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final exam Flashcards

1
Q

double bond consists of

A

sigma bond and pi bond

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2
Q

triple bond consists of

A

2 pi bonds, 1 sigman bond

each carbon is sp hybridized with a linear geometry (180 degrees)

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3
Q

acyclic alkene structure

A

CnH2n

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4
Q

unsaturated hydrocarbons

A

alkenes and alkynes

have fewer than the max # of H atoms per carbon

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5
Q

cycloalkanes have the general formula:

A

CnH2n

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6
Q

alkynes have the formula:

A

CnH 2n-2

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7
Q

one degree of unsaturation:

A

each pi bond or ring removes two H atoms from a molecule

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8
Q

how to calculate degrees of saturation

A

compare actual number of H atoms in a compound to the max number of H atoms possible for the number of carbons present (if molecular were a straight chain alkane)

this gives total number of rings and/or pi bonds in a molecule

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9
Q

formulaic changes needed to be made to find degrees of unsaturation for heteroatoms

A
  • for halogens: replace each ‘halogen’ with ‘H’
  • for O or S: delete each oxy and sulfur

for nitrogen: for each ‘N’, delete 1 ‘H’ and 1 ‘N’

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10
Q

C2H4Cl2 =

A

C2H4H2 =

C2H6 =

CnH2n+2

NO degrees of unsaturated
substituted alkane

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11
Q

C5H6OCl2 =

A

C5H8O =

C5H8 =

CnH2n-2

TWO degrees of unsat; 2 pi bonds, 2 rings or (1 ring and 1 pi bond)

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12
Q

naming alkenes

A

1) find longest chain containing both carbon atoms of double bond
2) change -ane to -ene
3) number chain, use 1st number assigned to C=C
4) name and number substituents

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13
Q

E isomer has

A

two higher priority groups on OPP sides

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14
Q

Z isomer has

A

two higher priority groups on SAME side

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15
Q

what denotes priorities for substituents in an alkene?

A

E or Z

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16
Q

how to name an alkyne

A

1) change -ane to -yne

2) choose longest continuous chain that contains both atoms of the triple bond

3) number the chain to give the triple bond the LOWER number

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17
Q

diynes:

A

compounds w/ two triple bonds

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18
Q

triynes:

A

compounds w/ three triple bonds

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19
Q

compounds w/ both a double and triple bond is a

A

enyne

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20
Q

in alkynes, the first site of unsaturation is given

A

the lower number

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21
Q

two carbon alkyl group derived from acetylene is called an

A

ethynyl group

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22
Q

physical properties of alkenes and alkynes resemble

A

those of hydrocarbons of similar shape and molecular weight

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23
Q

alkenes and alkynes have

A

low melting pts and boiling points

melting and boiling pt increase number of carbons increases

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24
Q

alkenes and alkynes are not soluble in

A

water

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25
Q

alkenes and alkynes are soluble in

A

organic solvents

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26
Q

cis and trans isomeric alkenes often have

A

somewhat diff physical properties as consequence of diff dipoles

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27
Q

more polar isomers have

A

higher boiling points

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28
Q

cis alkene is more (x) than a trans alkene

A

POLAR

making bp slightly higher and more soluble in polar solvents

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29
Q

increasing number of double bonds =

A

lower melting point

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30
Q

symmetry effect:

A

cis bonds make these molecules less symmetrical and have poorer packing which affects melting pts

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31
Q

alkenes can be prepared from

A

alkyl halides and alcohols via elimination rxns

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32
Q

in said elimination rxns that are stereoselective/regioselective, most stable (x) is the

A

ALKENE

is the major product

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33
Q

alkynes are prepared by

A

eliminiation rxns

strong base removes 2 equivalents of HX from a vicinal/geminal dihalide = alkyne through successive e2 rxns

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34
Q

carbon atoms of a double bond are both

A

trigonal planar

elements of X and Y can be added to them from the same side/opp sides

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35
Q

syn addition takes place

A

when both X and Y are added from the SAME side

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36
Q

anti addition takes place

A

when X and Y are added from OPP sides

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37
Q

alkynes undergo

A

addition rxns b/c they contain relatively weak pi bonds like alkenes

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38
Q

in an addition rxns, two sequential rxns can take place:

