[souro10]

Let us consider ring opening of Propylene oxide with conc. HCl --- Case (i)

At first the epoxide oxygen gets protonated, making it a good leaving group. Then, Cl- being a strong nucleophile attacks at the less substituted end, and opens up the ring. What we get is 1-Chloro-2-Propanol.
In this case the reaction is SN₂ and the outcome is SN₂-like too.

I appreciate that if the nucleophile would have been something weaker like water or methanol, instead of Cl- , then the reaction out-come would have been SN1-like. I have been told that the reason for this is that, Cl- being a strong nucleophile can directly make the ring rupture. While water being a nucleophile cannot make the ring rupture and hence attacks at the delta-plus carbon. Okay, accepted.

Now, let us consider acid catalyzed opening of Isobutylene oxide with conc. HCl. ---- Case (ii)

Here, after protonation of epoxide oxygen, Cl- attacks at the more substituted end.
If the previous rationalization is true, then in this case also, Cl- being a strong nucleophile should have attacked at the less-hindered end and should have made the ring rupture. But, it didn't.

That leaves me confused.

Any help?

[orgopete]

This is a little complicated, but I'll give it a try. All reactions are electronically neutral, so whether we chose to designate them as being driven by cations or anions can seem arbitrary. However, we may still discern that reactions are driven by acids or Lewis acids or bases/nucleophiles.

What is given is some data and we are to decide whether the reaction is driven by nucleophilic attack or formation of a cation, even if neither cation or anion may be completely formed. If protonation of propylene oxide gave a cation, then the product might have been 2-chloro-1-propanol. If the rate limiting step is ring opening by the chloride, then the alternate product may form.

If the same analysis were applied to isobutylene oxide, a different set of factors present themselves. Opening the epoxide can give a tertiary cation and alter the opening step. This seems plausible, even if no bona fide carbocation is actually formed.

[souro10]

Well, that is just another way to present the facts, isn't it?

What is not clear from your explanation is: why in one case the reaction is driven by attack of chloride on less-hindered carbon, while in another case the same logic can-not be applied?

However,  with the help of your inputs, I can now try to rationalize the products as follows:

We are to note that the nucleophile, in each case can attack at two places, both of which have partial positive charge. Of these two sites of attack, the more substituted carbon has greater positive charge density.

Now, a secondary carbocation-or a carbon atom closely resembling a secondary carbocation is not excellent for either SN₁ or SN₂.

Now, reaction conditions in each case is such that there is a high concentration of the nucleophile. This is essential for a good SN₂.

Given these, we try to rationalize the products in each case.

In the first case, an excellent SN₂ reaction can happen if the nucleophile attacks at the less hindered carbon- a primary carbon + good conc. of nucleophile. However, attack at secondary end might result in an SN1 (-like ) reaction - but such an SN₁ reaction should be slow. ¹

In the second case, an excellent SN₁ reaction can happen if the nucleophile attacks at the more substituted carbon- tertiary carbon, polar solvent. However, attack at primary end might result in an SN-₂(like) reaction - but SN₁ wins over SN₂ because a good SN₁ reaction is faster than the fastest SN₂ reaction

¹ Now, both the reactions are experimentally SN₂ in nature; Having said that, we may now appear to look at the situation closely- especially in the second case where the product seems to be SN₁ like. Now this may be subject to different explanations, but a plausible explanation I thought of waswith the help of ion-pair concept. In the tertiary carbon of the second case, the C-O bond is almost broken to form an intimate-ion-pair. The intimate-ion-pair, in this case proceeds till almost-but not yet- at the verge of forming a carbocation. This is the loose-ion-pair. At this stage, the nucleophile attacks at this carbon, and hence the rate expression involves concentration of the nucleophile because slight C-O bonding character is still present.

Now, why it prefers a loose ion-pair over a tertiary carbon, demands explanation.
A plausible explanation for this could be, in the bonded state, octet for all atoms is complete. The protonation of epoxide's oxygen makes the bonding unsymmetrical and weak, but a loose-ion-pair would still be more stable than a carbocation.

Inputs appreciated!

[opsomath]

For the curious, there's a JCE paper on the topic:

http://pubs.acs.org/doi/abs/10.1021/ed072p624

I, like others reading this thread, was surprised to learn that 1-chloro-2-propanol was the major product of this reaction. (propylene oxide + HCl). However, the ratio is only about 1.5:1, so it is not dramatic. Apparently a single methyl group is not enough to drop the barrier to the "Sn1-like" pathway to where it efficiently competes with the Sn2 pathway, but two methyl groups (isobutylene oxide) are sufficient.

Much of mechanistic organic chemistry is about understanding correctly what happens. True predictive power is not always a possibility.

[souro10]

For the curious, there's a JCE paper on the topic:

http://pubs.acs.org/doi/abs/10.1021/ed072p624

I, like others reading this thread, was surprised to learn that 1-chloro-2-propanol was the major product of this reaction. (propylene oxide + HCl). However, the ratio is only about 1.5:1, so it is not dramatic. Apparently a single methyl group is not enough to drop the barrier to the "Sn1-like" pathway to where it efficiently competes with the Sn2 pathway, but two methyl groups (isobutylene oxide) are sufficient.

