O Chem

Question 1

Molecular Bonds

Answer 1

Longer bonds are weaker bonds
Triple > Double > Single
Sigma bonds > Pi bonds

Question 2

sp3 Hybrid Orbitals

Answer 2

s+p+p+p
109.5 degree angle
Tetrahedral, Bent (2 lone pairs), Triganal Plane (one lone pair)
4 sigma bonds

Question 3

sp2 Hybrid Orbitals

Answer 3

s+p+p
120 degree angle
Triganal planar

Question 4

sp Hybrid Orbitals

Answer 4

s+p
180 degree angle
Linear

Question 5

Acid Strength by Orbital

Answer 5

The more s character in a hybrid orbital, the stronger the acid
sp > sp2 > sp3

Question 6

Resonance

Answer 6

e- density is shifted through regions of a molecule via pi bonds
Resonance reduces basicity (makes an atom more acidic)
Rules:
1) Atoms should have filled orbitals
2) Best structure minimizes formal charges
3) (-) charge is best placed on most electronegative atom
4) (+) charge is best placed on least electronegative atom

Question 7

Inductive Effect

Answer 7

Induces charge separation in a molecule b/c of delocalization of e- by electronegative atom

• Transfer of e- through sigma bond
• Dissipates over distance

Question 8

Steric Hindrance

Answer 8

Occurs when two atoms attempt to be in same place at same time
Large substituents should be equitorial on cyclohexane

Question 9

Aromaticity

Answer 9

Huckel Rule: 4n + 2 (n is integer)

Aromaticity increase stability

Question 10

Bronsted Lowry Acid-Base

Answer 10

Transfer of H+ from acid to a base
BL acid - compound w/ a H+ that can be lost
BL base - e- available to bind to an H+
As an acid gets stronger it has a weaker conjugate base

Question 11

Lewis Acid-Base

Answer 11

Transfer of e- pair from base to acid
Lewis bases are nucleophiles
Stength of an acid depends on intramolecular forces
-Acid is stronger w/ an e- withdrawing group

Question 12

Hydrogen Bonding

Answer 12

Weak bond
Bond btw lone pair of e- and hydrogen w/ partial (+) charge
No H bonding on H’s bonded to carbons

Question 13

Van der Waals Forces

Answer 13

Exist btw all compounds
Considered only when no other factors
btw temporary dipoles

Question 14

Constitutional Isomer

Answer 14

Same formula, different connectivity

Finite # of constitutional isomers for a formula

Question 15

Cycloalkanes

Answer 15

Farther from 109.5 degree angle, the greater the strain and reactivity

Question 16

Index of Hydrogen Deficiency

Answer 16

Alkanes - C(n)H(2n+2)

IHD = 2(#C) + (#N) - (#H) - (#X) +2 /2

Question 17

IR Spectroscopy

Answer 17

Finds functional groups
OH - broad at 3600
NH - 3400-3200 (medium)
Ketone - ~1700 (strong)
Aldehyde - ~1740 (strong)
Aromatic - 1600-1400

Question 18

NMR

Answer 18

Takes advantage of hydrogen nucleus magnetic spin

Determines unique hydrogens

Question 19

Meso Compounds

Answer 19

Have opposing chiral centers, therefore optically inactive

Contain mirror plane through an even # of chiral centers

Question 20

Stereoisomerism

Answer 20

# of stereoisomers = 2^n
-less if some are meso

Question 21

Nucleophilic Substitution

Answer 21

Substitution of one functional group for another
Nucleophile attacks the electrophile
Sn1 or Sn2

Question 22

Nucleophile

Answer 22

Smaller nucleophiles are better
Nuc strength is correlated to base strength
Therefore small strong bases are good nucleophiles (but OH and OR can induce elimination)

Question 23

Electrophile

Answer 23

Strength is correlated to stability of LG
Good LG is stable in solution
Good LG has weak bond w/ carbon
I > Br > Cl&raquo_space; F

Question 24

Sn2 Reaction

Answer 24

Nuc attacks backside and pushed LG out
Causes inversion at chiral carbon
One step reaction
rate = k[Nuc][Elec]
Sn2 favors primary > secondary > tertiary
Doesn't form a racemic mixture

