ORGANIC & BIOLOGICAL CHEMISTRY

Transcription

1 GARETH J ROWLANDS ORGANIC & BIOLOGICAL CHEMISTRY LECTURE FIVE Now we re on the homestretch......well for the first 6 lectures (section 1 if you like) 1 Now we will start looking at specific reactions...and a number of substitution reactions...we have braved it through nomenclature and started to discuss organic reaction mechanisms (yay)... 2 First SN2, as it is slightly simpler....and this is it. The reaction is just a single step...no intermediates. Nucleophile attacks carbon and the new bond starts to form and the leaving group starts to leave simultaneously. (yeah right, as if it was that easy...well for the purpose of this course it is but...) (Q. What film and why use it here? (one of the best 80s horror films in my opinion; then again I m showing my age)) 3 4

2 The nucleophile approaches from directly behind the leaving group....this has the wonderful name of backside attack The reason for this is found in molecular orbitals; the electrons of the nucleophile interact with the sigma* antibonding orbital, which is located behind the bond (see the first lecture). snigger This is a transition state...it does not have a finite lifetime; it cannot be isolated. It is not a real molecule. Carbon does not have five bonds; two of the bonds are partially formed / broken. The reaction proceeds with inversion of configuration; the backside attack can be thought of an umbrella blowing in the wind. This is one of bits of evidence for this mechanism... It is the highest point of an energy profile; it represents the energy maximum in passing from reactants to products. Any change in structure leads to a more stable arrangement. 5 6 This is the reaction profile, showing that the products are more stable than the starting materials (hence the reaction occurs). Because there is only one step, the rate of reaction depends on the concentrations of both the nucleophile and the substrate (SM). We can see that the transition state is the highest point...it is unstable with both the SM and product being lower in energy (hence it cannot be isolated) The reaction is bimolecular (hence the 2). 7 So lets take a look at what factors effect the reaction... Primary halides (or their equivalent) are the best substrates......first we will look at the substrate / starting material. The reason for this is that the nucleophile can easily approach the back of the electrophilic carbon; nothing hinders backside attack. 8

3 Secondary halides (or their equivalent) are ok...but they will be considerably slower...backside approach is much harder... (how tempted was I to put a picture of the choir scene from Face / Off...or even worse...) Tertiary halides (or their equivalent) do not undergo SN2 substitution. The three alkyl groups provide too much steric hindrance and the nucleophile cannot approach from behind the leaving group. Hopefully, a movie of this molecule (in the lecture) shows how hard backside approach is... So the structure of the substrate effects the rate...what else? 9 10 The better the leaving group the faster / better the reaction. Weak bases are better leaving groups. So iodide (HI pka = 10) better than fluoride (HF pka = 3.18). Note: inversion of stereochemistry 11 I guess a quick recap of pkas wouldn t go a miss at this point... Ka is the equilibrium constant of a reaction, in this case the dissociation of an acid. It is important to remember that when we talk about the pka scale we must be talking about an acid or a conjugate acid (so when we say pyridine has a pka of 5.2 we actually mean pyr H + has a pka of 5.2)...but what is this p anyway? So here is the maths... Ka is the equilibrium constant and pka is the log of it. Note how I keep stressing that we are dealing with acids or conjugate acids. For some reason people always forget this! 12 To avoid maths, all you have to know is... Small pka large dissociation and thus a high concentration (lots of) of the anion/ base. Or in other words we have a strong acid and...

4 ...a good leaving group Or to summarise it all on one slide... low pka = good acid, poor base, good leaving group. high pka = poor acid, good base, poor leaving group....and here for your edification is a list of the pkas. Alcohols make poor substrates for SN2 as hydroxide anion is a very poor leaving group (it is a reasonably strong base, hence we use KOH etc. in reactions). Note how HF is actually a poor acid (but a horribly insidious compound) Any nucleophile strong enough to kick out a hydroxide anion would certainly be strong enough to remove the proton from the alcohol first! For those of you who like numbers I recommend: pdf/evans_pka_table.pdf Text...to inspect the leaving group ability of water we talk about the conjugate acid (H3O+)...when we were looking at the leaving group ability of HO we were actually using the pka of water!? But, more importantly, it s a weak base so it is a good leaving group. As I said, we re out to confuse you (it certainly confuses me!) Please remember, chemists are out to confuse you So, here is an example of when we can convert OH to a good leaving group by protonation and the formation of water. So, how would we normally achieve the displacement of an alcohol? (and on a more philosophical tack, why are we so worried about trying to make alcohols into good leaving groups?) Of course, very few molecules are robust enough to survive treatment with such strong acids! 16 (because they are very easy to make) Water is a good leaving group. Lets face it, there is quite a lot of it about so we know it s a stable molecule.

