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Part 2.3
The chemistry of
ALKENES - unsaturated hydrocarbons
Doc Brown's
Chemistry Advanced Level Pre-University Chemistry Revision Study Notes for UK
KS5 A/AS GCE advanced level organic chemistry students US K12 grade 11 grade 12 organic chemistry
the nature of an alkene double bond explaining reactivity towards
electrophiles mechanisms
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Part 2.3
Bonding in alkenes, reactivity of alkenes compared
to alkanes, electrophilic addition reaction with
hydrogen chloride, hydrogen bromide
and hydrogen iodide
Sub-index
for this page
2.3.1
The covalent bonding in
alkenes (σ and π bonds)
2.3.2
The reactivity of alkenes (compared to alkanes)
2.3.3
The
addition of hydrogen halides to alkenes and using the
Markownikoff rule
2.3.4 The electrophilic
addition mechanism of HX addition to alkenes (non-aqueous
conditions), carbocations
and
explaining and using Markownikoff's rule
2.3.5 The electrophilic
addition mechanism of HX addition to alkene (aqueous conditions),
and using Markownikoff's rule to predict and explain the likely products
2.3.1 The
covalent bonding in alkenes
From the
quantum level rules,
carbon's electron configuration is
1s22s22p2
In terms of the separate orbitals you can express it as
1s2, 2s2,
2px1, 2py1, 2pz0
This can be further expressed as an electron box diagram:
1s 2s2p
with two unpaired electrons.
However, as you should know by now, carbon usually forms
four bonds (valency of 4) rather than two.
This is because it is energetically favourable to promote
one electron from the 2s orbital into the third empty 2pz
orbital, the energy required for this 'promotion' is far less than that
released when the carbon atom forms four bonds rather than two..
This gives a theoretical electron configuration of
1s2, 2s1,
2px1, 2py1, 2pz1
or 1s2s 2p
This gives four unpaired electrons, all of which can pair up with an
electron from another atom to form four covalent bonds, but they may be of
two varieties if a double or triple bond is involved in the carbon based
molecule - read on ...
Reminders (if needed): The diagrams below show the bonding situation in alkenes, that
conveniently involve two of the most important types of covalent bond
between atoms, including carbon.
Sigma and pi
covalent bonds in alkenes
In carbon based compounds a single bond (sigma bond, σ bond)
is formed by the overlap of two orbitals which can be either an s orbital
and p orbital, (illustrated above).
Two electrons (an electron pair) are mutually attracted to the
positive nuclei on either side.
A molecular orbital is formed and the
axis of the bonding orbital lies on a central line between the two nuclei.
So we have a
'central' C-C sigma bond shown by the black line and the other atoms C,
H or Cl etc. are also bonded to the C=C carbon atoms with sigma bonds
too i.e. a planar >C-C< sigma bond system of single covalent bonds.
So, we now have to
account for the 4th valency electron of the C=C carbon atoms.
In the case of
alkenes, the pi
bond (π bond) is formed by the overlap of
two 2p orbitals of the carbon atoms, but, due to repulsion with the bonded pairs of the sigma
bond electrons, they cannot form another sigma
bond molecular orbital along the same central axis.
Instead, two pi orbital
(in yellow, containing one electron each), are formed above and below the planar arrangement
of the sigma bond linking the two carbon atoms (and other C or H atoms),
giving the planar >C=C< bond arrangement.
Incidentally, the presence of the pi
orbitals inhibits rotation around the double bond in unsaturated
alkenes, because it requires a lot of energy to twist the orbitals
around and break the pi bond and this accounts for
the possibility of
E/Z isomerism in alkenes)
However, in alkanes everything is free to
rotate around the sigma bonds of the C-C bonds in saturated alkane
molecules.
TOP OF PAGE and
sub-index
2.3.2 The reactivity of alkenes (compared to alkanes)
 Alkenes are reactive molecules,
particularly when compared to alkanes.
They are
reactive towards electron pair accepting electrophiles
because of the high density of negative electron charge associated
with the pi
electrons of the double bond (diagram below)
π
and σ
bonds of the C=C double bond are shown in the diagram.
The sigma bond is present between
all the atoms in organic molecules. However in alkenes, the double
bond consists of a sigma bond (σ)
and a pi bond (π).