Much of mechanistic organic chemistry is about understanding correctly what happens. True predictive power is not always a possibility.

That is true but here the understanding is not clear, at least to me, hence the thread.

Nucleophile power definitely plays an important role. - During acid catalyzed ring-opening of propylene oxide in methanol or water, the solvent molecules attack at the more-substituted end. However, chloride attacks at less substituted end.

[orgopete]

As I suggested, this can be complicated.

We are to note that the nucleophile, in each case can attack at two places, both of which have partial positive charge. Of these two sites of attack, the more substituted carbon has greater positive charge density.

That is the conventional argument. I prefer a different rationale. I argue as follows. Carbon is electron donating, so a tertiary halide has the greatest electron density about its carbon. Hence, a nucleophile faces the greatest resistance to attack at that carbon. A methyl halide has the lowest electron density about its carbon and nucleophiles can most easily attack its carbon (the most positive). In order for a nucleophile to attack a tertiary carbon, the C-X bond must be broken or partially broken. This will flatten the groups attached to the carbon and expose it to attack. Conditions that enable bond breakage are required in order to allow this attack to occur.

If that were the case, then one could not simply argue attack is due to the effectiveness of a nucleophile. That is, even a very good nucleophile cannot overcome the electrostatic resistance of the electrons surrounding a tertiary carbon. Attack must be the combination of bond formation and bond cleavage. If this is described by kinetics, we would find that SN1 reactions have a high degree of bond cleavage such that the nucleophile does not enter the rate equation. Conversely, if bond breakage has not occurred to any appreciable extent, then a nucleophile must first attack a carbon before bond breakage can occur. I argue that attack under these conditions should occur at the most positive carbon or the carbon with the least electrostatic resistance, e.g., a primary or methyl halide. 

From what I have read, if a chiral tertiary halide were subjected to an SN1 reaction, the products may still retain optical activity. That is, even though we may write a discreet carbocation intermediate, bond formation may actually occur faster than diffusion of a freely rotating carbocation. That is, even SN1 reactions may retain some SN2-like inversion.

Okay, what might happen if the leaving group is tethered to the adjacent carbon? I should not be surprised that the products would appear to be SN2-like in their stereochemistry, but I don't think this reflects an improved nucleophilicity.

Now, reaction conditions in each case is such that there is a high concentration of the nucleophile. This is essential for a good SN₂.

I consider this a kinetic effect, not an atomic effect. Even at low concentrations, methyl iodide will undergo an SN2 reaction. It will not change mechanisms. Bond breakage leads to SN1 reactions and in its absence, SN2.

... a plausible explanation I thought of was with the help of ion-pair concept. In the tertiary carbon of the second case, the C-O bond is almost broken to form an intimate-ion-pair. The intimate-ion-pair, in this case proceeds till almost-but not yet- at the verge of forming a carbocation. This is the loose-ion-pair. At this stage, the nucleophile attacks at this carbon, and hence the rate expression involves concentration of the nucleophile because slight C-O bonding character is still present.

Now, why it prefers a loose ion-pair over a tertiary carbon, demands explanation.
A plausible explanation for this could be, in the bonded state, octet for all atoms is complete. The protonation of epoxide's oxygen makes the bonding unsymmetrical and weak, but a loose-ion-pair would still be more stable than a carbocation.

Now, I want to contradict myself. I referred to bond formation and cleavage. I don't necessarily think of bonds as bonds. I prefer to argue about forces. In this case, since single electron reactions are not implicated, I prefer to think about how the electrostatic forces and distances may contribute to the outcome. If that is the case, then ions and covalent bonding descriptions disappear. We need not refer to a loose ion pair. Attack can only occur at atoms in which the electrostatic repulsions are lowest. Conditions that increase a carbon-electron pair distance will have a weaker force. That is, an increase in a carbon-electron pair distance will flatten the carbon and increase its susceptibility to attack.

Nucleophile power definitely plays an important role. - During acid catalyzed ring-opening of propylene oxide in methanol or water, the solvent molecules attack at the more-substituted end. However, chloride attacks at less substituted end.

How does this fit in? In my classes, I used to argue that we should expect bromohydrin formation in the bromination of an alkene in water. I argued that we knew water was more basic than bromide as the equilibrium of HBr and water greatly favored hydronium ion, therefore we might expect a bromonium ion to be opened by water and not bromide ion.

If I extend this argument to the opening of an epoxide with HCl, then I might not expect chloride to be all that effective as a nucleophile, per se. It is a nucleophile, but not necessarily one that exerts a significant kinetic effect. (Someone might be able to verify this. If an epoxide is reacted with HCl in the presence of NaCl, then the rate should be increased if it is an SN2 controlled reaction. If the rate determining step is related to the concentration of protonated epoxide, then I might infer that even though the carbon attacked is typical of an SN2 reaction, the actual driver is the relative weakness of the bond breakage.)

