I got a 1st in my first year studying chemistry at the University of Birmingham using these revision notes that I have uploaded. They include detail on the structure and bonding of alkenes, consequences of the pi bond, E- and Z- isomers, stability of alkenes, bonding and geometry of alkynes, acidit...
Solution Manual For Organic Chemistry Second Edition Jonathan Clayden, Nick Greeves, and Stuart Warren
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Chemistry
Organic Chemistry I
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Alkenes and Alkynes
16 October 2017 18:01
HANDOUT 1 1. Alkenes - Structure and Bonding
Learning objectives for handout 1 • In ethene for example, assuming that the hybridisation in alkenes is sp3
1. Understand and be able to describe the hybridisation and bonding in alkenes would be incorrect as the overlap is very poor and the resulting bonds
2. Understand the implications of the π-bond in an alkene on the geometry of the alkene would be very weak.
3. Be able to describe the isomers of alkenes in terms of E and Z nomenclature • Assuming this would also suggest two equivalent carbon-carbon bonds
4. Understand the factors which contribute towards the stability of an alkene and know how when they are not the same.
to measure the stability of alkenes The bonding in ethene is sp2 hybridised:
5. Understand and be able to describe the hybridisation and bonding in alkynes
6. Understand the implications of the bonding in an alkyne on the geometry of the alkyne
7. Know the typical pKa of a terminal alkyne, and understand why alkynes are relatively acidic.
2. Consequences of the π-bond There is still a pz orbital left orthogonal to the trigonal plane of sp2 hybrids.
• In a σ-bond, rotation about C-C axis does not affect overlap - free rotation is permitted. Side-on overlap between the two pz orbitals results in a π-bond:
• In a π-bond, rotation destroys overlap - it requires a lot of energy to break a π-bond.
To achieve this side-on overlap the two p-orbitals must be parallel to one another. Free
rotation is not allowed. The consequence of this is that we can trans and cis (E and Z
stereoisomers)
3. E- and Z- isomers
• E- and Z- isomers are stereoisomers (the term geometric isomers is no longer used - they
are stereoisomers that are not enantiomers, hence they are diastereoisomers).
• The term configuration is used to describe the arrangement of atoms about a double
bond, not conformation. Therefore, σ-bond framework of alkenes formed
• Can interconvert between E and Z alkenes by heating to a high temperature (several from three sp2 hybrids, with remaining 2p orbital
hundred oC - barrier around 270kJ/mol. Light can also interconvert E- and Z-alkenes: forming π-bond. sp2-sp2 overlap results in σ-
bond, 2p-2p overlap results in π-bond.
1H NMR spectroscopy can
help us distinguish between E-
and Z-alkenes - the vicinal
coupling constants are
In diradical excited state there is one bonding and one antibonding electron = no pi bonding - different for E- and Z- protons:
free rotation about C-C sigma bond. (energy (E) of photon light given by hv, hv hence represents
light energy)
Substituted alkenes
4. Stability of alkenes • In general, more substituted alkenes are more stable. One of the reasons
• E-isomer generally more stable than Z-isomer due to steric interactions. Can determine for this is hyperconjugation (overlap between the filled C-H σ-bond and
the stability of an alkene by measuring its heat of hydrogenation: empty C=C π*-orbital gives a stabilising interaction):
H2C=CH2 ------> H3C-CH3 The more C-H bonds
available to overlap
Lose: C=C π-bond, H-H = greater
σ-bond stabilisation. Hence
Gain: 2xC-H σ-bonds more substituted =
Net result: exothermic more stable
Cyclic Alkenes
Z-but-2-ene releases 120kJ/mol of energy and E-but-2-ene releases 115kJ/mol, so E is more • Cyclohexene has bond angles close to optimum 120o, so has average
stable by 5kJ/mol. reactivity
• Z-isomer has steric clash of methyl groups making it less stable. • Cyclopropene has bond angles approx 60o, so very strained and highly
reactive.
In bicyclic alkenes we have to consider carefully where a double bond
is possible. When we attempt to constrain the double bond within a
ring it may be impossible to maintain efficient p-orbital overlap.
• Bredt's rule - cannot form a double bond at the bridgehead
position (unless the ring is large) as the geometry is unsuitable
for the p-orbitals to overlap.
The Newman projection along the
Bridgehead This double bond is at the "double bond" shows that the p-
position: bridgehead position therefore orbitals are almost at 90o to one
the isomer doesn't exist: another i.e. not a double bond at all
5./6. Alkynes - Bonding and Geometry
• The two carbon atoms forming a triple bond are sp hybridised, leaving The sigma framework: Two C-H σ bonds, formed by the overlap
two orthogonal p-orbitals on each carbon. C-C σ bond formed by the between C sp hybrid
• Alkynes are linear, which means there The overall picture: overlap of two sp hybrid orbitals orbital and H s orbital
isn't an issue of E/Z isomerism. The pi orbitals:
• The sp hybrids form the C-C and C-H
bonds, while side-on overlap of the p-
orbitals forms two orthogonal π-bonds
7. Acidity of Alkanes, Alkenes and Alkynes
• Acidity increases from alkanes to alkenes to alkynes (alkenes and alkynes are
not particularly acidic, but can both be deprotonated by strong bases (when Remember that conjugation arises when we have alternating single and double bonds, such
alkynes are deprotonated, negative charge is in the sp hybrid orbital)). that there is overlap between the π-systems (unbroken overlap between the p-orbitals)
• Why when there are several bonds in conjugation, this leads to absorption of light in
the visible region of the spectrum:
The acidity difference of each circled hydrogen can be understood in terms of
the hybridisation changes:
• An s-orbital is tightly held around the nucleus, whilst p-orbital is more
Organic Chemistry I Page 1
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