6 Reactions of Alkanes
Summary
TLDRThis educational video delves into the chemical properties of alkanes, a type of hydrocarbon, focusing on their reluctance to react due to high bond enthalpies and low bond polarity. It explains the concept of complete and incomplete combustion, providing equations for both. Additionally, the video explores alkanes' reactions with halogens under UV light, detailing the free radical mechanism involved in these substitution reactions, from initiation to termination, and emphasizing the importance of understanding homolytic fission in the context of these chemical processes.
Takeaways
- 🔍 Alkanes are part of the homologous series and are characterized by their saturation, meaning they have no double or triple bonds.
- 🔄 Alkanes are generally unreactive due to their high bond enthalpies and low bond polarity between carbon and hydrogen atoms.
- 🔥 All hydrocarbons, including alkanes, can undergo combustion, which is a reaction with oxygen, resulting in carbon dioxide and water as products.
- 🌐 Complete combustion of alkanes involves the formation of carbon dioxide and water, with the number of carbon dioxide molecules equal to the number of carbon atoms and the number of water molecules equal to half the number of hydrogen atoms.
- 🚫 Incomplete combustion occurs when there is a lack of oxygen, leading to the formation of carbon or carbon monoxide, which is a poisonous gas.
- 🌞 Alkanes can react with halogens under the influence of bright light, such as UV light, in a process known as a substitution reaction.
- 💡 The reaction of alkanes with halogens involves a free radical mechanism, starting with the formation of free radicals from the halogen molecules.
- 🔄 The propagation step in the free radical mechanism involves the reaction of a free radical with an alkane, leading to the formation of new radicals and products.
- 🔚 The termination step in the free radical mechanism occurs when two radicals collide and react, forming a stable molecule without unpaired electrons.
- 📚 It is important to understand the concept of homolytic fission, where a bond breaks and each atom retains one electron, leading to the formation of free radicals.
Q & A
What are homologous series in organic chemistry?
-Homologous series refer to a group of organic compounds that have similar chemical properties and a gradation in physical properties, differing by a constant unit, typically a -CH2- group.
Why are alkanes considered saturated hydrocarbons?
-Alkanes are considered saturated hydrocarbons because their molecules contain only single bonds between carbon and hydrogen atoms, meaning they have the maximum number of hydrogen atoms possible for the given number of carbon atoms.
What is the significance of bond enthalpy in the reactivity of alkanes?
-The bond enthalpy, or bond dissociation energy, indicates the strength of a bond. Alkanes have high bond enthalpies, meaning their carbon-hydrogen and carbon-carbon bonds are strong and difficult to break, which contributes to their unreactive nature.
How does bond polarity affect the reactivity of alkanes?
-Bond polarity arises from an electronegativity difference between atoms. Since carbon and hydrogen have similar electronegativities, the bonds in alkanes are nonpolar, and there is an even distribution of electron density, making alkanes less likely to participate in reactions.
What are the products of complete combustion of alkanes?
-In complete combustion, alkanes react with oxygen to produce carbon dioxide (CO2) and water (H2O) as the primary products.
How can you balance the chemical equation for the complete combustion of methane?
-The balanced equation for the complete combustion of methane (CH4) is CH4 + 2O2 → CO2 + 2H2O. The number of carbon dioxide molecules equals the number of carbon atoms, and the number of water molecules equals twice the number of hydrogen atoms in the alkane.
What is incomplete combustion and why does it occur?
-Incomplete combustion occurs when there is insufficient oxygen to fully react with the fuel. This can lead to the formation of carbon monoxide (CO) and carbon (C), which are dangerous due to their toxicity and potential to cause fires.
What are the conditions required for alkanes to react with halogens?
-Alkanes react with halogens under conditions of bright light, specifically ultraviolet (UV) light, which is necessary to initiate the reaction through the formation of free radicals.
What is a free radical mechanism and why is it important in the reaction of alkanes with halogens?
-A free radical mechanism is a chemical reaction that involves the participation of free radicals, which are molecules with unpaired electrons. This mechanism is important in the reaction of alkanes with halogens because it explains the initiation, propagation, and termination steps of the reaction, which would not easily occur due to the unreactive nature of alkanes.
What is the term used to describe the breaking of a bond in a free radical mechanism?
-The term used to describe the breaking of a bond in a free radical mechanism is 'homolytic fission', which results in two free radicals, each with an unpaired electron.
What are the possible products of the reaction between alkanes and halogens, and why are they significant?
