[BacksideAttack]

I hit a problem that seems simple but I don't know the answer and it's one of those things I'm sure would take forever to figure out, so I'll just ask:

A problem stated "which mechanistic step is least likely to occur?"

The answer was:


Can somebody give me a good reason why this reaction step is unreasonable? I've drawn it exactly as it was shown.

I can only come up with guesses, and that's not good enough.

My guess was that the molecule would be more likely to lose bromine and form a aromatic ring, but there could be a lot that I don't see.

[macman104]

We don't know what the original question was...

What was the mechanistic step in relation to?

[BacksideAttack]

The original question was "which mechanistic step is least likely to occur?"

I'd rather not post the other options, because it would take forever and because I can see why the other reactions might occur. Problem is, they all kind-of seemed likely.

The only other one I thought was questionable was:

Ph-CH2I ---(H2O, MeOH)---> Ph-CH2+

In fact, that was my answer, but it was wrong.

Literally, the only info I was given was what is in the picture. This was a multiple choice problem for a general exam.

[macman104]

Well, I'll be honest.  That mechanistic step is not that unlikely to occur.  That is a pretty standard SN1 type mechanism (formation of carbocation, attack of nucleophile).  True that if that compound can aromatize, then that would obviously be the most favored next step.  And I'm not sure how you plan to simply kick out the bromine.

But in a question that says "which step is least likely", it's hard to tell you why this is the correct answer, without seeing the other possibilities.  Are you able to possibly scan the question?

[James Newby]

Well you have two electrophilic sites on the intermediate and a bromide nucleophile:

1) the carbocation can be directly attacked to form the species you have shown

2) the bromine nucleophile can attack the H geminal to the bromine on the ring giving mono-bromo benzene.

Draw the mechanism for mono-bromo benzene formation and compare that mechanism to the one you already have.  Just for reference can you tell us how much benzene chemistry you have done?  I can give you a better hint if i know what you already understand

The other question you thought was questionable:

Carbocations next to phenyl rings are relatively stable compared to most other carbocations.  You can draw resonance forms where the positive charge is encorperated into the aromatic ring.  Iodide(-) is also very stable

Macman sorry if i repeated any of your points, this took ages to type!

[BacksideAttack]

Sorry, I had no idea I had my webcam handy. Boy would that have saved me some effort.

Here is the full problem:

[James Newby]

Well what do you think?  Can you explain which ones look correct and which appear peculiar to you?  I cant just give you the answer I'm afraid!  To me (B) is a difficult one as i didnt meet this chemistry until my 3rd year of uni so ill give you that one.  B is likely to occur

[BacksideAttack]

...And I'm not sure how you plan to simply kick out the bromine.

What bromine are you referring to? In the reaction bromine is not lost, but gained.

...Carbocations next to phenyl rings are relatively stable ...

I'm aware that iodide is a great leaving group, but I didn't know it would just up and leave if you dropped benzyl iodide in some wet alcohol. I suppose if I knew that benzyl cations were particularly stable, then that would have been a giveaway, now I know

Maybe I am supposed to consider how the intermediate was formed, could somebody guess how I would have formed that carbocation?

I would have thought that the bromide would attack the C+ right way, I just can't see why it wouldn't.

As for my experience, I'd say it is extremely limited. I have a BS chem degree that I may as well have picked up off the corner for $5. Also, it has been 3 years since I was in school, and 6 years since I finished Ochem.

Oh, and I know the answer, that is not the problem. I now I just need to know why.

A seems like a familiar reaction (I'm afraid I must rely on intuition)
B seems like an obvious hydride transfer to stabilize the cation.
C looks exactly like any free radical halogenation, although I couldn't back up why it would attack that site in particular.
D looked perfectly reasonable as the second step of an electrophilic addition (hydrohalogenation), except that it would have started with benzene, which wouldn't happen. If it was the second step of Sn1, then sure bromine would attack it, except what would have left in the first step?  I can see bromine leaving in this step in favor of forming an aromatic system, but I'm not sure if that is what I was supposed to see.
E I didn't think this would happen at all, in a neutral alcoholic solvent, but apparently benzylic carbocations are particularly stable.

[macman104]

James, do you not take organic chemistry until your 3rd year?  I would think that would be standard 1st semester organic knowledge, as it is an integral part of SN1 mechanisms.

Bromine Removal:  You were talking about forming the aromatic system by losing bromine.  I thought you were referring to the bromine still on the ring being lost.  This could happen, but I've only seen it as a dehalogenation, which usually follows an E2 type mechanism.

But I will agree with you BacksideAttack, looking at the rest of those, I think D is probably the least likely.  You'll notice that all of them COULD happen (none of them are impossible), which may be part of the difficulty in this question.

A, B, and C to me seem very plausible.  The first one is a nucleophile attacking a carbonyl, second is hydride shifts of a primary carbocation to a tertiary, and the third is allylic hydrogen abstraction.

As James said, benzylic carbocations are more stable (more than tertiary carbocations too), because of the delocalization throughout the ring.

Which really leaves D.  Like you said, formation of the aromatic system would be much more favorable than the addition of the bromine.

**And...some of this is repeat to what you just said, but I'm too lazy to type a new response **

And A is a familiar reaction probably:
http://www.organic-chemistry.org/namedreactions/stetter-reaction.shtm
But regardless, it's a reasonable strength nucleophilic attack on a carbonyl = reasonable.

[James Newby]

Yea i have done loads of SN1 but must have missed the bit on protons moving around! Had a course on stereochemistry last semester and it came up a lot.

The answer is D.  If you look into the mechanism of bromination of benzene you will see this is the intermediate before reforming the aromatic system.  The benefit of reforming the aromatic ring is huge, and some molecules will almost fragment in order to get to an aromatic ring system.  Its why such harsh/specialised conditions are needed to get benzene to do anything.

C attacks here because the radical formed is stabilised by the alkene adjacent to it.  Radicals will pretty much charge into anything they come into contact with.

E is a polar solvent (water) and the carbon iodine bond is so weak in this molecule it can almost be modelled as ionic

[BacksideAttack]

I thought that benzyl iodide would form benzyl alcohol in water, I suppose that is why I thought it was most likely to be wrong. By the way, why wouldn't that happen?

Thanks for the help everybody, it is nice to be able to discuss chemistry with real people. I'm the only sort-of chemist that I know, and it has been a lonely frustrating hobby for me by all means.

[macman104]

It probably will actually, or the ester (although, williamson is really more of SN2 reaction).  However, the mechanism is likely of an SN1 type, especially influenced by the stability of the carbocation and the solvents.

I have actually run a reaction of a propyl iodide, with AgNO3 and refluxing methanol/water mixture to make the alcohol.

[nj_bartel]

Yea i have done loads of SN1 but must have missed the bit on protons moving around! Had a course on stereochemistry last semester and it came up a lot.

Hydrides* =P