CHEM 51B FINAL STUDY

pKa of Hydronium Ion (H₃O⁺)

-2

pKA of Methane (CH₄)

50

pKa of Acetic Acid (CH₃COOH)

5

pKa of Methanol (CH₃OH)

16

pKa of Phenol (Aromatic Ring)

10

pKA of Sulfuric Acid (H₂SO₄)

-5

pKa of Ammonia (NH₃)

38

pKA of Propyne (CH ₃C≡CH)

25

pKa Pyridinium ion (C₅H₆N⁺C)

5

pKa of Hydrocyanic Acid (HCN)

9

pKa of Phenylacetylene (C₈H₆)

25

pKa of Hydrogen Gas (H₂)

40

pKa of Triethylammonium ion ((CH₃CH₂)₃NH⁺)

11

pKa of Hydroiodic Acid (HI)

-10

pKa of Hydrazoic Acid (HN₃)

5

pKa of Hydrofluoric Acid (HF)

14

pKa of H2

40

pKa of Water (H₂O)

14

pKa of HCl

-7

Hydroxybenzene/Phenol (C₆H₅OH)

10

pKa of HOH

15.5

Pka of HBr

-8

Sn1 Reaction

-two steps

-forms an intermediate carbocation

-unimolecular (substrate only)

-Uses weak nucleophile (generally neutral)

-check for carbocation stability

-Higher degrees the better (tertiary > secondary > primary)

-Stereochemistry: mix of retention and inversion

-All organic stuff has to be all neutral or positive charge

-Favors Protic Solvents:

-Alcohol Dehydration: uses any halide ex. HBr

Sn2 Reaction

-single step “attack of nucleophile on backside.”

-Bimolecular (substrate and nucleophile)

-check for steric hindrance and inversion/retention

-Lower degrees the better (primary > secondary > tertiary)

-Uses strong nucleophiles (generally negative charge)

-Stereochemistry: Inversion only

-Alcohol Dehydration: SOCL3 (use pyridine), and PBr3

E1 Reaction

-unimolecular transition state

-depends only on substrate

-needs a more stable carbocation to be fast

-does not require strong base

-No stereochemistry

-forms new C-C pi bond, breaks C-H bond and C-leaving group bond

-species acts as base to remove a proton, forming new pi bond

-Follows Zaitsev’s rule: removes hydrogen from carbon attached to fewest hydrogens

-favored by heat

-Alcohol Dehydration/Acid-Catalyzed Dehydration: Uses H2SO4

E2 Reaction

-bimolecular transition state

-depends on both substrate and base

-needs strong base

-Leaving group must be anti-periplanar to hydrogen to be removed

-FAVORS FORMATION OF STABLE ALKENES (MORE SUBSTITUTED)

-forms new C-C pi bond, breaks C-H bond and C-leaving group bond

-Fav base: KOtBu, NaNH2 for alkynes, likes bulky bases

-species acts as base to remove a proton, forming new pi bond

-Follows Zaitsev’s rule: removes hydrogen from carbon attached to fewest hydrogens

-favored by heat

-Alcohol Dehydration: uses POCl3 (with pyridine)

Zaitsev Rule

-Elimination reactions occur such that they remove a hydrogen from the carbon attached to the fewest hydrogens

-When forming alkene in elimination reaction make sure to form most substituted alkene (most carbon atoms directly attatched)

Anti Periplanar

180 degrees skewed/facing away from each other

Protic Solvents

solvents with large dipole movements and high dialectic constants

How to tell which reaction for which?

Alkene Hydrohalogenation & Hydration:

- Acidic Conditions

-go through carbon cation intermediate

-creates both sym addition (same face/added from same side) and anti addition (opp. faces added from opposite sides)

Ch 7: Alkyl Halides and Nucleophilic Substitution

SN1

-Reagent: H-Nuc (Hydrogen bonded with a nucleotide)

-Product: Racemic mixture = (two enantiomers, 50:50 mixture) + a H-X (X= I, Cl, Br), nucleotide replaces leaving group

SN2

-Reagent: Na+-Nuc (Na with a nucleotide)

-Product: one product with stereoinversion + X- (I, Cl, Br), nucleotide replaces leaving group,

Ch 8: Alkyl Halides and Elimination Reactions

E1

-Reagents: H-B (hydrogen bonded to base)

-creates carbocation intermediate

-Products: two diastomers (leaving group makes the double bond) plus H-X (Hydrogen bonded with Cl, Br, or I)

E2

-Reagents: Na+-Base- (Na bonded to a neg base)

-must be in anti-periplanar to do E2

-Products: leaving group bond used to make alkene also creates NaX (Na bonded to Cl, Br, or I) and H-B (Hydrogen bonded to the base)

Ch 9: Reactions of Alcohols

E1: Dehydration

-Reagents: H2SO4

-Products: creates a major contributor and minor contributor (diastereomers) plus H2O, OH leaves and becomes alkene

