H R' 1 Amine 2 Amine 3 Amine Quaternary Ammonium Salt R'' R'

Transcription

1 Amines rganic Bases amines are trivalent, but with a tetrahedral arrangement of electron pairs, although the geometry of a neutral amine is properly described as trigonal pyramidal, since unlike nuclei, the exact positions of an electron pair cannot be defined, an exception is an ammonium salt that is surrounded by 4 atoms (nuclei) trigonal pyramidal ' 1 Amine 2 Amine 3 Amine Quaternary Ammonium Salt 1 omenclature IUPA priority: acid > ester > amide > nitrile > aldehyde > ketone > alcohol > amine > alkene > alkyne > halide named as for alcohol, replace -ol with -amine, AMIE suffix when an anime is a SUBSTITUET it is an AMI substituent propanamine number the chain as usual to give the amine functional group the lowest number amines are named as alcohols except that "amine" is used in place of "ol" and so instead of propan-l we have propan-amie (above), instead of pentanl we have pentanamie (above) etc. '' ' (S) tetrahedral (2S)-pentanamine Me longest chain Et -ethyl-,3-dimethylbut-2-en-1-amine Me find the longest chain that contains the maximum number of functional groups (alkene and amine) = 4 number to give the highest priority functional group AMIE the lowest number comes BEFE 3 when specifying the two methyl substituents l #1 by definition 3-bromo-2-chloroaniline EMEMBE, the common name aniline has been incorporated into IUPA nomenclature some common names for amines diethylamine diisopropylamine piperidine * * pyridine 2,6-lutidine pyrrole 2 putrescine cadaverine (amines can stink!) * you EED T KW TESE TW, piperidine seen in the Stork reaction, pyridine a common organic base Amines : Page 1

2 2 Amines as Bases deciding the stronger of two acids, e.g. -A 1 and -A 2 weaker acid higher energy electrons A 1 A 1 stronger base A 2 A 2 the stronger acid has the more stable conjugate base anion, lower energy electrons in the base the stronger acid corresponds to the more favorable "reaction", left to right deciding the stronger of two bases, e.g. B 1 and B 2 higher energy electrons stronger base B 1 B 1 weaker conjugate acid B 2 B 2 there are no anions here, but the stronger base still has the higher energy, more chemically reactive electrons the stronger base is stronger because the energy of the electrons go down more by getting into a bond the stronger base has the weaker conjugate acid since it has the stronger bond Example D group, raises energy of electrons on, more reactive 3 W group, lowers energy of electrons on, less reactive 2 stronger base weaker base more favorable less favorable D group stabilizes positive charge on 3 W group destabilizes positive charge on 2 more stable cation less stable cation Acidity is Measured in terms of pka, ow is Basicity Measured? Basicity can be measured in terms of the corresponding pkb, however, in organic chemistry we often don't do that, we specify basicity in terms of the acidity of the conjugate acid Acidity of an Amine in terms pf pka. Amine acid strength 3 2 pka ~35 weak acid 2 strong base 3 ompare with (and don't confuse with) Basicity of an Amine in terms of pka of the conjugate acid Trends in Basicity? Amine base strength 4 2 pka ~9 moderate acid 3 3 moderate base 3 2 ( 3 ) 2 ( 3 ) pka = pka = ( 3 ) 2 2 pka = ( 3 ) 3 3 pka = increasing cation stability but decreasing solvation effects cancel!! Amines : Page 2

3 there are no clear trends among the aliphatic amines because of the competing effects of alkyl group donation and reduced solvation of the larger cations But other resonance contributors pka ~ 4 (minor) resonance delocalization (i.e. partial bonding) stabilizes (lowers energy) of the non-bonding electrons on, these electrons are less reactive aromatic amines are thus MU less basic than aliphatic amines sp 3 sp 2 3 sp "pka" ~9.8 "pka" ~5.3 "pka" ~ -10!!! decreasing basicity clear decrease in basicity with decreasing energy of non-bonding electrons as the hybridization of the orbitals gains more s character and less p character the protonated nitrile is about the strongest acid that we discuss this entire course 4 eactions of Amines 4.1 Amines a ucleophiles ecall: S2 3 I X reaction does T go, 2 not strong enough nucleophile 3 I nucleophilicity strong intermediate weak owever: 3 I S2 3 I 3 3 I slightly stronger nucleophile 3 3 "exhaustive" alkylation product etc. etc is a stronger enough nucleophile that this S2 reaction goes! Me 2 is an even stronger stronger nucleophile, reaction keeps going! clearly not very useful, due to "over-alkylation", except in the ofmann elimination (see next reaction) Amines : Page 3