5 Alcohols can be converted in to good leaving groups. The tosylate is a good leaving group because it is a weak base (it is the conjugate base of ptsa, which has a pka of 6). In this example the alcohol is reacted with tosyl chloride (toluenesulfonyl chloride; next slide) to make a good leaving group. We then react it with an acetate and finally hydrolyse the product. Note that by performing this sequence (with its SN2 reaction) we invert the alcohol stereochemistry. What about the nucelophile? How does it effect the rate of reaction? A strong nucleophile increases the rate of reaction. The strength of a nucleophile often follows the order of basicity (more basic = better nucleophile). As a result pka values can give us a reasonable estimate of nucleophilicity. H2O pka = 15.7 H3O+ pka = 1.7 ipr2nh pka = 36 (THF) iproh pka = pka only gives a rough estimate. It is a measure of a compounds ability to donate electrons to a proton. Nucleophilicity is the donation of electrons to anything else. Nucleophilicity is... So a compound can be highly basic but if it is sterically demanding then it will be a poor nucleophile (t-buo for instance). pka is not effected by size

6 Annoyingly, solvent can play a big role in the efficiency of SN2 reactions... Polar aprotic (non-protic) solvents (solvents that have a big dipole but no hydrogen atoms suitable for hydrogen bonding) are good! (eg. DMSO (above), DMF) Reason: They solvate the counter ion leaving the nucleophile naked and nonsolvated and therefore highly reactive. Polar protic solvents (solvents capable of hydrogen bonding; water, MeOH etc.) slow SN two steps. In this reaction, the leaving group departs first to give an intermediate that is then attacked by the nucleophile. As a result the reaction has... (other substitutions include SN and SNi amongst others) Such solvents are capable of hydrogen bonding to the nucleophile, solvating it (surrounding it with a cage of solvent) and thus stabilising it and making the nucleophile less reactive (and considerably larger/more sterically encumbered) and this is it... The other main substitution reaction is the SN1 reaction. The first step is the slow, rate determining step (RDS) and this is the dissociation of the leaving group from the substrate. To give a cationic intermediate. An intermediate is not only observable but in some cases can actually be isolated. It is a molecule or ion that represents a localised energy minimum; an energy barrier must be overcome before the intermediate forms a more stable species. (and your clue for todays picture... I ve seen things you people won t believe... the best sci-fi film ever...and whilst I prepared to enter a discussion about 23 this statement, I do so in the knowledge that I am right and anyone with a contradictory view is 24 wrong)

7 The second step is the fast combination of the cation and the nucleophile... Now to quickly return to our orbitals...a carbocation is in the vast majority of cases trigonal planar (the carbon will be attached to three groups and will have an empty p orbital). It is equally possible for the nucleophile to approach from either face. As a result stereochemical information will be lost....so if we start as a single stereoisomer we will end up with a 50 : 50 mixture of stereoisomers in the product....a racemic mixture... Such a mixture has no optical rotation (50% of it rotates light +x and 50% rotates the light x thus canceling each other out). If the compound had a single stereogenic centre that reacted then we will get two new enantiomers. Such a compound is known as a racemate or... (this is chemistry, so there are exceptions!) 25 If an optically active starting material reacts to give us a racemic mixture (no rotation) then an SN1 reaction occurred. 26 If the reaction occurs with inversion then the reaction was SN2. This is the energy profile for an SN1 reaction. As the nucleophile is not involved in the RDS the reaction is first order or unimolecular (hence the 1) and only the nature of the substrate is important... Once again the product is more stable than the starting materials. The highest energy bump (highest transition state) will be the rate determining step (as it is hardest to achieve). (this is chemistry, there are of course exceptions but lets keep it simple) (this is chemistry, there are of course exceptions but lets keep it simple) 27 You can see the intermediate sitting in its little energy minimum. 28

8 So lets have a look at how the substrate can effect the rate of reaction... The biggest factor is the stability of the newly formed cation. Tertiary cations are the best as they are the most stable......secondary cations are ok and primary cations are disfavoured / useless. Note: any group that can stabilise a cation (positive charge) will encourage cation formation and hence SN1 (or E1)... So heteroatoms, aromatic rings, double bonds etc. are all good....and yes, I couldn t be bothered to redo all the slides and numbering so I have doubled this one up just so I can continue as before... What other factors effect the rate? 31 32

9 As with SN2, the nature of leaving group is important. The better the leaving group the faster the reaction. Water is a good leaving group in SN1 reactions and so alcohols can be used in this form of substitution (as long as they are treated with an acid). Yet again, solvent can play an important role... Good leaving groups are weak bases (are the conjugate bases of strong acids) 33 Polar solvents are good as they encourage charge separation Polar protic solvents stabilise both the cation and the anion so are good, promoting SN1 over SN2. In this reaction the alkyl bromide ionises and the bromide ion reacts with the silver (I) nitrate to form an insoluble percipitate (helps drive the reaction to completion)....but aprotic (non-protic) solvents are not the best as they only stabilise the cation. The carbocation is then attacked by the ethanol to give a cationic intermediate, which losses its proton to more of the ethanol

