1559r.ch19_344-35710/17/0513:51Page354 EQA 354.chapter 19 CARBOXYUC ACIDS (d)CH,COOHBrCH-COOH 1GO →product (e)BrCH2COOH +(CH;CH2)2NH product occurs as written.The third s ourth step in the mechanism doesn t apply because only a 0 41.Acidity:CH,COH CH,CNH,CH,CCH.The m ns in Ch.CNH.are OH CH,CNH and on O.giving,C-NH2 Resonance stabilization causes protonation ontobe favored 42.See Problem 34 in Chapter 17. HOCH.CH.CHCHOH HOCH-CH.CH.CH- 43.(a)CHCH2OH +Cc 6: CH3CH2-C CH:CH2-0 -q-(d) (e) BrCH2COOH (CH3CH2)2NH n product (f) 40. No tricks here! The mechanism is almost exactly as shown in Section 19-12, with only minor differences. Because the method starts with an alkanoyl halide, step 1 is unnecessary. Step 2 (enolization) occurs as written. The third step uses Cl2, generated in low concentration from NCS, or Br2, formed in the same way from NBS, or I2. The fourth step in the mechanism doesn’t apply because only alkanoyl halides are present, not carboxylic acids. OO O O BB B B 41. Acidity: CH3COH  CH3CNH2  CH3CCH3. The most acidic hydrogens in CH3CNH2 are on the nitrogen. Acidity order is determined by electronegativity. Two protonation possibilities: on N, giving O B CH3CNG H3, and on O, giving Resonance stabilization causes protonation on O to be favored. 42. See Problem 34 in Chapter 17. 43. (a) CH3CH2 Cl Elimination O C Cl OH Tetrahedral intermediate CH3CH2 H O C H Cl CH3CH2 H O C Cl  O  O CH3CH2OH Addition Cl C O HOCH2CH2CH2CH2OH HOCH2CH2CH2CH CrO3 Initially O CrO3 O OH O Hemiacetal form O OH CH3C NH2 OH CH3C NH2 . CH3CH2COOH product 1. Br2, cat. P 2. (C6H5)3P, (CH3CH2)2O CH3COOH BrCH2COOH Br2, cat. P product 1. Excess KSH, CH3CH2OH 2. I2 (Section 9–10) 354 • Chapter 19 CARBOXYLIC ACIDS 1559T_ch19_344-357 10/17/05 13:51 Page 354