If (and this may seem more like a house of cards) water or methanol were the nucleophile, then it should be a better nucleophile than chloride. If the reaction were an SN2 reaction, then water or methanol should attack at the primary carbon. In the other example cited, this seems to not be the case. If I continue with my prior thought, then I may infer the conditions for opening in water or methanol may shift to a greater degree of bond breakage, that is the electrons are pulled further from the more substituted carbon than the HCl reaction. This could be considered consistent with the a typical SN1 reaction. Tertiary halides are more stable in SN2 conditions than SN1.

Can the opening of an epoxide be classified as strictly SN1 or SN2? I don't think it can.

[souro10]

I have a few questions before I accept your argument fully.

1. The first part of your argument deals with tertiary and primary halides. You are trying to put the same logic in case of protonated epoxides. My question is, for protonated epoxides aren't things quite different? The positive charge density is more on the tertiary carbon or secondary carbon than the primary carbon, because of the "resonance structures" of the protonated epoxides, and their relative stabilities, isn't it?

2. Considering things to be as clean-and-simple, your argument should hold true for Bromonium ions. How do you account for the difference, that Cl- attacks at the secondary carbon in case of the boromonium ion formed by reaction between propene and Br+ ?

[orgopete]

I have a few questions before I accept your argument fully.

1. The first part of your argument deals with tertiary and primary halides. You are trying to put the same logic in case of protonated epoxides. My question is, for protonated epoxides aren't things quite different? The positive charge density is more on the tertiary carbon or secondary carbon than the primary carbon, because of the "resonance structures" of the protonated epoxides, and their relative stabilities, isn't it?

I argue the conditions for any substitution reaction should be guided by the same principles. The electrons of a nucleophile should be attracted to the carbon whose nucleus is most exposed to attack. You can use reaction data to learn about how groups or conditions may contribute to that effect. For example, fluorine must only weakly pull electrons away from a carbon compared to bromine, hence attack is more facile if bromide is the leaving group.

Substitution reaction occur faster at primary carbons than secondary or tertiary carbons in SN2 reactions. However, the reverse is true for SN1 reactions. If electrons attack the most exposed nucleus, then bond cleavage must remove electron density from secondary and tertiary carbons in order to overcome the electrostatic repulsion of the more hindered carbons. This seems consistent with bond lengths and our notion of carbon as an electron donor.

I may not know the charge density for any given reaction, but if the principle I am arguing were to hold true, then the products must be indicating the ability of the nucleophile to donate electrons and the extent of bond cleavage that may be occurring in a reaction. If a protonated epoxide is giving the product of attack at the primary carbon, then bond cleavage must not be occurring to a significant extent under those conditions. Changing the nucleophile probably cannot be done without affecting the protonated epoxide. A loss of water may occur differently in water or alcohol than in HBr. Certainly loss of bromide from t-butyl bromide is different in aqueous acetone than in alcohol or acetone.

These reactions are barometers of electron density and movement. From the products, we learn about the electrons.

2. Considering things to be as clean-and-simple, your argument should hold true for Bromonium ions. How do you account for the difference, that Cl- attacks at the secondary carbon in case of the bromonium ion formed by reaction between propene and Br+ ?

If the reaction principle were to hold, then a greater amount of attack at a secondary carbon in preference to a primary carbon should only occur if the nucleus of the secondary carbon were more exposed. Certainly 1-bromobutane must react faster than 2-bromo. If that were to reverse, then the nucleus of the secondary carbon must be more exposed. If chloride prefers to attack at the secondary carbon, then a greater degree of bond cleavage must be occurring than in another completing reaction. I would accept that a bromonium ion may have a longer C-electron distance at the secondary carbon and expose the carbon to attack than an epoxide or protonated epoxide than at the primary carbon.

One must be cautious in thinking about charge. Ammonia, without a charge, is more basic than Br(-). The charge of two electrons is exactly the same despite the net number of protons. I think of electrons as being like a boxer, the longer the reach, the better they can hit you. The further electrons may extend beyond the positive field of its nucleus, the better they can attack another atom. A pair of electrons has the same charge irrespective of the net charge of the atom.

This should apply to a nucleus. As electrons are pulled from a nucleus, that may expose the nucleus to attack. An attack may occur without fully forming a cation. Bromine can react with bromide to form Br3(-) without forming a Br(+) or its mechanistic equivalent in the formation of a bromonium ion. It seems to me that we should seek to understand the forces (energy levels) in terms of charge and distance (inverse square force). I believe that these forces are not well understood in those terms. (I had recently argued a different model for atomic radii. I find it quite irrational that ionic distances should be much larger than the electrons of a neutral atom. A challenge for most chemists is to determine a reasonable atomic radius. If we cannot agree upon an atomic radius, using distance in a force argument can give variable results.)