Part 2

Question 2

Nucleophilic Substitutions

Answer 2

Occurs in Alkanes.

Nucleophiles are electron rich species that attracted to positively polarized atoms.

Basicity: Stronger the base, stronger the nucleophile.

Size/polarizability: dependson solvent. Protic = larger atoms. Aprotic = more basic atoms are better.

Question 3

Best Leaving Groups

Answer 3

I > Br > Cl > F

Weak bases make good leaving groups.

Can accept electron pair and dissociate to form stable species.

Question 4

Sn1 Reactions

Answer 4

Rate is dependent on ONE species.

Rate determining step is the carbocation.

Favored in POLAR protic solvents (water or acetone)

rate = k[RX]

Want stable carbocations: 3 > 2 > 1 > CH3 (favored with use of bulky nucleophiles)

Produces Racemic products (lose optical activity)

Leaving group = weak bases are best.

Question 5

Sn2 Reactions

Answer 5

ONE step (no carbocation)

rate = k[Nu][RX] (substrate and nucleophile)

Usually attacks from the backside.

Best Reactant: 1 > 2 > 3

Favored in polar APROTIC solvents.

Optically active and inverted products.

Need strong nucleophile and transition state.

Question 6

Elimination Reactions

Answer 6

Used in synthesis of alkenes.

Elimination reactions of either alcohols or alkyl halides.

In these reactions, the carbon skeleton loses HX (X = halide) or a molecule of water, to form double bond.

Two types E1 and E2

Question 7

Unimolecular Elimination (E1)

Answer 7

TWO step process, proceeding through a carbocation intermediate.

k = [RX] (substrate)

Elimination of leaving group plus proton = double bond.

1. Leaving group departs, producing carbocation.
2. Proton is removed by a base.

E1 is favored by same factors as SN1: highly polar solvents, Weak Nu, highly branched carbon chains (3), and good leaving groups.

However, HIGH temps favor E1.

Question 8

Bimolecular Elimination (E2)

Answer 8

ONE step process.

k = [RX][Base]

Strong base such as ethoxide ion (C2H5O-) removes a proton, while halide ion anti to proton leaves = double bond.

Often two possible products, but more substituted preferred.

Steric hindrance does NOT affect E2.

Strong base favors E2 over SN2.

SN2 favored over by E2 by weak Lewis bases. (Strong Nu)

Question 9

Catalytic Hydrogenation

Answer 9

Reductive process of adding hydrogen to a double bond with aid of a metal catalyst. (platinum, palladium, and nickel)

Product: Alkane with syn addition of H.

Reaction takes place on metal surface, thus H atoms are added to same face (syn addition).

Question 10

Addition of HX

Answer 10

An electrophilic addition that occurs in alkenes.

Electron of double bond acts as Lewis base and reacts with electrophilic HX molecules.

1. Yields carbocation after double bond reacts with H+
2. Halide ion combines with carbocation to give alkyl halide.

Follows Markovnikov rule (add to the most substituted carbon)

Product: Alkyl halide

Question 11

Addition of X2

Answer 11

Addition of halogens to double bond.

Rapid process.

Nucleophile is the double bond, which attacks an X2 molecule, displacing X-.

Forms intermediate cyclic halonium ion, which is then attacked by X- to make a dihalo compound.

Anti-addition (attacks SN2).

Product: dihalo alkane

Question 12

Addition of H20

Answer 12

Water can be added to alkenes under acidic conditions.

Protonated according to markovnikov.

Performed at LOW temperatures.

Product = alcohol

Question 13

Free radical Additions

Answer 13

Alternate mechanism for addition of HX to double bond.

Occurs when PEROXIDES, O2, or other impurities are present.

Disobeys Markovnikov rule.

Product: terminal alkyl halides.

Question 14

Hydroboration

Answer 14

Diborane (B2H6) adds readily to double bonds.

1. Boron atoms is the Lewis Acd and attaches to less sterically hindered C atom.
2. Oxidation-hydrolysis with PEROXIDE/Aq. base produces alcohol with ANTI-markovnikov, syn orientation.

Product: Alcohol with anti-Markovnikov.

Question 15

Potassium Permangate (KMnO4)

Answer 15

Involved in Oxidation.

1. Cold, dilute KMnO4 produces 1, 2 diols (vicinal) with syn orientation.
2. Hot, basic KMnO4 plus acid produces:

• Nonterminal alkenes = 2M COOH
• Terminal - 1M COOH and 1 CO2 (if disubsituted it makes a ketone).

Question 16

Ozonolysis

Answer 16

Involved in Oxidation.

Treatment of alkenes with OZONE, followed by reduction with Zn/H20 results in CLEAVAGE of double bond.

Product: Aldehyde.

If reduced with NaBH4 instead of Zn/H2) will result in Alcohols.

Question 17

Peroxycarboxylic Acids

Answer 17

Alkenes can be oxidized with peroxycarboxylic acids.

e.g. Peroxyacetic acid (CH3CO3H) OR m-chloroperoxybenzoic acid (mcpba).

Products = Epoxides or Oxiranes.

Question 18

Polymerization

Answer 18

Creation of long, MW chains composed of repeating subunits in alkenes.

Occurs through radical mechanism, requires HIGH TEMP and PRESSURE.

