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CHEM 2211: Final Exam

Sp3
4 single bonds 109.5 3 2p orbitals combined with 2S orbital
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Sp2
1 double bond 120 1 2S combined with 2 2P forming 3 SP2 and 1P unchanged
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SP
triple bond 180 1 2S overlaps 1 2P= 2 SP orbitals 2 P unchanged
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Formal Charge
FC= (# valence e) - (# of bonded e)/2 - (# non bonding e)
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Bronsted lowry
acid= donates H+--- forms conjugate base base= accepts H+ --- forms conjugate acid
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PKa
Pka= -logKa strong acid= small Pka Strong base= large PKa Ka= [products]/ [reactants]
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lewis acid/ base
acids= accept e - low energy orbitall or polar bond to H - metal cations/ halogen acids/ 3 compounds/ transition metals base= donates e or pair of non bonding e - most contain O or N
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dipole-dipole interaction
between polar molecules
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dispersion forces
constantly changing electron distribution van der wall force
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hydrogen bond
H- EN hydrophillic
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ant-conformation
180 apart
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gauche
60 apart 3.85 KJ more than anti - due to steric strain= atoms forced closer together
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eclipsed
highest energy/ least stable 12 KJ more than gauche -torisonal strain -C-H = 4.0 KJ
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ortho
2 substituents on adjacent carbons on a benzene ring
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aromatic rings
assume all are planar 4n + 2 = pi electrons
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weak acids
large Pka value OCH3 group contributes EN to ring= weak acid
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heat of combustion
amount of heat released when a compound is burned in O2 higher the strain= higher the heat of combustion
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specific rotation
SR= observed rotation/ pathlength x concentration
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chiral
not superimposable mirror image no plane of symmetry
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enantiomer
mirror images- not superimposable light cancels out
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# of sterioisomers
1 chiral center= 2 2 chiral centers= 4 3 chiral centers= 6
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meso compounds
2 chiral centers line of symmetry not optically active
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racemic mixture
50:50 mixture of 2 enatiomers 0 optical rotation
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racemic mixture
...
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prochirality
if a molecule can be converted from achiral to chiral in single chemical step
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exothermic
bonds formed are stronger than bonds broken releases energy to surroundings
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degree of unsaturation
# of bonds/ rings a molecule has CnH2n+2 subtract 1 for every N ignore O
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alkene
no free rotation around double bonds 350KJ to break
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hygrogenation
larger heat of hydration= higher energy= less stable
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hydrolysis
degradation of substance in water
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dehydrohalogenation
alkyl halide + strong base (KOH)---- loss of HX-- forms alkene - elimination
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dehydration
alcohol + strong acid (H2SO4)--- alkene
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polar reaction
unsymmetrical bond breaking and forming - between compounds w/ even number of electrons - most common type of reaction - 1 element keeps the pair of electrons -heterolytic - electron rich sites react with electron poor sites
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radical reactions
symmetrical bond breaking and making - 1 electron remains with each element - homlytic - take place in gas phase
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Keq
equilibrium constant that shows the rate of products to reactants - [products] / [reactants] - K > 1= product concentration > reactant concentration - K = 1 = fair amount of both - K < 1 = reactant > product---- does not appear as written; in reverse
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favorable reactions
exergonic: - favorable G= -RT ln Keq R= 8.314 T= kelvin
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bond dissociation energy
measures heat needed to break a bond exothermic= bonds formed are stronger than bonds broken
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stability of alkenes
cis alkenes are less stable than trans b/c of steric strain stability determined by: -cis trans equilibrium constants= product favored= more stable - hydrogenattion= treatment of alkene w/ H2 ---- more energy released= less stable - more subs on double bond= more stable= stronger sp2-sp3 bonds
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carbocation
planar increase with substitution
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the hammond postulate
the transition state resembles reactants or products= which ever closer in energy to exergonic resembles reactants endergonic transition state resembles products
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substitution vs elimation
major product: -sub= if leaving group is on primary -elimination= if leaving group is on tertairy Strong base: -OH Poor nuc: H2O or RHO 1= Sn2 favored 1= no reaction 2= mostly E2 2= Sn1/ E1 in polar protic solvents 3= E2 3= Sn1/ E1 in polar protic solvents weak base: Cl 1= Sn2 2= mostly Sn2 3= SN1/ E1 in polar protic solvents
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Sn2
rate depends on alkyl halide and leaving group single step= no intermediates attaches at opposite side favors: -small methyl/ primary molecules - vinylic and aryl halides are unreactive leaving group: F<Cl<Br<I < TosO aprotic solvents: CH3OH< H2O< DMSO<DMF<CH3CN< HMPA
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nucleophile
electron rich negative nuc= neutral product neutral nuc= positive product nuc increases going down periodic table
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Sn1
2 steps= intermdiate unimolecular produces racemic mixture polar protic solvents - -OH or -NH favors tertairy weak nucleophile= H2O RHO
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E2
Strong base -tertiary/ secondary primary=sn2
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Sn2
strong nucleophile/ base primary weak base- Cl primary/ secondary
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O-H/ N-H
3400
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C-H Sp3
2800-3000
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C-H Sp2
3000-3100
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C-H Sp
3300
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alkyne/ nitrile
2100
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C=O
1700
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C=C
1600
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aromatic
1500
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oxidation
increases O content/ decreases H content CrO3, K2Cr2O7, K2CrO4, Hot KMnO4 potent oxidizers chromium under acidic conditions KMnO4= basic conditions alkenes/ alkynes/ alkylbenzenes= cleavage product HIO4=vicinal diol= OH attached to adjacent carbons
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Reduction
loss of O content Carbonyl= Metal Hydrids alkenes= hydrogenation alkynes= hydrogenation or metal in NH3
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diels-Alder reaction
in heat Diene + Alkene---- cyclohexene stereospecific syn addition= same side endo forms major project= subs opp of chair
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