chemistry study guide exam2

Aaron Wood chem. 2050 study guide Exam 2

Benzene ChemistryWith cyclohexene, if you add Br2 you will get two Br’s attached at the double bond.

Benzene will only accept 1 Br and requires a Lewis Acid as a catalyst and will also get H-Br biproduct. Lewis Acid: An electron pair acceptor, Feribromide, i.e. FeBr3, Aluminumbromide, AlBr3.

Benzene is flat! Cyclohexane is in boat form.Benzene is thus a perfect hexagon.

Electrons in Benzene π bonds are 1.39 Angstrom(10-10 m), σ bonds are usually 1.54 and π are usually 1.34, Benzene is stable, despite electrons bouncing all around.

All the carbons have sp2 orbitals (double bonds) despite only 3 double bonds at a time.

Benzene functional groups and nomenclatureCommon name: Toluene (methyl group) IUPAC: MethylBenzene

Common name: Aniline (nitrile group) IUPAC: Amenobenzene

Common name: Styrene (vinyl group) IUPAC: Vinylbenzene Vinyl is an Alkene, C=C, connected to a benzene.

Common name: Phenol (hydroxy group, or alcohol)IUPAC: Hydroxybenzene

Any group attached will be named according to the atomic name.

Br Br

Br

CH3

NH2

OH

Br

Bromobenzene Propylbenzene


Benzene Nomenclature continued

Disubstituted (two or more functional groups)

Use the common name as the root: 4-chlorophenol

The number can be replaced by a word,Ortho -- 2 nd position Meta -- 3 rd position Para -- 4 th position OH CH3 OH

Br NO2 Cl

3 – nitrotoluene 2 – bromophenol 4 - ChlorophenolMetanitrotoluene Orthobromophenol Parachlorophenol

If the benzene is attached to a longer chain (nonane), the chain gets the base and the benzene is called a PHENYL, not phenol (with an alcohol group) unless it does have an OH attached. Br

5 – Phenylnonane 5 – (4 – bromophenyl) nonane

It is energetically favorable to have 3 pi bonds in a cycle. Lined P orbitals (especially 6 of them in benzene)

Electrophilic Aromatic Substitution (EAS)

Electrophiles (electron lovers) E+ are created when reacted with a lewis acid (it steals an electron).

Br – Br + AlBr3 Br – Br AlBr3 Br+ + AlBr4 - The Bromide is now an E +

After an E+ is formed, the Benzene will react with the E+ to form a functional group.Chlorination:

YX This will never

happen if Y is electronegative!


FeCl3 or AlCl3 H Cl H+ Cl + Cl – Cl σ complex Lewis acid +

The lewis acid creates a Cl+ as the E+. ChlorobenzeneNitrogenation:

H2SO4 H NO2 H+ NO2

+ HNO3 σ complex Sulfuric acid + Lewis acid

The lewis acid steals an Oxygen and Hydrogen, creating NO2+ as the E+. Nitrobenzene

SOHO2

Sulphination works in the same manner and creates

Meta and OP directorsDisubstituting benzene molecules and determining where the functional groups will attach.

Ortho-Para selectiveThese are functional groups with a group that can double bond to the benzene if the + charge in resonance can be put on the original group. Bromination of Phenol:

OH

+ Br-Br The bromine will attach to Ortho, Meta, or Para depending on what director it is. In this case the plus charge needs to be on the Oxygen to form a double bond with it.

OH OH OH OH OH OH

Para Meta H H H Br Br Br H Br H Br H Br

OH OH OH You will only get ortho and para groups out of the H H H reaction. Absolutely no meta. Thus this reaction is Ortho Br Br Br OP selective.

Meta directors


These are groups on benzene rings that have this formula:

A plus charge will never end up on the connection to x if y is electronegative (oxygen or fluorine)

O O O O O O OCH3 OCH3 OCH3 OCH3 OCH3 OCH3

H H HPara Ortho Br Br Br

H Br H BR H Br

O O O OCH3 OCH3 OCH3 This molecule is meta selective.Meta You will only get meta out of this H H H reaction! Br Br Br

Benzene reaction speeds

σ height of the hills determines the reaction speed. The lower the hills, the faster the reaction. The higher the hills, the slower the reaction. (more energy must be put in to react. σ complex is the energy that needs to be put in to form a sigma bond hop over of the hydrogen in the intermediate step of the EAS.