-The reaction between alkanes and halogens can produce haloalkanes (e.g., CH3Cl) and hydrogen halides (e.g., HCl). These products are significant because they demonstrate the substitution of hydrogen atoms in alkanes with halogen atoms, and the reaction can lead to various other products depending on the reaction conditions and the presence of radicals.
Outlines
🔍 Understanding Alkanes' Unreactivity
This paragraph introduces the topic of alkanes and their chemical reactions. It explains why alkanes are unreactive due to their high bond enthalpies and low bond polarity. The high bond enthalpy means strong bonds that are hard to break, and the low bond polarity between carbon and hydrogen atoms results in an even distribution of electrons, leading to a lack of regions with high or low electron density for reactions to occur. The paragraph sets the stage for further exploration of alkanes' reactions, particularly with halogens and their combustion processes.
🔥 Alkanes in Combustion: Complete and Incomplete
This section delves into the combustion of alkanes, a reaction with oxygen that is crucial for their use as fuels. The paragraph explains the concept of complete combustion, where alkanes react with sufficient oxygen to form carbon dioxide and water. It provides a step-by-step guide to writing balanced chemical equations for complete combustion, emphasizing the relationship between the number of carbon and hydrogen atoms in the alkane and the resulting carbon dioxide and water molecules. The paragraph also touches on incomplete combustion, which occurs when there is a lack of oxygen, potentially leading to the formation of carbon or carbon monoxide, the latter being a poisonous gas. The summary underscores the importance of balanced equations and the products formed in both types of combustion reactions.
⚛️ Free Radical Mechanism in Halogen Reactions
The final paragraph discusses the reaction of alkanes with halogens, which requires the presence of bright UV light. These reactions are substitution reactions, where a hydrogen atom in the alkane is replaced by a halogen atom. The paragraph introduces the concept of a free radical mechanism, detailing the initiation, propagation, and termination steps involved in these reactions. It explains how UV light causes the formation of free radicals from halogen molecules, which then react with alkanes to form new compounds and additional radicals. The summary highlights the importance of understanding the mechanism behind these reactions, including the role of homolytic fission in forming free radicals and the potential formation of various products beyond those represented in the simplified reaction equations.
Mindmap
Keywords
💡Homologous Series
💡Alkanes
💡Bond Enthalpy
💡Bond Polarity
💡Combustion
💡Complete Combustion
💡Incomplete Combustion
💡Halogens
💡Free Radical Mechanism
💡Homolytic Fission
💡Propagation Step
Highlights
Introduction to the sixth film in a series about the standard level organic topic.
Focus on alkanes and their chemical reactions.
Explanation of why alkanes are unreactive due to high bond enthalpies and low bond polarity.
Alkanes are saturated, meaning they have no double bonds, making them less reactive.
Bond enthalpies in alkanes are high, making them difficult to break.
Carbon-hydrogen bonds in alkanes are non-polar due to similar electronegativities of carbon and hydrogen.
All hydrocarbons, including alkanes, can burn well, making them suitable as fuels.
Complete combustion of alkanes involves oxygen and produces carbon dioxide and water.
Equations for complete combustion reactions can be balanced by matching carbon dioxide to carbon atoms and water to hydrogen atoms.
Incomplete combustion occurs when there is a lack of oxygen, leading to the formation of carbon or carbon monoxide.
Incomplete combustion reactions can be balanced with any combination of carbon and carbon monoxide, as long as water is formed.
Alkanes can react with halogens under bright light, a process called a substitution reaction.
The reaction of alkanes with halogens requires UV light to initiate.
Free radical mechanisms are involved in the reactions of alkanes with halogens.
Initiation in free radical mechanisms involves the formation of free radicals, such as chlorine radicals from chlorine molecules.
Propagation steps in free radical mechanisms involve radicals reacting with other molecules, such as alkanes, to form new radicals.
Termination steps in free radical mechanisms occur when two radicals collide and form a stable molecule.
Homolytic fission is the breaking of a bond into two identical parts, which is crucial in free radical mechanisms.
Understanding the free radical mechanism is important for contrasting with other types of reactions.
Encouragement to practice writing equations for combustion and reactions with halogens to solidify understanding.