4 4.2 ofmann Elimination evist a oncept we ave Seen Many Times ow: Base E2 X no reaction : poor leaving group Base E2 2 2 reaction goes : better leaving group This reaction ALS doesn't work (poor leaving group) Base 2 E2 X 2 no reaction : poor leaving group owever, this one does (better leaving group) Base E2 3 exhaustive methylation Me 3 excess Me-I reaction goes : better leaving group 2 Example 3 2 Excess 3 I exhaustive methylation 3 I Me 3 Ag converts from iodide to hydroxide anion = good base Me 3 AgI (ppt.) eat E2 2 verall XS 3 I 2. Ag 2 / 2 3. eat Major ofmann Minor Sayetzeff the reaction is overall E2 ELIMIATI, forming an alkene with a neutral amine as the leaving group Amines : Page 4

5 WEVE, the LEAST substituted alkene is the major product (ofmann product), not the more substituted Saytzeff (Zaitsev) alkene Why is this? There are TW reasons, the first is a STEI EFFET, the second is an ELETI EFFET eason #1: The steric effect of the bulky -3 leaving group S 2 excess 1. XS 3 I 2. Ag 2 / 2 3. eat lower 3 2 energy conformation Major ofmann bulky Me 3 and leaving 3 group are anticoplanar bulky Minor Sayetzeff S Me 3 3 gauche interaction the Me 3 group is LAGE (bulky), it is roughly the same size as a t-butyl group therefore the E2 conformation for ofmann elimination is lower in energy than the E2 conformation for Sayetzeff elimination for steric reasons (fewer gauche interactions) therefore elimination in the ofmann conformation is more likely because the structure spends more time in that lower energy conformation eason #2: electronic effect 3 is a good leaving group, but not as good as, I etc. at this point we need a more sophisticated and subtle look at the transition state that goes beyond conventional general organic chemistry, the transition state is not really "symmetrical", in that although all of the bonds are made and broken in one continuous process, as required for all concerted reactions, the bonds are not all broken and made at EXATLY the same time the proton "leaves" earlier than the 3, which means that in the transition state the partial bond to the departing hydrogen atom is LG and WEAK, but the leaving group leaves "later", which means that in the transition state the bond to the leaving group is still comparatively ST and STG, as illustrated below because the proton leaves "earlier", a partial negative charge starts to develop on the carbon atom that the proton is attached to, and in the ofmann elimination this carbon is less substituted: EALL: negative charges are LESS STABLE on more substituted carbons, thus the transition state for the ofmann elimination has the partial negative charge on the less substituted carbon, this transition state is lower in energy, it is thus formed faster (under kinetic control), the ofmann alkene is thus the major product lower energy leaving "later" Me 3 δ 2 3 Me 3 faster ofmann S leaving "earlier" partial negative developing on 1 carbon S Sayetzeff slower higher energy partial negative developing on 2 carbon δ leaving "later" Me S leaving "earlier" Amines : Page 5

6 this is a rather subtle advanced organic chemistry concept, but it is important that you are exposed to this thinking as we get towards the end of our general organic chemistry course Example Problem 1 1. Excess 3 I 2. Ag 2 / 2 eat Major, ofmann eat Minor, Sayetzeff there are TW possible elimination products once the ammonium ion has been form in the reaction with 3-I, the reaction that forms the least substituted (ofmann) alkene is fastest, the ofmann alkene is the major alkene product, together with its associated amine in some cases the amine leaving group will be a substantial major organic structure on its own Example Problem 2 1. Excess 3 I 2. Ag 2 / 2 heat mono-substituted alkene formed Example Problem 3 2 Me Me 1. Excess 3 I 3 Me 2. Ag 2 / 2 Me 3. eat 3 Synthesis of Amines Amines are made in EDUTI EATIS, several of which we have ALEADY SEE seen before 2 /Pd/ 2 amine nitro 1. LiAl 4 ' 2. ' 3 amine amide addition of, removal of addition of, removal of 1. LiAl 4 nitrile 2. 3 amine addition of Amines : Page 6

