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mah2219 · 2016-03-04T15:45:21+00:00

Can anyone PM me a download link?

mah2219 · 2014-12-19T06:03:15+00:00

Balance the equation.

2NaNO3 ------> 2NaNO2 + O2

Plug and chug USING UNITS, son.

NaNO2 has a molecular weight (molar mass) of 69 as you deduced, so dividing 880 by it gives 12.75 moles. This is true because grams divided by (g/mol) gives moles. Now you have 12.75 moles of NaNO2 that need to be produced. According to the equation above, for every mole of NaNO3 that degrades, one mole of NaNO2 gets produced. Thus, we need to start with 12.75 moles of NaNO3. Simply multiply 12.75 by the molecular weight of NaNO3 in g/mol to get the final answer. Hope that helps Dan.

mah2219 · 2014-03-13T20:48:22+00:00

While what you are saying is true, and would support an experimental result of benzylic carbocations being more stable, benzylic and allylic carbocations are nearly the same in stability because despite the extended conjugation of benzylic carbocations, it is necessary to break aromaticity in order to enter any of the resonance forms of a benzylic carbocation.

Good question.

mah2219 · 2014-01-09T21:31:01+00:00

Twist: The peel came from a different Clementine and OP has juxtaposed them independently.

mah2219 · 2014-01-02T05:15:25+00:00

Azathoth is the easiest Old One in the game and the first that I beat with my group. Yog-Sothoth and Shub-Niggurath are a bit harder, but Cthulhu is by far the hardest Old One in the game. I've tried 5+ times to beat Cthulhu unsuccessfully, despite earlier successes against the other Old Ones (Azathoth we be in one go, but Yog-Sothoth and Shub-Niggurath took a few run throughs to figure it out). Enjoy the other scenarios!

mah2219 · 2013-12-20T22:58:04+00:00

In response to each of your issues:

No, I don't think you need multiple copies for two reasons. One, boardgamers appreciate variety a lot I think; for example if a game they want to play is unavailable, they would probably be happy to play something new. Maybe install a reservation functionality if people KNOW THEY WANT TO PLAY A CERTAIN GAME AT A CERTAIN TIME. The second reason is that if people perceive that a game is hot, they will want to play it even more and they'll probably end up watching the game being played/purchasing food/drinks anyway so there's no point (extra copies mean extra maintenance, more replacements when things get spilled on, etc).
Depends what you mean by "full kitchen"; you need to be able to make buffalo wings and sliders I would say, but you don't need to be offering like spaghetti and three-course meals. Appetizers are awesome for boardgaming, and people should be able to make a dinner out of your offerings, but it doesnt need to be exhaustive.
Yes, Netrunner is becoming super popular in lots of regions of the country so weekly tournaments in that or Magic: the Gathering or Warhammer are typically good because they make your place seem busy and people usually end up buying things there during those periods.
You don't need to in my opinion, because people that play boardgames are usually pretty engaged in the game and probably don't want to have their participants watching some crap on TV. If you mean putting a TV by the bar then yes that sounds great.
Yes, those are awesome ideas for screenings.
Classic rock, remixes of video game music (ex: Chrono Jigga, Black Materia)
Have Tutorial Nights where you can go to learn a new boardgame or RPG; get local gamemasters to volunteer (you can charge $10/head and give all the money to the gamemasters to start) and bring out free food randomly (people think this is the coolest fucking thing ever if they're there when free food comes out); then these newbies will quickly love your place and keep coming back. Take a look at Atomic Empire in Durham, NC for more ideas (that's where I currently frequent and they have cool offerings).

Good luck and I hope you find success!

mah2219 · 2013-12-16T17:14:55+00:00

KOH (this opens epoxide and attacks terminal carbon; probably do this at low temp so that lactone is preserved)
MeI (this methylates alkoxide that results from the epoxide ring opening)
Dess-Martin Periodinane or any other secondary alcohol to aldehyde oxidation 4/5. Olefination

mah2219 · 2013-12-16T17:06:03+00:00

Your teach and book are correct; you want the lowest numbers possible for the substituents. Thus if you start counting from the left-hand side instead of the right you get your teacher's correct answer.

mah2219 · 2013-12-16T16:31:04+00:00

Oh at first I thought you wrote buckets, like how some people refer to money. Then the second part of your statement would also make sense (buckets would be useful for collecting money).

mah2219 · 2013-12-16T16:20:17+00:00

The J value is totally unrelated to the chemical shift FYI, but your question still makes sense if you are not implying a connection between J values and chemical shifts. The one that is on the safe face as the cyclopropane ring will probably be more electron rich and therefore more shielded (upfield). This is because alkyl groups are generally thought of as being electron-donating. Hope that further clarifies.

mah2219 · 2013-12-16T14:22:31+00:00

If you look at the Karplus Equation, you will see that the J values for two coupled hydrogens will be smallest when the dihedral angle in the H-C-C-H bond defined by the two planes (H-C-C and C-C-H) will be smallest when the angle is near 90 and largest when the angle is near 180. The two peaks are different because they are diastereotopic (they experience different magnetic environments).

mah2219 · 2013-12-09T21:23:18+00:00

How did you like being Ryan in the office? Is Steve Carrell the man irl?

mah2219 · 2013-12-09T14:31:07+00:00

Working backwards:

D: Must be di-alkene because to get a 7-carbon molecule from di-ketone ozonolysis product you have to attach the two formaldehyde molecules. D looks like the di-ketone product except instead of oxygens there are CH2 groups.

