SN1 Reaction Introduction ,Mechanism and Finally Characteristics of SN1 Reaction

1. SN1 Reaction
Saeed Ur Rehman
Roll NO : 180131
Class of 2022
Department of Chemistry Islamia College
University Peshawar

2. In today Lecture We will focus on ;
• Introduction to SN1 Reaction
• Mechanism Of SN1 Reaction
• Characteristics Of SN1 Reaction

3. Introduction
• Those reaction in which one molecule is involved in the rate determining
step or slow step of the reaction is called SN1 Reaction
• In the term SN1; ‘S’ represent that the reaction is Substitution
/replacement Reaction , ‘N’ represent that the reaction is Nucleophilic
reaction / in nature and The ‘1’ Represent that the reaction is
unimolecular which means that only one molecule is involved is the
transition state of rate limiting step. So its mean that in SN1 reaction
substitution and nucleophilic reactions is occurring in it
SN1
Substitution
Nucleophilic Show kinetic study
means tell us about
order of reaction

4. Component of Sn1 Reaction :
SN1 reaction have 3 parts and one is the material in which the
reaction is occurring,
• Substrate
• Leaving Group
• Attacking Nucleophile
• Solvent
If a reaction takes place in a series of steps, and if one step is intrinsically
slower than all the others, then the rate of the overall reaction will be
essentially the same as the rate of this slow step. This slow step,
consequently, is called the rate limiting step or the rate-determining step.
(CH3)3-C-Br + OH- ---H2O-- (CH3)3-C-OH + Br-
Leaving
group Attacking
Nucleophile
Solvent

5. Mechanism Of SN1 Reaction
SN1 reaction is a two step reaction in which one is bond breaking and other
one is bond forming.
Step 1: Here Bond breaking occur between Leaving group and substrate
(Heterolytic breakage), it is endothermic in nature because bond breaking is
occurring here. Before bond breaking a lot of energy is absorbed by the
substrate to equalize its energy with the activation energy of the reaction as
it is necessary for the reaction to occur and also this step is rate determining
step(show order of reaction) Because it is the slowest step of the reaction.
In this step:
• a leaving group is removed
• and a carbocation is formed ( group having Carbon is central element and have
positive charge ).

6. Substrate-leaving group [substrate]
++ [Leaving group]
-
Rate = [substrate ]
*The question that arises here is what motivates it? Are there external
reasons that cause the leaving group to leave easily?
* Why it’s Rate only depend upon the substrate ?
Step 2: Here bond forming is occur between the positive charge substrate
and Attacking Nucleophile. This step is exothermic in nature because here a
bond formation is occurring and we know when bond formation is occurring
then energy is released to the environment and here a lot of energy is
release . This step is the fast step of the reaction because here Ions are
involved which required very low activation energy to combine because they

7. have itself very high energy and unstable and tending to losing energy and
to gain stability.
In this step :
• Nucleophile addition is occurring
• Highly stable product is formed
[Substrate]
+ + Nu
-/Nu
.. Substrate Nu
This step is not the rate determining step because it’s fast step and we
only consider the slowest step is the rate determining step.
Fast
Step

8. Mechanism Of SN1 Reaction explanation with example :
Lets we have a reaction between tertial butyl bromide and methanol that is
(CH3 ) 3 C – Br + CH3- O H (CH3 ) 3 C – O–CH3 + HBr
The above reaction is occurring in two steps, there are explain as;
Step 1 : In this step the bond between C – Br is transferred to
Br in the presence of methanol as a solvent and formation of
carbocation is occurring as:
(CH3)3 C- Br (CH3)3 C+ + Br-
Methanol
boil

9. Step 2: In this step the Nucleophile attack on Carbocation and Form the
stable product that is;
(CH3)3 C+ + O CH3 (CH3)3 C-O+- CH3
Final step is attack of a methanol over the H of Oxygen and as a result the final
methyl t-butyl ether is produced .
(CH3)3 C-O+- CH3
H- O- CH3
..
H
.
.
H
H
(CH3)3 C-O- CH3 + CH3-O+-H
H

