L6 :Alkane preparation from alkyl halide - Hydrocarbons, Class 11 Video Lecture - Class 11 | Best Video for Class 11

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the preparation alkene from al-qaeda
right we have three methods to prepare
I'm paying to park at night the first is
you can do a reduction of Mecca light
the second you can use the mood reaction
third if you can use the brick not reach
let's start the first one via read with
the reduction of the highlights see all
the alkyl halides except fluorides and
you reduce them with zinc and s here it
gives alkane why because this guy is a
good reducing agent Jane plus s CL will
give you NASA and hydrogen Delta Q how
it gives zinc plus s it gives NASA and
hydrogen which is a good reducing agent
I'll take you through the reaction for
this and instead of jinkin l care you
can also use li LS for and nab a spoon
they can also be used as reducing engine
this is the required reaction we have
here ch3cl
oxidation number of carbon s or minus 2
here and oxidation number of carbon here
is minus 4 correct because this guy you
see if you assume discover as X plus 4
into hydrogen is 4 into 1 1 into
chlorine is minus 1 this has to be 0 so
if you see the value of x is minus 2
here carbon is X 4 into hydrogen is 4
into 1 is equal to 0 so oxidation number
carbon here is minus 4 so from minus 2
to minus 4 that means it got dude use
right - both no - 2 got reduced more
reaction if you want this is the
reaction if you have this guy C 2 H 5 CL
with il-28 when you react with hydrogen
so this will replace this chlorine and
you will get you think
similarly this Lawrence example then you
have profile alright you add hydrogen
and then it becomes propane too you see
the reaction of zinc and SCL this is
like this you have zinc and you have s
steel they react
it gives G and C n to n nascent hydrogen
the balanced reaction this is a nascent
hydrogen and this guy is a very strong
reducing agent cut and that's how this
reaction is the next reaction of the
booze reaction so here al Qaeda bites
similarly when you still using the alkyl
halides only they retreat with sodium
salts in right has to be free from
moisture it gives higher alkenes gives
higher in this reaction is called huge
reaction and as I told it has to be
performed under anhydrous condition
because if there even if there is small
amount of water right so the Alka
article which is formed in this reaction
right reacts the intermediate she
strongly basic and can react with proton
from the water so you see this is a
reaction I have my rx here this is my
metal it can be a sodium metal when you
react sodium with this metal what
happened is you get the free radical of
art and this MX the sodium bill takes
this chlorine or whatever halogen it had
and you get the free radical ah and this
guy is very reactive actually even if
you have a water here small amount of
water this will react with hydrogen from
or you call the water
correct so it has to happen in the NI
interest rate and then if you see this
is again my free radical will react with
sodium to form this component actually
what I am doing is I am making this to
this hour - emphasize okay I'm giving it
more ionic - when this reacts with rx
you get ah this is my reaction mechanism
this is my reaction mechanism what
happens and those it is preferred that
this has to be in hydros condition but
if it is if it is a if it has some water
this guy is a very reactive and will
react with photon from the water
and please note it is used to prepare
higher alkanes with even number of
carbon atoms only why because if you see
this is our X right even it - I suppose
three carbon atoms when combined it
becomes our Arts in laughs 3 into 2 so
when if it has n carbon atom it will
become n into n into 2 is what always a
even number correct so it is always
preferred for even number of comments I
take one example where in this point
will be cleared even if you take let
suppose see this guy the thigh bromide
and let's suppose you take the thigh
Roman together and then you do the
reaction so then in that case you should
you will be able to form compounds which
has Evod numbers but you will not be
able to separate it right because they
have similar physical chemical problem I
take this little examples understand is
called muj reaction Mary use zunium
right and then you use a dye ether to
make sure everything is dry and then
this sodium will form this free radical
it will ionize there's actually this
radical origami al qaeda and then this
alkyne will be formed the question is
find the product when is height I died
and if I die right under oppose whose
reaction so if you see here I have it
highlighted ite to h5 re methyl iodide
ch3 I write it underscored this reaction
with sodium when dry ether right so what
you get is one product you can have is
this reacts with this radial reaction
inside then you get c2h5 ends is ch3
that is one option
the second option can be ethyl the air