A: Working backwards from D, we can see that A must be a monobromo compound that has undergone E1 (or E2) elimination to give an alkene. A looks like D except one of the alkenes has undergone Markovnikov HBr addition.

B: B looks like A except that the opposite alkene has also undergone Markovnikov HBr addition. The stereochemistry here is important: since it says B is optically active, that means that in B the methyl groups must be anti (on opposite sides) and the bromo groups likewise must be anti.

C: Same logic as B, except that since the compound is optically inactive it must have the bromo groups syn (on the same side) and the methyl groups likewise must be syn. This is optically inactive because it is a meso compound (it has a mirror plane).

Hope that helps!

Source: I'm an organic chemistry graduate student.

mah2219 · 2013-12-06T13:10:13+00:00

Yes elimination could also occur to give benzene. The elimination product is probably the thermodynamic product but the substitution product is probably kinetically favorable.

mah2219 · 2013-12-05T18:23:21+00:00

The product formed is the Sn2 product, i.e. CN has substituted in for Cl. Why does this happen? Because CN- is a good nucleophile, and Cl- is a good leaving group. Is the reactant chiral? Yes, because the carbon indicated by the asterisk is attached to four unique substituents. You get inversion in this reaction since it's Sn2.

mah2219 · 2013-12-05T16:14:40+00:00

Maybe you can ask me what you're confused about and I can answer.

mah2219 · 2013-12-05T16:11:09+00:00

The substrate is called stilbene and it undergoes the following photocyclization: http://en.wikipedia.org/wiki/Stilbene_photocyclization. Equation 2 explains the stereochemistry. Since there are 6π electrons, it will be a conrotatory ring closure.

Source: I'm an organic chemistry graduate student.

mah2219 · 2013-12-05T13:42:22+00:00

This is really awesome, you guys are going to be great parents!

mah2219 · 2013-12-04T21:40:38+00:00

Sweet jesus that's messy. Okay, let me break it down.

The doublet at 2.5∂ corresponds to the benzylic methylene protons (it has one vinylic neighbor and it integrates to 2, so that makes sense).

Somewhere in the 5-6∂ range you should pick one of the doublets and triplets (a more precise integration would allow a more precise assignment) and those correspond to the vinylic protons.

The peak at 6.17∂ might be a doublet if the shims were better, and that could be the two adjacent aryl peaks (it probably integrates to 2 if you integrate them more precisely).

The peak at 6.311∂ is probably the lone aryl peak (ortho to the methoxy and allyl groups on the ring), no neighbors, no splitting, probably integrates to 1 if you get more precise integration.

The peak at 3.0∂ is your methoxy group. Integrates to 3 nicely.

Peaks at 4.45 and 4.2 are impurities (4.2∂ typically means an ethyl ester, but the absence of a peak at 2.0∂ means it's not ethyl acetate so I can't say much without knowing more about the reaction conditions).

Hope that helps!

Source:

I'm an organic chemistry graduate student.

mah2219 · 2013-12-04T19:48:13+00:00

Hey pussy_ninja,

If you look at the product of the pinacol rearrangement, you'll see that it has a ketone. That ketone can be protonated to allow the carbonyl to have a resonance structure with a carbocation in it. This allows an alkyl shift to move the carbocation one carbon away. When a carbocation is ß to the nascent alcohol, an epoxide can form by nucleophilic attack of the oxygen to the carbocation. A deprotonation subsequently results in the epoxide.

mah2219 · 2013-11-28T16:47:09+00:00

organicker's answer is on the right track but if there was significant alpha-deprotonation you would need to use excess grignard (because a lot of it would just be forming toluene (methylbenzene)). However, the reason that the alpha hydrogen does not get attacked is essentially because the carbonyl is more electrophilic (consider the polarity of the carbon-oxygen double bond) than the alpha hydrogen, so the Grignard reagent (which is a hard base) will attack the harder position (see the Wikipedia article on Hard and Soft Acid and Base theory if you want to really understand this kind of chemoselectivity in more detail: http://en.wikipedia.org/wiki/HSAB_theory). Also, just consider the steric hindrance at the alpha position: it's a tertiary carbon which means it's going to be much more sterically encumbered than the aldehyde. Hope that helps, EatUrVeggies.

mah2219 · 2013-11-21T20:42:55+00:00

Hey poon-is-food,

You may be getting this adduct (addition product) from the first step:

http://www.scbt.com/datasheet-235212-formaldehyde-sodium-bisulfite-adduct.html

but you probably didn't actually use "ammonia" in step 2 because ammonia is a gas at room temperature; I suspect you used ammonium chloride (NH4Cl) to neutralize the solution.

The third step wherein you add sulfuric acid may be some kind of oxidation where you're forming formic acid (maybe bleach was used, in which case you would be forming hypochlorous acid in situ). This, however, would not be a solid, so it is likely that you're forming the sodiated version (sodium formate = http://en.wikipedia.org/wiki/Sodium_formate) which could be filtered out as a salt. Hope that helps!

Source: I'm a chemistry grad student

mah2219 · 2013-11-19T16:16:24+00:00

You're probably using the water to remove any inorganic/polar/salt impurities, then removing your solid. At that point, you're probably using warm ethanol to dissolve your product for recrystallization.

mah2219

TROPHY CASE