Two electrons are in the molecular orbital of the sigma bond which
is directed linearly between the two carbon atoms.
BUT, the other two electrons of
the C=C double bond are in the two pi orbitals which lie above and
below the plane of the sigma bonds of the >C-C< system (1 electron
per pi electron
cloud).
It is the electrons of these
pi orbitals that are susceptible to electrophilic attack by an
electron pair acceptor - an
electrophile.
The electrophile (electrophilic
reagent) can be a positive ion (e.g.
H3O+),
a permanently polarised molecule (i.e.
one with a partial
positive charge e.g. Hδ+Brδ-)
or a neutral molecule that become polarised on collision with the
substrate molecule (e.g.
Brδ+Brδ-)
The pi bond orbitals give the C=C
bond a high electron density region for the electrophile to attack.
The pi bond makes unsaturated alkenes much more reactive than saturated
alkanes, hence most of alkene chemistry involves addition reactions, and
many of these involve an electrophilic addition mechanism.
It is the pi bond that breaks to form two new single
sigma bonds with two other atoms - often (but not necessarily) both from the
attacking reagent.
Consider the bond enthalpies
(bond energies)
|
Bond |
Bond
enthalpy kJ/mol |
|
σ bond of C-C |
(i)
348 |
|
σ +
π bond in C=C |
(ii)
612 |
|
π bond only |
(ii)
- (i) 264 |
The pi bond is considerable weaker than the
sigma bond in a carbon - carbon double bond in alkenes.
This fact, combined with the pi orbital regions
of high electron density, makes alkenes much more reactive than
alkanes with their strong bond enthalpies - 348 kJ/mol for C-C and 412
kJ/mol for C-H (both sigma bonds).
Note that a higher bond enthalpy means a higher
activation energy inhibiting reactivity.
Irrespective of whether the reaction is
electrophilic in nature, ignoring other atoms/bonds involved, the overall addition reaction of alkenes is
expressed as:
C=C + a-b
===> a-C-C-b
e.g.
 + HBr
===>
or
The pi bond of the C=C bond is broken and
two new sigma bonds (C-Br and C-H) are formed.
In these reactions you are going from an
unsaturated alkene molecule to a saturated alkane molecule, and apart
from addition of hydrogen, the product is a substituted alkane.
With some alkenes, there is the
possibility of isomeric products depending on which way round the HX adds to
an unsymmetrical alkene like propene.
Note that alkenes can
also readily undergo free radical reactions e.g. their
peroxide catalysed polymerisation to form a poly(alkene) and these
reactions also involve the interaction of free radicals with the
π
(pi) electrons.
TOP OF PAGE and
sub-index
2.3.3 The addition of hydrogen halides to alkenes
Examples of the
addition of hydrogen bromide/hydrogen chloride/hydrogen iodide to alkenes, which readily occurs under
non-aqueous, gaseous conditions HBr/HCl(g) or aqueous conditions with concentrated
hydrobromic acid/hydrochloric acid/hydriodic acid HBr/HCl/HI(aq), and usually at room temperature and pressure.
(i) ethene + hydrogen
bromide ===> bromoethane
+ HBr ===>
This is a symmetrical alkene, so this
reaction can only give one product.
Symmetrical here means the
atoms/groups attached to each carbon of the double bond are identical.
Hex-3-ene
is another symmetrical alkene and will give only one possible product
e.g.
CH3-CH2-CH=CH-CH2-CH3
+ HCl ===> CH3-CH2-CHCl-CH2-CH2-CH3
(3-chlorohexane)
(ii) propene +
hydrogen bromide ==> 2-bromopropane
+ HBr
===>
Propene is an unsymmetrical alkene -
giving rise to two possible products depending on which way round the
hydrogen bromide molecule adds to the carbon - carbon double bond..
Some of the other isomer of C3H7Br, 1-bromopropane, is also formed in this reaction.
The minority product
is
from a con-current reaction, when the H-Br molecule adds the
'other way round' - more on this later when discussing the mechanism.
This reaction involves addition of an
unsymmetrical reagent (H-X) to an unsymmetrical alkene, where
different atoms/groupings differ on either side of the double bond e.g. H2C=CH-CH3.
The major product is predicted from
Markownikoff's Rule which can be stated in different ways e.g.
