Introduction to reaction mechanisms
Introduction to Reaction Mechanisms . Addition reaction to an alkene. Created by Sal Khan.
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- I'm a little confused. Even if bromine is a little negative, isn't it still satisfied with a stable octet? Why would it bond with the carbon?(26 votes)
- Bromine has a negative charge when it has a full octet; this makes it an negative anion so it wants to bond with a positive carbocation.(Unlike charges attract) This is because, as posted above, a bromine atom is neutral when it has 7 valence electrons. If it has 8 v.e then it will have a charge of 1-. You're right that it is 'stable' with a full octet, but as it has a charge, it will still bond with another oppositely charged ion. Hope this helps.(85 votes)
- i have a little confusion about this tutorial..... why the hydrogen attached to the first carbon and bromine to second carbon and not vice-versa? i want to ask if it can possible?(8 votes)
- It depends on the position of the positive charge in the carbocation. A positive charge is pretty unpleasant for a C-atom, so it likes to share the burden with neighbouring C-atoms. The more carbon neighbours a C-atom has, the less inconvenient the positive charge. As the second C has more carbon neighbours, it cares less about giving away its electron (and thereby getting a positive charge) than the first C. And yes, the first carbon can give away it's electron, get a positive charge and react with the bromide-ion, but as this is energetically less favourable and therefore it happens less frequent.(13 votes)
- I know this isnt the most relevant question, but i was wondering if aomebody could give me tips on how to study organic chemistry.. I have trouble remembering the reactions, preparations of different compounds etc etc.. Thank you!(4 votes)
- In order to truly understand organic chemistry, you must be able to visualize on a 3D scale what is happening for each reaction. This may be difficult to most but it is completely necessary for learning and knowing exactly what is going on for each reaction. I advise for those who are having a hard time, to follow these Khan Academy videos while making 3D structures from models to better visualize spacing of the atoms. At the same time, you must also be thinking of acid/base chemistry. What makes a strong base? What makes a strong nucleophile and what is the difference? These are all important questions to ask when looking at any reaction mechanism. OK ok enough. How do you remember all this?.... Watch the videos and take notes. Categorize the reactions based on what they need to react and practice. As much as you can, find reaction mechanism problems and solve them. This will help the most.(6 votes)
- at5:00i didn't understand how the hydrogen can take electron from carbon ?(5 votes)
- The H has a δ⁺ charge.
The π electrons in the double bond are weakly held and move to form a bond to the H.
At the same time, the electrons on the other side of the H move onto the Br.(8 votes)
- wait why would BR have a negiative charge when it gained the electron. When it gained it then had 8 electrons and was neutral ... right? or am i confusing that with something else(3 votes)
- Charge arises from electrons and protons only, not from filled valence shells. Since elements are defined by the number of protons they have, Bromine always will have 35 protons, which is on the periodic table. So when all orbitals are filled, you end up with a neutral charge when bromine has 7 electrons in its valence shell. This is one less than an octet!(13 votes)
- I got lost at4:51. Are the two equations close to each other when the carbon loses its electron?(4 votes)
- As Sal says, the 1-pentene molecule and the HBr molecule must come in collision to start this reaction, so I suppose they must come very close. At that moment a electron from the secondary carbon goes to the H, and so forming a bond with the primary carbon.(2 votes)
- After the pent-1-ene has bonded with the Hydrogen, wouldn't it form an ionic bond with the Bromide anion?(2 votes)
- Certainly the bromide ion is attracted to the 1-methylbutyl cation, and a bond will form between them. The Br- wants its electrons, but the carbocation also wants electrons. Although the Br wants to hog the electrons, it is not strong enough to prevent the carbon from getting some of them (the electronegativity difference is not great enough). So we end up with a polar covalent bond.(4 votes)
- does it matter which carbon the H from H-Br bonds to? Why did H from H-Br bond to the left carbon, on the alkene? is this dependent on hyper conjugation?(2 votes)
- Yes, it does matter which carbon of the alkene the H bonds to. You're right, it does have to do with hyperconjugation. The left-hand carbon has 2 hydrogen substituents. The right-hand carbon has a hydrogen and an alkyl group. When the H from HBr attaches to the left carbon, it leaves a carbocation on the right carbon. The alkyl substituent on the right carbon is able to donate electron density (through hyperconjugation) to the carbocation, thus stabilizing it. Had the H from HBr added to the right carbon, the carbocation would have formed on the left carbon where no such hyperconjugative stabilization is possible. Thus, the more stable/more energetically favourable mechanism involves putting the H on the left carbon.(3 votes)
- There are so many different reactions, ie halogenation, acid catalyst and so forth, my problem is how do you know which reaction to use, how do you know what source is used in the reaciton?(3 votes)