Transcripts
hello and welcome to the sixth in a
series of films about the standard level
organic topic in the first five films
we've looked at what we mean by
homologous series and we've looked at
two different homologous series the
alkanes and alkenes here we're going to
be focusing on the alkanes once again
and in particular on their chemical
reactions so hopefully by the end of
this film you will understand first of
all what it is about alkanes that means
they don't particularly like reacting
with other things and then for the
things that they do like reacting with
you're going to be able to write some
equations so you'll be able to write
equations for complete and incomplete
combustion reactions and also for the
reactions of alkanes with halogens and
what is more for the reactions with
halogens you know what we mean by a free
radical mechanism okay so if we start
off by looking at why it is that alkanes
don't really like reacting with other
things you might remember that we termed
alkanes and saturated when we were
introducing this analogous series and
that means that they're already full if
you like that we have to break some
bonds and remove some atoms before we
can put new atoms into the molecule so
one way of explaining why they're so
unreactive would be to look at the bond
enthalpies in the molecules right
because if you've got a high bond
enthalpy that means you're a bond that
is difficult to break and the bonds
between carbon single bonds between
carbons and hydrogen's and carbons and
carbons tend to have quite a high bond
enthalpy so they're hard to break so
before we can add anything here we've
got to break quite strong bonds and
that's a difficult thing to do we could
also look at this in terms of bond
polarity and this is a little bit more
complicated but we don't have to
understand it fully okay bond polarity
arises as a result of an
electronegativity difference between
atoms and this can lead to a kind of
imbalance of electron density around the
molecule and quite often reactions will
happen because things will attack
regions of high or low electron density
however if you've got atoms that have
quite similar electronegativities then
the bonds won't be very polar
so because carbon and hydrogen are quite
similar in their electronegativity
carbon hydrogen bonds aren't very polar
the electrons are quite evenly spread
spread throughout the molecule and so
what these molecules don't react very
much okay so you don't have to explain
that in a great deal of detail but you
do need to be able to say that because
their bond enthalpies are high and
because the bond polarity is low alkanes
tend to be unreactive molecules now then
let's have a look at some of the
reactions that they do take part in now
all hydrocarbons not just alkanes but
all hydrocarbons can burn quite well
which is why we use them as fuels so
combustion is a reaction with oxygen
combustion just being another name for
burning so if we think about how we're
going to write equations for this well
we can think of any particular alkane
like methane for example and if we're
going to write an equation for it then
we're going to need to include oxygen as
one of our reactants because combustion
is a reaction with oxygen now
we're talking here about complete
combustion complete combustion means
that the carbons get as much oxygen as
they can and so the hydrogen so do the
hydrogen's when we come to deciding the
products so potentially if carbon were
to combine with oxygen you could imagine
it forming carbon monoxide or carbon
dioxide but if we're to give it its
maximum amount of oxygen it forms carbon
dioxide there is only one oxide of
hydrogen and that's water so in any
complete combustion reaction these are
going to be our products carbon dioxide
and water writing the equation now is
just simple that's left really is to
balance it now the number of carbon
dioxide's will always be the same as the
number of carbons in your hydrocarbon
and the number of water's will always be
the same as half the number of
hydrogen's in your hydrocarbon so
there'll be two water molecules here
right one carbon dioxide now all that's
left is to do the oxygens and we've got
two oxygen atoms here another two there
so we need four on the left
so we need two O twos now you could say
this is a little bit easier when the
number of carbon atoms in the
hydrocarbon or in the alkane is an odd
number and we'll see why in just a
moment because I'll just do one quickly
for ethane where we've got an even
number of carbon atoms and once again
combustion involves oxygen and if it's
complete combustion then we're going to
be making carbon dioxide and water and
this time we're going to make two carbon
dioxides and we're going to make three
waters now what you'll see with any even
at any alkane that contains an even
number of carbons you'll end up with an
odd number of oxygens on this side okay
so we've got four oxygen atoms here in a
seven sorry three there so seven in
total so we need three and a half o2 s
now that equation can be left just like
that if you like but some people get a
little bit upset by the presence of
these half numbers so if we wanted to we
could double everything in this equation
we could make that two key things seven
oxygens makes four carbon dioxide's and
six water's either one of those
equations being just as good as the
other the key point about complete
combustion equations is not only that
they balance but they contain carbon
dioxide and water as their products now
if we can trust that with incomplete
combustion this is when there's a lack
of oxygen so often gas heaters for
example get tested to see if they're
working properly which means is there a
sufficient supply of oxygen to the fuel
because if there isn't then the carbon
in the alkane could just be left as
carbon or more importantly or more
dangerously it could turn into carbon
monoxide instead of carbon dioxide and
the danger with that is it's a very
poisonous gas but it doesn't smell of
anything so if you're producing it in
your home it's not really a very good
thing okay so if we think about writing
balanced equations now what we need to
make sure we do is that we include at
least one of those things or any
combination of them
