[x1101x]

Alright I'm in organic chemistry right. I'm trying to study for an exam so I do some practice problems. In this one problem I'm given an equation with a reactant, reagent, and solvent, and I'm told to find the product. I'm really f&#^$*@ positive that it is supposed to undergo an E2 reaction, but when I check the answer it evidently is supposed to undergo an Sn2 reaction. This is extremely frustrating because one of the few things that I thought I knew turns out to be false and adds to my growing suspicion that it really is not possible to determine how a reaction will occur without actually running it in a lab.

I'm not entirely sure what the name of the reactant is but I'll try my best to describe it to you,
its a benzene ring with a single bond to another carbon, which is bonded to a chlorine, a hydrogen and then a ethyl group (CH₂CH₃)

So basically something like this
           Cl    H
             \  /
              C
            /   \
 benzene     CH₂CH₃
The Cl is facing OUT, and the H is facing IN

The reagent (the thing above the arrow) is a carbon double bonded to an oxygen, single bonded to a methyl group, and single bonded to an oxygen with a negative charge, balanced by a sodium, or something like this
         O
        ||
        C
      /   \
Na O     CH₃


Lastly the solvent (or at least I think it is the solvent, I don't entirely know my chem teacher sucks, its below the arrow of the reaction so maybe it just means its taking place in acidic conditions) is this:
         O
         ||
         C
       /   \
H - O     CH₃

Lastly it notes that I need to find the single major product leading me to believe that an Sn2 or E2 reaction must occur

So there are my crude drawings and explanations, here's what I THINK should happen. Obviously chlorine is the leaving group so that's gonna leave. Its bonded to a 2nd degree substrated carbon and the solvent is Protic (has an OH bond) so that means it should either be E2 or a combination of Sn1 and E1. Lastly the reactant is a good nucleophile so it should under go the E2 reaction which means the final product should be:

              H
              |
              C
            /   \\
 benzene     CHCH₃

With no stereo chemistry to consider (trigonal planar)

The \\ is supposed to  mean double bond.

HOWEVER, when I check the answer it says the product should be:
                     O
                    ||
                    C
                   /  \
          H    O      CH₃
           \  /
            C
           /  \
benzene     CH₂CH₃

with the H facing out and carboxylic acid group facing in
Obviously the reaction must have undergone the Sn2 reaction (or maybe not what the hell do i know)

So, to sum up my question why Sn2 and not E2???

EDIT: Just remembered that E2 requires the anti stereo chem (and also the acidic conditions might make it impossible to give up an H), but even still that means the reaction would have to undergo SN1 and or E1, but its not E1, and if it was SN1 there would be two major products (a racemic mixture) so how in the hell can there be a single major product (as said in the original question)

[Dan]

Hmm, to me it looks like anything but E2.

1. There are no strong bases around to facilitate E2, so that's out.
2. The Nu is reasonable, but not great, S_N2 is a maybe.
3. The LG is pretty good, doesn't narrow it down much.
4. The solvent is polar-protic - stabilises Nu (making it less reactive), LG, and potential carbocation - favours S_N1/E1.
5. If you generate a carbocation, it will be secondary but also benzylic - a secondary carbocation stabilised by resonance with the benzene ring. That looks viable.
6. In acidic conditions, the olefin formed by E1 could protonate and regenerate the carbocation, which could be trapped by acetate. Acetate is not a very good LG, so the ester you get should be quite stable.

Overall, I would predict mainly S_N1 under these conditions, with perhaps some S_N2 as well (which is frequently observed for benzylic secondary chlorides). If there is some S_N2 as well as S_N1, you will get incomplete racemisation (or if you prefer, incomplete inversion) - overall, most of your product will be inverted, but not all.

For example, even if the rate of S_N1 is four times the rate of S_N2 we might expect 80% racemisation and 20% inversion - this is overall 60% inversion and 40% retention of stereochemistry - so the major product is the "S_N2 product" even though S_N2 is not the dominant pathway.