- I think it's largely based on the reagents and catalysts. For example, with hydrohalogenation, you have an alkene react with HBr or HCl (usually) and add a Br to one end of the broken double bond, then a H added to the other end. Different solvents go further to determine which end, so with peroxides, you get the anti-Markovnikov product. Memorizing reagents and solvents is the key here.(1 vote)
- Is there anywhere I can get more examples of these basic reaction mechanisms that I can use for practice? Just to make sure I understand the general concept before moving onto SN2, SN1, etc. reactions. Like websites and such. :) Thanks!(2 votes)
Let's think about how hydrogen bromide might react with this thing right here. Let's think about what this would be called. We have one, two, three, four, five carbons. It has a double bond right here, if we start numbering at this end, because that's where the double bond is. Then this would be pent. The double bond starts at the number one carbon, pent-1, and it's obviously an alkene. It has a double bond, so it's pent-1-ene. Sometimes this is called 1-pentene, either way. So let's think about how these two characters might react with each other in some type of solvent. Usually when a solvent's not specified, it's usually water or alcohol. If this was water, then we would have a solution of hydrobromic acid, but let's not worry about that right now. Let's just worry about how these two characters might react with each other. Now, the first thing we might look at is this hydrogen bromide right there. And as you get more experience there, you'll say, well, you know, bromine is much more electronegative than hydrogen. It likes to hog electrons much more. If that's a completely foreign concept to you, I'll do a little bit of a review. This is the Periodic Table. Electronegativity increases from the bottom left to the top right of the Periodic Table. Electronegativity. And really, that's just a fancy way of saying how much does an atom like to hog electrons. So, for example, fluorine loves to hog electrons and calcium doesn't like to hog electrons so much. So if we think about hydrogen bromide, hydrogen's way up here on the left side of the Periodic Table. Bromine's all the way to the right. Bromine is much more electronegative. So in this situation, since bromine is more electronegative, it will hog the electrons in this bond. And since it's hogging the electrons in this bond, you'll have a partial negative charge on the bromine end, and you'll have a partial positive charge on the hydrogen end. And, whenever you do these reactions, actually, it's useful to draw all the valence electrons. So bromine right here has an atomic number of 35, which means it has 35 electrons in its neutral state. And we don't have any charge here, so it's in a neutral state. But you can look at its group. It's in Group Number 7. If you just count from here, one, two, three, four, five, six, seven, or has seven valence electrons. Let me draw that. Seven electrons in its outermost shell. Let me draw them right here. So one, two, three, four, five, six, and then the seventh is in the bond with the hydrogen. Hydrogen obviously has one electron. It's right there. So these two guys are bonding with each other and they both allow each other to pretend that they're part of a pair. And that's what gives it a lower energy level or makes it a little more stable. That's why the bond forms in the first place. But anyway, let's think about what might happen here. This guy is really electronegative, so maybe he wants to take this electron away, this green hydrogen electron. And maybe that happens. Let me just draw it out. So maybe that happens. So he takes that electron, that green electron, away. It's just getting closer and closer. If there was only some other place that this hydrogen could get an electron from, then this guy could just go to the bromine, which is what bromine really wants to happen, because it's so electronegative. So where can the hydrogen get an electron to replace the one that it's about to lose? Well, we have a double bond here. And maybe one of these carbons lose an electron. And in future videos, we'll talk more about which one is more likely to give up the electron at this stage of the reaction. But just for simplicity, let's assume that this carbon right here gives up an electron. Where are its valence electrons? And just as a bit a review, here's carbon on the Periodic Table. it has six protons and six electrons in its stable state. Two are sitting in its first shell and then the other four are its valence electrons. And you can see that. It's in Group 4, right? One, two, three, four. We can ignore these for now, and usually you're not going to be dealing with-- well, we'll ignore those for now. It has four valence electrons. Most of what you're going to deal with, you just have to count the group numbers like this. We won't worry too much about all of the metals and all of those right now. And that's four valence electrons and you see it right here. It has one, two, three, four valence electrons. It has two in its one shell, so it actually has six, but you only draw the four out there. So what we're saying is this green electron can go to the bromine as long as the hydrogen can take an electron maybe from this carbon right here. So let's draw that. So this electron right here is going to go to the hydrogen. And when that electron goes to the hydrogen, simultaneously that will allow this electron to go to bromine. And, obviously, they won't be this far when it happens. There would be some type of collision that would have to occur in just the right way with just the right level of energy for this to occur. Maybe the hydrogen is getting really close to this part right at the right moment when this electron is being