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30 Cards in this Set
- Front
- Back
aq. HONO2 ------------------ aq. H2SO4 |
Nitration. Gives NO2 or OH Substituent aq. H2SO4 uses H3O+ as protonating group. Protonate the HONO2 with H3O, then remove the OH2 and move lone pairs from O- to make = bond. This NO2 can now add |
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H2SO4 ------------ CH3OH |
Gives OCH3 substituent that is added to pi bond |
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HBR ------- HOOH |
HOOH will split into 2, creating radical 2HO The radical from one HO will be used to create a bond with the H from HBR. We now have H2O with Br radical. The product will have a Br and result in anti-markovnikov reaction. Syn/anti mix |
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H3O+ |
Parallels HBr. Can capture nucleophile as H2O because it's most prevalent |
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BH3 --------- NaOH HOOH |
Anti-markovnikov Hydroboration-oxidation. Results in OH & H addition to a pi bond. Syn reaction. |
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Br2 |
Results in addition of 1 Br if no pi bond. Results in addition of 2 Br if pi bond. Markovnikov's is irrelevant. Anti addition |
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O2 ----------- (CH3)2S |
Breakage of pi bond to add a =O to the ends connected to pi bond |
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SO3 -------- H2SO4 |
Sulfanation. Results in SO3H substituent. Similar to nitration. |
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aq. H2SO4 |
Results in =O addition. Markovnikov. |
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HX (Includes H-OH2) to alkene |
Hydrogen Halide Addition. H will add to less substituted C. X will add to more substituted C. Markovnikov + Syn/Anti mix |
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HX (Includes H-OH2) to alkyne |
Hydrogen Halide Addition. Same steps as alkene, but we will end up with an alkene, so we can continue reaction to get rid of the pi bond. Repeat steps and will end up with 2 of X attached. |
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HOH2 |
Makovnikov Hydration. Similar to hydrogen halide addition but will need to get rid of H2O. |
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Alkyne Hydration of HOH2 |
Results in C=O. Markovnikov. |
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H2 ----- Pt |
Catalytic Hydrogenation. Markovnikov doesn't apply. Syn. No electrophile. |
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Catalytic Hydrogenation Br2 Hydroboration/oxidation HBr, H3O+ |
Syn, Markovnikov doesn't apply Anti, Markovnikov doesn't apply Syn, Anti-Markovnikov Syn/Anti mix, Markovnikov |
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Br2 ------- FeBr3 |
Bromination. Results in a Br substituent. Make electrophile by joining Br2 and FeBr3. Br2 will separate and pi bond will take one of them. Will have a carbocation, but can get rid of it by forming pi bond with H attached to Carbocation. |
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Ortho/Para directors |
Alkyl Groups (CH3, CH2CH3 etc.) Pi bonds, aromatic rings Lone Pairs (electron donating outweighs electron withdrawing): OH, OR, NH2, NHR, NR2, F, Cl, Br, I |
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Activators |
Alkyl Groups (CH3, CH2CH3 etc.)Pi bonds, aromatic ringsLone Pairs (electron donating outweighs electron withdrawing): OH, OR, NH2, NHR, NR2 SAME AS ORTHO/PARA DIRECTORS EXCEPT FOR HALOGENS. THEY ARE DEACTIVATORS IN THIS CASE |
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Meta Director |
NO2 (Also deactivator) NH3, NH2R, NR3 C=O C=-N CF3 SO3H |
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Cl2 ------ AlCl3 |
Chlorination. Same as bromination. |
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(CH3)3CCl -------------- AlCl3 |
Friedel-Crafts Alkylation. R-Cl will bond to AlCl3. R breaks off. C(CH3)3 group is the substituent. |
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CClCH3O ------------- AlCl3 |
Friedel-Crafts Alcylation. COC(CH3)3 is substituent. |
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H2SO4 ---------- H2O |
Syn/anti mixture. Markovnikov addition of water. Electrophile is H3O+ and results in OH or =O substituent. |
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Br2 ---------- in hexane, not benzene hv |
Br will take place of H. H will bond with the other Br. 1. H on most stable CC will have half of it leave a radical on the CC, and the other half will make a sigma bond with Br. 2. Radical on CC will form a sigma bond with one of the Brs. The other Br will break. |
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Radical fate #1: |
Add to pi bond |
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Radical fate #2 |
Atom transfer |
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Radical fate #3 |
Radical combination |
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Biological Oxidation Steps |
O2, ----> Superoxide -----> (OH)2 ----> 2HO radical e- 2H |
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O3 ------ (CH3)2S |
Ozoneolysis. Break pi bond and replace with =O on both ends |
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H2SO4 |
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