Chapter 7 – Intro to Alkenes

  • Sigma vs. pi bonds
  • Sigma bonds are stronger than pi bonds
  • But remember that a double bond is the sum of the two, so it is stronger than a single bond
  • Elements of unsaturation
  • 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.
  • How do different elements affect the number of hydrogens on a molecule?
  • Oxygen has 0 impact!
  • So you would still expect 2n+2 hydrogens if there were no elements of unsaturation
  • Halogens replace hydrogens
  • 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
  • Now you would expect 2n+2+(# of nitrogens) if the molecule had no unsaturations.
  • 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”
  • 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”
  • 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.
  • 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.
  • How many pi bonds?
  • (hydrogens after-hydrogens before)/2
  • How many rings?
  • The number of unsaturations present in the “after”
  • Nomenclature
  • 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.
  • Change “ane” to “ene” at the end of the name
  • Instead of 10,10,10-trifluoro-3-isopropyldecane, we now have 10,10,10-trifluoro-3-isopropyldecene.
  • If there are two, three, etc double bonds, use “di,” “tri,” etc
  • 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.
  • Stereochemistry (E/Z nomenclature)
  • Assign priorities to the substituents on each side of the double bond just as you would with R/S
  • 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
  • Tetrasubstituted>tri>di>mono>unsubstituted
  • More stable = lower heat of hydrogenation
  • More stable = lower energy
  • Acid-Catalyzed dehydration of alcohols
  • Just an E1 where water is the leaving group
  • Remember that E1’s allow rearrangement, so dehydration of alcohols does too
  • Alcohol + Concentrated Acid (often H2SO4) yields alkene
  • 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
  • First you protonate the oxygen, generating a better leaving group
  • What makes a good leaving group? A weak base
  • Hydroxide is a strong base, so poor leaving group
  • Water is a weak base, so good leaving group
  • 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.