7 new BUT related! ' ' 2 /Pd/ LiAl amine IMIE 2 /Pd/ amine AZIDE 1. LiAl addition of addition of In reality there are lots of different ways of doing these reductions (many different sets of reagents/conditions work), but here we use reagents/conditions that we already know for simplicity., no need to learn even more new reagents at this point 3.1 Amine Synthesis by eduction of Azides: Synthesis of PIMAY amines AGAI, the reactions we already know in a synthetic context, AD A EW E acid Sl 2 ' 2 l acid chloride ' amide 1. LiAl ' makes a 1, 2 or 3 alkyl amine does not add a carbon a X alkyl halide S 2 nitrile 1. LiAl amine adds 1 carbon EW X alkyl halide a 3 S 2 1. LiAl does not add carbon azide amine 1, 2 or 3 amines can be made by reducing an amide, which in turn comes from an acid chloride, which in turn comes form a carboxylic acid a 1 amine can be made from a nitrile, which can be made in an S2 reaction from an appropriate alkyl bromide, E AB ATM IS ADDED MPAED T TE STATIG ALKYL BMIDE a 1 amine can be made from an azide, which can be made in an S2 reaction from an appropriate alkyl bromide, ZE AB ATMS AE ADDED MPAED T TE STATIG ALKYL BMIDE A ew eagent sodium azide = a 3 = a azide anion Very good nucleophile small "bullet" shaped similar to the cyanide anion cyanide anion Examples Synthesize these two PIMAY amines BS/hν a 3 S LiAl 4 2 no additional 2. 3 carbon atoms a 1. LiAl must add TIS adds carbon atom carbon atom Amines : Page 7

8 3.2 Synthesis of Substituted Amines by eduction of an Imine: eductive Amination Addition of nitrogen to a carbonyl to form an imine, FLLWED BY reduction to form an amine 2 /Pd/ LiAl / (cat.) ' imine 2 /Pd/ LiAl ' 2 amine reductive amination So now we have a new reaction name: eductive Amination, but is basically chemistry we already know We already knew that we could react an amine with an aldehyde/ketone to form an imine (also known as a Schiff base), and it is no great surprise that the = bond of the imine (Schiff base) can be reduced by addition of 2 addition similarly to a = bond, we can even use the SAME EDUIG EAGETS we already know! reduction of = is actually somewhat easier than reduction of =, is less electronegative and the - pibond is thus a bit weaker than the - p-bond IDIET eductive Amination combines these reactions in sequence: Example (cat.) 3 2 /Pd/ amine Example / cat. 2. LiAl DIET eductive Amination MBIES formation of the imine and the reduction in one reaction organic acid LA LB 3 1 amine ammonia ab 3 new!! reduced in situ "in situ" means formed in the reaction, reaction "in situ" means that the imine reacts as it is formed EW EDUIG AGET, ab 3 ecall ab 4 : stronger π-bond more electronegative slower faster Amines : Page 8 B LB/ucleophile a B 4

9 ompare ab 3 : a B 3 withdrawing B LESS nucleophilic, less reactive X weaker π-bond less electronegative The - withdrawing group makes the -B3 anion less nucleophilic, it reacts slower with carbonyls (=) and imines (=), but the reaction with = is slow enough that it does not happen under reductive amination conditions, a B3 is a selective reducing agent for imines remember, "no reaction" doesn t really mean no reaction at all, reactivity is not black and white, it means that the reaction with the = is now slow enough and the reaction with = fast enough to be useful in the reaction another subtle point, Et may not be required as a solvent to provide the protons that go with the hydride, since the acid that is used to catalyze the reaction may also supply these protons LA LB 1 amine or other organic acid TFA ab 3 Et imine reduced in situ 2 amine LA LB 1 amine or other organic acid Ts ab 3 Et iminium reduced in situ 2 amine Example of an IDIET reductive amination Et LA LB 1. l cat /Pd/ Et Example of a DIET reductive amination note that the selective reducing agent ab 3 is included in the reagent/conditions in a "one pot" reaction Example in everse how can we make the provided amine using a reductive amination? 2 LB A make bond B ab 3 Et LA A eak bond "backwards", the benzene ring cannot carry a =, the carbonyl must be the fragment on the "right" You can use EITE the DIET or the IDIET reductive amination methods in a synthesis problem Amines : Page 9 Ts ab 3 Et X 2 can't make bond A B obviously can't have this 2

10 5 Some Final Synthesis Problems 5.1 Synthesis of Amines ommon Methods of Synthesizing Amines (this is not all of the methods, see Section 3 of these notes 3 2 / (cat.) ab 3 reductive amination 1, 2 or 3 1. LiAl amide reduction 1, 2 or 3 1. a 3 2. LiAl azide reduction 1 only 1 1. a 2. LiAl nitrile reduction 1 only 1 carbon atom Example Problem 1 (±) MDMA 1. 2 /g(ac) reductive 2 2. ab amination Ts at ab 3 P same thing MDMA =,a-dimethyl-1,3-benzodioxole-5-ethenamine = Ecstacy : 5g ~ 20 years!! this is a secondary amine, which can in principle be synthesized either by reduction of an amide or by reductive amination, however, in this case only reductive amination would work Trying to make the amine via amide reduction in this case obviously doesn't work because the relevant amide can't exist can't exist 1. LiAl 4 amide reduction 2. 3 can't work here (±) Amines : Page 10