On the addition of HX to an alkene,
the hydrogen atom attaches itself to the carbon that already has the
most hydrogen atoms already directly attached.
(The rule is defined in
other ways later in this section and only fully explained when the
mechanism is considered in
section 2.3.4)
Reminder of the
Markownikoff
(Markovnikov)
rule which predicts which
isomer is likely to predominate for adding a non-symmetrical
reagent to a non-symmetrical alkene and the rule can be
stated in various ways:
For the heterolytic addition of a polar molecule to an alkene
(or alkyne), the more electronegative (the most nucleophilic like OH-,
H2O or Br-
etc.) atom (or part)
of the polar molecule becomes attached to the carbon atom
bearing the smaller number of hydrogen atoms.
This is usually step 2 in the
electrophilic addition mechanisms described below.
Or, you can
say, the least electronegative (the most electrophilic like Br+
or H+ etc.) will attach to the carbon atom bonded
with the most H atoms.
This is usually step 1 in the
electrophilic addition mechanisms, noting which carbocation is
the most stable in the mechanisms described below.
So in these reactions you think
of it as a sort of 'H+Br-' addition.
(iii) The symmetrical E/Z isomers of but-2-ene can only give one product
on addition of hydrogen chloride.
or
+ HCl ===>
(2-chlorobutane)
(iv)
Skeletal formulae to show addition of hydrogen chloride to but-1-ene to
give 1-chlorobutane or 2-chlorobutane.
+
HCl ===>
or
(v) The addition of hydrogen iodide to
cyclohexene to give iodocyclohexane.
+ HI ===>
Some general comments.
As you can see, hydrogen chloride (HCl), hydrogen bromide
(HBr) and hydrogen iodide (HI) all add to alkenes.
The reactivity order is HI >
HBr > HCl, because of the order of decreasing bond enthalpies
(bond energy).
The smaller the bond energy of 'HX', the lower
will be the activation energy for the reaction.
Bond enthalpies (kJ/mol):
HI 299 < HBr 366
< HCl 431; C-C (σ
bond) 348; C-C (π
bond)
264; C-H (412).
Apart from the HX bond enthalpy order,
also note the low value to break the pi bond of the alkene functional group
- the most reactive site on the molecule, it is also the weakest bond
involved in the addition of a hydrogen halide to an alkene.
These are mechanistically described as
electrophilic addition
reactions (mechanism discussed in section next).
Hydrogen chloride (HCl) and hydrogen iodide (HI) also add
via the same electrophilic addition mechanism, either mixing the gases, HX
in a non-polar solvent or bubbling the alkene into a concentrated aqueous
solution of HX.
The
Markownikoff
(Markovnikov)
rule predicts which
isomer is likely to predominate for adding a non-symmetrical
reagent to a non-symmetrical alkene and the rule can be
stated in various ways e.g.:
For the heterolytic addition of an
unsymmetrical polar molecule to an unsymmetrical alkene
(or alkyne), the more electronegative atom of the
electrophilic polar molecule, becomes attached to the carbon atom
bearing the smaller number of hydrogen atoms (with
respect to the >C=C< bond grouping).
[or you can
say the least electronegative (most electrophilic like Br+
or H+ etc.) will attach to the carbon atom bonded
with the most H atoms to actually form an intermediate
carbocation.
BUT the
'rule' only applies to the ionic mechanism described in the
next sections 2.3.4 and 2.3.5.
(but you can get
the opposite effect in free radical addition in the presence
of peroxides!)
The Markownikoff rule is NOT based on
the stability of the molecule, they are effectively ~ equally stable.
the rule is based on the relative stability of the intermediate
carbocations that can be formed -
see later in section 2.3.4.
Examples of using the Markownikoff
Rule
can see the major and minor product outcome in terms of the Markownikov
rule?
(i) methylpropene +
hydrogen chloride ===> 2-chloro-2-methylpropane
(major product)or
1-chloro-2-methylpropane
(minor product)
+ HCl ===>
or
+ HCl ===>
or
(ii)
pent-1-ene + hydrogen
bromide ===> 2-bromopentane (major product) or
1-bromopentane
(minor
product)
CH3-CH2-CH2-CH=CH2
+ HBr ===> CH3-CH2-CH2-CHBr-CH3
or
CH3-CH2-CH2-CH2-CH2Br
(iii)
2-methylbut-2-ene + hydrogen iodide ===>
2-iodo-2-methylbutane (major
product) or
2-iodo-3-methylbutane
(CH3)2C=CHCH3
+ HI ===> (CH3)2CICH2CH3
or
(CH3)2CHCHICH3
TOP OF PAGE and
sub-index
2.3.4
Electrophilic
addition mechanism of HX addition to alkenes
(non-aqueous conditions)
and
explaining Markownikoff's rule
I've already written a detailed
'theoretical' page on this reaction, so I'm just repeating the
essential points here.
