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flexastexas
 one year ago
Can someone help me balance this chemical equation?
flexastexas
 one year ago
Can someone help me balance this chemical equation?

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flexastexas
 one year ago
Best ResponseYou've already chosen the best response.0I 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

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Just refer this and try to solve it : Basic Tutorial on balancing chemical equations : http://openstudy.com/study#/updates/559f06b5e4b05bcb08a14634

flexastexas
 one year ago
Best ResponseYou've already chosen the best response.0Will do. How would do the second part of the question? Label the the energies of each step of the reaction and activation energy.

Photon336
 one year ago
Best ResponseYou've already chosen the best response.2Well, I think A would be your reactants, B would be the transition state, and C would be your products.. but do you know why?

flexastexas
 one year ago
Best ResponseYou've already chosen the best response.0Ohhh, 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.

nincompoop
 one year ago
Best ResponseYou've already chosen the best response.1dw:1436679303437:dw

Photon336
 one year ago
Best ResponseYou've already chosen the best response.2Catalyst 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.

nincompoop
 one year ago
Best ResponseYou've already chosen the best response.1that 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

nincompoop
 one year ago
Best ResponseYou've already chosen the best response.1I am just assuming that it would be Chlorine, but it can be any halogen with either F or Cl

nincompoop
 one year ago
Best ResponseYou've already chosen the best response.1F, Cl, Br, I name your big ball

Photon336
 one year ago
Best ResponseYou've already chosen the best response.2Seems like you have SN2 reaction going on there..

nincompoop
 one year ago
Best ResponseYou've already chosen the best response.1since the ball is much much bigger than C, it may be a halide on the third or 4th row

Photon336
 one year ago
Best ResponseYou've already chosen the best response.2i see now , but that carbon at the bottom has 5 bonds.. that must be your Transition state.

nincompoop
 one year ago
Best ResponseYou've already chosen the best response.1it 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.

nincompoop
 one year ago
Best ResponseYou've already chosen the best response.1the carbon in methyl is slightly partially positively and that can provide a little attraction with the halide.

Photon336
 one year ago
Best ResponseYou've already chosen the best response.2ok.. It looked like that was a bond there makes sense!

nincompoop
 one year ago
Best ResponseYou've already chosen the best response.1I didn't draw a curved arrow from halide to the partial positive carbon :)

nincompoop
 one year ago
Best ResponseYou've already chosen the best response.1in any case I will say that the products are \(\sf MeOH + Cl^\)

nincompoop
 one year ago
Best ResponseYou've already chosen the best response.1our charge was preserved ()

Photon336
 one year ago
Best ResponseYou've already chosen the best response.2I mean if that's a concerted mechanism.. practically happens in one step so .. like Cl would be able to hold  charge good LG..

nincompoop
 one year ago
Best ResponseYou've already chosen the best response.1oh that did not look right let me type it again \(\sf Me–OH + Cl^ \) dw:1436680471791:dw

Photon336
 one year ago
Best ResponseYou've already chosen the best response.2just curious.. how would you justify the opposite reaction

Photon336
 one year ago
Best ResponseYou've already chosen the best response.2because i see two arrows, which implies the opposite reaction, I mean chlorine's pretty happy with that  charge

nincompoop
 one year ago
Best ResponseYou've already chosen the best response.1I shouldn't probably draw the last reversible arrow in the end because OH wouldn't be a good leaving group for a reversible reaction

Photon336
 one year ago
Best ResponseYou've already chosen the best response.2OK. 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?

nincompoop
 one year ago
Best ResponseYou've already chosen the best response.1dw:1436680676160:dw

nincompoop
 one year ago
Best ResponseYou've already chosen the best response.1\(\sf Cl^ \) is a pretty stable base (weak), and initially \(\sf OH^ \) was a strong nucleophile

Photon336
 one year ago
Best ResponseYou've already chosen the best response.2Ah.. I see.. that double dagger I think that's what they call it.. HOC bond breaking, while the Cl C bond is forming.

Photon336
 one year ago
Best ResponseYou've already chosen the best response.2I guess if you think about it the opposite way you can't actually just kick off OH unless you introduced some proton source.

nincompoop
 one year ago
Best ResponseYou've already chosen the best response.1and you are correct, the product is most likely more stable and our graph would look like this dw:1436680983637:dw

Photon336
 one year ago
Best ResponseYou've already chosen the best response.2YOU sir NAILED this one

nincompoop
 one year ago
Best ResponseYou've already chosen the best response.1you 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

nincompoop
 one year ago
Best ResponseYou've already chosen the best response.1my organic chemistry is pretty crappy so I am refreshing with some problems.

Photon336
 one year ago
Best ResponseYou've already chosen the best response.2crappy is a relative. that was well explained

nincompoop
 one year ago
Best ResponseYou've already chosen the best response.1the double dagger that you mention is the opposite it is the concerted \(OH^ \) forming a bond to C whilst Cl breaking

Photon336
 one year ago
Best ResponseYou've already chosen the best response.2mixed that up nucleophile was OH so it's forming a bond haha

flexastexas
 one year ago
Best ResponseYou've already chosen the best response.0Sorry Im late. It was really late when I posted. Question, how did you determine what kind of chemical this was from the picture?

flexastexas
 one year ago
Best ResponseYou've already chosen the best response.0And also, that last graph you posted would be what they asked for ?

flexastexas
 one year ago
Best ResponseYou've already chosen the best response.0And the first would be the balanced equation?
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