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4.2 Hydration of alkenes
Hydration of alkenes
Synthesis of Alcohols
o By organometallic reaction (organlithium and Grignard
reagents)
o Alcohols can be prepared by the hydration of alkenes in the
presence of acid catalyst.
 Electrophilic hydration is the act of adding electrophilic
hydrogen to alkene
o By the reduction of aldehydes, ketones, acids, and esters
Hydration of alkenes
o Markovnikov’s addition
o Acid catalyzed reaction
Markovnikov's rule (Markovnikov addition)
 In an addition reaction of alkyl halides (HX; such as hydrogen chloride,
hydrogen bromide, or hydrogen iodide) to an alkene or alkyne,
 The hydrogen atom of HX becomes bonded to the carbon atom
that had the greatest number of hydrogen atoms in the starting
alkene or alkyne (Markovnikov's rule)
How Does Electrophilic Hydration of alkene Work?
o Markovnikov’s addition
o Acid catalyzed reaction
 Mechanism for 3º Alcohol (1º and 2º mechanisms are similar):
Addition of hydrogen bromide to propene follows Markovnikov's rule.
Regioselectivity and the Markovnikov Rule
 Only one product is possible from the addition of these strong acids to
symmetrical alkenes such as ethene and cyclohexene.
 However, if the double bond carbon atoms are not structurally equivalent, as in
molecules of 2-methyl-2-butene, the reagent conceivably may add in two
different ways..
 When addition reactions to such unsymmetrical alkenes are carried out, we
find that one of the two possible constitutionally isomeric products is formed
preferentially. Selectivity of this sort is termed regioselectivity.
 In the example below , 2-chloro-2-methylbutane is nearly the exclusive product.
 Similarly, 1-butene forms 2-bromobutane as the predominant product on
treatment with HBr.
Electrophilic Addition Reactions of Alkenes
o All alkenes undergo electrophilic addition reactions with the hydrogen
halides
o As substituents increases, the reactivity also increases
There are two ways of looking at the reasons for this - both of which need you to
know about the mechanism for the reactions.
1. In the reaction, the carbocation intermediate is important to determine the rate
of the reaction
2. Alkenes react because the electrons in the pi bond attract things with any
degree of positive charge. Anything which increases the electron density
around the double bond will help this.
 Alkyl groups have a tendency to "push" electrons away from themselves
towards the double bond. The more alkyl groups you have, the more
negative the area around the double bonds becomes.
Orientation of addition
 If HCl adds to an unsymmetrical alkene like propene, there are two possible
ways it could add. However, in practice, there is only one major product.
 This is in line with Markovnikov's Rule which says: in this case, the hydrogen
becomes attached to the CH2 group, because the CH2 group has more
hydrogens than the CH group.
 Notice that only the hydrogens directly attached to the carbon atoms at either
end of the double bond count.
Dehydration reaction of alcohols
 The dehydration reaction of alcohols to generate alkene proceeds by heating the
alcohols in the presence of a strong acid, such as sulfuric or phosphoric acid, at
high temperatures.
 The required range of reaction temperature decreases with increasing substitution
of the hydroxy-containing carbon:
o 1° alcohols: 170° - 180°C
o 2° alcohols: 100°– 140 °C
o 3° alcohols: 25°– 80°C
Is this a Reversible Synthesis (hydration of alkene and dehydration of alcohol
Rxn)?
Electrophilic hydration is reversible because an alkene in water is in equilibrium with
the alcohol product. To sway the equilibrium one way or another, the temperature or
the concentration of the non-nucleophilic strong acid can be changed.
For example:
o Lower temperatures help synthesize more alcohol product.
o Less sulfuric or phosphoric acid and an excess of water help synthesize more
alcohol product.
Primary alcohol dehydrates through the E2 mechanism
 Oxygen donates two electrons to a proton from sulfuric acid
H2SO4, forming an alkyloxonium ion.
 Then the nucleophile hydrogen sulphate anion (HSO4– ) backside attacks on adjacent hydrogen and the alkyloxonium ion
leaves in a concerted process, making a double bond.
Secondary and tertiary alcohols dehydrate through the E1 mechanism
 Similarly to the primarily alcohol Rxn, secondary and tertiary –OH first
protonate to form alkyloxonium ions.
 However, in this case the ion leaves first and forms a carbocation as the reaction
intermediate.
 The water molecule (which is a stronger base than the HSO4- ion) then abstracts
a proton from an adjacent carbon, forming a double bond.
 If the reaction is not sufficiently heated, the alcohols do not dehydrate to
form alkenes, but react with one another to form ethers.
 Alcohols are amphoteric; they can act both as acid or base.
o The lone pair of electrons on oxygen atom makes the –OH group
weakly basic, and the hydrogen atom in OH makes a weak acidic
The Williamson ether synthesis
 The Williamson ether synthesis is the most widely used method to produce ethers .
 The Williamson ether synthesis is an organic reaction used to convert an alcohol and an
alkyl halide to an ether using a base such as NaOH.
