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Questions About Aamc 6 (ps & Bs)

Discussion in 'MCAT Discussions' started by phatty925, Aug 12, 2002.

  1. phatty925

    phatty925 Senior Member
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    i have a lot of questions about aamc 6, so if you guys could help me out with some of these, that would be great:

    PS:
    107) how come linear momentum is not conserved? is it b/c the velocity is constantly changing?

    142) the solutions manual says that power is lost as I^2R.
    how did they come to this conclusion?

    BS:
    153) i can see how this reaction is bimolecular from the rate chart, but how can you tell the reaction is E2 and not E1 without looking at the chart, but just by looking at the compounds involved and the description in the passage?

    also, in this reaction, what does sodium ethoxide do as a strong base? the solutions manual says that if a strong base is present, an E2 reaction is likely....why?

    168) how come compound 5 is not conjugated whereas compound 1 is when the double bonds in compound 5 are at the same location as the double bonds in compound 1?

    207) why is an sn1 reaction unlikely under basic conditions?

    i know there are a lot of questions here, but these were the ones that i couldn't figure out just by reading the solutions manual. if some of you could help with ANY of these, that'd be great.
    short, but thorough and to-the-point explanations would be appreciated!!! =)
     
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  3. 107) Yes velocity is constantly changing so Momentum can not remain constant.

    153) The passage says that the reactions involve a transition state so it has to be either E1 or SN1. You know that it's an E reaction because a DB is formed so the answer has to be E1.

    207) I don't understand what you are asking here.

    I don't know about the other questions. Hopefully someone else will post them.
     
  4. lola

    lola Bovine Member
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    153 -- E2's & SN2's form transition states

    168 -- don't have the test or answers, but doesn't a conjugated system have to have alternating double bonds around a ring? i'm confused about this and would love to have this clarified. maybe compound 1 satisfies this but compound 5 doesn't?
     
  5. No I really think that SN1 and E1 form transition states. SN2 and [email protected] occur in 1 step with no transition state.
     
  6. I to A

    I to A Member
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    Conditions for Elimination reactions
    You have a base which abstracts the protons.
    (Acids are proton donors and bases are proton acceptors).

    Sn1 reactions are where nucleophiles attack the carbocation.

    In short:

    A Base-Elimination reaction
    A Nucleophile-Substitution reaction

    Phys. Sci:
    I^2R losses or Joule's heating effects. A concept by itself. Current as it is moved from one location to another-through those transmission lines- is lost as heat. This is known as I^2R losses.
     
  7. lola

    lola Bovine Member
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    i think it's the other way around, although i haven't taken ochem in 7 or 8 years :eek: sn1 and e1 form carbocations, not transition states and are done in 2 steps. sn2 and e2 form transition states (not a real step or product) and are done in one step.
     
  8. I to A

    I to A Member
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    E1 and Sn1 both form carbocations. The end product is where the difference is. E1 gives you an alkene.

    In E1, the proton leaves-gets eliminated.

    Sn1-substitution of one group for a good leaving group.
     
  9. Mudd

    Mudd Charlatan & Trouble Maker
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    107) how come linear momentum is not conserved? is it b/c the velocity is constantly changing?

    The system is subjected to a force. Conservation of momentum only applies in the absence of an external force. So, you are correct that because the velocity is changing, linear momentum is NOT conserved.

    142) the solutions manual says that power is lost as I^2R.
    how did they come to this conclusion?


    There is always some inherent resistance in any wire. As a result, current passing through a wire experiences some power drain. This energy per unit time is being released as thermal energy from the system, which explains why a wire heats up when current is passed through it. They choose P = i^2R, because the two values we know are current and resistance. Power can be found other ways, but they use the formula that contains the appropriate terms.

    153) i can see how this reaction is bimolecular from the rate chart, but how can you tell the reaction is E2 and not E1 without looking at the chart, but just by looking at the compounds involved and the description in the passage?

    also, in this reaction, what does sodium ethoxide do as a strong base? the solutions manual says that if a strong base is present, an E2 reaction is likely....why?



    An E2 reaction involves the deprotonation of a hydrogen on the carbon adjacent to the carbon with the leaving group. You may recall that the leaving group and hydrogen must be anti-paraplanar. The presence of a strong base should be an indicator that reaction proceeds by way of an E2 mechanism. This is just a common feature of an E2 reaction. The strong base pulls off the hydrogen, from which the electrons push into the molecule to form the pi-bond, forcing the leaving group off. The reaction is concerted, thus it has no intermediate and only a transition state.

    168) how come compound 5 is not conjugated whereas compound 1 is when the double bonds in compound 5 are at the same location as the double bonds in compound 1?

    Compound 1 is a conjugated ketone (carbons 3, 4, 18, and the carbonyl O make up the system) which results in an intense UV absorbance around 240 nm. Compound 5 is not a ketone and simply has independent pi-bonds. Single pi-bonds result in a weak UV absorbance around 180 nm. This question is tricky, because technically speaking, they are correct that conjugation is necessary. But one could make a viable argument that a carbonyl was lost. The question does not implicitly state that the carbonyl is conjugated, so choice D is not as good as choice C.

    207) why is an sn1 reaction unlikely under basic conditions?

    An SN1 involves a carbocation intermediate, which is not stable under basic conditions. If you could even form a carbocation, it would readily undergo elimination due to the base. The likelihood of forming a carbocation in a basic medium is minimal, because the basic and nucelophilic nature of hydroxide will take place faster than a leaving group could leave.
     

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