I can apply a knowledge of the rate equations for the hydrolysis of halogenoalkanes to deduce the mechanisms for primary and tertiary halogenoalkane hydrolysis and to deduce the mechanism for the reaction between propanone and iodine (4.3h)

primary and tertiary halogenoalkane hydrolysis = nucleophilic substitution reactions

Nucleophilic substitution mechanisms of halogenoalkanes Two substitution mechanisms are possible when a bromoalkane reacts with aqueous alkali: R- Br + OH- → R-OH + Br Only the experimental rate data can show which mechanism actually takes place.

d- d+ d- : d+ Nucleophiles Nucleophiles are species that have a lone pair of electrons available to form a covalent bond with electrophiles. They will be found with either a full or a partial negative charge on one atom. Ammonia as a nucleophile reacting with chloromethane This bond will break d- d+ Lone pair forming a bond d- : d+ Both molecules are polar because of the high electronegativity of N and Cl compared to H and C respectively. Nucleophiles are species that have a lone pair of electrons available to form a covalent bond with electrophiles. They will be found with either a full or a partial negative charge on one atom. In this example, ammonia is attacking chloromethane. Ammonia can act as a nucleophile because it has a d- nitrogen atom and a lone pair of electrons. One curly arrow shows the formation of a bond to the carbon atom which itself is d+ due to the electonegativity of the neighbouring chlorine atom. This bond will be further polarised by the approaching ammonia so the polarity is not that relevant as I'll mention later. As the new bond is formed, the C-Cl bond will break at the same time. The Cl atom is substituted out so the process is known as nucleophilic substitution. The Cl atom here is acting as a leaving group and although this may not have been mentioned at A-level, how good your leaving group is is really what infers nucleophilicity on our electrophile, not the partial positive charge. You will have seen at A-level that Br2 can attack an alkene without being polar because it is readily polarised as a consequence of the way in which the two molecules meet. This kind of behavior goes beyond neutral species though. Although it seems counter-intuitive, you can in fact have electrophiles with a negative charge as long as they have an effective leaving group. A glance at the molecular orbitals shows the HOMO of the ammonia is the lone pair and that it can overlap with the LUMO on the chloromethane. Observe the lobe behind the carbon atom ready to form a bond. Note that this orbital is antibonding with respect to the C-Cl bond which explains why that bond breaks when electrons go in here. Because the Ammonia can only attack from this side, there is a reversal in the stereochemistry around the carbon like an umbrella being turned inside-out. This is known as a Walden Inversion. This type of reaction is known as nucleophilic substitution. A nucleophile attacks an electrophile forming a bond. The carbon atom can only support 4 bonds so another is broken and a leaving group (here Cl) is substituted out. © The Royal Society of Chemistry 2010-11

Nucleophilic Substitution: SN2 There are two major pathways a nucleophilic substitution can take… Nucleophile Lone pair(s) Negative charge Electrophile Partial positive charge Br leaving group Rate = k[H3CBr][OH-] Rate determining step contains both molecules* SN2 Bimolecular Nucleophilic Substitution *only one step (correctly) assuming there are no faster steps after this one! Transition State Nucleophilic substitution can follow two main pathways which involve everything happening in one step or over two separate steps. We call these pathways “mechanisms”. In the example we just saw, the whole process occurred together. If we were to measure the rate of the reaction whilst changing the concentrations of each of the reactants we would find that they both affect the speed at which the reaction takes place. This tells us they are both involved in the slowest, “rate determining” step and as there are only two reactants there must only be one step! The reaction proceeds via a transition state where there is a bond being formed and broken. This transition state is not a stable molecule you could store in a jar although we do have techniques for trapping and identifying high-energy intermediate species in reactions. Imagine the reaction profile as a hill and the chemicals as a ball. On the plateau to the left, our reactants will sit quite happily. If we input some energy, we can push our imaginary ball to the top of the hill but once there it will always slide off one side or the other. As two molecules are involved in the rate determining step, we call this “bimolecular nucleophilic substitution” or “SN2”. E Animation is © 2001 by Daniel J. Berger

Nucleophilic Substitution: SN1 There are two major pathways a nucleophilic substitution can take… slow fast Compare this animation to the previous one. Which type of mechanism will produce a pure product and which could produce a mixture of products? Carbocation Positively charged carbon Rate = k[(H3C)3CBr] Slow first step involving only (H3C)3CBr followed by fast addition of nucleophile SN1 Unimolecular Nucleophilic Substitution Vacant p-orbital Another pathway a nucleophilic substitution can follow involves two different steps. The rate equation shows that only the haloalkane is involved in the rate determining step so something must happen to that molecule which has a large activation energy and is therefore slow before a second, faster step with a lower activation energy can proceed. In this case, the slow step with the large activation energy involves the breaking of the C-Br bond and happens independently of the hydroxide ion. This leads to the formation of a reactive carbocation which the hydroxide ion can attack to form the product. Notice that the chemical in this example forms a tertiary carbocation (i.e. there are three other carbons surrounding the carbon with the positive charge) whereas if this mechanism had occurred in the last example, there would have been a primary carbocation (as there was only one adjoining carbon atom). This is no accident and will be explained on the next slide. The carbocation is planar and has a p-orbital with no electrons in it. Notice also that there is a reactive intermediate here so our imaginary ball can settle for a short period in this hump but it only takes the slightest nudge to push it to the products. As you might guess, because only one molecule is involved in the rate determining step, we call this mechanism "SN1" E Two steps, the first has a larger Ea so happens much more slowly Animation is © 2001 by Daniel J. Berger