-favors stable alkenes, favors more stable carbocations 3°>2°>1°

E2

-Reagents: POCl3 + pyridine

-Products: creates one product where OH leaves and becomes alkene

Sn1

-Reagents: HX (Hydrogen bonded to Br, I or Cl)

-Products: creates two hydrohalogenation diastereomers (OH is replaced by X) plus H2O

SN2

-Reagents: PBr3 or SOCL2 + Pyridine

-Products: one product with stereoinversion where X replaces OH (X= Br when PBr3 and X= Cl when SOCl2)

Ch 9: Reactions of Alcohols Part 2

Tosylation:

-good for replacing OH to an OTs group in order to make it easier to convert it to an alkyl halide (BEST used with 2 ∘ chiral carbon)

• Used with secondary chiral carbon bc wants to avoid racemic product of the chiral center (want complete inversion for product)

• Commonly two steps: 1st Step) TsCl + pyridine 2nd Step) X Nucleophile (Cl, Br, or I) for Sn2

-Reagents: TsCl + pyridine

-Products: OH is replaced with OTs (p-toluenesulfonate) when replaced O-C bond does not break so it is not invereted

Williamson Ether Synthesis from Alcohol:

-creates ether from alcohol

-Reagents: 1st Step) NaH, 2nd Step) Br-R (Br connected to carbons in other words an alkyl halide)

• First step treats alcohol with NaH bc alchol is a bad leaving group (deprotonate alcohol)

• Second Step is an Sn2 reaction with an alkyl halide (ideal alkyl halide would be a 1° for this reaction, more sterically bulky = bad)

-Products: 1st product) H deprotonates OH, leaving O neg and Na pos. 2nd product) Br attaches to Na while the rest of the alkyl halide attaches to O stabilizaing the charge

Epoxide Opening

-epoxide opens up to create alcohol

-acidic conditions: attack the more substituted side, can protonate

• Reagents: EtOH + H2SO4

• Products: one product epoxide opens up, OH goes to the more substituted side

-basic conditions: bases attack the less substituted side

• Reagents: 1st step) Na+Oet- 2nd step) H2O

• Products: one product epoxide opens up, OH goes to the less substituted side

Ch 10: Alkenes and Addition Reactions

Syn and Anti Additions (Non-Selective)

-Syn Addition: additions added to the same face/side (ex: from the top or bottom at the same time)

-Anti Addition: additions add from different side

• Hydration: Markovnikov reaction (attached to more substituted carbon)!

  • Reagents: H2O + H2SO4

  • Products: OH added by Markovnikov

• Hydrohalogenation: Markovnikov reaction (attached to more substituted carbon!

  • Reagents: H-X (Hydrogen bonded to I, Cl, or Br)

  • Products: added by Markovnikov

Anti-Addition Reactions (ALWAYS )

-Halohydrin: Markovnikov reaction

• Reagents: X2 + H2O

• Products: makes two enantiomers, OH added Markovnikov

-Halogenation: NOT Markonikov

• Reagents: X2

• Products: two enantiomers

Syn Addition Reaction (ALWAYS)

-Hydroboration-oxidation: Anti-Markovnikov (added to less substituted carbon)

• Regeants: 1st Step: BH3, 2nd Step) H202 + OH-

• Product: Two enantiomers, OH added to anti-Markovnikov

Ch 11: Alkynes

Hydration: Markovnikov reaction, Syn and Anti Additions (Non-Selective)!

-Reagents: Internal Alkyne (triple bond in between two carbons): H2O + H2SO4

• Terminal Alkyne (triple bond at end of carbon chain): H2O + H2SO4 + HgSO3

-Products: creates a ketone

Hydrohalogenation: Markovnikov reaction, Syn and Anti Additions (Non-Selective)!

-Reagents: 2HX (2 Hydrogen bonded to X)

-Product: makes geminal dihalide (Halogen bonded to the same carbon)

Halogenation: NOT Markonikov, Anti-Addition!

-Reagents: 2X₂ (4 Halogens)

-Product: makes a tetrahalide (known dead end for synthesis)

Hydroboration-oxidation: Anti-Markovnikov, Syn Addition!

-Reagents: 1st Step: R₂BH 2nd Step: H2O2, OH-

-Product: Creates an aldehyde

Ch 11: Alkynes Part 2

Acetylide Anion Formation:

-useful for Sn2 reactions and epoxide openings

-Reagents: NaNH2(more common) or NaH

-Products: creates acetylide ion and Na+ (strong nucleophile and base)

• FOR SN2: with the acetylide ion

• Reagents: alkyl halide (primary bettter)

• Product: halogen leaves attatches rest of structre to the neg carbon (creates carbon carbon bond, makes larger carbon chains)

Alkene to Alkyne Synthesis:

-Reagents: 1st Step) Br2 (takes alkene, makes it a single bond) 2nd Step) 2 NaHH2 (turns single bond to alkyne)

Ch 12: Oxidation and Reduction

Reductions

-Alkenes: syn addition!