See
Electrophilic
addition of hydrogen bromide to form
halogenoalkanes
(aqueous and non-aqueous media)
The hydrogen bromide molecule attacks the
electron-rich pi bond of the alkene molecule (e.g. ethene shown below)
and forms sigma bonds with the carbon atoms of the double bond of the alkene
functional group.
The hydrogen bromide is an
electrophile, which is a molecule/ion that can bond to an electron rich
site by accepting a pair of electrons to form a new covalent bond.
Hence the reaction of hydrogen bromide with an
alkene is described as an
electrophilic addition.
Diagram mechanism 3 illustrates the general
mechanism for adding an hydrogen halide electrophile HX (e.g. HBr) to an alkene R2C=CR2
(R = H, alkyl or aryl).
Mechanism diagram 53 shows the
electrophilic addition of hydrogen bromide to the alkene ethene.
The pi bond orbitals give the
C=C bond a high electron density region for the electrophile to attack.
In this case, for step
(1),
the attacking electrophile is the already polarised hydrogen bromide
molecule, Hδ+Brδ-,
which splits heterolytically to protonate the alkene,
forming the carbocation and leaving a free bromide ion.
The HBr molecule is an electrophile
because it accepts a pair of electrons from the ethene π
bond to form the new C-H bond.
It is the positive end of the
hydrogen bromide molecule that attracts the electrons of ethene's pi
bond.
The two pi electrons of the C=C
bond move to form a covalent bond with the δ+
hydrogen atom of the electrophile Hδ+Brδ-
which undergoes heterolytic bond fission.
Simultaneously the hydrogen - bromine
bond of the hydrogen bromide molecule is broken and a bromide ion formed
- this is an example of heterolytic bond fission, where the bonding pair
of electrons moves onto only one atom of the original bond (H-Br).
The full curly arrow indicates the
movement of a pair of electrons - it must start from the pi bond and go
towards the atom forming the new bond (H in this case).
Note there are two electron pair
shifts in step (1).
(i) The pair of pi electrons of
the alkene move onto the proton (C-H bond formed).
(ii) The H-Br bond pair shift
completely
onto the bromine atom to form the bromide ion.
In step
(2) the negative bromide ion formed in step (1) rapidly combines
with the positive carbocation to form the bromoalkane
(bromoethane).
The bromide ion
donates a pair of electrons to form the new C-Br bond - shown by
the full curly arrow indicating the shift of a pair of electrons from
the bromide ion to the positive carbon atom.
Note that ethene is a symmetrical
ethene, where the atoms/groups on either side of the double bond are
identical, so only one product is possible (at least under non-aqueous
conditions).
Mechanism diagram 64 for electrophilic
addition: >C=C<
+ H-X ===> >CH-CX<
The
reaction profile
progress diagram for the addition of a hydrogen halide to an
alkene is shown above.
Ea1 is the activation
energy for step 1, the formation of the carbocation after the
electrophile 'attack'.
This is the highest activation
energy because it initially involves bond breaking processes.
This is also the reason why step
1 will be the slowest of the two steps.
Ea2 is the activation
energy for step 2, the formation of the final product when
the carbocation combines with an anion (or any other electron pair
donating species e.g. it can be a water molecule in aqueous media).
This is the lower activation
energy because it only involves a bond making progress.
This will be the fastest of the
two steps, especially as it involves two oppositely charged ions
coming together.
ΔH is the overall
enthalpy change for the reaction and not to be confused with
either of the activation energies.
Note there are complications in this
reaction if aqueous hydrogen
bromide HBr(aq) is used, discussed further down.
The next reaction described is the
addition of hydrogen bromide to propene.
Diagram mechanism 54 shows the
electrophilic addition of hydrogen bromide to the alkene propene.
