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Organic Chemistry > Organic Test 2 > Flashcards

Organic Test 2 Flashcards

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1
Q

Activation energy (Ea)

A

the minimum kinetic energy the molecules must have to overcome the repulsions between their electron clouds when they collide.

Represents the energy difference between the reactands and the transition state

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2
Q

Transition State

A

The highest-energy state in a molecular collision that leads to reaction.

This configuration is the transition between the reactants and products, and the molecules can either go on to products or return to reactants.

A transition state is unstable and cannot be isolated

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3
Q

Intermediate

A

A species that exists for some finite length of time, even if it is very short. An intermediate has at least some stability.

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4
Q

Catalyst

A

Creates a transition state of lower energy, thereby lowering the activation energy and increasing the reaction rate.

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5
Q

Rate-Limiting Step (Rate-determining step)

A

Controls the overall reaction rate. The highest energy step of a multistep reaction is the “bottleneck,” and it determines the overall reaction rate.

If we have the reaction-energy diagram, the highest point in the energy diagram is the transition state with the highest energy– generally the transition state for the rate-limiting step.

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6
Q

Halogenation reaction with methane:

Cl vs. Br vs. F vs I

A

fluorine reacts explosively with methane, and chlorine reacts at a moderate rate. A mixture of bromine and methane must be heated to react, and iodine does not react at all.

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7
Q

Order of stability of Radicals

A
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8
Q

Hammond postulate

A
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9
Q

Selectivity of Cl, Br, and F in free-radical halogenation

A
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10
Q

Different types of reactive intermediates

A
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11
Q

What is the order of stability of carbocations?

A
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12
Q

What is the order of stability of carbanions?

A
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13
Q

What are carbenes?

What is the stability of carbenes?

A

Carbenes are uncharged reactive intermediates containing a divalent carbon atom. The simplest carbene has the formula :CH2 and is called methylene, just as a - CH2 - group in a molecule is called a methylene group.

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14
Q

What are the properties and order of stabilities of the reactive intermediates?

A
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15
Q

Constitutional Isomers

A

(structural isomers) differ in their bonding sequence, their atoms are connected differently.

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16
Q

Stereoisomers

A

have the same bonding sequence, but they differ in the orientation of their atoms in space

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17
Q

Chiral

A

Non superimposable mirror image

(“handed”) different from mirror image, having an enantiomer

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18
Q

Achiral

A

(not chiral)

superimposable mirror image

(“not handed”) identical with its mirror image; not chiral

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19
Q

enantiomers

A

mirror-image isomers; pairs of compounds that are nonsuperimposable mirror images

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20
Q

The Cahn-Ingold-Prelog convention

A
  1. Assign a “priority” to each group bonded to the asymmetric carbon. Atoms with higher atomic numbers receive higher priorities
    a) Atoms with higher atomic numbers receive higher priorities
    b) In case of ties, use the next atoms along the chain of each group as tiebreakers
    c) Treat double and triple bonds as if each were a bond to a separate atom.
  2. Put the fourth-priority group away from you. Draw an arrow from the first-priority group to the 2nd. Clockwise= R, counter-clockwise=S.
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21
Q

Optical Activity

A

Rotation of the plane of polarized light

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22
Q

The difference between two enantiomers

A
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23
Q

Using IUPAC notation, what denotes the direction of rotation of plane-polarized light

