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49 Cards in this Set
- Front
- Back
saturated hydrocarbons contain no...
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double/triple bonds
CH(n+2) |
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primary carbon
secondary carbon tertiary carbon quaternary carbon |
CH3R
CH2R2 CHR3 CR4 |
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primary hydrogen
secondary hydrogen tertiary hydrogen |
CH3R
CH2R2 CHR3 |
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as molecular weight increases the ___ increase/decreases
-MP -BP -Density |
increase
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branched molecules have reduced surface area and therefore lower/higher BP
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lower
-lower van der waals forces |
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combustion
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alkanes + O2 --> CO2 and H2O + heat
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free radical halogenation
-diatomic halogens(Cl2) -can be cleaved using... |
UV light or heat
aka initiation |
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the halogen radicals become extremely reactive...
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begin propagation...
then end with termination when two radicals combine and end process |
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the more stable the intermediate is....
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the more likely the reaction is to occur
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3>2>1>methyl for...
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stability of radicals
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bromine attacks to form more stable radicals while chlorine will attack to form less stable radicals
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know
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free radical chlorination yields mixtures of products...it is dependent on the number of hydrogens present
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know
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pyrolysis
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occurs when a molecule is broken down by heat
-commonly used to reduce the average molecular weight or heavy oils and increase the production of the more desirable volatile compounds |
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disproportionation
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when radical donates a hydrogen to another radical and forms an alkene
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nucleophiles are ____ charged
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negatively (e- rich)
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electrophiles are ___ charged
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positively
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nucleophiles and basicity
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nucleophilic strength decreases
OR- > HO- > RCO2- > ROH > H2O |
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nucleophile and size
protic solvents aprotic solvents |
protic solvents:
large atoms tend to be better nucleophiles since they are more polarizable CN- > I- > RO- > Br- > Cl- >F- > H2O aprotic solvents : EN - F- > Cl- > Br- > I- |
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leaving groups
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best leaving groups are weak bases
-can accommodate and e- pair |
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halogen leaving group order...
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I- > Br- > Cl- > F-
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Sn1 rxns
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unimolecular substitution
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unimolecular = rate of rxn depends only on ___ specie/s
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one
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SN1 rate determining step =
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dissociation of specie to form a stable carbocation
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SN1 involves __ step/s
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two
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SN1 require a/protic solvents
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polar protic solvents
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SN1 carbocation stability
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3 > 2 > 1 > methyl
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original substituent must be a ____ leaving group than the nucleophile
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better
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SN1 rxns DO/DO NOT require strong nucleophiles
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DO NOT
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carbocation acts as a strong electrophile
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know SN1
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rate of SN1 rxns
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slowest Rate determining step is formation of the carbocation
-dependent only on the concentration of the original molecule (1st order rxn) |
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the rate of the rxn for SN1 does/does not depend on the concentration/nature of the nucleophile
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does not
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rate of rxn for SN1 can be increased by anything that accelerates the formation of the....
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carbocation
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structural factors: highly substituted alkyl halides are more/less stable
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more stable carbocation
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solvent effects: highly polar solvents better with ions
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know
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leaving group: weak bases dissociate more easily; make better leaving groups
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increase carbocation formation
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SN2 rxns
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bimolecular substitution
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SN2 involves a strong/weak nucleophile
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strong
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SN2 occurs in ___ step(s)
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one
-this one step is the rate determining step |
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bimolecular means it involves 2 molecules
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know
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nucleophile attacks and displaces leaving group...aka...
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backside attack
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SN2 requires less hindered substrates
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methyl > 1 > 2 > 3
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transition state of SN2 =
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trigonal bipyramidal
SP2 |
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rate of SN2
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rate determining step involves substrate and nucleophile
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concentrations of both substrate and nucleophile will have an impact on the rate of the rxn
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know
-increasing concentrations increases likelihood of the rxn |
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SN2 = 2nd order kinetics
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know
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SN1 stereochemistry
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carbocation intermediate = 120degree planar
-sp2 hybrid -nucleophile can attack either top/bottom |
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since it can attack from top/bottom...it will result in 2 products....
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loss of stereochemistry for SN1
-if original compound was optically active... the product will be a racemic mixture and not be optically active :( |
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SN2 stereochemistry
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inversion of direction (r-->s) etc
if chiral... |
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it will only lead to an inversion if the ranking of the nucleophile and leaving group are the SAME...
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if different...no necessarily so!
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