(Oe) Chapter 5: Alkenes, Alkynes, And Elimination Reactions

Front 1

How are Pi bonds normally formed?

Back 1

Through elimination reactions in which a hydrogen is removed from the alkane which leads to a double bond.

Front 2

Olefins

Back 2

Another name for alkenes

Front 3

Boiling Point vis a vis cis / trans configuration

Back 3

Trans will have a higher melting point than than the cis alkene due to the higher symmetry, and greater packing capacity. The cis alkene will have the higher boiling since the electrons are more concentrated leading to greater polarity and thus greater attraction between the molecules

Front 4

What occurs as the molecular weight of the alkene goes up?

Back 4

As you increase the molecular weight, the density, melting point, and boiling point increase.

Front 5

How do you draw a dipole moment?

Back 5

It is an arrow pointing the in the direction of electron density.

Front 6

Why is the trans alkene non polar while the cis alkene polar?

Back 6

The dipole moments of the trans alkene cancel each other while in the cis alkene the dipole is concentrated in one region of the molecule.

Front 7

What occurs during an elimination?

Back 7

A hydrogen or a water molecule is removed from the carbon back bone and a double bond is created.

Front 8

Feature of the E1 Elimination.

Back 8

--Unimolecular in that the rate of the reaction only depends on the concentration of the SUBSTRATE
--CARBOCATIONformation is the first step of the reaction
--WEAK BASE will pull off hydrogen from the beta carbon to form the double bond
--Polar Protic Solvent
--Good Leaving Group
--Highly branched
--

Front 9

High temperatures and E1 mechanisms

Back 9

A high temperature will favor an E1 over an Sn1

Front 10

Sn1 vs E1 Competition

Back 10

Sn1 and E1 mechanisms will often compete but heat will favor the elimination reaction

Front 11

Features of the E2 mechanism

Back 11

--Bimolecular in that rate depends on both the concentration of the substrate and the concentration of the nucleophile
--Relies on STRONG BASES
--Will remove a proton in an ANTI manner

Front 12

Which isomer will predominate in an E2 mechanism?

Back 12

In an E2 mechanism, the double bond can form on either side of the departing group. The double bond can be either cis or trans. The trans configuration will predominate since it is more stable and less sterically hindered.

Front 13

How does substitution affect the E2 reaction?

Back 13

The Zaitzev product is much more favored than a Hoffman product in which the latter will end up forming a double bond with a carbon that has less carbon attached to it.

Front 14

C2H5O- (ethoxide) is an example of what?

Back 14

It is a strong base that will favor an E2 mechanism.

Front 15

A big bulky base will favor which reaction?

Back 15

It will favor an E2 reaction since the bigger the base the more susceptible to steric hinderance.

Front 16

What are two factors that will favor a E2 over an Sn2 reaction?

Back 16

1) A big bulky base will favor an E2 reaction since it will have a harder time coming in for an in line attack
2) A strong base will favor an E2 mechanism over an Sn2 mechanism.

Front 17

Two example of good nucleophiles that are poor bases which are used for Sn2 reactions.

Back 17

Cyanide -- CN-
Iodide -- I-

Front 18

Catalytic Reduction

Back 18

--Using Pt, Pd, of Ni to turn a double bond into a single bond
--This process involves a SYN addition in that the reaction takes place in the surface of the metal so the hydrogens add to the same side.

Front 19

Stereospecific Reaction

Back 19

A reaction in which only one stereoisomer is formed

Front 20

Markovnikov's Rule

Back 20

The fact that when adding to a double bond, substituents will add to side of the bond with more substituents.

Front 21

Addition of X2

Back 21

--Results in an anti addition of the X2 molecule to the double bond
--Will end up forming a dihalo compound
--The halogens will end up adding in an ANTI fashion given the fact that cyclic halonium ion will form and the other negatively charged halogen will attack from the opposite side.