it with anti theoretical but if you get
c2h5 against eth 5 other is this with
air we are reading a react with the
methyl radical itself correct three
possible scenarios curry because
intermediate if you see
this guy when it reacts with sodium we
will get c2h5 free radical we can
correct because c2h5 Iwan react with
sodium it you give c2h5 free radical and
nai correct similarly ch3 i and you
react with sodium it'll get ch3 free
radical plus nai so now if you see we
have two here leaders now they get
combined among themselves become c2h5
c2h5 CSTA csv or they both can be here
so you'll get three protons now say
phonetic this product is going to figure
because they have similar she's kind
chemical properties and cellular boiling
points so we generally don't use or
don't recommend this booze reaction for
this kind of thing because if you see
here it has three carbon atoms in s4 and
it has to see if you see this has odd
number of carbon atoms so chain of
scenario we generally avoid see we
generating only a signal compound here
so you can do if I light it is occur
instead where you are good to take only
the highlights with that case you made
only this guy you want to take with high
riding a goalie this you get only this
number correctly
why is boost reaction not preferred from
regulation of alkanes containing odd
number of carbon atoms this is the same
thing the same questions but in that
question earlier you were given that you
have to react it highlighted it is
highlighted here that is not given right
you have to think more you have to think
of a scenario where you can have a
reaction between a two carbon and one
carbon halides is that way in this case
you will have the same thing you can see
that C 2 H 5 and CH 3 I that is with
highlighted inside it reacts with sodium
imprudence or either you get these kind
of words alright so this is what you
have to say and then you have to say
that these product is really difficult
separate and so we don't use muj
reaction to form odd number of carbons
amounts so generally we use only this
guy it's a puzzle so in this case you
need only one product
correct or we use this game in this way
this is easy so let us understand now
the preparation of Alki alkanes from
alkyl halides using grignard reagent so
what is the trigger operation it is
nothing but rmgx format we will discuss
this so if you see this my alkyl halide
when you react with magnesium metal in
diethyl ether it forms a lock I'll
magnesium halogen and this is nothing
but mine quick knowledge and here you
see my carbon this will become carbon
here carbon and mg this bond is very
very weak RMG bond is very very is
actually very very polar actually is
very highly polar can be broken easily
so if you see this why it is broken
easily so if we see the
electronegativity of carbon is two point
five five and for magnesium it is one
point three one
so you believe there is a huge
difference in electronegativity right
and that's why it's polar this guy will
get negative charge carbon get negative
charge and this way we get positive sir
correct so we see this this is my
reaction I have my alkyl halide right
this is my ethyl bromide it reacts with
magnesium in presence of dry ether what
you get is this like c3 cs3 cs2 MGB are
correct there is nothing but it high in
magnesium provide now in this case if
you see this bond is very weak why
they told carbon will be slightly
negative and this guy will be slightly
positive because the difference in
electronegativity
correct so when this thing this whole
thing reacts with water actually now
this bond will break and you'll get ch3
C so let me show you the reaction
actually so I have C
it's three I have ch2 mg that I have we
are correct now water so let us assume I
have h plus nos - I love it presume that
make things little simple since this guy
is slightly positive ending it is a
slightly positive this H will attack
here right and with this this bond will
break and this OS - will attack here so
what you get here is out of this bleep
this is the SVA time Jesus second carbon
and here you will have this hydrogen
here plus this mg magnesium you get Oh H
here then this is what you get same
thing correct since this bond is polar
high H+ will attack this guy this carbon
and since this is a time this carbon
since this was slightly negative
slightly positive this bond will break
and this is a bond will break and this
resin is a negative charge here positive
charge rate this TD of course in form 1
and this positive and this h- negatively
correct so this is example of electro
philic addition reaction so this is a
chart where you say you have rmgx you
react with water there is we done this
the sharp we say that this is a very
critical region that you react is very
with water you get something that he
told you get something if you act with
earlier you get something this is the
whole chart thank you
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