Straumanis Memorization

Strong Acids

HCl, HBr, HI, HNO3, H2SO4, OH3

Very Good Leaving Groups

RSO-, I-

Good Leaving Groups

Cl-, Br-, H2O

(Very) good leaving groups are...

very weak bases

Poor leaving groups are...

strong bases

Poor Leaving Groups

F-, RO-, R2N-, R3C-

Very good nucleophiles

RS-, NC-, I-, PR3, R3C-, R2N-, RC---C-, RO-

Good nucleophiles

Br-, R2S, NR3, Cl-, RCO2-, N3-

(very) good nucleophiles are...

good bases

poor nucleophiles

F-
HCO3-
R2O (water, alcohol, or ether)

poor nucleophiles are...

weak bases

examples of soft nucleophiles (also are good leaving groups)

I-, Br-, PR3, R2S, & Cl-

Nucleophicity increases...

as you go down the periodic table

13.1: R-CH2-OH => R-CH2-X

reagents: HX (X= Cl, Br, or I)
OH is replaced with an X

13.2: R-CH2-OH => R-CH2-Cl

reagents: SOCl3
OH is replaced with a Cl

13.3: R-CH2-OH => R-CH2-Br

reagents: PBr3
OH is replaced with a Br

13.4a: R-CH2-LG => R-CH2-Nuc

reagent: any nucleophile
LG is replaced with nucleophile

13.4b: R-CH2-LG => R-CH2-OH

reagents: NaOH or KOH
LG replaced with OH

13.4c: R-CH2-LG => R-CH2-OR

reagents: Na (metal; same one u use for alkyne to alkene reduction trans) + ROH (an alcohol) = NaOR
LG replaced with OR

13.4d: R-C---C-H => R-C---C-CH2-R

reagents = 1) strong base so NaNH2
2) R-CH2-LG

H is replaced with everything in step 2 except the leaving group

8.1: Markov Addition of H & OH

reagents: 1) Hg(OAc)2, H2O & THF (solvent)
2) NaBH4
H2O creates OH and it gets added onto the Markov. carbon. Remember the double bond gets broken.

8.2: Markov Addition of H & OR

reagents: 1) Hg(OAc)2, HOR & THF (solvent)
2) NaBH4
OR gets added on Markov. carbon.

8.3 Anti-Markov. Addition of H & OH

reagents: 1) BH3 & THF
2) NaOH, H2O2, H2O
Double bond gets broken & OH is added to Anti-Markov carbon.

8.4: Anti-Markov. Addition of H & Br

reagents: HBr & peroxides like light & heat
Br is an exception from the usual HX unless its no peroxides.
Double bond breaks & Br is added to the Anti-Markov. carbon.
Needs more heat/energy to add to Anti-Markov.

8.5: Markov. Addition of H & X (X = Cl, Br, or I)

reagents: HX & no peroxides (dark & cold)
X is added to Markov. carbon. HX is a strong acid & acids prefer Markov.

9.1: Addition of Br2 to an Alkene (a ring)

Reagents: Br2
Double bond in a ring breaks. Creates 2 Br additions.
Has trans configuration.

9.2: Oxidation of an Alkene to an Epoxide

Important points:
reagent - MCPBA = a peroxy (carboxylic acid with an addition O)
Double bond between two C's break & creates a triangle with an O in the middle.

9.3: Anti-Markov. Epoxide Ring Opening (Basic Conditions)

reagents: 1) strong base (ex: RO-)
2) dilute acid (H gets added to O-)

The base causes the triangle to split. Only one side gets the O- & the other gets the H & the base. The acid job is to make the O- to a neutral value.

9.4: Markov. Epoxide Ring Opening (Acidic Conditions)

reagents: H+ & HOR

9.5: Markov. Addition of Br & OR to an Alkene

reagents: Br2 & HOR

OR ends up on the more subs. carbon & Br on Anti-Markov.

10.1: Hydroboration-Oxidation

reagents: 1) BH3 & THF
2) NaOH, H2O2, H2O
Anti-Markov. cuz OH ends up on less substituted carbon.
H & OH = cis/syn
H3C & OH = trans

H2 & Pt/Pd can...

completely reduce any bond to a sigma bond. the two H's are syn of each other.

H2 / Ni2B
or
H2 / Lindlar Catalyst

alkyne => alkene in cis/syn

Na0 (metal) NH3 (ammonia)

alkyne => alkene in trans

D2

Double bond will reduce but 2 D's will form in a syn configuration touching each side of the bond.