10 So, which of the following substrates will react faster?? Hopefully, you think that these two could be the best substrates for the SN1 reaction...they are both tertiary bromides so should form a relatively stable carbocation. 37 The secondary halide will be ok...it will react by an SN1 mechanism but will be much slower than the tertiary halides and vice-versa, this is the worst as it is a simple primary halide and can t form a stable cation. 38 What about these two?...but you would be wrong! They look like primary halides so you might think they were poor substrates, unable to form a stable cation... If we look at allyl bromide you should be able to see that the cation, the empty p orbital, can be stabilised by the alkene (the pi bond, the adjacent p orbitals)......resonance rears its ugly head again 40

11 Remember that any adjacent p orbitals can overlap (as long as they are parallel). So the two electrons can be shared over three atoms......or in terms of resonance structures and hybrids we see that the positive charge or cation is spread over the entire molecule...the more we spread a charge out the more stable it is. So what about the benzylic cation?...it is resonance stabilised... Well, hopefully you can see that it is a very similar situation. We have a collection of parallel p orbitals and so... Therefore allyl bromide is a good substrate for SN or pictorially...the charge can be spread over seven carbon atoms So benzylic cations, like allylic cations, are very good in SN1 reactions. 43 Lets take a few minutes just to look at a few examples and see if we can decide which mechanism is in action... So does this simple displacement occur via SN1 or SN2? 44

12 Two arguments suggest that this reaction is SN2. Firstly, formation of a cation next to a carbonyl group is disfavoured; the carbonyl group is electron withdrawing and would destabilise a cation (something that already lacks electrons). Secondly, the carbonyl group activates the halide by creating a new LUMO from the interaction of the!* and!*. Again, what mechanism does this reaction occur by, SN1 or SN2? The molecule contains two leaving groups, both are secondary chlorides (so, so far no help). The one that is displaced is next to the oxygen. This can be explained if we invoke a cation (or oxonium species) with the lone pair on the oxygen allowing delocalisation of the positive charge. This suggest that the reaction proceeds via SN1. Once again, what is the mechanism? Trevor may have covered the 2 nd point (or, as I am now doing my lectures before Trevor...he will cover this point...) The reactions proceeds via the protonation of the oxygen to create a good leaving group. The chloride is the nucleophile and it could attack at either the secondary or primary position. It attacks at the primary position so the reaction must be SN2 (remember: primary cations are unstable). And the last one is a nasty bit of work. Again, the question is, which mechanism operates? The problem is we either have to have SN2 at a tertiary position (disfavoured) or we have to have SN1 next a carbonyl group (disfavoured)... It turns out that these types of reaction are about the only known example of SN2 at a tertiary carbon. They can proceed because the!-carbonyl group accelerates SN2 so much that it overcomes steric problems. We know this reaction is SN2 as there is inversion of stereochemistry, Hopefully this diagram shows how the orbitals of the carbonyl group and the halide align to cause the activation. Furthermore, it should be noted that the azide anion is an excellent nucleophile being a sharp, narrow cylinder of negative charge that is barely effected by steric hindrance

13 The other reaction class I want to cover in detail are the eliminations... If we treat 2-bromobutane with a base it can undergo an elimination reaction giving us hydrogen bromide (HBr) and an alkene...in fact it can give us three alkenes. It depends on the mechanism? What about this one? If we treat 1-methylcyclohexanol with an acid we get an elimination of water (H2O) and the possible formation of two different alkenes. Which one (if any) is favoured? Which one is preferred and why? 49 Again, it is all dependent on the mechanism of the reaction. 50 The first we ll cover is E1... So lets start looking at some of the elimination reactions... In this reaction the leaving group dissociates from the substrate to give a carbocation intermediate in a slow, rate determining step. The intermediate then losses a proton to give the alkene. Like the SN1 reaction, E1 is a two step process... The first step is identical to the SN1 reaction

14 Like the SN1 reaction the slow step of E1 is the initial dissociation to give the cationic intermediate. An intermediate is not only observable but in some cases can actually be isolated. It is a molecule or ion that represents a localised energy minimum; an energy barrier must be overcome before the intermediate forms a more stable species. The second step is fast and is deprotonation of the cation by a base to give an alkene. Virtually any electron donor can behave as a base in this reaction! The reaction profile is the same as the SN1 reaction, showing that the formation of the intermediate has the highest energy barrier so is the slowest step. It also reinforces the relative stability of the intermediate. (in case any of you are wondering...the film is Blade Runner ) 53 As a result the rate of reaction depends on the substrate only and not the nature of the base. 54 This means the stability of the cation is the most important factor. 55