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I am supposed to balance this equation to reprsent the reaction aswell as draw and energy diagram which labels the energies of each step in the reaction and activation energy
Will do. How would do the second part of the question? Label the the energies of each step of the reaction and activation energy.
Well, I think A would be your reactants, B would be the transition state, and C would be your products.. but do you know why?
Ohhh, OK I didn't look at the question that way. So what you are essentially saying is, A would be the process of using the catalyst to kick start the reaction. B would then be the peak of the reaction and C would be the reactions end results.
Catalyst lowers activation energy Ea of both the forward and reverse reactions. I don't remember what the colors/atoms represent for the ball stick model.
that is what essentially occurred, but what you really need to focus on are the reactants from A and the product in C balance those two
I am just assuming that it would be Chlorine, but it can be any halogen with either F or Cl
F, Cl, Br, I name your big ball
Seems like you have SN2 reaction going on there..
since the ball is much much bigger than C, it may be a halide on the third or 4th row
i see now , but that carbon at the bottom has 5 bonds.. that must be your Transition state.
it would be an \(\sf SN_2 \) and creating a true bond between the halide will violate the octet rule, so I am positing that the interaction was due to induction.
the carbon in methyl is slightly partially positively and that can provide a little attraction with the halide.
ok.. It looked like that was a bond there makes sense!
I didn't draw a curved arrow from halide to the partial positive carbon :)
in any case I will say that the products are \(\sf Me-OH + Cl^-\)
our charge was preserved (-)
I mean if that's a concerted mechanism.. practically happens in one step so .. like Cl- would be able to hold - charge good LG..
oh that did not look right let me type it again \(\sf Me–OH + Cl^- \) |dw:1436680471791:dw|
just curious.. how would you justify the opposite reaction
because i see two arrows, which implies the opposite reaction, I mean chlorine's pretty happy with that - charge
I shouldn't probably draw the last reversible arrow in the end because OH wouldn't be a good leaving group for a reversible reaction
OK. he asked about energy of the reactants vs products... technically if this process is favorable the nucleophilic attack on the on Methylchloride, then wouldnt this mean that the product is more stable, at a lower energy, than the reactant?
\(\sf Cl^- \) is a pretty stable base (weak), and initially \(\sf OH^- \) was a strong nucleophile
Ah.. I see.. that double dagger I think that's what they call it.. HO-C bond breaking, while the Cl- C bond is forming.
I guess if you think about it the opposite way you can't actually just kick off OH- unless you introduced some proton source.
and you are correct, the product is most likely more stable and our graph would look like this |dw:1436680983637:dw|
YOU sir NAILED this one
you are correct. OH is a bad leaving group so we're going to need a good source of energy to overcome the barrier, which translates to a good proton donor
my organic chemistry is pretty crappy so I am refreshing with some problems.
crappy is a relative. that was well explained
the double dagger that you mention is the opposite it is the concerted \(OH^- \) forming a bond to C whilst Cl breaking
mixed that up nucleophile was OH- so it's forming a bond haha
Sorry Im late. It was really late when I posted. Question, how did you determine what kind of chemical this was from the picture?
And also, that last graph you posted would be what they asked for ?
And the first would be the balanced equation?