The description of the steps and reaction
profile are identical to that of ethene + HBr, so no repeats!
However, complications arise when an asymmetrical
reagent (H-Br) adds to an asymmetrical alkene (where different
atoms/groupings differ on either side of the double bond) e.g. H2C=CH-CH3.
The products can be 1-bromopropane or
2-bromopropane.
Note that bromine (Br-Br) and ethene
(H2C=CH2) are both symmetrical molecules and give
only one product under non-aqueous reaction conditions.
However, before we further discuss the
possible products from the hydrogen bromide - propene reaction, the
major product AND understand the Markownikoff Rule, you
need to know more about the relative stability of carbocations -
illustrated below.
A
carbocation has three sigma bonds (via
three single covalent bond pairs) to other atoms/groups e.g. H, alkyl etc.
A carbocation has empty 2p orbital,
therefore it is electron deficient by one electron and so carries
overall a single (but whole) positive charge.
The empty 2p orbital accepts the pair of electrons
from the electron donating electrophile to form the new C-Br bond.
The three C-R bonds are actually in a trigonal planar
arrangement, bond angles are all ~120o (R = H, alkyl).
Note on some structural terms used:
Tertiary means 3 alkyl
groups (and no H's) attached to the relevant positive carbon atom (in this
case the carbon atom carrying the positive charge) e.g.
(CH3)3C+
Secondary
means two alkyl groups (and 1 H) attached to the relevant positive carbon
atom e.g. (CH3)2CH+
Primary means one alkyl group (and 2 H's
e.g. propyl or ethyl carbocation) attached to the relevant carbon atom OR no
alkyl group and 3 H's for the methyl carbocation e.g. CH3CH2CH2+,
CH3CH2+
or CH3+
The orientation of the products from non-symmetrical
addition (HX or Br2(aq) see later) is governed by the stability of the carbocation
intermediate formed by the protonation of the alkene by
the attacking H-X electrophile, and explains the
Markownikoff rule.
The
order of carbocation stability is tertiary >
secondary > primary, because alkyl groups give a
slight electron donating inductive effect
via the attraction of the positively charged carbon
atom.
This +I effect of the
electron shift and its direction, is shown by the brown arrow heads
in the diagram above and note from left to right the number of brown
arrow heads goes 3, 2, 1 and then 0 as the inductive effect
decreases, as does the stability of the carbocation.
The alkyl groups involve
bigger electron clouds than that of a hydrogen atoms and even lone
pairs on bromine atom can have the same stabilising effect (see
addition of bromine to alkenes).
This inductive electron cloud shift spreads the positive charge of the
carbocation beyond the single carbon atom and gives the carbocation more stability
by lowering its potential energy. It is possible that
the vacant 2p orbital of the positive carbon accepts some of the electron
cloud of the adjacent carbon atoms.
Whatever, the true full explanation,
it is a general rule
of physics, that spreading out electric charge lowers the
potential energy and increases the stability of a
situation - here the context is the stability of a
carbocation involved in an organic chemistry reaction mechanism.
The
most stable carbocation will be the one most likely to
exist with a sufficient life-time to be hit by the
electron pair donating ion (e.g. :X-, :OH-,
:Br- or H2O:) or any
other electron pair donor that acts as an electrophile.
The Markownikoff (Markovnikov) rule is
NOT based on the stability of the molecule, the orientation of the major
isomeric product is governed by the relative stabilities of the possible
carbocations that could be formed.
e.g. for the following reaction
between but-1ene and hydrogen chloride ...
+ HCl ===>
or
... you would expect
2-chlorobutane to be the major product, and 1-chlorobutane the minor
product.
... and the mechanism possibilities
are similar to reaction mechanism diagram 58 for addition of HBr.
Mechanism diagram 56 shows the
electrophilic addition of hydrogen bromide to cyclohexene to give
bromocyclohexane, just one product (if non-aqueous), no need for
Markownikoff Rule, its a symmetrical alkene.
Mechanism diagram 57 shows the
electrophilic addition of hydrogen bromide to but-2-ene to give
2-bromobutane, just one product (if non-aqueous), no need for
Markownikoff Rule, its a symmetrical alkene.
Mechanism diagram 58 shows the
electrophilic addition of hydrogen bromide to but-1-ene to give
1-bromobutane or 2-bromobutane.
