(Od) Chapter 4: Alkane, Alkyl Halides And Substitution Reactions

Front 1

Saturated Hydrocarbon

Back 1

A saturated hydrocarbon means a carbon that has the maximum number of hydrogens. Saturation refers to how many hydrogens the carbons have. A hydrocarbon that is unsaturated will have a double bond and less than the max number of hydrogens

Front 2

Primary, Secondary, Tertiary, and Quaternary Carbons

Back 2

Refer to the number of carbons a carbon atom is bonded to

Front 3

Effect of increasing molecular weight of the hydrocarbon.

Back 3

As you increase the molecular weight of the hydrocarbon, you also INCREASE the BOILING POINT, MELTING POINT, and DENSITY

Front 4

Effect of branching on a hydrocarbon

Back 4

Branching will decrease the boiling point and the melting point due to the fact that there is less surface area for van der wall interaction.

Front 5

Combustion

Back 5

The reaction of a hydrocarbon and oxygen

Front 6

Equation for combustion

Back 6

C3H8 + 5O2 →3 CO2 + 4 H2O + heat

Front 7

Halogenation

Back 7

The process by a halogen is added to an alkane. This process occurs though FREE RADICAL SUBSTITUTION

Front 8

Three steps of free radical substitution

Back 8

Initiation, propagation, termination

Front 9

Initiation

Back 9

DIATOMIC halogens are cleaved via light or heat to create free radicals which have have unpaired electrons that will react with anything

Front 10

Propagation

Back 10

Radical will pull off hydrogen from alkane. The alkane now a radical. the radical alkane will now react with the diatomic halogen. It will attack the halogen creating an alkyl halide and a halogen radical

Front 11

Termination

Back 11

The reaction will stop when two radicals come together and bond.

Front 12

How do you determine how likely a carbon will engage in a radical reaction?

Back 12

The more substituted the carbon is the more stable it is and more capable it is bearing a radical.

Front 13

Pyrolysis

Back 13

Breaking down a molecule via heat. One example is CRACKING in which C-C bonds are broken down to create alkyl radicals which combing to to make a variety of alkanes.

Front 14

Dispropornation

Back 14

When a alkyl radical gives its electron to another alkyl radical to from an alkene and an alkane.

Front 15

In an aprotic solvent, how are nucleophilicity and basicity related?

Back 15

In an aprotic solvent, since there are not protons to coat the nucleophile, it is more capable of attacking the electrophile since there are no protons blocking it. In this case, there is a direct correlation between how basic a molecule and how good of a nucleophile it is. The better base it is the better nucleophile it is.
Trend: F-> Cl->Br-> I-

Front 16

In a protic solvent, how are nucleophilicity and basicity related?

Back 16

In a protic solvent, since the protons surround the nucleophile, it will have a harder time getting to the electrophile. In this case, the bigger the molecule, the nucleophile it will be since it will have an easier time breaking through the proton shell and getting to the electrophile. The feature of a molecule in which its size will allow it to shed the protons that surround it is called POLARIZABILITY.

Front 17

What makes a good leaving group?

Back 17

A good leaving group is a group that is the conjugate base of a strong acid. The stronger the acid, the weaker the conjugate base and the better leaving group it is. The stronger the acid, the better electron acceptor it will be which means that weak bases are better at taking electrons.

Front 18

Why are weak bases make good leaving groups?

Back 18

They are better able to spread out negative charge thus making the molecule more stable.

Front 19

Why is Sn1 unimolecular?

Back 19

Because it only relies on the level of one species. The substrate, not the nucleophile, is the only molecule that affects the rate of the reaction.

Front 20

Features of the Sn1 reaction.

Back 20

- The sn1 reaction is unimolecular and only relies on the levels of the substrate
This first step of the reaction is the formation of a positively charged carbocation.
- Does not need a strong nucleophile since the carbocation is such a good electrophile
-The likelihood of the Sn1 reaction occurring is based on how substituted the carbon atom is: the greater substitution the more stable the carbocation the more likely the Sn1 reaction to occur.

Front 21

Rate determining step of the Sn1 reaction

Back 21

The rate determining step of the reaction is based on the speed at which the leaving group dissociates from the molecule and forms the carbocation

Front 22

Solvent effect on Sn1 reaction

Back 22

A polar protic solvent works best for Sn1 reactions since they are better able to surround and stabilize the carbocation

Front 23

Pentacoordinate transition state

Back 23

In the Sn2 reaction, one bond is forming as another one is breaking so the transition state is said to have five intermediate bonds. this is the hallmark of the sn2 reaction

Front 24

Why is the Sn2 reaction bimolecular?

Back 24

It is bimolecular because the rate of reaction is based on the concentration of the substrate and the concentration of the nucleophile

Front 25

Feature of the Sn2 reaction

Back 25

-Rate is bimolecular in that it depends on the concentration of the substrate and the nucleophile
-Results in an inversion of the chiral orientation given the back side attack
-Will form a trigonal bipyramidal transition state
-Kinetics are second order

Front 26

In an Sn2 reaction, how did the alpha carbon affect the type of reaction?

Back 26

The alpha carbon would be a primary carbon since in an Sn2 reaction the nucleophile is attacking the carbon. The more substituents there are on the carbon the more blocked off it is. A carbon that only has one bond is much more likely to undergo an Sn2 reaction.

Front 27

What is the difference between an intermediate and a transition state?

Back 27

An intermediate is real. It is a real entity. A transition state is a theoretical structure.

Front 28

Stereochem of Sn1 vs Sn2 reaction.

Back 28

In an Sn1 reaction, the stereochemistry will be lost while in an Sn2 reaction there will be an inversion of the stereochemistry owing to the backside attack.

Front 29

In an Sn1 reaction, why will the product be a racemic mixture?

Back 29

In an Sn1 reaction, the nucleophile can attack from the top or the bottom of the molecule. In the reaction, both scenarios will play out so you will get a mixture.

Front 30

Which solvent will favor which reaction?

Back 30

Polar protic solvents will favor Sn1 reactions while aprotic solvents will favor Sn2 reactions.

Front 31

A bulky nucleophile will favor which type of reaction and why.

Back 31

A bulky nucleophile will favor an Sn1 reaction since it cannot get in to the carbon in order to attack the carbon will dissociate forming a carbocation.