Chapter 5

E

H+

H E+

OH CH3 Cl NO2

1000* 24.5 1 0.037 0.0000001~ Phenol Toluene Benzene chlorobenzene Nitrobenzene*(1000 time faster than benzene) ~(10-7 time the speed of benzene)

Chiral Achiral Mirror images But cannot superimpose Not mirror image But can superimpose


Chirality: The condition of non-superimposable mirror image. i.e. Hands are chiral.

They are mirror images of each other.But if you put them on top of each other (superimpose) they don’t match up. Thus these are chiral.

If a carbon has 4 different groups attached, the carbon is called a chiral center, the molecule is called an enatiomer.

Achiral: any molecule that is super imposable but not a mirror image. Simply change any of the two groups of a chiral carbon to get Achiral.

CIP rules

2 CH3 Br 1 2 CH3 CH3 2

C C C C

1 F CH3 2 1 F Br 1 E – isomerization Z – isomerization (E) 2 – Bromo – 3 – Fluoro – 2 – Butene (Z) 2 – Bromo – 3 – Fluoro – 2 – Butene

4 1 Since the rotation is clockwise from 1-2-3, This is R configuration.

(R) – 2 – Iodobutane 3 2

H Iodine Iodine = 1 Ethane = 2 C Methane = 3 Hydrogen = 4 C C

H H

CO2H

H2N H

CH3

4Same as 1 S - config

2 3


R & S rules (chirality)

#1. Rank the substituents by atomic number. (highest number = 1, lowest atomic number = 4) #2. If there is a tie, look for the first point of difference. C-C-C is higher than C-C. #3. Hold the #4 atom pointing away from you. And count from 1-2-3. If clockwise, R configuration. If counterclockwise, S configuration.

R and S configurations are enatiomers of each other.

Enantiomers have the same chemical properties in achiral environments. Different in biological systems.

R and S Fisher projections

Horizontal lines are coming out at me!Vertical lines are going into the page!

The model can be rotated 180˚ but NOT 90˚. 90˚ will make the enatiomer. It will turn and S into an R or vice versa

E and Z isomerization (cis and trans for alkenes) This is another way of expressing cis and trans for alkenes. E is trans, Z is cis. You number from 1 to 2 on each side. You give the halide group the 1 and the carbon groups a 2.

Molecules with two or more stereocentersEnantiomers: Stereoisomers that are non-superimposable mirror images.

R CH3 CH3 R

H Br Br H

H Br Br H line of symmetry

S CH3 CH3 S

They can mirror and if flipped, they can superimpose. These compounds are Identical.The # of possible configurations decreases by one if they are mesocompounds. 2 stereocenters. 22 – 1 identical configuration = 3 possible configurations.


Diastereomers: stereoisomers that are not enantiomers. Diastereomerism occurs when two or more stereoisomers of a compound have different configurations at one or more (but not all) of the related stereocenters and are not mirror images of each other, they can or cannot be superimposable.

Diastereomers. They only differ at one stereocenter, and are not mirror images.

Diastereomers have different physical & chemical properties! Enantiomers have the same physical and chemical properties in Achiral environments!

How many stereoisomers are possible?

N is the number of stereocenters.2N = the # of stereoisomers. Thus is a molecule has 2 stereocenters, there are at max 4 configurations. Unless they are Meso compounds.

Meso compounds are molecules that are mirror images AND and superimposable. This happens if there is a line of symmetry between the stereocenters.

Stereochemistry in ReactionsWhen an atom wants to connect to a stereocenter (an Alkene), it will choose a side to go to one side of the double bond.

If it doesn’t matter which side, it will make a 1 : 1 mixture (racemic mixture) of being on one side and on the other. Racemic mixture (A mixture of 1 : 1 R & S configurations of a molecule)

If it does prefer one side of the molecule, most of the products will be in that configuration.

