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Transcript
Nucleophilic Substitution
Swapping
Generic Equation: Swap
R-X + Nu:  R-Nu + :XThe problem lies in the mechanism.
Importance of Alkyl Halides
Precursor to many others
NUCLEOPHILIC SUBSTITUTION
Theory - nucleophile means ‘liking positive’
- a dipole is induced (by a Nu:) in the C-X bond and it is polar
polar bond, why is it polar with chlorine?
Answer: ∆E=0.4, electronegativity difference
OH¯
CN¯
NH3
H 2O
These are the 4 Nu: NaOH, NaCN the 2 moleculars are as is
SN2 :NUCLEOPHILIC SUBSTITUTION
MECHANISM
MECHANISM: 2 Steps
1) The NU: electrons attacks the slightly positive carbon atom
2) The polar bond breaks unevenly (heterolytic) and Br- released
Back
attack
Polar Bond
Both eTo halogen
2 products
Note/Points:
-attack from back (electronegative halogen will forces this)
-show the polar bond
SN2
All 10 (primary)halides react as SN2 reactions.
The 2 does NOT mean steps
S = substitution
N = Nucleophile
2 = second order (both reactants determine rate)
Note: this is for primary halides, tertiary will under go SN1 as
we will see later, secondary will do both – but we are not
responsible to know about secondary reactions
1O Haloalkanes
SN2
• https://www.youtube.com/watch?v=h5xvaP
6bIZI
There are 4 Substitutions
1) WITH HYDROXIDE
C2H5Br(l) + OH- (aq) -->C2H5OH(l) +Br-(aq)
2) WITH A CYANIDE
C2H5Br(l) + CN- (aq/alc) —>C2H5CN + Br-(aq)
3) WITH 2 AMMONIAS
C2H5Br(l) + 2NH3(aq/alc)—> C2H5NH2 + NH4Br
4) WITH H2O (solvolysis) solvent and reactant at same time
C2H5Br(l) + H2O (l) —> C2H5OH(l) + HBr (aq)
SN2 Energy Diagram
• The SN2 reaction is a one-step reaction.
• Transition state is highest in energy.
Chapter 6
8
Substitution Reactions : The Nu: will be one of these 4 reactants
OH¯
CN¯
NH3
H2O
These four reactants will be similar in result and mechanism
They will ALL be refluxed
Note: NaOH or KOH, NaCN etc is the source of
The OH¯ , and CN¯ above as they are ionic
But we can ignore the metals as spectators
NaOH ---> Na⁺ + OH⁻
NUCLEOPHILIC SUBSTITUTION
AQUEOUS HYDROXIDE ALCOHOLS
Reagent
Conditions
Product
Aqueous* NaOH or KOH (aq) (OH¯)
Reflux in aqueous solution (SOLVENT IS IMPORTANT)
Alcohol
Nucleophile
hydroxide ion ( : OH¯)
Equation
e.g. C2H5Br(l) +
Na OH-(aq)
Spectator
——>
C2H5OH(l) + Na Br-(aq)
Spectator
Mechanism
STEP 1
Nu: attack
STEP 2
:Br- forms and leaves
* WARNING
It is important to quote the solvent when answering questions.
Elimination takes place when ethanol is the solvent - SEE LATER
NUCLEOPHILIC SUBSTITUTION
CYANIDE makes Nitriles AND makes chain LONGER
Reagent
Conditions
Product
Nucleophile
Equation
Alcohol/ Aqueous, NaCN / KCN
Reflux in aqueous , alcoholic solution
Nitrile (cyanide)
cyanide ion (CN¯)
e.g. C2H5Br + Na CN (aq/alc) ——> C2H5CN + Na CN-(aq)
Mechanism
1-bromoethane
Importance
propanenitrile
Br-
extends the carbon chain by one carbon atom,
called propanenitrile .
NUCLEOPHILIC SUBSTITUTION
AMMONIA makes an amine
Reagent
Conditions
Product
Nucleophile
Equation
Aqueous, alcoholic ammonia (in EXCESS, or 2 Ammonia’s)
Reflux in aqueous , alcoholic solution under pressure
Amine
Ammonia (NH3 (aq/alc) )
e.g. C2H5Br + 2NH3 (aq / alc) ——> C2H5NH2 + NH4Br
(i) C2H5Br + NH3 (aq / alc)
(ii) C2H5NH3 + NH3 (aq / alc)
——> C2H5NH3 + HBr
——> C2H5NH2 + NH4Br
_______________________________________________
C2H5Br + 2NH3 (aq / alc) ——> C2H5NH2 (l) + NH4Br (aq/alc)
Mechanism
Ethaneamine
Bromide
NUCLEOPHILIC SUBSTITUTION
AMMONIA
Why excess ammonia?
The second ammonia molecule ensures the removal of the H to make an amine. A
large excess ammonia ensures that further substitution doesn’t take place.
Problem
Amines are also nucleophiles (lone pair on N) and can attack another molecule of
haloalkane to produce a 2° amine. This too is a nucleophile and can react further
producing a 3° amine and, eventually an ionic quarternary ammonium salt.
..
C2H5NH2 + C2H5Br
——> HBr + (C2H5)2NH
diethylamine, a 2° amine
..
(C2H5)2NH + C2H5Br ——> HBr + (C2H5)3N
(C2H5)3N +
C2H5Br ——> (C2H5)4N+ Br¯
triethylamine, a 3° amine
tetraethylammonium bromide
a quaternary (4°) salt
NUCLEOPHILIC SUBSTITUTION
WATER Makes Alcohols
Details
A similar reaction to that with OH¯ takes place with water.