Oxidation of 1* alcohol to aldehyde

PCC
reduces an OH => double bond O

Oxidation of 2* alcohol to ketone

KMnO4 or Na2Cr2O7/acid/H2O or CrO3/acid/H2O
reduces an OH => double bond O

10:9 Syn Addition of Two OH Groups to a PI Bond

reagents: 1)OsO4
2)H2O2 or NaHSO3

the two OH added Syn/Cis

Trans Addition of Two OH Groups to a PI Bond

acidic: 1) MCPBA
2) H3O+ & H2O
OH and methyl Markov. carbon

basic: 1) MCPBA
2)KOH
3) neutralizes with dilute acid
methyl group on other carbon

OH & a methyl shown

acid: OH is on anti-Markov. & methyl on other
that means: 1) MCPBA
2) Na + (methyl)
2) H+

base: OH is on Markov, & methyl is on anti-Markov
1) MCPBA
2) (methyl) + OH
2)H2SO4 or H+

10.10: Ozonolysis

reagents: 1) O3 2) Zn/acetic acid
cleaves double bond in half, it only oxidizes the carbon to an aldehyde under reducing conditions. if ozidizing make same product as KMNO4

10.11: Permanganate

reagents: 1) KMnO4 (hot)
2) H3O+, H2O
cleaves C=C into making 2 O=C. OH is attached to one of the sides.

11.1: Markov. Addition of H & X (X = Cl, Br, or I)

reagents: 1) HX so no peroxides
2) excess HX
goes from triple bond to sigma & creates germinal dihalide (2 X on one side Markov)

11.2 Addition of X2 (X = Cl, Br, or I)

reagents: 1) X2
2) X2 (excess)
2 X's on each side. They are trans to each other.

11.3: Markov. Addition of H2O to an Alkyne

reagents: H2O2, H2SO4, HgSo4

O added to Markov & triple bond becomes a double bond with O connected to it

11.4: Anti-Markov. Addition of H2O to an Alkyne

reagents: 1) BH3 & THF
2) NaOH, H2O2, H2O

double bonded O ends up on Anti-Markov. carbon

11.5: Deprotonation of an Alkyne

reagents: NaNH2

The H on the right of the triple bonded C disappears. The C is left with a lone pair of electrons and a - charge. Ends up creating an acetylide anion.

polar aprotic solvents

- polar solvents without an H on N, O, or X
- faster in SN2

polar protic solvents

- polar solvents with an H on N, O, or X
- faster in SN1 (since it lowers the PE of the intermediate, lowers E act)

Substitution of allylic & benzylic

Can undergo either SN2 or SN1 depending on the solvent; although, in tertiary groups, they cannot undergo SN2 (too hindered)

steric factors

the ability of groups, because of their size, to hinder access to a reaction site within a molecule
overcrowding & depends on the degree of carbocation

electronic factors

factors having to do with where the electrons are found (ex: electronegativity, resonance, hybridization, or inductive effects)

methyl leaving group

SN2 only possibility

1 degree leaving group

good nucleophile (both weak & strong bases work) => SN2
strong base & poor nuc (& hot) => E2

2 degree leaving group (or 1 degree with allylic or benzylic leaving group)

good nucleophile => weak base & polar protic solvent (cold) => SN1

good nucleophile => weak base & polar aprotic solvent => Sn2
good nucleophile => strong base & cool => SN2

poor nucleophile => weak base (& polar protic solvent hot) => E1

poor nucleophile => strong base => E2
good nucleophile => strong base & hot => E2

3 degree leaving group

weak base => cool or good nucleophiles => SN1

SN2 not possible

weak base => hot and poor nucleophile => E1

strong base => E2

Zaitsev's Rule

an elimination usually gives the most substituted alkene product.

Hoffman Rule

When an elimination reaction yields the alkene with the less substituted double bond.