A

addition of one equivalent of reagent forms an alkene

then add a second equivalent of reagent to = a product having 4 new bonds

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39
Q

alkenes’ oxidizing reactions

A

epoxidation, dihydroxylation, oxidative cleavage

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40
Q

alkynes oxidizing rxn

A

oxidative cleavage

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41
Q

hydrohalogenation is

A

the electrophilic addition of HX

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42
Q

to draw products of hydrohalogenation

A

1) locate C-C double bond

2) identify sigma bond of reagent that breaks H-X bond in hydrohalogenation

3) break pi bond of alkene and sigma bond of reagent

4) form two new sigma bonds to the C atoms of the double bond

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43
Q

mechanism of electrophilic addition consists of

A

two successive lewis acid-base rxns

step 1: alkene is the lewis base that donates an electron pair to H-BR (lewis acid)

step 2: Br- (lewis base) that donates an electron pair to the carbocation (lewis acid)

44
Q

hydrohalogenation (markovnikov’s rule)

A

applies with an unsymmetrical alkene, HX can add to the double bond to = 2 constitutional isomers (only one is formed)

45
Q

markovnikov’s rule states that

A

in the addition of HX to an unsymmetrical alkene, the H atom adds to the less substituted carbon atom (carbon w/ greater number of H atoms)

46
Q

basis of markovnikov’s rule is the

A

formation of a carbocation in the rate-determining step of the mechanism

47
Q

in the addition of HX to an unsymmetrical alkene,

A

the H atom is added to the less substituted carbon to form the more stable/more substituted carbocation

48
Q

trigonal planar atoms reacts w/ reagents from

A

two directions w/ equal probability

49
Q

achiral starting materials yield

A

achiral products or new stereogenic centers are formed

50
Q

halogenation (rxn stereochemistry)

A

slide 29

51
Q

hydration (electrophilic addition of water) –>

A

addition of water to an alkene to form an alcohol

52
Q

in a hydration rxn, alcohols add to alkenes –>

A

forming ethers by the same mechanism

i.e. addition of CH3OH to 2-methylpropene forms tert-butyl methyl ether (MTBE)

53
Q

three consequences to the formation of carbocation intermediates

A

1) markovnikov’s rule holds

2) addition of H and OH occurs in both syn and anti fashion

3) carbocation rearrangements can occur

54
Q

internal alkynes undergo

A

hydration w/ concentrated acid

55
Q

terminal alkynes require presence of

A

HgSO4 to yield methyl ketones by addition of water

56
Q

tautomers:

A

enol form
keto form

57
Q

tautomers are

A

constitutional isomers that differ in the location of a double bond and H atom

two tautomers are in equilibrium w/ each other

58
Q

tautomerization =

A

equilibrium w/ isomers

59
Q

enol tautomer

A

has an OH group bonded to C=C

60
Q

keto tautomer has

A

C=O and additional C-H bond

favored by equilibrium (C=O is much stronger than C=C)

