Example 14 Example 6, Application in total synthesis 11 picture to explain. Example 14 Please indicate with an arrow and draw a three-dimensional OH Ph CH3C(OME)2NMe2 PhH, 100 °C, 2 h, 75% Ph. Example 6, Application in total synthesis" xylene MeO OMe 200 °C, MW P x = } OH Me NMe2 10 min, 89% CF3 O NMe2 NMe2
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- PMB-0OTBS H2 Pd OMe `Ph) Physical Properties of Alcohols and Phenols pt Unknown # Choose... Fill in the table with data collected BIIIU X2 | X² | → EEE fx | ®I D Normal Alcohol pH Soluble in Water? Ethanol 2-Propanol 2-Methyl-2-propanol Cyclohexanol Phenol UnknownDrive-Google X Logout Successhil X O- Mail- Mariin, And X e Kegistration-Lifer X Sign pw.com/iln/takeAssignment/takeCovalentActivity.do?locator=assignment-take Clear All (C,H3),NH, Stronger Bronsted-Lowry acid Weaker Bronsted-Lowry acid NO, (C;H3)½NH Stronger Bronsted-Lowry base HNO2 Weaker Bronsted-Lowry base Cengage Learning | Cengage Technical Support a W
- CH30 carboxylic acid CH3C-OH ester 0 0 CH3COČCH, anhydride CH3CH2CHËI amide CH2CH3 CH3CH2CHCOCI none of the above ČH2CH3 Cengage Learning | Cengage Technical Support e y-axis and pH on the x-axis. If the points stv N IVUL 4IL VL LLIV VM tBn 1) Mel 2) NaH a) b) -NH 1) NANO2/HCI| Cul/MeLi SH COCl2 pyridine -10 °C, THF 2) LIALH4 ОН Yegt MeO2C- -CO,Me TICL, Zn THE NaH, THF reflux OH N, Bn h) Sn, HCI 1) nBuLi, 2) O i) Zn/Cu CH212 S. NO21) CO2 (s) g) Ph(CH:)MgBr 2) 11 (aq) H'(aq)/A h) 6-Bromohexanoic Acid + NACN- 1) NaOEt/ELOH i) PhCOHC-CH2 + CH:(CO Et) 2) H (aq), A,-CO. Butanoic Acid 1 equivalent Bry/P H.NEt k) Methyl Benzoate NaOMe/MEOHI I) 2 (Ph) CHCOOME -
- Which of the following can inhibit nitrification? 1. Low HNO2 2. High NH3 3. High NH4+ 4. pH < 5 5. Low HCO3- 6. High HNO2 7. High NO2a) " -78 °C, THE 2) H₂0 Phyp THF, 0'C 8 TICH MICH reflujoaund= ney, churge Na O NH2 attack Tonic c) CH3CH2CH2CH,Li + NH3 (acid CH 3 CH2 CH2-CHz t WH2 (Conj acud) (base) (ca bu 2. The equilibrium idea means that if the reaction is reversible there will be four species in solution at one time, the acid, the base, the conjugate acid, and the conjugate base. Sometimes this is what is required, but at other times we need to choose bases that will completely deprotonate every molecule of acid, i.e. send the reaction completely to the right. These bases will include CH3CH,CH2CH;Li, NaNH2, and LiN(i-Pr)2. Weaker bases will include NaOH, NAOCH3, KOtBu, and NaOCH2CH3. For each of these bases, give the products formed when they react with H2O, then use pKa values to get an idea of the relative base strengths of these compounds. You will provided 7 separate reactions. 3. In CHEM 3112, we will study reactions based on the deprotonation of ketones such as acetone, (CH3)2C=0, which has a pKa of 19. Given the bases LiN(i-Pr)2 and NaOCH3, decide which one…
- 40 mg of sodium hydroxide is required to saponity Igm sample, this means that (MW of NaOH= 40 g/mole, MW of KOH is 56g/mol) O a acid value equals 40 O b saponitication value equals 40 OG. saponification value equals 56 O d. Acid value equals 562. Predict the structure of the products (A-E) of the following reactions. BugSnH OMe AIBN Toluene, 80 "C OH 1) NaH, CS,, BnBr Ph B HO" 2) BuzSnH, AIBN H Toluene, 80 °C BuşSnH SPh AIBN Toluene, 80 "c 1) LDA, THF, -78 °C then PhSeBr D 2) H202 1) AIBN, SnBu, Me. Br E 2) O, then PPhg33. In the reaction CH3COOH + H₂O ⇒ H3O+ + CH3COO- the base constant K₂ of acetic acid CH3COOH is approximately 10-5 at 25 °C. What is the pk of the acetate ion, CH3COO™? (A) 10-⁹ (B) 9 (C) -5 (D) 5 (E) 10-5