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Chapter 7 – Intro to Alkenes
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
Sigma vs. pi bonds
o Sigma bonds are stronger than pi bonds
o But remember that a double bond is the sum of the two, so it is stronger than a
single bond
Elements of unsaturation
o For every pi bond or ring you lose two hydrogens.
 These are called elements of unsaturation because now the molecule
doesn’t have as many hydrogens on it as possible
 It isn’t “saturated” with hydrogen.
 When an acyclic hydrocarbon has no elements of unsaturation then its
formula is CnH2n+2
 You can determine how many elements of unsaturation there are based
upon the number of hydrogens in a chemical’s formula.
 For every two missing hydrogens you have one element of
unsaturation.
 Remember that a triple bond has two pi bonds.
o How do different elements affect the number of hydrogens on a molecule?
 Oxygen has 0 impact!
H
H
H
H
C
C
C
H
H
H
H
H
H
H
H
C
C
C
H
H
H
O
H
H
H
H
C
C
H
H
O


H
So you would still expect 2n+2 hydrogens if there were no
elements of unsaturation
Halogens replace hydrogens
H
H
C
C
C
H
H
H
C3H8
H
H
H
H
Br H
C
C
C
H
H
H
C3H7Br
H
H
H
H
H
C
C
C
H
H
H
C3H7I
C
H
C3H8O
C3H8O
C3H8
H
I
H


H
Thus for each halogen on the molecule you have one fewer
hydrogen.
 Now you would expect 2n+2-(# of halogens) if there are no
unsaturations.
For every nitrogen you add an extra hydrogen
H
H
H
C
C
C
H
H
H
H
H
C3H8
H
H
H
H
C
C
C
N
H
H
H
H
H
H
H
C
C
N
C
H
H
H
H
H
C3H9N
C3H9N

Now you would expect 2n+2+(# of nitrogens) if the molecule had no
unsaturations.
o Putting this all together to figure out how many elements of unsaturation a
molecule has.
 Find this out by figuring out how many hydrogens are “missing”
 How to find out how many elements of unsaturation a molecule has
 1) Calculate the “expected hydrogen”
o 2n+2+(number of N)-(number of X)
 Note: This is different than the way Mr. Baker
and the book do it, so be careful not to blend the
two methods.
 I just think this way is easier.
 2) Subtract the “actual hydrogen”
o Note: If you’re using the formula in step one, you do not
add the halogens to the number of actual hydrogens.
 3) Divide by 2 to get the number of unsaturations
 What if you are given a “before and after”
 We can figure out how many pi bonds and rings by hydrogenating
the molecule.
H2
Pt
C8H12
C8H16
H


o In this reaction, hydrogen is added across all pi bonds but
the rings are left alone.
Often you will be given a before and after type problem where
you need to figure out the number of pi bonds and the number of
rings.
o How many pi bonds?
 (hydrogens after-hydrogens before)/2
o How many rings?
 The number of unsaturations present in the “after”
Nomenclature
o The double bond has priority over halogens or alkyl groups, so get to it first
 So for the molecule above we will count from the right.
o Change “ane” to “ene” at the end of the name
 Instead of 10,10,10-trifluoro-3-isopropyldecane, we now have 10,10,10trifluoro-3-isopropyldecene.
o If there are two, three, etc double bonds, use “di,” “tri,” etc
o Say where the double bond(s) is(are)
 We just say the first number of where the double bond is.
 This double bond is between carbons 2 and 3 of the molecule so
we’ll say it’s a dec-2-ene or 2-decene.
o Stereochemistry (E/Z nomenclature)
 Assign priorities to the substituents on each side of the double bond just
as you would with R/S
1
1
2
2



If the high priority groups are on the “zame zide,” then it’s Z
If the high priority groups are on the “epposite side” then it’s E
 So this molecule is (Z) 10,10,10-trifluoro-3-isopropyldec-2-ene.
Stability of Alkenes

o Tetrasubstituted>tri>di>mono>unsubstituted
o More stable = lower heat of hydrogenation
o More stable = lower energy
Acid-Catalyzed dehydration of alcohols
o Just an E1 where water is the leaving group
 Remember that E1’s allow rearrangement, so dehydration of alcohols
does too
o Alcohol + Concentrated Acid (often H2SO4) yields alkene
o The acid needs to be concentrated to drive the reaction forward
 Review of LeChatelier’s Principle
 If you remove a product of a reaction, you drive the reaction
forward
o First you protonate the oxygen, generating a better leaving group
 What makes a good leaving group? A weak base
o Hydroxide is a strong base, so poor leaving group
o Water is a weak base, so good leaving group
o Why no SN1?
 In these conditions, there is no nucleophile present to attack the
carbocation.
 The water that fell off could choose to add to the carbocation, but it
would just fall back off.