• Reagents: Pd/C + H2

• Products: creates one product and a enantiomer, H2 attaches, alkene is removed

-Carbonyls

• To reduce an Aldehyde:

  • Reagents: Pd/C + H2

  • Products: makes an alcohol

• To reduce a Ketone:

  • Reagents: Pd/C + H2

  • Products; makes an alcohol + enantiomer

-Alkynes

• To reduce to a single bond:

  • Reagents: Pd/C + H2

  • Product: product without an alkene

• To make an alkene:

  • Reagents: Lindlar’s Catalyst + H2

  • Product: syn addition, Z, Cis alkene

  • Reagents: Na + NH3

  • Product: anti-addition, E, trans alkene

-Epoxides

• To reduce epoxide:

  • Under basic conditions

  • Reagents: LiAlH₄, H₂O

  • Product: Alcohol forms, Alkane forms

    • moves to more substituted carbon,

    • breaks bond at less substituted carbon,

Ch 12: Oxidation and Reduction Part 2

Reductions

-Alkyl Halides and Tosylates

• To reduce an alkyl halide:

  • Reagents: LiAlH₄, H₂O

  • Product: Halogen is lost, Stays as alkane

• To reduce a tosylate (Ts):

  • Reagents: LiAlH₄, H₂O

  • Product: OTs is lost, Stays as alkane

Oxidations

-Epoxidation

• To epoxidate:

  • Reagents: mCPBA

  • Product: Epoxide forms, alkene breaks, inverted

-Dihydroxylation

• To dioxidize in a cis fashion:

  • Reagents: OsO₄,NaHSO₃, H₂O

  • Product: syn-diol, alkene breaks, 2 OH bonds form

• To dioxidize in a trans fashion:

  • Reagents: (1) mCPBA, (2) KOH

  • Products: (1) Epoxide forms, alkene breaks, (2) Epoxide breaks, 2 OH bonds form, anti-diol

-Alcohols

• To oxidize an alcohol → aldehyde → carboxylic acid

  • Reagents: (1) PCC, (2) CrO₃, H₂SO₄, H₂O

  • Product: (1) Aldehyde, (2) Carboxylic Acid

• To oxidize an alcohol → carboxylic acid

  • Reagents: CrO₃, H₂SO₄, H₂O

  • Products: Carboxylic Acid

• To oxidize an alcohol → ketone

  • Reagents: PCC

  • Products: Acetone (Ketone)

  • Reagents: CrO₃, H₂SO₄, H₂O

  • Products: Acetone (Ketone)

Ch 12: Oxidation and Reduction Part 3

Oxidative Cleavage

-Alkenes

• To cleave an alkene:

  • Reagents: O₃, Me₂S

  • Products: 2 double-bonded oxygens replace alkene, two ketones

-Alkynes

• To cleave an alkyne:

  • Reagents: O₃, H₂O

  • Products: 2 double-bonded oxygens/2 alcohols replace alkyne’s cleavage, 2 carboxylic acids

• To cleave an alkyne with H end:

  • Reagents: O₃,H₂O

  • Products: Carboxylic acid, carbon dioxide

Ch 13: Radical Reactions

Chlorination

• Reagents: Cl₂, hv/Δ (heat)

• Products: 4 constitutional isomers, methyl chloride (1) or a methyl/chlorine (3)

Bromination

• Reagents: Br₂, hv/Δ (heat)

• Product: Stable tertiary radical (selective) → major product - Methyl/Bromine on same carbon

Allylic Bromination

• Reagent: NBS, hv/ROOR

• Product: Alkene does not react, Br replaces H

Radical Hydrobromination

• Reagents: HBr, hv/ROOR

• Products: Br added anti-Markovnikov

Ch 14: Conjugation, Resonance, and Dienes

Generic Diels-Alder

• Reagent: Diene, Dienophile

• Product: Aromatic ring with pi bond

Trans Product

• Reagent: E configured dienophile

• Products: Aromatic ring with trans branches

Cis Product

• Reagent: Z configured dienophile

• Products: Aromatic ring with cis branches

No Reaction Case

• No reaction because the diene can not be rotated to the cis config

Hydrohalogenation of Conjugated Dienes

• Reagents: HBr

• Products: (1) 1(H),2(Br) addition, (2) 1(H),4(Br) addition product made from a resonance structure of the carbocation intermediate

Ch 15: Benzene and Aromatic Compounds

Labeling Disubstituted Benzenes (Only Aromatic Rings)

• 1,2 - Ortho

• 1,3 - Meta

• 1,4 - Para

How to determine aromatic, antiaromatic, or non aromatic

1. Is it cyclic?

2. Are all atoms sp2 hybridized?

3. Number of π electrons (Huckel’s Rule)

• Aromatic: (4n+2)π electrons

• Anti-aromatic: (4n)π electrons

Orbital of nitrogen lone pair

• N lone pair is in the p orbital since it can participate in aromaticity

• N lone pair is in the sp2 orbital

• If the nitrogen in a pi conjugated system is double-bonded to another atom, the lone pair does not participate in aromaticity, therefore it is sp₂ hybridized