Question 25

Sn1 Reaction

Answer 25

``` LG leaves before Nuc attacks Does not depend on Nuc concentration Forms carbocation intermediate Favored tertiary > secondary > primary (no methyl) Favored in poor nucleophile rate = k[Elec] Slow two step reaction Forms a racemic mixture Carbocations can rearrange to form tertiary ```

Question 26

Free Radical Halogenation

Answer 26

Halogen bond is cleaved to form halogen radicals Hydrocarbon radical stability: tertiary > secondary > primary > methyl Mech: 1) initiation; 2) propagation; 3) termination Selective forming tertiary alkyl halogens (favors substituted product)

Question 27

E2 Reaction

Answer 27

Carried out under basic conditions at high temperature Occurs in one step Requires a bulky base; prevents nuc addition Substituents must be anti Yields most substituted alkene

Question 28

E1 Reaction

Answer 28

Carried out under acidic condition at high temp | Forms a carbocation intermediate (therefore possible rearrangement)

Question 29

Hofmann Elimination

Answer 29

Forms least substituted alkene | Uses a quaternary amine

Question 30

Alcohols

Answer 30

Higher boiling points and miscibility in water Become less hydrophilic as carbons are added As branching increases, boiling point decreases As molecular mass increases, boiling point increases Not good as Nuc, must be deprotonated to an alkoxide

Question 31

Aldehydes and Ketones

Answer 31

``` Don't form H-bonds Only slightly miscible in water Reactivity: at electrophilic carbon Ald: IR - 1720-1740; NMR - 9-10 Ket: IR - 1710-1725; NMR - 2.0-2.5 ```

Question 32

Carbonyl Reactivity

Answer 32

Attack at carbonyl group C=O, carbon is good electrophile Reactivity: Acid halide > Acid anhydride > ester > amide

Question 33

Aldol Condensation

Answer 33

Kinetic Product - major product minimizes steric hindrance of TS -chooses least substituted alpha carbon (low T, bulky base) Thermo Product - maximizes stability of intermediate or product -choose more substituted alpha carbon (high T, small base)

Question 34

Epimers

Answer 34

Vary in chirality at most oxidized C | Sugar diastereomers at one chiral center

Question 35

Anomers

Answer 35

Vary in chirality of least oxidized C Diastereomers of the cyclic form (alpha or beta) Beta - Up; alpha - down

Question 36

Osazome Test

Answer 36

Epimers yield the same osazome | Helps determine identity of sugar

Question 37

Tollens Test

Answer 37

Reduction of silver by oxidizable sugar | (+) test if silver mirror forms quickly, indicative of an aldose

Question 38

Amine Basicity

Answer 38

Secondary > Primary > Tertiary > NH3 Amines are increasingly nucleophilic w/ more alkyl groups Therefore primary amines are hard to synthesize

Question 39

Distillation

Answer 39

Removes a liquid from another liquid by explooiting differences in boiling point Fractional Distillation is used when liquids have small boiling point differences

Question 40

Thin Layer Chromatography

Answer 40

Spot sample on silica gel Solvent moves up plate interacting w/ samples Use Rf values for spots of the compounds -High Rf indicates most solubility in solvent (most like solvent)

Question 41

Column Chromatography

Answer 41

Separate bulk quantities of product | If a compound travels quickly, then minimal attraction to stationary phase

Question 42

Gas Chromatography

Answer 42

Vaporize sample into gas phase Heavier gases move slower and have longer elution times -b/c heavy and polar gases have higher point and also have longer elution times (therefore lower boiling point compounds come off GC first)

Question 43

Extraction

Answer 43

Works on solubility -Takes advantages of drastic solubility differences Acid-Base: Separate components on pH -Loss of charge favors organic (uncharged polar); gain of charge favors water (protonation) -Brine solution is used to increase differences in aqueous and organic layers

Question 44

Recrystallization

Answer 44

Solid product separated from solid impurities through selective precipitation Dissolve solid into hot solvent, filtering out insolubles, then cool solution to precipitate purified crystals Use minimal solvent so soluble at high temps and insoluble at low temps

Question 45

Mixed Melting Temperature

Answer 45

A mixture of compounds has a broader melting point indicating impurities