SN1 – These reactions involve only one of the reactants in the rate determining step (SUBSTITUTION NUCLEOPHILIC FIRST ORDER) This applies to tertiary haloalkanes SN2 – These reactions involve two reactants in the rate determining step (SUBSTITUTION NUCLEOPHILIC SECOND ORDER) This applies to primary haloalkanes

Nucleophilic Substitution: Why does SN1 happen? As we move from a primary to a tertiary carbon two things happen... The carbon atom that is to be attacked becomes more sterically hindered by the surrounding atoms. The carbocation becomes more stable due to inductive effects and hyperconjugation Although not usually referred to as a polar bond, there is a difference in electronegativity between C + H which results in charge being pushed towards the vacant p-orbital to stabilise it. This is an inductive effect Electrons in adjoining s-bonds to hydrogens that are b to the carbocation can stabilise it by forming an extended molecular orbital. This is called hyperconjugation We can explain why primary haloalkanes tend to undergo SN2 and tertiary haloalkanes SN1 by comparing the stability of the respective carbocations and the ease with which a nucleophile can attack the haloalkane. As long as pop-ups aren't blocked, you should see a new window appear now which will give you a 3D look at our tertiary haloalkane. Consider first where the nucleophile needs to attack. The lobe of the molecular orbital indicated in the main slide is where the nucleophile needs to get to but it's protected by the surrounding hydrogen atoms on the methyl groups – we say it is sterically hindered – there's no way the HOMO of a nucleophile will achieve effective overlap with this lobe of this orbital. You can now close the pop-up window. Once the carbocation has been formed it can be stabilised in two ways. Although there is not a significant difference in electronegativity between carbon and hydrogen, there is a slight difference. Methyl groups as a result can weakly push charge towards adjoining atoms. Will call this an inductive effect. Also, adjoining s-bonds (highlighted here in red) can stabilise the vacant p-orbital by forming an extended molecular orbital – this is known as hyperconjugation. You could imagine hyperconjugation as being a little like 2 resonance forms which both tell us a little about the whole picture of what's going on. To some extent, the C-H bonding electrons can move towards the positive charge to stabilise it and this latter picture is an extreme form of that. In reality what we see is a blend of the two resonsance forms that will have more in common with the left hand picture. © The Royal Society of Chemistry 2010-11

What factors affect the speed of nucleophilic substitution? How attractive will the carbon be to the nucleophile? Bond Polarity – increases up the group Image by: Lupo AgCl precipitation How easy will it be to break the carbon-halogen bond? Bond Strength – increases up the group. i.e. how effective is our leaving group? A simple experiment can be performed to determine which of these factors is the most important using silver nitrate solution which will form silver halide precipitates in the presence of the halide ions released in the reaction. There are two things to consider that will affect the speed of nucleophilic substitution – one is the attractiveness of the + carbon to the nucleophile and one is how easily the carbon-halogen bond is broken. As we move towards the top of the halogens, the strength of the bond will increase as the bond will be shorter this would suggest the higher halogens would form less reactive haloalkanes – however they will also create a more polar bond having the reverse effect. One way of seeing which effect wins out is by adding some silver nitrate solution to the reacting mixture. As the halogen is substituted out as a halide ion, it will form a silver halide precipitate. Experiments show that AgI precipitates form first suggesting that the strength of the bond to be broken is more important than its polarity. This echoes what I suggested in an earlier slide. The existence of the partial positive charge is a bit of a red herring - it is the existence of an effective leaving group which infers electrophilicity on our nucleophile. In fact this point can be stressed even further if you look back at the electronegativity chart I included earlier. Take a close look at carbon and iodine. Although iodomethane is the fastest of the haloalkanes to react, carbon and iodine have identical electronegativities, so the bond isn’t even polar! A precipitate will be formed from the reaction of CH3I before it will from CH3Cl indicating that bond strength is more important to the rate of the reaction. Image by: Dr. T

How would they expect you to do this? I can use kinetic data as evidence for SN1 or SN2 mechanisms in the nucleophilic substitution reactions of halogenoalkanes (4.3j) How would they expect you to do this?

I can demonstrate that a proposed mechanism for the reaction between propanone and iodine is consistent with the data from the experiment in 4.3e (4.3i)

rate = k[CH3COCH3 ][H+] What mechanism would fit this reaction? Look back at the results from the experiment for the propanone and iodine reaction that we carried out. rate = k[CH3COCH3 ][H+] (zero order with respect to iodine) What mechanism would fit this reaction? The mechanism must be consistent with the evidence: • if the reaction is second order overall it must involve two different species The rate determining (slow) step must contain both chemical species. These are CH3COCH3(aq) and H+(aq). The mechanism will contain a step with these two species forming a new species. E.g. CH3COCH3 + H+  CH3COH+CH3

Proposed mechanism for the reaction between iodine and propanone

From a rate equation you can decide on one step in the mechanism From a rate equation you can decide on one step in the mechanism. This will be the slow, rate determining step and must contain all species in the rate equation. Any sensible species can be chosen as possible products, only further experiment can show if they really are present in the reaction