Ch23 - Chapter 23: Amines Solutions 616 Chapter 23: Amines...

Info iconThis preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon
Chapter 23: Amines Solutions 616 CHAPTER 23 . Solutions to the Problems Problem ~3.1 Identify all carbon chiral centers in coniine, nicotine, and cocaine. CH yo / 3 0 N " . . ~D CH 3 .('l. H * C. ... ~ * H.' 1 (a) (c) (b) O~""" ~ . ~ ~N;J<:/~ ..-9 0i 3 ~ . N H * I . H o (5)-(-)-Nicotine (5)-(+)-Coniine Cocaine Problem23 2 Write structural formulas for these amines. (c) (R)-2-Butanamine (a) 2-Methyl-l-propanamine (b) Cyclohexanamine CH 3 H/'•. ~ CH 3 CHt" 'NH 2 Problem 23.3 Write structural formulas for these amines. (c) Diisopropylamine (a) Isobutylamine . (b) Triphenylamine ( )-N-{) 6 ~N~ H Problem 23.4 Write IUPAC and, where possible, coromon names for each compound. . 4-Aminobutanoic acid 2,2·Dimethylpropanamine (S)-2-Amino-3-phenyIpropanoic acid (y-Aminobutyric acid) (Neopentylamine) (L-Phenylalanine) Problem 23.5 Predict the position of equilibrium for this acid-base reaction. CH 3 NH 3 + +~O -- CH 3 NH 2 + H:30+ pK 10.64 a -1.74 a . (weaker (stronger . acid) acid) Because equilibrium favors formation of the weaker acid, equilibrium will be to the left as shown. Problem 23.6 Select the stronger acid from each pair of compounds. (a) o,,~NH; or CH 3 · <. )-NH; (A) (B) 4-Nitroaniline (pKb 13.0) is a weaker base than 4:m:ethylaniline (pKb 8.92). The decrease~ b~sic.i~y o~ 4- nitroaniline is due to the electron-withdrawing effect of the para nitro group. Because 4-mtroamhne IS the weaker base, its conjugate acid (A) is the stronger acid. Chapter 23: Amines Solutions 617 o-~H; (0) Pyridine (pKb 8.75) is a much weaker base than cyclohexanamine (pKb 3.34). The lone pair of electrons in the sp2 orbital on the nitrogen atom of pyridine (C) has more s character, so these electrons are less available for bonding to a proton. Because pyridine is a weaker base, its conjugate acid, (C), is the stronger acid Problem 23.7 Complete each acid-base reaction and name the salt formed. (a) (EthN + Hel .. (EthNtf' CI- Triethylammonium chloride ---. ... . CN~H -CH 3 COO- . H Piperidinium acetate Problem 23.8 FolJowingare structural formulas for I?ropanoic acid and the conjugate acids of isopropylamine and alanine, . along with pK a values for each functional group:' . pK 4.78 . a ) . ~ CH 3 CH 2 COH Conjugate acid of Propanoic acid Conjugate acid isopropylam ine of alanine (a) How do you account for the fact that the -NH 3 + group of the conjugate acid of alanine is a stronger acid than the - -NH 3 + group of the conjugate acid of isopropylamine? I The electron-withdrawing properties of the carboxyl group adjacent to the amine of alanine mak'e the conjugate acid of alanine more acidic than the conjugate acid of isopropylamine. (b) How do you account for the fact that the -COOH group of the conjugate acid of alanine i's a stronger acid than the -COOH group of propanoic acid? " The -NH 3 + group is electron-withdrawing, so the adjacent -COOH group is made more acidic by an inductive effect. This situation is analogous to the electron:withdrawing effects of halogens adjacent to carboxylic acids in molecules such ,as chloroacetic acid, which has a a of 2.86. In addition, deprotonation ofthe carboxylic acid function of alanine results in formation of an overall neutral zwitterion. Thus, the carboxylate form of alanine can be thought of as being neutralized by the adjacent positively-charged ammonium ion.
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Image of page 2
This is the end of the preview. Sign up to access the rest of the document.

This note was uploaded on 07/09/2011 for the course CHE 301 taught by Professor Diver during the Spring '08 term at SUNY Buffalo.

Page1 / 23

Ch23 - Chapter 23: Amines Solutions 616 Chapter 23: Amines...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online