Question 19

Synthesis of Alkynes

Answer 19

1. Elimination of HX from geminal and vicinal dihalides with HEAT and BASE.
2. Add existing Triple bond to Nucleophile by removing acidic proton with STRONG BASE.

Question 20

Reduction of Alkynes

Answer 20

Can be hydrogenated with a catalyst.

Two Ways:
1. Lindlar’s catalyst (H2, Pd/BaSO4, Quinoline)

Product: Alkane (cis)

2. Na, NH3 (liquid) -33 degrees C.

Product: Trans alkene.

Question 21

Electrophilic addition of alkynes.

Answer 21

Electrophilic addition to alkynes occurs in same manner as it does alkenes. (X2 added)

Occurs according to Markovnikov.

Can be stopped at alkene or proceed further.

Product: Dihalo alkene or alkane

Question 22

Free Radical Addition of Alkynes

Answer 22

Radicals add to triple bonds as with double bonds with ANTI-markovnikov (X-).

Product: trans isomer.

Because intermediate vinyl radical can isomerize to its more stable form.

Question 23

Hydroboration Addition of Alkynes

Answer 23

Addition to triple bonds is same as double bonds.

Addition is syn and BORON adds first.

Boron can be replaced with proton from acetic acid = cis alkene.

Product: syn alkene.

Question 24

Oxidation Addition

Answer 24

Alkynes can be oxidatively cleaved with either basic KMnO4 followed by acidification or Ozone.

KMnO4 Product: 2 M COOH

Ozone Product: 2 M COOH

Question 25

Huckel’s Rule

Answer 25

4n + 2 pi electrons

Important indicator of aromaticity.

n = any nonnegative integer; thus can be 2, 6, 10, 14, 18, etc.

Question 26

Aromatic Compounds

Answer 26

Aryl compounds or arenes (Ar)

1,2 substituted = Ortho- or 0-
1,3 subst = meta or m-
1,4 subst = para or p-

Physical properties are similar to hydrocarbons.

All 6C atoms are sp2 hybridized.

has delocalized pi electron system.

Question 27

Electrophilic Aromatic Substitution

Answer 27

Most important reaction of aromatic compounds.

Reaction: electrophile replaces a proton on ring, producing substituted aromatic compound.

Common e.g.:

1. Halogenation
2. Sulfonation
3. Nitration
4. Acylation

Question 28

Halogenation of Aromatic Compounds

Answer 28

An electrophilic aromatic substitution

Aromatic ring reacts with Br2 or Cl2 in presence of FeCl3, FeBr3, or AlCl3 to produce a monosubstituted product.

Product: aromatic ring with X.

Question 29

Aromatic Rings
\+
Br2 or Cl2
\+
FeBr3, FeCl3, or AlCl3

(name product)

Answer 29

Electrophilic Aromatic Substitution of Br or Cl to aromatic ring.

Question 30

Sulfonation of Aromatics

Answer 30

An electrophilic aromatic substitution.

Aromatic reacts with fuming H2SO4 (mixture of SO3 and H2SO4) to form sulfonic acids.

Question 31

Aromatic Rings + Fuming H2SO4 (SO3/H2SO4)

name product

Answer 31

An electrophilic aromatic substitution.

Formation of Sulfonic acids.

Aromatic ring + SO3H

Question 32

Nitration of Aromatic Rings

Answer 32

An electrophilic substitution.

Mixture of nitric and sulfuric acids is used to create the nitronium ion, NO2+ (strong electrophile).

This reacts with Ar. rings to produce nitro compounds.

Question 33

Aromatic Rings
+
Nitric (HNO3)/Sulfuric acids

(name product)

Answer 33

Product: Nitro Compounds

Question 34

Friedel-Crafts Acylation

Answer 34

A carbocation electrophile, usually an acyl group is incorporated into Aromatic.

Usually catalyzed by Lewis Acids such as AlCl3.

Product = Ar-Acyl group

Question 35

CH3C(=O)Cl + Aromatic + AlCl3 –> ?

Answer 35

Product: Aromatic Ring-Acyl group + HCl

Question 36

Substituent Effects

Answer 36

1. Activating, ORTHO/PARA-directing substituents: (electron donating) NH2, NR2, OH, NHCOR, OR, OCOR, R
2. Deactivating, ORTHO/PARA-directing subs (weakly electron withdrawing): F, Cl, Br, I
3. Deactivating, META-directing substituents (electron withdrawing): NO2, SO3H, Carbonyl compounds (COOH, COOR, COR, CHO)

Question 37

NO2, SO3H, Carbonyl compounds (COOH, COOR, COR, CHO)

Answer 37

Deactivating, META-directing substituents (electron withdrawing)

Question 38

Catalytic Reduction of Aromatics

Answer 38

Benzene rings can be reduced under vigourous conditions (elevated Temperature and Pressure) to yield cyclohexane.

Ruthenium and Rhodium are most common catalysts.

Product: cyclohexane

Question 39

Aromatic ring + H2/Rh/C and high T/P –> ?

Answer 39

Benzene –> cyclohexane

Question 40

Br2, Hv

Answer 40

electrophilic addition

occurs according to Markovnikov most substituted””

Question 41

1) CH3MgBr

2) H3O+

Answer 41

Converts carbonyl compounds to alcohol

Addition reaction by Markovnikov.