[victor]

I just wondering about which mechanism that will be run...
If I react a primary alkyl halide with a weak nucleophile reagent like methanol, which mechanism will occur? S_N1 or S_N2?

The thing that makes me confuse is that, primary alkyl halides tends to run S_N2 mechanism just if they're reacted with strong nucleophile..but weak nucleophile like methanol tends to substitute alkyl halides only if the carbocation is formed...so?which one of nucleophilic substitution will occur if I react between primary alkyl halide with a weak nucleophilic reagent?

[wereworm73]

It depends on the primary alkyl halide.  If it was chloroethane, then you wouldn't have much of a reaction for either mechanism (only a primary carbocation is possible and it's unstable).  Allyl chloride, on the other hand, forms a reasonance-stabilized carbocation and undergoes S_N1 at a moderate rate (about the same as a secondary alkyl halide). 

[movies]

With allyl electrophiles you tend to get a lot of S_N2' reaction as well, so that's a complicated case.

[victor]

Hmm...from one side, allyl carbocation is stable enough to run S_N1 reaction but on the other side its primary structure also tends to run S_N2 mechanism..
Arggh, it makes me confused.... How can I face my organic chemistry final exam if I like this...hope you all help me guys..

Oh, I also wanna ask question about ether. Let's say that I react ethyl methyl ether with HCl. It's said that it'll run S_N2 mechanism, where the insertion of Cl^- ion is happen altogether with the breaking up of C-O bond. My question is, are those 2 processes which happen altogether occurs when O atom is protonated?
I think like this because S_N2 mechanism doesn't need any protonation...or the protonation is just an additional process which occurs because HCl releases H⁺?

[wereworm73]

For S_N2 (and S_N1) to occur, the electrophilic molecule needs a good leaving group.  With alkyl halides, there is already a good leaving group so nothing else needs to be done to set up an S_N2 reaction.  However, ethers & alcohols don't have good leaving groups, though you can create good leaving groups if you protonate them (ROH and HOH are much better leaving groups than -OR & -OH).  Once the ether or alcohol is protonated, they can undergo a S_N2 reaction the same way as alkyl halides do. 

[wereworm73]

Definitely feel free to ask if you have any questions about S_N2.  As someone who has taken organic chemistry classes already, I can tell you it's especially important to have a solid understanding of S_N2 reactions because you'll be seeing examples of them over and over.  The Williamson ether synthesis is really an S_N2 reaction involving an alkoxide and a primary alkyl halide, for instance. 

[victor]

  Thanks for the offer..um, okay..next question..
It's said that S_N1 is oftenly run together with E1 (80% : 20% for the rough measurement). From the book I see that in order to yield E1 product, a molecule which has become carbocation will remain like that until one of its beta hydrogen deattaches from that molecule and resulting a dehydrogenated molecule.
Is there any ways to yield a higher product of E1 compared to S_N1??

[wereworm73]

Sure, if you do this reaction at higher temperatures, then you can get a better yield of the E1 product.

[Donaldson Tan]

You can promote E1 by restricting nucleophilic behavior. This can be achieved by using a "bulky" protic solvent such as propanol. This is because protic solvent can form H bonds with the nucleophile. Another means of promoting E1 is to use a strong base that is more likely to abstract hydrogen than to act as a nucleophile. High temperature favours the formation of a more stable product, which in our case is the elimination product.

[victor]

Thanks..and, next question..
the book sez: aryl and vinyl halide can't undergo S_N2 process.

Then I'm just curious why vinyl can't undergo that thing? (I've already known for aryl halide.) Then, the book answered that it's because of two major factors. One is, because double bond makes a stronger bonding effect. Then the second is because of vinyl's molecular geometry (I don't understand this one).
What I can say is, double bond makes a 120º geometry which 'prevent' it from nucleophile attack...With 120º? then what? isn't it's just make vinyl are easier to be attacked from 'behind'? Or it has a corelation with pi electron bonding??

[Donaldson Tan]

Subjecting a vinyl halide to nucleophilic attack requires the C=C bond to break. This is definitely not energetically favourable. The geometry of the SN2 mechanism requires the nucleophile to attack alpha carbon from the "backside", which is not possible in the case of vinyl halide because the backside is coplanar with the halogen atom and the C=C group.

[wereworm73]

Using strong, nonnucleophilic bases actually favors the E2 reaction because they deprotonate the electrophilic molecules before they have the chance to form carbocations.

[victor]

Um, I have another question...can we make the benzene get substituted with nucleophile? I mean, in example, how can we make a phenyl methyl ether? At first, I thought Grignard reagent would solve the problem, but after trying to write the mechanism, I don't think so....

What I think now is, by substitute the benzene become nitrobenzene which has a positive charge in N atom, so that the electron in the benzene ring would get attracted towards N atom. Then, because of that, C atom from benzene ring (especially in the para position from nitro groups) will  get a more positive charge compared to others and thus, can be reacted with nucleophile. What do you think about that?

[wereworm73]

For a standard nucleophilic aromatic substitution, you'll need an electron withdrawing group (like -NO₂) that's ortho or para to a good leaving group.  The positive charge on the nitrogen (along with resonance) stabilizes the sigma complex so the nucleophilic substitution reaction goes forward.   So, you could make bromobenzene, add a nitro group (this will be ortho or para to the bromine atom) and then use sodium methoxide as a nucleophile, but the nitro group will still be on the aromatic ring.

Another way you could make the ether with benzene as the starting material is to first make bromobenzene (through electrophilic aromatic substitution) and then use the Dow process (a rather extreme type of nucleophilic aromatic substitution)  by adding two equivalents of NaOH at 350^oC.  Under these harsh conditions, the bromobenzene gets deprotonated and forms a carbanion.  The carbanion then expels the bromine and creates a benzyne intermediate.  This intermediate easily reacts with hydroxide, a strong nucleophile, to from phenoxide.  You can then use the phenoxide as a nucleophile to react with chloromethane to form phenyl methyl ether.