10. *The question that arises here is what motivates it? Are there
external reasons that cause the leaving group to leave easily?
Answer : the reason to this Problem is….
• Recall that SN1 Reaction which is highly favors by Polar protic solvents
(polar/H+). In such cases, the nucleophile that is we called as leaving-group (to be
Nu-) is going to be attracted to the positively charged molecules of the polar
protic solvent (...H+) by the dipole force ( S-) which cause increase in the electron
affinity of leaving group so its greatly packing electron from the partially
charged carbon of carbocation . The power of the attraction comes from the
dipole between C-Nu, where Nu has the negative bit of the dipole. At the end of
this step, when the nucleophile group leaves, you often see "Nu-" hanging around,
but really it's with the solvent, e.g. H3O+Br-

11. Characteristic Of SN1 Reaction
1. A two step and 01 Order reaction:
SN1 reaction is a 2- step reaction in which one is bond breaking
and another one is bond making step. The 1st one is slow and rate
determining step and 2nd one is fast reaction and it does not paly any role
in rate of the reaction .
Yes! SN1 reaction is 1st order reaction because its rate is only depend upon
the concentration of Substrate and it does not depend upon the
concentration of Nucleophile
Normally, rate of a reaction depend upon the concentration of reactant ,
when concentration of reactants is greater than it’s take less time to reach
the completion and similarly , if the concentration of the

12. reactant is less than its takes greater time to reach the completion and
similarly, if the rate of a reaction does not depend upon the concentration
of reactant then the time during which the reaction is completing does not
increase or decrease with decrease or increase of concentration of that
reactant, Keeping this concept in mind we have to proof that SN1 reaction
only depend upon concentration of Substrate by a practical work .
Practical work :
Let suppose ! We have a reaction ,
(CH3 ) 3 C – Br + CH3- O H (CH3 ) 3 C – O–CH3 + HBr
The rate equation for the above reaction is;
Rate = [ (CH3)3C- Br] [CH3-OH]
Methanol
boil
*Why it’s Rate only depend
upon the substrate ?

13. Experiments:
From the above experiment it is clear that rate does not change with change
in concentration of CH3-OH( Nucleophile).
So its proof that rate is only depends upon concentration of Substrate that
is (CH3)3 C- Br.
S.N
O
[(CH3)3C-
Br]
[CH3 –
OH]
Time of
completion of
reaction
Rate of reaction
01 0.01ml 0.01 5 seconds X
02 0.02ml 0.01ml 2.5 seconds 2x
03 0.01ml 0.02ml 5 seconds x
04 0.02ml 0.02ml 2.5 seconds 2x

14. 2.Nu attacks from the top and bottom sides of the C+ intermediate:
Yes ! In SN1 reaction in step 2, Nu attacks from the top and
bottom sides
of the C+ intermediate.
Reason : In SN1 reaction Nu only attack from top and bottom
side of C+ intermediate ,because the molecule of Intermediate
C+ is planner and all the C+ is sp2 hybridized and the bond
angle is 120 degree . If we look at to the Intermediate C+
complex structure al the sides are blocked from the Nu by the
three bulky group attached to the C+ center ..so for Nu there is
only these two side from which the nu attack can attack so
that’s why in SN1 reaction the NU is attacking from top and
bottom sides.

16. 3. Reactant Undergoes Racemization :
Yes ! Racemization is occurring in SN1 reaction , but it depend upon the
type of substrate used .
Explanation: SN1 reaction, The Nu attack from the top and bottom side
and produce two types of product in case of ,when the intermediate C+ is
produced from a chiral molecule . In one product the Nu is attached from
the top and another one the Nu is attached from below . The chances of
formation of these product is 50% which means that if we have 100
molecules of product is formed then 50 will be the one in which the Nu is
attached from the top and the 50 molecule will be like the in which the Nu
is attached from the Below . Which is actually called racemization .This
percentage may be changed means that the two isomers which is produced
may have different percentage probability which is called partial
racemization .Now if the substrate is not chiral then there will be no concept
of racemization phenomena because only type of product will be formed .

