Term
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Definition
| the relationship between the concentration of reactants and the rate of reaction. |
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Term
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Definition
substitution, nucleophilic, bimolecular (depends on the concentrations of TWO molecules)
- takes place in one step, no intermediate
- the nucleophile comes from the side opposite the leaving group, flipping its stereochemistry |
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Term
| as a halide increases in substitution, its SN2 reactivity... |
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Definition
DECREASES. The more highly sub'd the carbon in question is, the more crowding there is around it, and the harder it is for the nucleophile to attack the carbon.
- vinylic halides and aryle halides [i.e. on a double bond] don't react with SN2 at ALL. |
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Term
| regarding SN2 reactions: if the nucleophile is negatively charged vs. if the nucleophile is neutral. |
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Definition
neutral nucleophile: product is negatively charged.
negatively charged nucleophile: product is neutral. |
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Term
| what increases the effectiveness of a nucleophile? (nucleophilicity) |
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Definition
1. the larger, the better - electrons are further away and looser held, so more reactive
2. usually strong bacisity helps, although a REALLY strong base might pull off an H instead of attacking the electrophile
3. lone pairs are a must; a negative charge is even better |
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Term
| protic solvents and SN2 reactions |
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Definition
| contain OH or NH group; they are BAD for SN2 reactions because they SOLVATE the reacting nucleophile, stabilizing it and keeping it from reacting. |
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Term
| what is a good solvent to use for a SN2 reaction and why? |
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Definition
| polar deprotonated solvent is best; these dissolve salts, but RAISE the ground-state energy of the nucleophile, increasing the reactivity and therefore the reaction rate |
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Term
| to stabilize the TRANSITION STATE of an SN2 reaction, you would use a _________ solvent. To PREVENT the stabilization of the reacting electrophile, you would use a __________ solvent. |
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Definition
| POLAR is good for SN2 because it stabilizes the transition state of the reaction. PROTIC is bad for SN2 because it stabilizes (by hydrogen bonding) the reacting electrophile. |
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Term
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Definition
substitution, nucleophilic, UNIMOLECULAR - only depends on electrophile, NOT on nucleophile
- two steps, the first of which is the rate-limiting step, and has a CARBOCATION INTERMEDIATE |
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Term
| why is the SN1 reaction faster than the SN2 reaction? |
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Definition
| SN1 is first order - i.e. its rate limiting step depends only on the concentration of ONE molecule. However, SN2 is second order, so it has to wait for a collision between its two important molecules instead of reacting spontaneously. |
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Term
| more highly substituted halohydrins are _________ reactive with SN2 and ________ reactive with SN1. Why? |
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Definition
| more highly substituted halohydrins are LESS reactive in SN2, because the nucleophile is blocked from attacking the carbon with more groups on it. More highly substituted halohydrins are MORE reactive in SN1 because SN1 uses a carbocation intermediate, which is more stable when highly substituted (hyperconjugation) |
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Term
| in an SN1 reaction, the stereochemistry will come out ____________; in an SN2 reaction, it will __________ |
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Definition
SN1: because carbocation intermediate is planar, stereochemistry is equally likely to reverse or not to reverse and comes out NEARLY as a racemic mixture (not quite, will have slightly more inverted product, because the leaving group hangs around on its original side and repels the nucleophile slightly.)
SN2: reversed stereochemistry every time |
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Term
| will allylic and benzylic halohydrin substrates react slowly or quickly with a nucleophile via SN1? |
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Definition
| quickly; because SN1 uses a carbocation intermediate and is exothermic, the Hammond postulate says that a low energy intermediate (more substituted) will stabilize a transition state and make the reaction faster. allylic and benzylic halohydrins are lower energy, as well, because they are stabilized by RESONANCE. |
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Term
| how do we know that in E2 reactions the C-H bond is broken in the rate-limiting step (which supports a one-step process?) |
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Definition
| the reaction of a deuterium containing molecule is much slower than a normal H-containing molecule; because the effects of the stronger D bond show up in the rate, the rate-limiting step must include breaking of the C-D/C-H bond, which supports the one-step process |
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Term
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Definition
| in elimination reactions, you get a mixture of products, but the more-substituted alkene product will be the most common. |
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Term
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Definition
- occurs in one step
- rate determined by both reactants
- must be periplanar to happen; anti-periplanar is lower in energy because of staggered conformation
--- periplanality is important because the p orbitals are lined up so forming a double bond is easier |
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Term
| what happens if periplanar geometry physically can't form? (like in anything but 1,2 diaxial trans stereochemistry) |
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Definition
| then you'll get a non-Zaitzev product! the double bond will be LESS substituted, because E2 physically can't happen without periplanar geometry |
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Term
| why is there no geometric requirement for E1 reactions? |
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Definition
| H and leaving group leave in two seperate steps; therefore they don't have to be in periplanar arrangement |
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Term
| why does E1 show no deuterium isotope effect? |
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Definition
| it has a two-step mechanism, and the breaking of the C-H/C-D bond isn't the rate-limiting step |
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Term
| if you add a whole lot of base, which reaction will happen, E2 or E1? Why? |
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Definition
| E1 doesn't rely on the amount of base because it's first-order; E2 needs more base to react because it depends on base concentration also (second-order). therefore with lots of base E2 will happen. |
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Term
| why do high temperatures favor E1 reactions? |
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Definition
| E1 must be in periplanar confirmation to react, but this is energetically unfavorable. The heat allows it to assume this conformation and then revert back when the reaction is finished (ringflip) into a more stable conformation. |
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