15.1: Radical Chlorination of an Alkane
start: C is attached to 2 R's & H & a methyl group (CH3)
result: it creates 3 products = 1) Cl attaches to the methyl group, 2) Cl replaces the H as the fourth bond to the central C, 3) HCl

*remember why it creates more products than Br = Cl products equally divided & doesn't prefer a certain form

reagents = Cl2 & light (hv)

Synthetic Transformation 15.2: Radical Bromination of an Alkane
start: C is attached to 2 R's & H & a methyl group (CH3)
end: 2 products = 1) Br replaces the H as the fourth bond to the central C 2) HBr

*Br is more selective => tertiary bromide product is most favored

reagents:
Br2 & light OR NBS (creates Br2 & radicals quickly) & light

radical

species with an unpaired electron (result from nonpolar bond breakage)

shown with a half arrow ("single barbed" arrow)

radical potential energies (highest P.E. to lowest P.E.)

1. H
2. CH3
3. 1 degree
4. 2 degree
5. 3 degree
6. allylic
7. benzylic
8. X (halogen)

radicals in synthetic transformation:
Br2 => ???
radical initiator + HBr =>
HBr =>

look at pg 231 for visual reference

1. Anti-Markov.
2. Anti-Markov
3. Markov.

bond angles & shapes

1. count the number of electron domains
=> 4 - 109.5 degrees, 3 - 120 degrees, 2 = 180 degrees

2. Find the molecular shape (you can figure this out by knowing the electron domain & how many bonds/lone pairs in each molecule)

Linear

- 2 electron domains
- 180 degrees
- no lone pairs, all bonds

Bent (2 types)

- 3 electron domains
- 120 degrees
- 2 bond domains (the 3 bonds are usually split as 1 double bond and 1 single bond) & 1 lone pair

- 4 electron domains
- 109.5 degrees
- 2 bonds & 2 lone pairs
- common ex: water (H2O)

Trigonal planar

- 3 electron domains
- 120 degrees
- all bonds, no lone pairs

Pyramidal

- 4 electron domains
- 109.5 degrees
- 3 bonds & 1 lone pair

Tetrahedral

- 4 electron domains
- 109.5 degrees
- 4 bonds, 0 (no) lone pairs

Formal charge

group # - # lines - # dots

hybrid orbitals

sp3 => 109.5 degrees
sp2 => 120 degrees
sp => 180 degrees

Strong base

molecule with a lone pair & -1 formal charge on H, C, N, or O

phenol

10

carboxylic acid

5

pKa table relation to acids and bases

More acidic molecules will have a smaller (or even negative) pKa. pKa below -2 are considered strong acids, which dissociate in aqueous solution

More basic molecule will have larger pKa.

endothermic

PE change from low to high
positive (+) or uphill
unfavorable
breaking a bond; energy must be added

exothermic

PE change from high to low
negative (-) or downhill
favorable
making a bond: energy be released by molecule

inductive effect

polarization caused by electronegative atom that can induce minor polarization in neighboring bonds
very weak

In one of the CTQ it talks about an H connect to an N or O is more acidic than H on C, F, or Cl. I am not sure tho.

Newman projections (lowest to highest PE)
remember lowest PE is always more favored

lowest: staggered
highest: eclipsed

staggered anti (180 degrees apart) > staggered gauche (60 degrees apart) > eclipsed (when same groups are right hidden by each other, the PE becomes even higher)

degree of unsaturation

no. pi bonds + no. rings

constitutional isomers

same molecular formula, different connectivity

conformers

identical
alternate way of showing the same molecule

stereoisomers

Compounds with the same structural formula but with a different arrangement of the atoms in space. (ex: one could be trans & other cis)

distereomers

stereoisomers that are not mirror images

2 types of orientation

Z (cis) = largest groups on same side of a double bond
E (trans) = largest groups on different sides of double bond

why is it favorable to put large groups in equatorial position?

- points away from the ring where there is more space
- lower PE conformation

isopropyl

isobutyl

sec-butyl

CH3 - CH- [CH2CH3 (ethyl)]

tert-butyl

Chapter 14 (& 15 a) memorization recap

Br2 hv => adds to most sub C

KOH heat => usually E2; since OH- is a small base, it obeys Zaitsev rule (double bond to most sub)

Large base E2 => Hoffman product

E1 reactions always follow Zaitsev's rule

NBS => Adds Br to most sub carbon. Next a small base or large base will follow. You add it according the whether it is an E2 or E1.

Chapter 15b recap

initiation = always breaks a bond to create 2 radicals
propagation step = usually 2 steps; and involves 2 breaks of excess bonds (H-Br, Br-Br, etc)
termination = combine leftover products

resonance structures acid & base

multiple resonance structures in acid = less acid - base = (-) downhill

multiple resonance structures in base = acid - less base = (+) uphill

M1 Q13