18. 4. Mechanism is favored by stable carbocation :
Yes! SN1 reaction is favored by stable carbocation
Reason : Stable carbocation
• Easily attain transition state
• Easily activated
• Proceed reaction easily
Explanation: The rate of a reaction depend upon the Rate determinant step
or we can say that reactivity of a reaction depend upon its rate
determinant step. In SN1 reaction the 1st step is the rate determinant step
now if the carbocation produced is stable then the rate will be high , As we
know that the bond between leaving group-C is polar , Now when positive
part of a solvent comes near the Leaving group then Leaving group show
some interaction with it as result of that the leaving group will attract the
electron pair from the Carbon more and the carbon will get more partial

19. charge, Now if the Carbon atom is attached with electron donating group
then these group will donate electron to it and the Partial charge of carbon
will be reduced more as a result of which the leaving group will get more
power to attract electron more toward itself , so greater the electron
donating group attached to carbon of substrate then partial positive charge
of carbon will be reduced more and the leaving group will get high and high
power to attract electron toward itself and the bond breaking will occur
easily,
From the whole above discussion its clear that if the carbon have greater no
of electron donating group then less time will be required to the bond
between C-Leaving .g to break .
Now as we know that bond breaking process is totally a transition state so
less time will be required for the transition state to achieve (bond breaking)
if the carbon have high number of electron donating group to stabilize its
partial positive charge and vice versa . And as a result the rate will be high
respectively . So due to this the SN1 reaction is favored by Stable
carbocation

20. Now question is arising that which Alkyl halide is best for SN1
reaction and why? And what is the order ?
Answer : Tertiary alkyl halide is best for SN1 reaction
Reason:
• Contain greater no of alkyl groups
• It’s carbocation is stable
Explanation ;
As we know that in SN1 reaction if carbocation is stable than
the reaction will proceed quickly means rate of reaction will be fast. The
newly produced carbocation in 1st step of SN1 reaction is in need of
stability and there stability comes by the sigma donation of electron from
the electron donating group (alkyl group ) and also by the hyperconjugation
phenomena in which filled orbital from the electron donation group overlap

21. with Empty p-orbital of C+ center of carbocation. Now greater the no alkyl
group, greater no of electron will be donated toward the C+ center and its
positive charge will be reduced and also greater hyperconjugation will occur
and vice versa . Now In methyl halide there is no alkyl group and in
primary alkyl halide there is one electron donating group and 2 in
secondary alkyl halide and there is 3 electron donating group in tertiary
alkyl halide.. As the tertiary alkyl halide have greater no of Electron
donating group so it will stabilize the carbocation more then the other alkyl
halide by the sigma donation and hyperconjugation phenomena.
Order: The order is given below which is on the basis of stability of
Carbocation;
Tertiary > secondary > primary > Methyl Halide

22. 5. Mechanism is best in polar protic solvent
Polar Protic Solvents :
The solvents which can give an H+ ion or a proton in the solution is known as protic solvent
.Similarly the polar protic solvents are those solvents , which have an acidic H-atom, but
they have a high dielectric constant and high polarity and so they form hydrogen bonding
and dissolve the salts. They prefer SN1 reactions.
The common polar protic solvents are: water, alcohols , formic acid, ammonia etc.
Formation of carbocation intermediate is the rate determining step.
Since for the formation of stable intermediate carbocation highly polar
solvent is required. The negative part of the polar protic solvent interacts
with the highly positive part and minimizes the energy and makes it stable.
And secondly Polar protic solvent makes nucleophile less nucleophilic and stabilizes anionic
leaving group.

23. Our Next lecture will
be regarding ;
“Factor affecting the SN1
Reaction along with Energy
diagram study ”