final exam Flashcards
1
Q
double bond consists of
A
sigma bond and pi bond
2
Q
triple bond consists of
A
2 pi bonds, 1 sigman bond
each carbon is sp hybridized with a linear geometry (180 degrees)
3
Q
acyclic alkene structure
A
CnH2n
4
Q
unsaturated hydrocarbons
A
alkenes and alkynes
have fewer than the max # of H atoms per carbon
5
Q
cycloalkanes have the general formula:
A
CnH2n
6
Q
alkynes have the formula:
A
CnH 2n-2
7
Q
one degree of unsaturation:
A
each pi bond or ring removes two H atoms from a molecule
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
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’
10
Q
C2H4Cl2 =
A
C2H4H2 =
C2H6 =
CnH2n+2
NO degrees of unsaturated
substituted alkane
11
Q
C5H6OCl2 =
A
C5H8O =
C5H8 =
CnH2n-2
TWO degrees of unsat; 2 pi bonds, 2 rings or (1 ring and 1 pi bond)
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
13
Q
E isomer has
A
two higher priority groups on OPP sides
14
Q
Z isomer has
A
two higher priority groups on SAME side
15
Q
what denotes priorities for substituents in an alkene?
A
E or Z
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
17
Q
diynes:
A
compounds w/ two triple bonds
18
Q
triynes:
A
compounds w/ three triple bonds
19
Q
compounds w/ both a double and triple bond is a
A
enyne
20
Q
in alkynes, the first site of unsaturation is given
A
the lower number
21
Q
two carbon alkyl group derived from acetylene is called an
A
ethynyl group
22
Q
physical properties of alkenes and alkynes resemble
A
those of hydrocarbons of similar shape and molecular weight
23
Q
alkenes and alkynes have
A
low melting pts and boiling points
melting and boiling pt increase number of carbons increases
24
Q
alkenes and alkynes are not soluble in
A
water
25
alkenes and alkynes are soluble in
organic solvents
26
cis and trans isomeric alkenes often have
somewhat diff physical properties as consequence of diff dipoles
27
more polar isomers have
higher boiling points
28
cis alkene is more (x) than a trans alkene
POLAR
making bp slightly higher and more soluble in polar solvents
29
increasing number of double bonds =
lower melting point
30
symmetry effect:
cis bonds make these molecules less symmetrical and have poorer packing which affects melting pts
31
alkenes can be prepared from
alkyl halides and alcohols via elimination rxns
32
in said elimination rxns that are stereoselective/regioselective, most stable (x) is the
ALKENE
is the major product
33
alkynes are prepared by
eliminiation rxns
strong base removes 2 equivalents of HX from a vicinal/geminal dihalide = alkyne through successive e2 rxns
34
carbon atoms of a double bond are both
trigonal planar
elements of X and Y can be added to them from the same side/opp sides
35
syn addition takes place
when both X and Y are added from the SAME side
36
anti addition takes place
when X and Y are added from OPP sides
37
alkynes undergo
addition rxns b/c they contain relatively weak pi bonds like alkenes
38
in an addition rxns, two sequential rxns can take place:
addition of one equivalent of reagent forms an alkene
then add a second equivalent of reagent to = a product having 4 new bonds
39
alkenes' oxidizing reactions
epoxidation, dihydroxylation, oxidative cleavage
40
alkynes oxidizing rxn
oxidative cleavage
41
hydrohalogenation is
the electrophilic addition of HX
42
to draw products of hydrohalogenation
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
43
mechanism of electrophilic addition consists of
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
hydrohalogenation (markovnikov's rule)
applies with an unsymmetrical alkene, HX can add to the double bond to = 2 constitutional isomers (only one is formed)
45
markovnikov's rule states that
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
basis of markovnikov's rule is the
formation of a carbocation in the rate-determining step of the mechanism
47
in the addition of HX to an unsymmetrical alkene,
the H atom is added to the less substituted carbon to form the more stable/more substituted carbocation
48
trigonal planar atoms reacts w/ reagents from
two directions w/ equal probability
49
achiral starting materials yield
achiral products or new stereogenic centers are formed
50
halogenation (rxn stereochemistry)
slide 29
51
hydration (electrophilic addition of water) -->
addition of water to an alkene to form an alcohol
52
in a hydration rxn, alcohols add to alkenes -->
forming ethers by the same mechanism
i.e. addition of CH3OH to 2-methylpropene forms tert-butyl methyl ether (MTBE)
53
three consequences to the formation of carbocation intermediates
1) markovnikov's rule holds
