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Organic chemistry
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)
- ati didn't understand how the hydrogen can take electron from carbon ? 5:00(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 at. Are the two equations close to each other when the carbon loses its electron? 4:51(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)
Video transcript
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.