4.3 Hydroboration-oxidation of
Alkenes
Hydroboration-oxidation of Alkenes
Hydroboration is the treatment of alkene or alkynes
with borane (BH3), followed by H2O2
―To produce alcohol (for the case of alkene)
― In which aldehydes and ketones are obtained in
case of alkynes.
 In all asymmetric addition reactions to carbon,
regioselectivity is important and often determined by
Markovnikov’s rule.
 Organoborane compounds give anti-Markovnikov
additions
Anti-Markovnikov addition
 In an addition reaction of a generic electrophile HX to
an alkene or alkyne, the hydrogen atom of HX
becomes bonded to the carbon atom that had the least number
of hydrogen atoms in the starting alkene or alkyne.
 The 'anti' in 'anti-Markovnikov' refers to the fact the reaction
does not follow Markovnikov's rule, instead of being a reference
to anti addition stereochemistry.
 Anti-Markovnikov addition of hydrogen bromide to propene, illustrating
by the peroxide effect.
Mechanism: Peroxide effect follows free radical mechanism.
 Peroxy linkage in peroxides is weaker.
o Hence they decompose to form free radicals which attack on
HBr to form Bromine free radicals.
 These bromine free radicals attack on alkene to give alkyl free
radicals, which attack on next mole of HBr to form alkyl bromide
 Organic reactions can be generally classified into two broad classes of
mechanisms
1. Polar reaction (heterolytic cleavage) and
2. Radical reactions (homolytic cleavage)
Heterolytic bond cleavage:
o It is initiated by polar bond (polar reaction)
o Bond breaking in which the bonding electron
pair is split unevenly between the products
(often produces at least one ion).
Homolytic bond cleavage
o It is initiated by radical reactions
o A covalent bond breaking in such a way that each
fragment gets one of the shared electrons.
 Free radicals have the potential to be both
 extremely powerful chemical tools and
 extremely harmful contaminants because of their high reactivity.
 Much of the power of free radical species stems from the natural tendency of
radical processes to occur in a chain reaction fashion.
 Radical chain reactions have three distinct phases:
1. initiation,
2. propagation, and
3. termination.
 In borane (BH3), this situation flips around.
o Hydrogen is more electronegative than boron:
Hydroboration of Alkenes: The Mechanism
 After hydroboration, organoborane is formed.
 Treatment of the organoborane with basic hydrogen
peroxide leads to replacement of C-B with C-OH.
 Note that there is no change in stereochemistry; it has
occurred with retention!
 Step1: The first step here is deprotonation of hydrogen
peroxide to give NaO-OH. Since the conjugate base is a better
nucleophile, this speeds up the rate of the subsequent step.
 Step 2: The next step is a simple Lewis acid-base reaction.
o The deprotonated peroxide anion then adds to the empty
orbital of boron, forming a negatively charged boron species:
 Step 3: It is a rearrangement reaction
o the pair of electrons in the C–B bond migrates to oxygen,
leading to breakage of C–B and formation of C–O, along with
rupture of the O–O bond.
Note: how the charge on boron goes from negative to neutral.
 In step 4: Hydroxide ion attacks the empty p orbital of boron,
and the O–B bond breaks. Although drawn here as a
“concerted” step, where bond formation accompanies bond
breakage, it need not be so, since addition of hydroxide to
boron does not violate the octet rule.
 In step 5: Finally, the negatively charged oxygen is then
protonated by water (the solvent).
 Other halogen containing reagents which add to double
bonds include
o hypohalous acids, HOX, and
o sulfenyl chlorides, RSCl.
 Hypohalous acid is an oxyacid consisting of a hydroxyl group
single-bonded to any halogen.
 The electrophilic moiety of these reagents is the
halogen.
 In practice, these addition reactions are regioselective,
with one of the two possible constitutionally isomeric
products being favored.
o HOBr= Hypobromous acid
o HOCl = Hypochlorous acid
Hydroboration of Alkynes
 Hydroboration is the treatment of alkynes with BH3, followed by
H2O2 in which aldehydes and ketones are obtained.
 Hydroboration of alkynes is a two-step reaction sequence which
proceeds via an anti-Markovnikov addition where installing a hydroxyl
group at the less substituted position.
 Step 1: In the first step comes to hydroboration and oxidation of
alkyne.
o Addition of borane forms an organoborane. When we have an
asymmetric alkyne, the addition of BH2 will be achieved on a less
substituted carbon atom (anti-Markovnikov rule).
o And oxidation of this organoborane with basic H2O2 forms an enol.
o Enol is an unstable intermediate which we cannot isolate.
 Step 2: In the second step, enol that we obtained is subject to
tautomerization which cannot be prevented. Depending on which alkyne
is started (terminal or internal), aldehydes or ketones are obtained.
 Tautomerization can be catalyzed by either acid or base.
 Tautomerization can be catalyzed by either acid or base.
STEP 1. The hydroxyl group of the enol is deprotonated and
resonance-stabilized enolate ion is formed.
STEP 2. The enolate ion is protonated to give an aldehyde.