Lecture Outline 2. Amines as nucleophiles

Lecture outline 2. Amines as nucleophiles — a. Formation of imines and enamines H+ + R–NH 2 1° (R = H, alkyl, aryl, NH2, OH) O H C C H+ + R 2NH 2° H+ + R 3N 3° Equilibrium can be driven either way — imines and enamines can be hydrolyzed to carbonyl compounds with aqueous acid (i.e. excess H2O) 1° amines form imines (Achtung! You must add charges or this will make no sense!) O mech: + H+ H2 N—R H H2 N + –H R–NH2 O R N N H R R – H2O +H –H HO H N H H2O R a carbinolamine or "hemiaminal" H N R N R H2 O The imine mechanism might look frighteningly complex at first, but it's even simpler than the acetal mechanism. It's essentially two steps — nucleophilic add'n of the amine to the carbonyl group, and acid-catalyzed dehydration. Oh yeah, and a bunch of H+s scurrying around. 2° amines form enamines. The mechanism is the same, except that the dehydration forms a C=C because it can't form a C=N. HO R N R + H+ R – H2O N R R N R – H+ Now let's briefly return to formation of imines from 1° amines. That reaction could produce either tautomer — the imine or the enamine. Why do we get just the imine? 3° amines — this is even simpler 'cause nuthin' happens. O + – H+ NR3 HO As they say Out West NR3 "That thar's th' end o' th' road, pardner." That hydroxy-ammonio-thing is thermodynamically unstable compared to the reactants, so it just falls apart and goes back where it came from. b. some other imine-forming reactions — draw the products! C O + H2N OH H+ hydroxylamine C O + H2N NH Ph phenylhydrazine H+ c. Wolff-Kishner reduction goes via a hydrazone — H2N–NH2 C O CH2 HO–, H2O, 200°C (ethylene glycol or other highboiling alcohol solvent is often used instead of water) C N NH2 — recall that Clemmensen reduction is an alternative way to accomplish the same transformation under acidic conditions ... C O Zn(Hg) aq HCl CH2 3. Hydride as a nucleophile — i.e. reduction of aldehydes and ketones to alcohols Two commonly used hydride reagents — a. Lithium Aluminum Hydride — LiAlH4 — very strong base e.g., LiAlH4 reacts violently with water — write the acid-base rxn — good nucleophile e.g., besides adding to carbonyl groups (below), LiAlH4 serves as a source of nucleophilic hydride in SN2 rxns — write the SN2 rxn of LiAlD4 with (bromomethyl)cyclopropane b. Sodium borohydride — NaBH4 — less reactive than LiAlH4 NaBH4 reacts slowly with water and alcohols. NaBH4 reductions are typically run in alcohol solvent, but they can also be run in water, although the reagent dies within a few hours in water. Both reagents reduce aldehydes and ketones to alcohols by nucleophilic addition of hydride to the carbonyl C. C H—AlH3 O Et2O H C H+ O (note the aprotic solvent and separate protonation step) C O H—BH3 EtOH (note here the use of a protic solvent that acts as the proton source) NaH does not reduce aldehydes and ketones because it's insoluble in most organic solvents. NaH can act as a base, but hydride embedded in the solid surface is not nucleophilic — the surrounding solid presents too much steric hindrance. Draw the products of the following rxns. NaBD4 O O MeOH LiAlH4 H+ Et2O Aldehyde and ketone carbonyl groups can also be reduced by catalytic hydrogenation, but these are harder to hydrogenate than CC double bonds. Hydrogenation of a carbonyl group typically requires higher temperatures and higher H2 pressures than the easy hydrogenation of a C=C. This is useful because it provides a way to do a selective reduction that's complementary to the 2nd example above.