diy ochem exam

polar reaction

movement of electron pairs

radical reactions

movement of single electrons

electrophile

accepts electron pairs, can be positive or negative

nucleophile

donates an electron pair, can be negative or neutral

polar reactions

move two electrons together, involve nucleophiles and electrophiles

radical reactions

move one electron at a time, involves radicals with unpaired electrons

heterolytic bond cleavage

both bonding electrons go to one fragment and produce ions

homolytic bond cleavage

each fragment receives one bonding electron, producing radicals

curved arrows show

where electrons move during a reaction

electrons moving between atoms can

change their formal charges

radical reaction

species containing an unpaired electron

radical reactions are often formed when

a bond breaks homolytically

free energy/ΔG∘

describes whether a reaction is energetically favorable under stated conditions

negative ΔG∘

exergonic, products are favored

positive ΔG∘

endergonic, reactants are favored

equilibrium constant/Keq

tells you which side is favored at equilibrium

Keq > 1

products are favored

Keq < 1

reactants are favored

ΔG∘ in relation to Keq

ΔG∘ = −RT lnKeq

negative ΔG∘ corresponds to

Keq greater than 1

thermodynamics/gibbs

is the reaction energetically favorable?

kinetics

how fast does the reaction happen?

enthalpy/ΔH∘

heat change of a reaction

negative ΔH∘

exothermic

positive ΔH∘

endothermic

breaking bonds requires energy just as

forming bonds releases energy

entropy/ΔS∘

dispersla of energy aand the number if osssuble molecular arrangments

posiive ΔS∘

entropy increases

negative ΔS∘

entropy decreases

ΔG∘ in relation to ΔH∘ and TΔS∘

ΔG∘=ΔH∘−TΔS∘

bond disassociation energy/D

energy needed to break a specific bond homolytically in the gas phase

calculating a reaction’s approximate enthalpy change

ΔH∘ ≈ energy to break bonds - energy from forming bonds

energy diagrams show

how energy changes as a reaction proceeds

activation energy/ΔG‡

energy barrier the reaction must overcome

transition state

high energy arrangement at top of an energy barrier

one step reaction

one transition state

two step reaction

two transition states with an intermediate between them

for multiple step reactions, overall activation energy is

measured from reactants to highest transition state along the way

lower ΔG‡

faster reaction

intermediate

exists briefly during a reaction and appears at a local energy minimum

transition state

exists only at the instant of crossing an energy barrier, appears at an energy maximum, and cannot be isolated as an orginary compound

example of intermediate

carbocation formed during alkene addition

transition state is

not a stable molecule; represents arrangement of atoms and electrons as bonds are breaking and forming

catalysts increase the rate of a reaction by

providing a pathway with lower activation energy

3 key facts about catalysys

they do not change the overall ΔG‡, they do not change the equilibrium constant, they help the reaction reach equilibrium faster

enzymes are

biological catalysts

biological reactions generally

occur in aqueous environments, operate at the organism’s temperature, have high substrate and reaction specificity

alkene

contains C=C

alkene carbons are

sp² hybridized and have a roughly trigonal planar arrangment

in order to freely rotate, the ______ in a double bond would need to be broken

pi bond

degree of unsaturation

tells how many rings and/or pi bonds are present in

1 degree of unsaturation

1 ring or 1 double bond

2 degrees of unsaturation

2 double bonds, 2 rings, 1 ring + 1 double bond, 1 triple bond

every ring or double bond removes

2H compared with the corresponding alkane

steps to name alkenes

find longest carbon chain containing double bond and change -ane to -ene, start numbering from end closest to double bond, add substiuents

e/z is used itstead of cis/trans for

more complicated alkenes

steps for e/z naming

determine priority on each alkene carbon, use cahn-ingold-prelog rules, look at 2 highest priotity groups

alkenes become more stable as

they become more substituted because more alkyl groups attached to double bond = greater stability

trans tends to be more stable than cis because

cis isomers cause more steric strain

alkenes are electron rich because

of their pi bond

alkene pi bond attacks

electrophile

markovnikovs rule

when HX adds to an unsymmetrical alkene, H goes to carbon that already has more H’s/more alkyl substitutes

carbocation

carbon with a positive charge, sp² hybridization, planar, empty p orbitals

why are tertiary carbons more stable

they have more alkyl groups which leads to more electron donation and positive charge being stabilized

hyperconjugation

nearby C-H bonds can interact with the empty p orbital

hammond postulate

transition state resembles nearby stable species

hammond postulate for endergonic step

transition state resembles the product

hammond postulate for exergonic step

transition state resembles the reactant

more stable transition state leads to

lower activation energy and faster reaction

two carbocation rearrangments

hydride and alkyl

hydride shift

hydrogen and its bonding electron pair move to an adjacent carbon

alkyl shift

alkyl group and its bonding electron pair move to an adjacent carbon

4 steps to recognize rearrangement problem

add H according to the pathway that gives the more stable carbocation, look at carbocation and ask: can the positive charge become more stable if something moves from the neighobring carbon, check for hydride shift and alkyl shift, if rearrangement produces a more stable carbocation expect the rearrangement

alkene addition

atoms or groups are added across the double bond

alkene elimination

groups are removed to form a double bond

oxidation reaction

often adds bonds between carbon and oxygen or breaks the double bond

alkene reduction

often adds hydrogen to carbon

alkene polymerization

many alkene molecules join to form a long chain

in many addition reactions

the double bond breaks and the molecule splits into fragments

elimination reactions

form alkenes by removing groups from negihboirng atoms

addition to an alkene

adds groups across C=C

elimination of alkene

removes groups to create C=C

elimination reactions can produce

mixtures of alkenes

hydration

adding water across a double bond to form an alcohik

oxymercuration puts

OH on the more substitued carbon

hydroboration puts

OH on less substituted carbon