Front 22

Addition of HX

Back 22

--This reaction will add a halogen to a double bond
--hydrogen will protonate the double bond creating a carbocation at the most substituted carbon
--The negatively charged halogen will attack the positively charged atom to form a halo alkane

Front 23

Addition of Water

Back 23

--An H will protonate the double bond making it an alkane with a carbocation at the the most substituted carbon
--The negative dipole of the water molecule will then attack the positive carbocation and attach to the alkane
--A hydrogen can dissociate from the water molecule forming an alcohol

Front 24

When you see a free radical reaction on which a halogen is added to an alkene, which compound is most likely to be used give the energetically favorable reaction?

Back 24

HBr is the only halogen that will undergo a free radical reaction to add to a double bond

Front 25

Free Radical Addition

Back 25

--This will result in a NON MARKOVNIKOV ADDITION to the alkene. The reason being that that the radical takes up the position of the carbon with the most substituents because it is much more stable. The halogen will go the carbon that is less substituted.
--Radical Br will attack the double bond
--Br will attach to the less substituted side of the double bond and the radical will attach to the more substituted
--The radical on the carbon will then remove a hydrogen from another HBr thus turning the molecule into a normal haloalkane and creating another Br radical

Front 26

BH3/THF and H2O2/OH

Back 26

--This is a HYDROBORATION REACTION
--In this reaction, an alcohol will add to the less sterically hindered carbon on the alkene in an anti markovnikov fashion
--the orientation will be syn

Front 27

Cold, Dilute KMnO4

Back 27

--Oxidation Reaction
--Will create a vicinol diol in which two alcohols add to both sides of the double bond in a syn fashion.

Front 28

Hot KMnO4 with Acid

Back 28

--Oxidation Reaction
--Will cleave a NON TERMINAL double bond into two molar equivalents of carboxylic acid. One a TERMINAL double bond, the double bond will become a ketone

Front 29

O3, CH2Cl2/(Zn/H20)

Back 29

--Ozonolysis Reaction
--Will cleave a double bond into two ketones
--You can take the reaction one step further by adding NaBH4, CH3OH to the reaction which will oxidize the ketone creating two alcohols

Front 30

mcpba

Back 30

Will take a double bond a create an epoxide

Front 31

What should you think when you see heat?

Back 31

radical mechanism

Front 32

Polmerization

Back 32

Occurs via heat and radical mechanism in which a double bond will break and fuse with other broken double bonds to make a longer chain

Front 33

What is acetylene?

Back 33

It is officially called ethyne in which two carbon atoms are triple bonded to each other

Front 34

Boiling temperature of internal alkenes vs terminal alkenes.

Back 34

Internal alkenes will boil at a higher temperature than terminal alkenes given greater stability.

Front 35

Physical properties of Alkynes.

Back 35

--Very acidic. The hydrogen molecule will pop off much easier
--The trend that follow molecular weight are the same
--Given the triple bond they are much more polar and will boil at a higher temp than other hydrocarbons

Front 36

Dihaloalkane + Heat + Base

Back 36

If you have an alkane with two halogens attached to it and add a base and heat, the halogens will be removed from the alkane and create a triple bond

Front 37

n-BuLi or CuLi and alkyne

Back 37

These two reagent will initiate an ion exchange reaction in which the the acidic proton of the alkyne is removed and a Li ion in put in its place. Then adding CH3Cl, the CH3 will displace the Li atom and attach to the alkyne thus making the chain longer

Front 38

Na / NH3

Back 38

This reagent will take a triple bond and reduce it to a double bond. The key point here is that the substituents will be in the TRANS position

Front 39

H2 / (Pd/BaSO4)

Back 39

This reagent will take a triple bond and reduce it to double bond. The key point is that that substituents will be in the CIS position

Front 40

Br2 + Alkyne

Back 40

The Br2 will add to the triple bond in an anti manner and if another mole is added, another two Br2 will add creating the alkane

Front 41

Alkyne Radical Reaction

Back 41

Alkynes will undergo radical reactions in the same manner to add a halogen that alkenes did.

Front 42

Alkyne Hydroboration

Back 42

This will follow that same path as the alkene hydroboration. The only difference is that at the end of the reaction, the product will tautermize between the ketone and the enol form. The reagent for this reaction is B2H6.

Front 43

O3 + Alkyne

Back 43

O3 will cleave an alkyne into two caboxylic acids

Front 44

KMnO4 + Acid

Back 44

Will cleave an alkyne to make two carboxylic acids.