11 Example Problem 2 1. KMn 4 / /boil Sl 2 l 1. LiAl cat. ab 3 a (S2) P in this case the strategy again is to note that we need to make a tertiary amine, which requires either a reductive amination or reduction of a tertiary amide, in this case either of these two routes would work, as shown Example Final Exam Synthesis Problems Involving Amines Problem 1: Difficulty = "moderate": Synthesize methamphetamine (meth) from benzene (±) meth 2 /Fe Ts at ab 3 Mg.TF Mg 1. (±) P 2. 3 (±) again, the strategy again is to note that we need to make a tertiary amine, which requires either a reductive amination or reduction of a tertiary amide, in this case only the reductive amination works This is not the normal method of synthesizing meth, this is just an organic chemistry exercise! Problem 2: Difficulty = "easy" 2 BS/hν 1. LiAl a Amines : Page 11

12 in this case we need to make a primary amine AD we need to add one carbon atom compared to the starting structure, the nitrile reduction method works well here Problem 3: Difficulty = "difficult" Ph BS/hν l Ph ignore stereochemistry cat. XaB3 Mg.TF Mg P cat. ab 3 Ph does not work for an amide this one is tricky because you have to construct the amine from which to make the amide it is also tricky because the "order" of the reactants to make the amide are reversed, usually we make the acid chloride and react it with an amine on the reagent arrow, here we make an amine and have the acid chloride on the reagent arrow Amines : Page 12

13 5.2 Different Types of Synthesis problems for the Final Exam Synthesis is hard because the strategy is often difficult and there are simply a lot of reactions to know. Synthesis problems on the final exam will be of two kinds. The first includes all of the major reactions we have learned this semester. The second kind of synthesis problem will include only a "minimal set" of reactions, that everybody should know. The idea here is these problems test your synthesis problem solving skills rather than your ability to learn a lot of reactions. There are two parts to this "minimal set: of reactions, the first are a set of basic functional group interconversions, the second includes the most basic carbon-carbon bond forming reactions, i.e. the Grignard and acetylide reactions. These two sets of reactions are summarized below, and also on the class website. Example Synthesis Problems that Include nly eactions from the "Minimal Set" Problem 1 BS/hν P Mg.TF Mg Amines : Page 13

14 Problem 2 Ph BS/hν 1. PhMg 2. 3 a MPBA Problem 3 2 a/ 3 (l) 1. Excess a 2 /heat 2. 2 a 2 a Amines : Page 14

15 6 Summary of eactions Involving Amines Do T start studying by trying to memorize the reactions here! Work as many problems as you can, with this list of reactions in front of you if necessary, so that you can get through as many problems as you can without getting stuck on the reagents/conditions, and so that you can learn and practice solving reaction problems. Use this list AFTE you have worked all of the problems, and just before an exam. By then you will have learned a lot of the reagents/conditions just by using them and you will only have to memorize what you haven't learned yet. Then do the following: over the entire page of reagents/conditions with a long vertical strip of paper, see if you can write down the reagents/conditions for each reaction, check to see which you get correct, if MPLETELY correct, circle Y, if incorrect or even slightly incorrect, circle. In this way you keep track of what you know and what you don't know. Keep coming back to this list and so the same thing only for those reactions you circled, until all are circled Y. Knowing the reagents/conditions on this page is ISUFFIIET to do well on an exam since you will ALS need to recognize how to use and solve reaction problems in different contexts, this page LY helps you to learn the reagents/conditions that you have not YET learned by working problems. You have seen PATS of some of these reactions in earlier sections! amine synthesis 1. LiAl Y / 2 1. a 3 Y / 2 2. LiAl a 2. LiAl /Pd/ IDIET reductive amination 1. 3 / (cat.) 2. 2 /Pd/ Y / 1 2 amphetamine DIET reductive amination 3 2 / (cat.) Y / 2 ab 3 /Et ( 3 ) 2 / (cat.) 3 Y / ab 3 /Et ofmann elimination 1. XS 3 I Y / 2. Ag 2 / 2 /eat Y / Y / 2 2 Amines : Page 15