sucked away from the bromine, so this has a partial positive charge attracted to the electron. That electron goes over there. It won't always happen, but this is a potential reaction mechanism. But this whole step happens at once. This electron goes from this character to the hydrogen at the same time as hydrogen loses its electron to the bromine. So what's going to happen right after this step? So right after that happens, what will everything look like? Well, the bromine will have gained the electron, so it's now a bromide ion. So let me draw it like this. So it had its original seven valence electrons, one, two, three, four, five, six, seven, and now it just stole an electron from the hydrogen. It was able to swipe an electron off the hydrogen. That's the electron it swiped off the hydrogen. And now what will this thing that was 1-pentene or pent-1-ene, what will this look like now? Let me draw it. So we have a carbon and a hydrogen and a hydrogen, and then you have a carbon, and then a hydrogen, and then you have the rest of the chain right over here. But this double bond was broken. This carbon lost an electron, went to the hydrogen. So this bond right here now forms between this carbon and that hydrogen. So let me draw that bond. You have this electron right there. That electron will now be with the hydrogen. So let me draw the electron. And now we have that orange hydrogen. I'll try to keep the colors consistent so that we know where things came from. And then we have this bond now to the hydrogen. Now, this carbon now only has three valence electrons. One, two, three. It has two sitting in its first shell, so it actually has a total of five electrons. It has six protons, so it has a positive charge. This carbon right here has a positive charge. And another way to think about it is it was completely neutral and then it lost an electron. So now it will have a positive charge right over there. So this is what we are left after that step of the reaction. Oh, and of course, we can't forget. Bromine over here was neutral. It had seven valence electrons, and that's when bromine is neutral. But now it has eight, so now this will have a negative charge. This will have a negative charge, because it gained an electron. And in general, your total charge-- over here our total charge is zero, So? Our total charge will still be zero. We have a negative and we have a positive. They would cancel out, so our total charge is still zero. So what's likely to happen for the next step of our reaction? Well, we have this positive thing here. Maybe bromine just bumped, just the right way, to let go of this guy and steal his electron. But now you have this guy who's negative and this guy who's positive. Maybe they'll be attracted to each other. Maybe they'll just bump into each other at the exact right way. And if they bump in the exact right way, maybe this guy can swipe the electron from the bromide ion, from this negative ion right here. And you might say, hey, isn't bromine more electronegative than carbon? Well, it might be, but this guy's electron rich. It's not just a regular bromine atom. This is a bromine plus an extra electron. So he's already hogged an electron, so he's electron rich. So in this situation, he's negative. He's positive. He can give this guy an electron. So if they bump in just the right way, this electron can be swiped by this carbon right over here. And this positive carbon, just so it gives you a little terminology, and we'll go over it in more detail in future videos, is actually called a carbocation. It's a positive ion of carbon. That's where the word comes from. But anyway, if this electron gets swiped by this carbon, it will then form a bond. Because remember, this was the electron that was originally in a bond with this hydrogen. It's still going to be, you could imagine, paired up with this other purple or magenta electron right over there. So if that happens, then we're going to be left with-- so the next step is going to be-- so on this end of the molecule we have C, carbon, hydrogen, hydrogen. Then we have this orange hydrogen that we stole from the hydrogen bromide. Then we have this carbon right here. It has a hydrogen. You have the rest of the chain, CH2-CH2-CH3. And now, since this guy stole an electron, a bond will form with the bromine. Let me draw it. So he's going to steal this-- let me draw it this way. So a bond will form. He's stealing this electron, so now this electron is with this carbon, so I can do it as one end of the bond. The other end of the bond will be that magenta electron right there on the bromine. And now the bromine lost an electron, so it's neutral. It once again has one, two, three, four, five, six, seven, valence electrons. Carbon is now neutral, because it gained an electron. It once again has one, two three, four electrons. So now everyone is happy again. In this video, we were able to get a mechanism. We can talk in the future about how likely it is to happen, how quickly it might happen. We were able to start with hydrogen bromide and 1-pentene, or pent-1-ene, and get to this thing, just by pushing around electrons and just thinking about what is likely to happen based on what's electronegative and what might be able to gain or lose an electron. And just a bit of a review, what is this thing right here? This might be-- This is one, two, three, four, five carbons still. So it's going to be a pent. It's now an alkane, no double bond. So it's pentane. And it has one group on it, so we'll start numbering closer to the group. So one, two. It's 2-bromopentane. So we were able to figure out a reaction mechanism to get us from hydrogen bromide and 1-pentene to 2-bromopentane.