It is slower as water is a poor nucleophile.
Water may be looked at as HOH THE OH AND BR SWAP
Equation
C2H5Br(l) + H2O(l) ——> C2H5OH(l) +
HBr(aq)
We show the product as HBr and not just Br- because it is molecular and the H in
the water is not an ion like Na+ or K+. As well, H3O+ forms, which we rewrite as
H+ and the water from the H3O+ is part of the (aq)
SN1 V SN2
There is very little difference in these reactions.
Both have 2 steps and the same product , the
same reactants. Hard to see any difference
S = substitution (both will swap)
N= Nucleophile (both will use the same Nu:
1 =Rate of reaction (depends on 1 reactant)
called 1st order
2 =Rate of reaction (depends on 2 REACTANTS)
called 2nd order
Steric Hinderance
• With a tertiary halogenoalkane, NU: attack
is impossible. The back of the molecule is
completely cluttered with CH3 groups.
• It has to go by an alternative mechanism.
Tertiary Haloalkenes SN1
Steric hindrance means the Nu: cannot approach and
has no effect on the reactions first step
The [Nu:] nucleophile has no effect. Only the [haloalkane]
affect the reaction rate. Thus it is first order.
rate = k [haloalkane] 1
Mechanism for
3o Halides SN1
SLOW (RDS)
STEP
Carbocation Intermediate
FAST STEP
http://ibchem.com/IB16/10.34.htm
SN1 Energy Diagram
• Forming the
carbocation is an
endothermic step.
• Step 2 is fast with a
low activation energy.
Chapter 6
19
Film Clip 30 Haloalkanes
SN2 (single step)
https://www.youtube.com/watch?v=JmcVgE2
WKBE
Secondary Halides
• They undergo both SN1 and SN2 reactions
so on an exam if you are given a secondary
halide, either mechanism will receive full
credit in marks
LAST REACTION
ELIMINATION
3 WAYS to Favor Elimination
To favor elimination rather than substitution use:
1) More heat
2) More concentrated hydroxide in ALCOHOL
3) pure ethanol as the solvent
Elimination
• The :OH- is a BASE, in alcohol and takes a hydrogen (H+) to
form water. This sets off a cascade where a C=C forms and
the halide ion is lost. (as always Reflux)
24
Elimination Via OH- BASE
Step 1: The :OH- is NOT a nucleophile, it is a BASE, and attacks H
Step 2: The H+ leave and e- pair make C=C
Step 3: The halogen (here as :Br-) leaves
GENERIC CASCADING REACTION
+ H2O + :Br-
An Example
+ H2O + :Br-
ELIMINATION
Reagent
Alcoholic sodium (or potassium) hydroxide
Conditions
Product
Mechanism
Equation
Reflux in alcoholic solution, ALL alcohol
Alkene
Elimination
C3H7Br + NaOH(alc) ——> C3H6 + H2O + NaBr
Mechanism
the OH¯ ion acts as a base and picks up a proton
the proton comes from a carbon atom next to that bonded to the halogen
the electron pair left moves to form a second bond between the carbon atoms
the halogen is displaced
overall there is ELIMINATION of HBr.
Complication:Cis / Trans with unsymmetrical halalkanes, you get mixture of products
ELIMINATION
Complication
If the haloalkane is unsymmetrical a mixture of isomeric alkene
products is obtained.
but-1-ene (Minor Product)
2-bromobutane
but-2-ene (Major Product)
2-bromobutane
can exist as cis and trans isomers
ELIMINATION v. SUBSTITUTION
The products of reactions between haloalkanes and OH¯ are influenced by the solvent
SOLVENT
ROLE OF OH–
MECHANISM
PRODUCT
WATER
NUCLEOPHILE
SUBSTITUTION
ALCOHOL
ALCOHOL
BASE
ELIMINATION
ALKENE
Modes of attack
Aqueous soln
OH¯ acts as a nucleophile
Alcoholic soln
OH¯ acts as a base (A BASE IS A PROTON ACCEPTOR)
Both reactions take place at the same time but by varying
the solvent you can influence which mechanism dominates.
Elimination Clips
• https://www.learnnext.com/CBSE/Class12/Chemistry/Haloalkanes-andHaloarenes/Chemical-Reactions-Of-AlkylHalides-Elimination-Reactions/L1804203.htm
• Franklychem
• https://www.youtube.com/watch?v=o3lk77s
dFHY
The ability of solvents to stabilize ions through solvation is directly associated
with their polarity. Polar solvents, such as water, can stabilize the ions 3X more
than alcohol (see chart)
Methanol can stabilize ions through solvation, but water is best at controlling
ions. The :OH- is a super strong ion and needs controlling. Alcohol controls less
and is better to dissolve weak :Nu’s such as NH3 and CN, as they do not need
to be controlled like OH- as they are weak ions
Aprotic solvents
ε
Protic solvents
ε
Hexane
1.9
Acetic acid
6.2
Benzene
2.3
1-Methyl-2-propanol
11
Diethyl ether
4.3
Ethanol
34.3
Chloroform
4.8
Methanol
33.6
Hexamethylphosphora
mide (HMPT)
30
Formic acid
58.0
Dimethyl formamide
(DMF)
38
Water
80.4
Dimethyl sulfoxide
(DMSO)
48