Bimolecular Nucleophilic Substitution Video Lecture | Chemistry Class 12 - NEET

laya hair from labor sicom and in this
video i will show you the reaction rate
and mechanism for the sn2 reaction sn2
stands for substitution nucleophilic by
molecular meaning it's a by molecular
nucleophilic substitution reaction
substitution means that when you have X
attached to a molecule Y will substitute
for X and B attached to the molecule
instead nucleophilic comes from the word
nucleophile which means nucleus loving
and translates to positive seeking the
sn2 reaction happens when a positive
seeking molecule meaning something that
is negative or highly electronegative
attacks a molecule that is partially
positive and uses that attack to create
a substitution and lastly by molecular
tells me that this reaction involves two
molecules reacting at the same time by
doesn't mean that it's two steps because
this is actually a one-step reaction
however because I have two molecules
reacting at the same time it's an
overall second-order reaction giving me
the two in sn2 let's take a look at a
generic sn2 reaction that starts out
with a leaving group attached to a
chiral molecule I have a strong
nucleophile attacking the carbon that
holds the leaving group and forming a
bond since carbon now has too many bonds
the leaving group is forced out along
with its bonding electrons and in one
step I get my product then now has the
nucleophile attached where the leaving
group used to be notice that the leaving
group started out as a wedge but the
nucleophile is now attached as a dash
and that's because an sn2 reaction
occurs via a backside attack and this
gives you an inversion in chirality
because if the leaving group was forward
the nucleophile had to come from the
back and when it formed that bond that
bond is towards the rear of the molecule
looking at the kinetics of this reaction
the rate of this reaction is dependent
on some constant K and now because we
have two molecules reacting together the
rate is going to depend on the
concentration of both so we write alkyl
to refer to the carbon chain holding the
leaving group and we'll write nu
referring to the concentration of the
new
clea file the reaction is first order
for the alkyl group first order for the
nucleophile giving me an overall second
order reaction and the two in sn2 this
tells you that if you double the
concentration of the alkyl group the
rate of the reaction will double or if
you double the concentration of the
nucleophile the reaction rate will
double however if you double the
concentration of both the alkyl group
and the nucleophile the reaction will
quadruple because it'll be two times the
reaction for the alkyl two times the
reaction for the nucleophile that's two
times two giving you four times the
overall reaction rate let's analyze the
reaction that takes place when the our
version of 2-bromobutane reacts with
sodium cyanide in the presence of DMSO
we'll use the for item checklist that we
analyzed all substitution and
elimination reactions I have explained
each item in detail in videos that are
found on my website layer four sicom
plus substitution - elimination we'll
start by analyzing the alkyl chain in an
sn2 reaction you want your leaving group
position on a carbon that is easy to
attack meaning you want as little
interference from side groups as
possible ideally you prefer the leaving
group on a methyl or primary carbon
however an sn2 reaction can take place
when a leaving group is attached to a
secondary carbon as long as the molecule
isn't too bulky getting in the way of
the attacking molecule in this case we
have a secondary carbon meaning that we
can have both a one type meaning sn1 e1
or a two type sn2 or e2 reaction bromine
as a leaving group is considered a good
leaving group because it forms a large
halide in solution where the charge is
nice and distributed however in an sn2
reaction it's not as important to have a
really good leaving group because the
strong nucleophile forces the leaving
group out in an sn2 reaction so what
you're looking for is not so much a very
good leaving group but rather a leaving
group that is capable of being kicked
out next we look at the attacking
nucleophile or base in this case na CN
in an sn2 reaction we're looking for a
strong nucleophile potentially with a
negative charge so that it
strong enough to attack the molecule
that holds the leaving group while NECN
appears to be neutral we actually have a
negative nucleophile anytime you see a
group one metal such as sodium at the
start of your molecule it's simply a
positive spectator ion balancing the
negative charge on your counter molecule
in this case we have CN minus which is
negative and therefore a strong
nucleophile this gives me a two-type
reaction because a negative charge will
go straight for the attack C triple
bound and because of its hybridization
and high polarizability makes a good
nucleophile but not as good of a base
this tells me that the SN meaning
substitution reaction will be favored
over E or the elimination reaction and
last we have the solvent DMSO discuss in
detail in my solvent video this is a
polar aprotic solvent which is not ideal
for stabilizing strong charges in
solution but favors a polarizable
nucleophile and therefore favors an sn2
reaction once again we have substitution
favored over elimination looking at all
these clues together we know that
substitution is favored over elimination
and we know that we want a two-type
reaction and we can have a two-type
reaction meaning we will have the sn2
reaction take place one more thing to
consider before we start the reaction is
that bromine being highly
electronegative will pull electron
density away from the carbonyl attached
to making the carbon partially positive
and this is what starts the reaction the
cyanide will use its lone pair of
electrons to attack the partially
positive carbon from behind given that
the bromine is facing forward when the
cyanide attacks it forms a new bonds to
the carbon chain kicking out bromine at
the same time I get my product in one
single step and have to show the cyanide
on a dashed line given that it attaches
from the back and of course we have a br
minus and solution the sum of the
charges on the reactants is negative one
given that we have a negative sign ID
and a neutral starting molecule the sum
of the charges and the products is also
negative one so we have a neutral
molecule and a negative bromide and
lastly recognize that because the
stereochemistry of the starting molecule
is our
and that the sn2 reaction results in an
inversion of configuration we get an S
chirality for the final product be sure
to join me in part two of this video
where I take you through additional sn2
reactions and mechanisms are you
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