2) addition of H and OH occurs in both syn and anti fashion
3) carbocation rearrangements can occur
54
internal alkynes undergo
hydration w/ concentrated acid
55
terminal alkynes require presence of
HgSO4 to yield methyl ketones by addition of water
56
tautomers:
enol form
keto form
57
tautomers are
constitutional isomers that differ in the location of a double bond and H atom
two tautomers are in equilibrium w/ each other
58
tautomerization =
equilibrium w/ isomers
59
enol tautomer
has an OH group bonded to C=C
60
keto tautomer has
C=O and additional C-H bond
favored by equilibrium (C=O is much stronger than C=C)
61
hydroboration-oxidation is a
two step rxn sequence that converts an alkene into an alcohol
62
first step in hydroboration-oxidation
addition of elements (H and BH2 to pi bond of alkene) = intermediate alkylborane
63
proposed mechanism of hydroboration-oxidation involves
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
with unsymmetrical alkenes,
boron atom bonds to the less substituted carbon atom
65
alkylboranes are oxidized without isolation w/
hydrogen peroxide
66
oxidation replaces the C-B bond w
C--O bond
forming new OH group w/ retention of configuration
67
syn addition of H and OH to a double bond =
anti Markovnikov
68
hydroboration-oxidation is a
two step rxn sequence that converts an alkyne to a carbonyl compound
69
addition of borane =
organoborane
70
oxidation with basic H2O2 =
an enol
71
tautomerization of the enol =
carbonyl compound
72
results of hydroboration-oxidation
addition of H2n to a 3x bond
73
hydroboration of an internal alkyne forms
a ketone
74
hydroboration of a terminal alkyne adds
BH2 to the less substituted, terminal carbon
after oxidation to the enol, tautomerization = aldehyde, carbonyl compounds
75
internal alkyne forms
ketone
76
terminal alkyne forms
aldehyde
77
sp hybridized C-H bonds are more acidic than
sp2 and sp3 hybridized C-H bonds
terminal alkynes are deprotanated w/ strong base
resulting ion = acetylide ion
78
acetylide anions react w/
unhindered alkyl halides = products of nucleophilic substitution
79
acetylides are
strong nucs
SN2 favored
fastest w/ CH3X and primary alkyl halides
80
steric hindrance around LG causes
2ndary and tertiary alkyl halides to undergo E2 mechanism
81
nuc substitution w/ acetylide anions form
new c-c bonds in high yield only w/ unhindered CH3X and primary alkyl halides
82
steric hindrance prevents
SN2 rxn
83
acetylide anions are
strong nucs that open epoxide rings by SN2
84
backside attacks occurs at the less
substituted end of the epioxide
85
markovnikov's rule:
in addition of HX to an unsymmetrical alkene, the H atom bonds to the less substituted carbon
86
syn addition:
reagent is added from the same side
HYDROBORATION
87
anti addition
X and X are added from opp sides
HALOGENATION
88
halohydrin formation also involves
anti addition where X and OH are added from opp sides
89
syn and anti addition occur when
carbocations are formed
H and OH are added from same and opp sides
90
hydrohalogenation also involves both
syn and anti addition
H and X are added from the same and opp sides
91
ALL reactions of alkenes involve addition
weak pi bond is broken and two new sigma bonds are formed
92
calculate degrees of unsaturation
1) calculate max number of Hs = 2n +2
2) subtract actual # from max number and divide by 2
93
assign priorities in naming an alkene
E or Z (assign to substituents on each end of C=C)
94
E-isomer
two higher priority groups on opp sides
95
Z-isomer
two high priority groups on same side
96
when drawing products of an addition rxn, in an unsymmetrical alkene:
determine the regioselectivity
97
you can use reagents to identify the two groups added to
the C=C
98
carbocation intermediates gives
syn and anti products
99
concerted addition gives
syn products
100
converting an alkene to an alkyne
1) add halogen (X2) to an alkene
2) eliminate two equivalents of HX by treatment w/ strong base
101
draw product of an addition rxn
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
convert enol to keto tautomer in acid
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
convert keto tautomer to an enol in acid
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
comparing products of hydration of alkyne
1) use reagents to determine regioselectivity and add h20 to form enol
2) convert enol to its keto tautomer
105
comparing rxns of acetylide anions
1) classify alkyl halide
2) draw products
106
devise synthesis
1) compare carbon skeletons and functional groups
2) work backwards
3) work forwards
4) complete synthesis