Here you need the Markownikoff Rule
to predict the major product will be 2-bromobutane.
In equation 58b, the secondary
carbocation is CH3CH2CH+CH3,
and is more stable than the primary carbocation CH3CH2CH2CH2+
in equation 58a.
So 2-bromobutane will be the major
product from the more stable secondary carbocation (pathway 58b),
and 1-bromobutane will be the
minor product from the less stable primary carbocation (pathway
58a).
2.3.5 The electrophilic
addition mechanism of HX addition to alkenes
(aqueous conditions),
and
using Markownikoff's rule to predict and explain the likely products.
If a liquid/gas alkene is mixed with
concentrated hydrobromic acid, HBr(aq)
(hydrogen bromide dissolved in water) a bromoalkane
is the expected product.
Expected overall reaction from section 2.3.4:
R2C=CR2
+ HBr ===> R2CH-CBrR2
but quite the
full story by a long way!
In water,
hydrogen bromide is a strong acid i.e. completely ionises to
give the oxonium ion and bromide ion.
HBr(g)
+ H2O(l) ===> H3O+(aq)
+ Br-(aq)
This means another attacking electrophile is more
likely to be the proton donating oxonium ion H3O+.
rather than a HBr molecule, which does NOT exist in aqueous solution.
Mechanism diagram 39
- electrophilic addition of hydrogen bromide to an alkene in aqueous
media (R = H or alkyl)
In the acid
solution via step (1)
the H3O+ or oxonium ion
(hydrated proton) is the 'attacking electrophile' and protonates the alkene to form the intermediate positive
carbocation R2CHCR2+. The
oxonium ion is an electrophile because it accepts a pair of
electrons from the alkene π
bond to form the new C-H bond.
In step (2) the (already present) negative bromide ion rapidly
combines with the carbocation to form the bromoalkane product.
The bromide ion donates a pair of electrons to form the new C-Br
bond.
BUT .... with the
high concentration of water present, a water molecule could
also interact with the carbocation to eventually form a
small amount of the alcohol R2CHCR2OH,
this again provides evidence of an ionic mechanism with a
carbocation intermediate.
Mechanism diagram 39a
- electrophilic addition of water to an alkene in acidic aqueous
media (R = H or alkyl)
Step2 in mechanism diagram 39a:
R2CHCR2+
+ H2O === > R2CHCR2OH +
H+
Using abbreviated structural formula, I've set
out the products that can be formed, indicating whether they are major
or minor (if known?) for several alkenes.
(a)
Ethene (symmetrical alkene)
(i) H2C=CH2
+ HBr ===> CH3CH2Br
(bromoethane)
(ii) H2C=CH2
+ H2O ===> CH3CH2OH
(ethanol)
Not sure which is the major product?
A high concentration of bromide ion favours reaction
(i), a low concentration (ii).
(b) Propene
(unsymmetrical alkene)
(i)
CH3CH=CH2
+ HBr ===> CH3CH2CH2Br
(1-bromopropane)
(ii)
CH3CH=CH2
+ H2O ===> CH3CH2CH2OH
(propan-1-ol)
(iii)
CH3CH=CH2
+ HBr ===> CH3CHBrCH3
(2-bromopropane)
(iv)
CH3CH=CH2
+ H2O ===> CH3CH(OH)CH3
(propan-2-ol)
Not sure which is the major product?
A high concentration of bromide ion favours reaction (i)
and (iii), but the Markownikov rule would make reaction (iii) give the
major product - the secondary haloalkane, rather than the primary
haloalkane.
A low concentration of bromide ion would favour
reactions (ii) and (iv) (ii) but again, the Markownikov rule would make
reaction (iv) give the major product - the secondary haloalkane, rather
than the primary haloalkane.
Comment - evidence of the ionic mechanism in aqueous solution
If other ions present from e.g. from dissolved salts
like sodium chloride, NaCl, subsequence analysis shows as well as the
products indicated by the equations above, you would also get chloroalkanes
formed as well as bromoalkanes.
e.g. reaction (a) would give some chloroethane
and (b) would give 2-chloropropane >> 1-chloropropane (again
using the Markownikov rule)
Future
developments
Produce more
skeletal formula
mechanism diagrams
More on
HCl and mixed ion medium
[SEARCH
BOX]
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