Stereochemistry in Reactions (continued)Take these reactions for example;


The Cl doesn’t care which side it joins, so it will The Cl does care which side it goes to in this molecule. form a 1 : 1 mixture. (doesn’t matter how much to us in this class which side)

+ HCl H CH3 + HCl H + 50% 80% C CH3 . Cl .

C CH3 Cl - 20% 50% H + CH3 H

Separating enantiomersIf you react a 1 : 1 mixture of R & S (racemic) with an S solvent. You will get R connected to the S solvent and S connected to the S solvent. You can then break these bonds to get S solvent, R, and S individually.

1 : 1 R/S mixture + S* R S* + S S* 2 S* + R + S React separate S* is solvent

Chapter 6 Organic Halides Primary carbocation 1 carbon attached to the carbon in question.Secondary carbocation 2 carbons attached to the carbon in question.Tertiary carbocation 3 carbons attached to the carbon in question.

4 Mechanisms of Halide reactions Nucleophilic substitution (nucleus lover, electron giver, A benzene ring is a nucleotide in the first step of a reaction.)

Elimination

2 types of substitution


SN2 Requires a good nucleophile. Negatives are better nucleophiles than neutral but it cannot be an O- or N- negative unless it can resonate through the rest of the molecule.

Down the column and from right to left (bottom and far left are best) are better nucleophiles.

Requires a WEAK base to form, if it has oxygen it must be OCH2CH3 to form SN2. Any Nu that has Sulphur is a weak base. A 3˚ carbon with an Oxygen nucleophile (OCH3CH3CH3) will not do SN2, The methyl’s will get in the way of elimination.

All bonding happens in one step, no cation formed. 1 ̊ and 2 ̊ work for these reactions. 3 ̊ do NOT work . The nucleophile (negative or neutral charge) will come from the back and push out the halide.

The configuration of the bond will be reversed when this happens, turning S into R or R into S. 3˚ do not work with SN2 because the starting molecule’s other carbon groups repel the nucleophile.

SN1 A Carbocation is formed (intermediate step)! Requires a strong base! NaOH, NaOCH3, NaOCCH3CH3CH3, NaNH2.Will work with 3 ̊ and 2 ̊ , but NOT 1 ̊ .

2 types of eliminationE2 no intermediate step is formed.

E1An intermediate step is needed.


Weak base:

Is the reactant a primary, secondary, or tertiary

carbon?

If Primary, Is the solvent a strong or weak base?

If a weak base, Sulpher, or Oxygen with resonance or attached to a 3˚

carbon. The reaction will be SN2 and will push out the halide and join

where it left, inverting the orbitals. R to S or S to R. 1 step process!

If a strong base, Nitrogen, or oxygen without resoance and not attached to 3˚ carbon. The reaction will be E2 and will eliminate the halide and will form a double bond where the elimination

occured. 1 step process!

If secondary, Is the solvent a strong or

weak base?

If a weak base or Nu, The reaction will favor SN2.

Is it a weak nucleophile? nucleophiles get stronger as

you go down a column and to the right. Most of the time it

will prefer E2.

If a strong nucleophile, the reaction will favor

SN2 over SN1.

If it is a weak nucleophile, the

reaction will favor SN1 over SN2.

If tertiary, Is the solvent a strong base or weak

base?

If a weak base, nothing will happen. The weak base (-S-H) will be repulsed and will do anything to the halide. But will favor E1 at

high temps.

If a strong base, Nitrogen, or oxygen without resonance and not attached to a 3˚ carbon.

The reaction will be SN1, it will form a carbocation with a plus charge where the

halide left as an intermediate step. the nucleophile will then join the carbocation to

form a product. 2 step process!


OCH2CH3 . SN2 (1˚) SCH3. SN2 (1˚)CN, cyanide, SN2 (1˚)

Strong Base:OH. E2 (1˚) or SN1 (3˚)OCH3. E2 (1˚) or E1 (3˚)O-CCH3(CH3)CH3 O-3˚ carbon (carbons get in the way of substitution). E2

Reaction speedsIf the leaving group is larger, the speed will be higher. Down a column and to the right of a row. = faster

chemistry study guide exam2

Documents