flying fish Posted July 30, 2009 Posted July 30, 2009 So yeah...I'm in organic 1 now... And as some of you know, I'm NOT a chemistry student. Frankly, I should not be in this class. But that's besides the point. I've been having trouble figuring out how to draw reaction mechanisms. I get that the curved arrow notation shows the movement of electrons...but the decision of how to arrange them just seems like magic to me. for instance, here's the first example we got in class: http://i22.photobucket.com/albums/b330/WhyAreAllUserNamesTaken/org01.jpg I'm not sure I understand why it was chosen, which way to move the electrons... And then here's another one that's not completed. I didn't write the whole thing down 'cause he said "OK, I really want you guys to understand this, so put your pens down and just listen." Of course, I didn't understand it, and I don't have the notes either. So I have no idea how the mechanism went... Here's the example: http://i22.photobucket.com/albums/b330/WhyAreAllUserNamesTaken/org02.jpg Of course, I'm going to consult the professor, but I thought I'd also tap into this forum and see if it helped me. By the way, I have frickin' quiz in two hours from now! HALP! Much appreciated if you can.
justanotherpyro Posted July 30, 2009 Posted July 30, 2009 (edited) They are both common acid base reactions. Acids will protonate bases and commonly favor the elimination of H2O. So the arrow pushing basically shows the formation of water on the propane and the new charged species(H2O) is the leaving group which then forms the carbo cation. A + charge and the free Br- group are favorable to react thus forming the 2-bromopropane. The second one is a bit trickier in that it has a 1,2 alkyl shift. Basically when the OH group gets protonated and breaks off as water the positive charged carbo cation is in close proximity to a methyl group and in this case it is more favorable for the molecule to rearrange and move one of the methyl groups to the nreformed carbo cation and counter balance the + charge on the molecule ionically with the negatively charged bromine vs just having a covalent interaction between the newly formed carbo cation and the negatively charged Br. When it comes to this the energetics and math of it explain why it is more favorable for the 1,2 alkyl shift to occur. Maybe someone else can shed some more light on it since Organic was never my strongpoint and my focus is inorganic chemistry Just memorize the mechanisms thoroughly and hopefully you can pull through. quick edit - The 1,2 alkyl shift occurs because it forms a more stable tertiary carbo cation on the C at the top of the ring. Edited July 30, 2009 by justanotherpyro
flying fish Posted July 30, 2009 Author Posted July 30, 2009 Thanks man for the explaination! I'll be sure to work through the problem again using that methodology...I think the book doesn't do very many examples like this (we are using Solomons & Fryle). Any reccomended resources for learning to write reaction mechanisms? I pretty much failed the quiz...but really only the test matter. I have one week to learn it before the "fear of god" test! There is this goofy kid from the Bahamas that called it that...said that it "taught him to fear god..." whatever that means. Yeah, he's in the class because he failed it the first time. I heard last term there were 6 people who failed (out of the 20 out of 25 students that didn't drop the class), and 4 that had a nervous breakdown! One of my co-workers at the company I intern for, who graduated from the same university I'm going to...said that "ahh, don't worry about it...it's only the people who try to attach 5 bonds to carbon who fail." But statistics say otherwise...
Mumbles Posted July 30, 2009 Posted July 30, 2009 My o-chem professor calls the 5 bond carbon a "Texas carbon". I found that funny. We used the same book for my first 2 org chem classes. It was ok, but I didn't like it all that much. For my 3rd we used a book of reaction called "Advanced Organic Chemistry" by Carey. It's like an encyclopedia. Lots of examples, lots of mechanisms, lots of the weird reactions that don't make much sense. Electrons are just like water. They flow from areas of high density to low density. Lone pairs, such as on oxygen and nitrogen, and double bonds are the normal "high density" sources. And remember, if a teacher ever questions a reaction you did, call it the "Corey reaction". There are like 2 dozen of them, chances are one of them will do what you want.
asilentbob Posted July 31, 2009 Posted July 31, 2009 And remember, if a teacher ever questions a reaction you did, call it the "Corey reaction". There are like 2 dozen of them, chances are one of them will do what you want. Haha, good times.
flying fish Posted July 31, 2009 Author Posted July 31, 2009 Hehe... I went to the "hang-out/ study room" in the chemistry department...hoping that some of the students would be able to help me out. None of the were studying...they were just ranting about the instructor the whole time! No wonder they fail the class! "Man, that guy is insane. He knows everybody by name and calls them out on their weeknesses during lecture until they cry""IDK Man, I think he's gone soft this term, since his wife is back from Iran""Not a chance...""Hmmm...Iranian chemist = terrorist?""LOL, the government is probably watching him" Me "So does anyone want to study for the quiz?"Someone "No, were all going to fail it anyway" *Prof walks in*"What are you guys doing, you're working hard, right?" Some kid with Jamaican accent who was doing some lab work for him: "Todaaay were just chiiilin man, you're giving me the day off, right?""No, I think you should be in the lab today!" Hehehe I guess that's enough for now
flying fish Posted August 30, 2009 Author Posted August 30, 2009 (edited) So wats the difference between cold, concentrated H2SO4 and warm, dilute as far as Alkene additions go? Can't seem to figure it out... The exact problem is...you take some 1-butene, react some of it with warm dilute, and some with cold concentrated. What are the products for each case? The book says that cold concentrated is a regioselective Markovnikov addition. Simple enough...but it says nothing about warm, dilute. And yet it asks that frickin question! Hehe... My test is on monday. I don't need to do great...I just need to pass! Oh Crap Oh Crap Oh Crap. By the way, I'm going to draw my invention, the "super-carbon" on the exam. Super carbon forms 8 bonds. Edited August 30, 2009 by flying fish
Mumbles Posted August 30, 2009 Posted August 30, 2009 Your super carbon gives new meaning to the term "texas carbon". The cold concentrated gives the 2-hydrogensulfate. I've never seen it done this way, but the warm dilute would likely give the 2-hydroxyl. All normal situations I've seen react with concentrated first, and then dilute it and distill/separate off the product.
flying fish Posted August 31, 2009 Author Posted August 31, 2009 Thanks! Turns out that particular question wasn't on the exam, but yeah, it still helps. I made some really stupid mistakes...like forgetting to count the carbons for a couple of reactions...And for many other questions I was just too incompetent to answer them correctly. So yeah, chances are I failed it...but the real question is how badly? Lol. This is allegedly the hardest of the four exams, so I'm hoping it'll get better from here. I didn't quite have the nerve to write my super carbon in one of the answer spaces, so I drew it on the attached periodic table...but large enough that he should see it!
Swede Posted September 2, 2009 Posted September 2, 2009 (edited) Can anyone else write the periodic table from memory? I used to be able to do this... Not anymore. I've forgotten 98% of what I once knew. Use it or lose it. FF, draw a Bucky-Ball. That will impress him. Or claim that the test does not apply, as in reality we are all part of the Matrix, and our silicon-based overlords don't believe in carbon-based life. It is all an illusion. Best of all - refer to silicon as silicone, which 99.999999% of the population does these days. Edited September 2, 2009 by Swede
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