okay because how much carbon and carbon
monoxide you produce depends on how much
oxygen you have and if someone tells you
that you're incompletely combusting of
fuel they're not often going to tell you
about the amount of oxygen you've got so
you're kind of free to write a number of
different equations here so for example
I could write carbon as my only carbon
containing product and two h2 O's
because I've still got four hydrogen's
and we have to make water because it's
the only oxide of hydrogen and that
equation is balanced and you could
imagine this reaction happening in a
Bunsen burner where the air hole was
closed because there wouldn't be very
much oxygen and so all the methane would
turn into water and carbon perhaps and
this carbon would make things so
this would make a beaker that you were
heating it would cover it in a black
substance which would be the carbon and
also this carbon would glow very
brightly in the flame so your flame
would be yellow okay so anyway that's a
little bit of a sidetrack the key point
about incomplete combustion reactions is
that they balance but how you make them
balance is kind of up to you because you
can include any amount of carbons and
carbon monoxides in the products that
you want just make sure that you're also
forming water
okay so alkanes are quite good at
burning other reactions they're not so
good at and in particular they're not
very good at reacting with halogens they
might wonder why we're doing this in
that case well because alkenes are quite
good and there's a contrast to be made
but anyway let's stick with our Canes
for the moment alkanes will react with
halogens
but we need bright light to do this so
often you'll see in the equation above
the arrow UV meaning there has to be
bright UV light in order to make these
reactions happen they're called
substitution reactions why they call
that because we're swapping one of the
hydrogen's in the hydrocarbon for a
halogen atom so what you'll see here is
that the CH 4 loses a hydrogen but gains
a bromine atom and then we're left with
a CH
our if Ethan reacts with chlorine we're
going to swap one of the hydrogen's for
chlorine but we'll be left with HCl so
writing the equations is actually quite
simple here what's a little bit more
tricky is the fact that we need to know
something called a mechanism for these
reactions so in that means we need to
know how the reactions happen in other
words we need to be able to describe the
stages that take place now with free
radical mechanisms the first step
involves the formation of a free radical
okay this is called initiation because
it's what gets everything started and
it's why we need the UV light so for
example a chlorine molecule could turn
into two chlorine radicals we're
representing a radical as having a dot
that means it's got an unpaired electron
the UV light causes this split to happen
and if you think about what's holding
two chlorine atoms together in a
chlorine molecule there's a pair of
electrons okay and this type of bond
breaking breaking being fission is
called homo lytic it's called homo lytic
because when we break this bond so lytic
means meaning breaking we end up with
two things that are the same in other
words the electrons in the bond are
shared between the two atoms and we form
two of these so-called free radicals
they're called free radicals because
they have an unpaired electron and this
makes them extremely reactive because
they want to find a partner for this
electron so they'll react with just
about anything including alkanes which
aren't very very reactive so if we
imagine that we had some methane and it
was reacting with chlorine that could
react with a chlorine radical now this
would be a difference that once we'd
form the rear the radicals in the
initiation step we could take part in
what is called a propagation step where
one molecule reacts with another this
takes one of the electrons from a carbon
hydrogen bond and makes the HCL that we
saw on the previous slide however what's
left now
is ch3 because it's lost a hydrogen atom
but it has now got an unpaired electron
so it's called propagation because
although we use up one radical we've
made another and now this radical that
we've just made could potentially react
with something else like perhaps another
chlorine molecule and if it did that
then this electron will pair up with one
of the electrons from this bond and
would make ch3cl and we'd produce yet
another chlorine radical so this
chlorine radical can now take part in
another reaction okay so these are this
is why these are called propagation
steps because although we use up a
radical we make another one the last
step of reaction notice here that we
have made HCl and ch3cl is what is
called a termination step and this is
where two radicals collide because if
two things with an unpaired electron
come together like for example these two
radicals that we've already seen then
they won't produce another radical
they'll produce something that has all
paired electrons and that's the end so
to speak right you could also end up
with two chlorine radicals colliding
together and that would just reform a
chlorine molecule or you could even end
up with two ch3 radicals colliding
together because that would make a
totally different product c2h6 which
would be ether so it's also important to
realize that although we can write quite
nice neat tidy reaction equations for
these like ch4 plus CL 2 makes ch3cl and
HCL it's important to realize that there
will be lots of other products that form
even though we don't necessarily have to
put them into our equation okay and it's
definitely important to know the meaning
of this term homolytic fission because
we'll contrast that with a different
type of fish and later on okay so
hopefully now that you've watched this
film you know why alkenes are unreactive
you can write equations for combustion
reactions and reactions with halogens
and you know what we mean by a free
radical mechanism it's important to
practice practice this stuff obviously
to get it stuck into your memory but if
you've got any questions or comments
please feel free to come and see me
or to post a comment on the YouTube
channel
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