peracid) to transfer oxygen atom to the alkene
2) Watch out coming from top or bottom - possibility for enantiomers.
1) Symmetrical Ether Reagents
4) Asymmetrical Ether Reagents
7) How to favor W vs. E2 reaction?
- 1) Two alcohols --> Ether (peeing mechanism!)
- 2) Use lower temperature to prevent dehydration of alcohol to alkene.
- 3) protonation of alcohol O, attack by other OH, deprotonation.
4) Use Williamson Synthesis:
- Pot 1: Create the alkoxide anion that will attack (NaH + ROH)
- Pot 2: Add SO2ClMe to ROH.
- 5) O first attacks S, kicking out Cl. Then, alkoxide attacks the R group to kick out the entire leaving group.
- 6) SN2 so we have inversion.
- 7) Apply the strong base NaH on the more substituted carbon substrate (we don't want THAT substrate attacked since it has lots of H available for pickings ... real reason = less likely to form the alkene)
Alcohol into Alkyl Halides:
1) Reaction with HX
- why not use?
2) Reaction with PBr3
3) Reaction with SOCl2
- 1) Protonation to form good LG, then SN1-like attack by the halide anion.
- - primary alcohol goes SN2
- - danger of rearrangement through this mechanism since we have free carbocation.
- 2) No rearrangements
- - Mechanism:
- 1) Alcohol O attacks the P and displaces three Br from the PBr3.
- 2) Br- attacks three times to form final alkyl halide product.
- Overall: 3 R-OH + PBr3 --> 3 R-Br + H3PO3
- 1) 3 R-OH + PBr3 --> P (OR)3 + 3 HBr
- 2) P (OR)3 + 3 HBr --> 3 RBr + H3PO3
- 3) R-OH + SOCl2 --> RCl + SO2 + Cl- + H+
- - Main thing is that we create alkyl chlorides.
- - Mechanism: NP --> EP attack (move carbonyl pi up to lone pair on O), reform pi bond + kick out LG + neutralize charge, Cl- attack to kick out LG.
Chlorination of Alkanes:
1) Reactants and products
4) 'Higher' Alkane reactions
- 1) Alkane + light + Cl2 --> RCl + HCl
- - Replace H atoms with Cl
- 2) When we have an excess of Cl2, the Cl* will continue to react with all H until all are substituted.
- 3) Remember: Initiation -> net increase in radicals, propagation -> no net change, termination -> net decrease
- - First, initiation with breakage of Cl-Cl.
- - Propagation with homolytic breakage of C-H to produce alkyl radical.
- - Propagation again with abstraction of Cl-Cl. This Cl* can go on and continue the reaction.
- - Termination: Any two radicals forming a regular molecule, like Cl* and alkyl*
- 4) Look for all the possible H's we can replace and get a new molecule.
- - More likely to break the C-H that is weakest - i.e., it forms the most stable radical intermediate.
2) Explanation for selectivity
- Bro, you're late!
- 1) Bromine is selective to form the most stable radical intermediate. We pull the hydrogen from the most substituted carbon.
- 2) HCl BDE: 432 kJ/mol, H-Br: 366 kJ/mol
- There is more radical character on carbon when we're doing bromination because the H-Br bond is easier to make.
- Therefore, stability of radical is more important for bromination - the transition state is closer to the products (finished radical product). Larger difference in Ea.
- - This is in contrast to chlorination, where the TS is early (harder to make the H-Cl bond). Smaller difference in Ea.
Radical Polymerization of Alkenes:
- Step 1: Initiation.
- - Break weak O-O bond to form two RO*
- - Break down more, if necessary, to form the 'barebone' R* that is used in propagation.
- - Example: Make 2 CO2 and 2 R* with the diacyl peroxide starting material.
- - R* attacks the monomer and attaches to form the most stable radical.
- - Propagation: The above radical attacks the original alkene to form a longer radical, etc. repeat.
- - Branching: Radical can abstract H atom from itself, 'moving' the radical. The next monomer attaches to where that radical is, etc.