F B - N + H. B-L A B-L B CB CA pka ~15 wkr A ~9. Lewis A Lewis B Lewis A-B complex. N O B-L A B-L B CB CA pka ~5 ~9 wkr A

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1 CEM 109A 1. Predict the products of the following reactions. Label the acids, bases (Lewis or Bronsted-Lowry) and conjugates (if applicable). Determine the side that is favored at equilibrium for the reactions between Bronsted-Lowry acids and bases). a. - + B-L A B-L B CB CA pka ~15 wkr A ~9 Keq = 10^(9-15) = 10-6 or Keq = Ka(C 3 ) 1 (pka prot amine pka alcohol) so R Ka ( + 4 ) favored b. B B B - + Lewis A Lewis B Lewis A-B complex c. - + B-L A B-L B CB CA pka ~5 ~9 wkr A Keq = 10^(9-5) = r Keq = Ka(C 3 C) > 1 (pka prot amine > pka carbox. acid) Ka( + 3 C 3 ) so P favored d. Al Al + Al - Lewis B Lewis A Lewis A-B complex Page 1 of 6

2 CEM 109A e B-L B B-L A CA CB pka ~ ~16 wkr A Keq = 10^(16-10) = r Keq = Ka( + 3 C 3 ) > 1 (pka prot amine pka water) Ka( 2 ) so P favored 2. Put the following in order of relative acidity and list the major effect(s) a. 2,, 2, 3, Br Br tart by grouping according to atom attached to the proton and look for element effects element effect E element effect size element effect E Br element effect size Br Br > > > > confirm w/ pkas b. is attached to an in each compound, so element effects o resonance structures possible, but there is an E atom attached to some of the compounds, let s look at the inductive effect Page 2 of 6

3 CEM 109A is most E element, so will stabilized the CB the most, making the compounds with the more acidic Compound w/ closest to the acidic proton will be the more acidic b/c can better stabilize the - of the CB. c. is attached to a C in each compound, so element effects o resonance structures possible and no E atoms. nly effect left is hybridization and these compounds have single, double and triple bonds! If attached to C-C single bond was removed have lone pair in an sp 3 orbital, the anion would If attached to C-C double bond was removed have lone pair in an sp 2 orbital, the anion would If attached to C-C triple bond was removed, the anion would have lone pair in an sp orbital The anion with the lone pair in an sp orbital would be the most stable since the e-s would be held closer to the nucleus due to the large s-character of the hybrid orbital, making it the most stable CB and the acid with the triple bond the strongest. > > 3. Put the following in order of relative stability Br In each compound there is a C - group, so while there is resonance stabilizing these anions, it will not allow us to distinguish between them. There is an E atom/group attached to some of the compounds, let s look at the inductive effect E groups inductively stabilize the anion, so the Page 3 of 6

4 CEM 109A more E the group, the more stabilizing (and the closer to the negative charge). E of E of 3.5 E of E of Br 2.8 Br all E groups are the same - distance from the C-, so the only difference is in the E. > > > - - Br Determine the favored structure at equilibrium AD the relative ratio or percentage for each of the following a. C 3 C at p = 6.75 pka = 4.75, p > pka so A - favored by 10 2 (b/c p pka = 2) A - :A is 100:1 or 99%:1% b. C 2 C at p 6.76 pka = 2.85, p > pka so A - favored by 10 4 (b/c p pka = 3.9 = 4) A - :A is 10000:1 or 99.99%:0.01% c. C at p = 7 pka = 10, p pka so A favored by 10 3 (b/c p pka = 3) A:A - is 1000:1 or 99.9%:0.1% d. C at p = 9.1 pka = 9.1, p = pka so [A]=[A - ], 1:1 or 50% ea. e. 2 at p = 9.1 pka = 7.04, p > pka so A - favored by 10 2 (b/c p pka = 2) A - :A is 100:1 or 99%:1% Additional Information: Definitions of A/Bs Bronsted-Lowry: + donor (any species w/ a proton)/ + acceptor (any species w/ lone pair of e- (i.e.,,, anion)) Lewis: e- pair acceptor (species w/ unfilled orbitals, usu a metal atom)/ e- pair donor (species w/ lone pair of e-, usu a nonmetal/halogen) use this definition more later in CEM 109B. A/B strength Page 4 of 6

5 CEM 109A rom Gen Chem: 7 strong A, all others weak. Acid strength p Ka pka vry strng A 1 med strng A 1-3 wk A 3-5 vry wk A 5-15 extrm wk A >15 actors affecting A strength* Element Effects when the atom that the proton is bound to is changing 1. ize when atoms are related vertically in the PT As the size of the atom bound to the proton inc, the length of the atom- bond inc making it weaker and easier to break and therefore a stronger A. 2. Electronegativity when atoms are related horizontally in the PT As the E of the atom bound to the proton inc, the e- being shared between the atom and the proton spend more time around the atom and make the atom- bond weaker and easier to break and therefore a stronger A. tructural Effects have to do with the structure of the compound 3. Inductive effect occurs when there is an E element somewhere on the compound (T bound to the acidic proton) The E element withdraws electrons from adjacent atoms and creates a Coulombic interaction that stabilizes the charge on the CB, making the A stronger. Must take into account the E of the element (more E substituent, more stabilization of CB, stronger the A) AD its proximity to the acidic proton (closer E substituent is to the negative charge of CB, more stabilizing, stronger A). 4. Resonance effect occurs when you can draw resonance structures for the CB If the CB of the A is stabilized by resonance, it makes the A stronger. 5. ybridization when atom bound to the proton has different hybridization The CB of the A is more stable when a lone pair of electrons is in a hybrid orbital that has more s-character (sp 3 sp 2 sp) b/c the e-s are closer to the nucleus and therefore held tighter and therefore the stronger the A. * Using the relationships: tronger the A, weaker the CB (or stronger the B, weaker the CA) and vice versa the relative B strength can also be determined using the above factors. Table of pkas in Appendix II of text or my website Aue s prescription for solving A/B problems Calculate the equilibrium constant and Gibb s free energy for the reaction of acetic acid and ammonia. Page 5 of 6

6 CEM 109A 1. Write a balanced chemical equation (You may ask you to use Kekule-Lewis structures) acetic acid 3 ammonia - acetate ion 3 + ammonium ion 2. Label A B CB CA 3. Label pkas ~5 ~9 4. ID relative strength: stronger A (smaller pka) wkr A (larger pka) 5. Determine direction favored at equilibrium Meek shall inherit the Earth, weaker acid wins 6. ind Keq: Keq = 10 ± pka [if Ps favored sign on pka term should be + = Keq very large, if Rs favored sign on pka term should be - = Keq very small] Keq = 10^(9-5) = ind G o 298K: G o 298K = -1.4log(Keq) kcal/mol** G o 298K = -1.4log(Keq) = -1.4log(10 +4 ) = -1.4 (+4) = -5.6 = -6 kcal/mol ** Comes from G o = -RTln(K) [from Gen Chem] G o = -RTln(K) R = kcal/mol K, room temp = K G o 298K = ln(K) ln = log G o 298K = -1.4log(K) Acid form (A) or base form (A - ) in solution? Depends on p of solution and pka of the acid: pka = p + log([a]/[a - ]) better known as p = pka + log([a - ]/[A]) enderson-asselbach [from Gen chem] [A or A - ] = 10 p-pka p pka: A form p > pka: A - form p = pka: [A] = [A - ], 50:50 Page 6 of 6