61
Q

hydroboration-oxidation is a

A

two step rxn sequence that converts an alkene into an alcohol

62
Q

first step in hydroboration-oxidation

A

addition of elements (H and BH2 to pi bond of alkene) = intermediate alkylborane

63
Q

proposed mechanism of hydroboration-oxidation involves

A

concerted addition of H and BH2 from same side of planar double bond

pi bond and h-bh2 bond are broken as two new sigma bonds are forms

trans state is four centered

64
Q

with unsymmetrical alkenes,

A

boron atom bonds to the less substituted carbon atom

65
Q

alkylboranes are oxidized without isolation w/

A

hydrogen peroxide

66
Q

oxidation replaces the C-B bond w

A

C–O bond

forming new OH group w/ retention of configuration

67
Q

syn addition of H and OH to a double bond =

A

anti Markovnikov

68
Q

hydroboration-oxidation is a

A

two step rxn sequence that converts an alkyne to a carbonyl compound

69
Q

addition of borane =

A

organoborane

70
Q

oxidation with basic H2O2 =

A

an enol

71
Q

tautomerization of the enol =

A

carbonyl compound

72
Q

results of hydroboration-oxidation

A

addition of H2n to a 3x bond

73
Q

hydroboration of an internal alkyne forms

A

a ketone

74
Q

hydroboration of a terminal alkyne adds

A

BH2 to the less substituted, terminal carbon

after oxidation to the enol, tautomerization = aldehyde, carbonyl compounds

75
Q

internal alkyne forms

A

ketone

76
Q

terminal alkyne forms

A

aldehyde

77
Q

sp hybridized C-H bonds are more acidic than

A

sp2 and sp3 hybridized C-H bonds

terminal alkynes are deprotanated w/ strong base

resulting ion = acetylide ion

78
Q

acetylide anions react w/

A

unhindered alkyl halides = products of nucleophilic substitution

79
Q

acetylides are

A

strong nucs

SN2 favored

fastest w/ CH3X and primary alkyl halides

80
Q

steric hindrance around LG causes

A

2ndary and tertiary alkyl halides to undergo E2 mechanism

81
Q

nuc substitution w/ acetylide anions form

A

new c-c bonds in high yield only w/ unhindered CH3X and primary alkyl halides

82
Q

steric hindrance prevents

A

SN2 rxn

83
Q

acetylide anions are

A

strong nucs that open epoxide rings by SN2

84
Q

backside attacks occurs at the less

A

substituted end of the epioxide

85
Q

markovnikov’s rule:

A

in addition of HX to an unsymmetrical alkene, the H atom bonds to the less substituted carbon

86
Q

syn addition:

A

reagent is added from the same side

HYDROBORATION

87
Q

anti addition

A

X and X are added from opp sides

HALOGENATION

88
Q

halohydrin formation also involves

A

anti addition where X and OH are added from opp sides

89
Q

syn and anti addition occur when

A

carbocations are formed

H and OH are added from same and opp sides

90
Q

hydrohalogenation also involves both

A

syn and anti addition

H and X are added from the same and opp sides

91
Q

ALL reactions of alkenes involve addition

A

weak pi bond is broken and two new sigma bonds are formed

92
Q

calculate degrees of unsaturation

A

1) calculate max number of Hs = 2n +2

2) subtract actual # from max number and divide by 2

93
Q

assign priorities in naming an alkene

A

E or Z (assign to substituents on each end of C=C)

94
Q

E-isomer

A

two higher priority groups on opp sides

95
Q

Z-isomer

A

two high priority groups on same side

96
Q

when drawing products of an addition rxn, in an unsymmetrical alkene:

A

determine the regioselectivity

97
Q

you can use reagents to identify the two groups added to

A

the C=C

98
Q

carbocation intermediates gives

A

syn and anti products

99
Q

concerted addition gives

A

syn products

100
Q

converting an alkene to an alkyne

A

1) add halogen (X2) to an alkene

2) eliminate two equivalents of HX by treatment w/ strong base

101
Q

draw product of an addition rxn

A

1) use reagents to ID the two groups added to C (3x bond) C

2) in an unsymmetrical alkyne, determine regioselectivity

3) draw products by breaking pi bonds and adding reagents

102
Q

convert enol to keto tautomer in acid

A

1) locate C=C and H atom on OH group

2) add proton to C=C, draw two resonance structures

3) remove proton from OH group

103
Q

convert keto tautomer to an enol in acid

A

1) locate C=o and H on alpha-carbon

2) add proton to C=O and draw two resonance structures

3) remove proton from alpha-carbon and draw C=C

104
Q

comparing products of hydration of alkyne

A

1) use reagents to determine regioselectivity and add h20 to form enol

2) convert enol to its keto tautomer

105
Q

comparing rxns of acetylide anions

A

1) classify alkyl halide

2) draw products

106
Q

devise synthesis

A

1) compare carbon skeletons and functional groups

2) work backwards

3) work forwards

4) complete synthesis

orgo (6 decks)

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