A
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24
Q

Racemic mixture

A

Optically inactive, a solution of equal amounts of two enantiomers

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25
Chiral molecules with a plane of symmetry
Some molecules are so bulky or so highly strained that they cannot easily convert from one chiral conformation to the mirror-image conformation. Ex-Three conformations of a biphenyl. This biphenyl cannot pass through its symmetric conformation because there is too much crowding of the iodine and bromine atoms. The molecule is "locked" into one of the two chiral, enantiomeric, staggered conformations . Allenes: Allenes are compounds that contain the C=C=C unit
26
Fischer Projections
27
Rules for Fischer Projections
Rules for Fischer projections: 1 . Interchanging any two groups an odd number of times (once, three times, etc.) makes an enantiomer. Interchanging any two groups an even number of times (e.g. twice) returns to the original stereoisomer. 2. Rotating the structure by 90 makes the enantiomer. Rotating by 180 returns to the original stereoisomer.
28
Rules for Drawing Fischer Projections `
The final rule for drawing Fischer projections helps to ensure that we do not rotate the drawing by 90°. This rule is that the carbon chain is drawn along the vertical line of the Fischer projection, usually with the IUPAC numbering from top to bottom. In most cases, this numbering places the most highly oxidized carbon substituent at the top. For example, to represent (R)-1,2-propanediol with a Fischer projection, we should arrange the three carbon atoms along the vertical. C1 is placed at the top, and C3 at the bottom.
29
Meso Compound
An achiral compound that has chirality centers (usually asymmetric carbons).
30
Diastereomers
stereoisomers that are not mirror images
31
absolute configuration
The detailed stereochemical picture of a molecule, including how the atoms are arranged in space. Alternatively, the (R) or (S) configuration at each chirality center.
32
Relative Configuration
The experimentally determined relation- ship between the configurations of two molecules, even though we may not know the absolute configuration of either.
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34
alkyl halide
has a halogen atom bonded to one of the sp3 hybrid carbon atoms of an alkyl group.
35
vinyl halide
36
aryl halide
has a halogen atom bonded to one of the sp2 hybrid carbon atoms of an aromatic ring.
37
What are primary, secondary, and tertiary halides?
38
Geminal Dihalide
(Latin, geminus "twin") has the two halogen atoms bonded to the same carbon atom.
39
Vicinal Dihalide
has the two halogens bonded to adjacent carbon atoms.
40
Carbon-halogen bond length order
Bond length increases as the halogen gets larger I\>Br\>Cl\>F
41
What is allylyic bromination?
An allylic position is a carbon atom next to a carbon-carbon double bond. Allylic intermediates (cations, radicals, and anions) are stabilized by resonance with the double bond, allow- ing the charge or radical to be delocalized.
42
allylic shift
The other product results from reaction at the carbon atom that bears the rad- ical in the second resonance form of the allylic radical. This second compound is said to be the product of an allylic shift.
43
good leaving group
The halogen atom can leave with its bonding pair of electrons to form a stable halide ion; we say that a halide is a good leaving group.
44
nucleophilic substitution
a nucleophile (Nu:) replaces a leaving group (X-) from a carbon atom, using its lone pair of electrons to form a new bond to the carbon atom.
45
elimination
In an elimination, both the halide ion and another substituent are lost. A new pi bond is formed.
46
substrate
the compound that is attacked by the reagent.
47
SN2 Mechanism Key Facts: steps reaction coordinate etc.
Stands for Bimolecular Nucleophilic Substitution Concerted reaction, takes place in a single step One transition state (no intermediates)
48
SN2 Mechanism
49
What is the order of reactivity for substrates in an SN2 reaction? (greatest to least)
50
What are halogen exchange reactions?
Reactions in which halides can be converted to other halides. Iodide is a good nucleophile. Many alkyl chlorides react with sodium iodide to give alkyl iodides. Alkyl flourides are difficult to synthesize directly. Often 18-crown-6 is used with an aprotic solvent to enhance the normally weak nucleophilicity.
51
Trends in nucleophilicity
52
Steric effects on nucleophilicity
An ion or molecule must get close to a Carbon atom to attack it.
53
Protic Solvent
A solvent with acidic protons (H-O or H-N bond)
54
Nucleophilicity in Protic Solvents
Generally increases down a column in the periodic table, as long as we compare similar species with similar charges.
55
Nucleophilicity in Aprotic Solvents
An anion is more reactive in aprotic solvent The relatively weak solvating ability of aprotic solvents is also a disadvan- tage: Most polar, ionic reagents are insoluble in simple aprotic solvents such as alkanes.
56
Nucleophilicity in polar aprotic solvent
have strong dipole moments to enhance solubility, yet they have no O-H or N-H groups to form hydrogen bonds with anions.
57
Good leaving group for an SN2 reaction should be: (3 things)
1. electron withdrawing, to polarize the carbon atom 2. stable (not a strong base) once it has left, and (conjugate bases of strong acids) 3. polarizable, to stabilize the transition state.
58
Steric effects on the substrate in SN2
59
Stereochemistry of the SN2 Reaction
100% inversion of stereochemistry. (back-attack)
60
SN1 Mechanism Key Facts:
Unimolecular Nucleophilic Substitution Unimolecular means there is only one molecule involved in the TS of the rate-limiting step. Multi-step reaction
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Nucleophilicty in an SN1 Mechanism
SN1 Reaction involves a weak nucleophile
62
SN1 Reaction mechanism
63
solvolysis
64
Reaction Coordinate SN1 vs SN2
65
Order of reactivity in an SN1 reaction
66
Which halide types do not undergo SN1 or SN2?
67
Leaving Group in an SN1 Reaction
The leaving group should be a weak base, very stable after it leaves with the pair of electrons that bonded it to carbon. Similar leaving groups are effective in SN1 and SN2 processes
68
Solvent effects in an SN1 Reaction
SN1 reactions go more readily in polar solvents. BEST in protic solvents.
69
Stereochemistry in an SN1 reaction
50% inversion, 50% retention -Back and front attack
70
Rearrangements
(Not seen in SN2) Carbocations frequently undergo structural changes, called rearrangements, to form more stable ions. A rearrangement may occur after a carbocation has formed or it may occur as the leaving group is leaving. Rearrangements are not seen in SN2 reactions, where no carbocation is formed and the one-step mechanism allows no opportunity for rearrangement.
71
hydride shift
the movement of a hydrogen atom with its bonding pair of electrons. A hydride shift is rep- resented by the symbol ~ H. In this case, the hydride shift converts the initially formed secondary carbocation to a more stable tertiary carbocation. Attack by the solvent gives the rearranged product.
72
methyl shift
the migration of a methyl group together with its pair of electrons. Without rearrangement, ionization of neopentyl bromide would give a very unstable primary carbocation.
73
Primary and methyl carbocations
Highly unstable and will rarely ionize into carbocations in solution. If a primary halide ionizes, it will likely ionize with rearrangement
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Effect of the nucleophile SN1 v SN2
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Effect of the substrate SN1 vs SN2
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Effect of the solvent SN1 vs SN2
77
Kinetics SN1 vs SN2
78
Stereochemistry SN1 VS SN2
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Rearrangements SN1 vs SN2
80
Order of reactivity in an E1 mechanism
80
Zaitsev's Rule
81
E2
Most strong nucleophiles are also strong bases, and elimination commonly results when a strong base/nucleophile is used with a poor SN2 substrate such as a 3° or hindered 2° alkyl halide. Like the SN2 reaction, the E2 is a concerted reaction in which bonds break and new bonds form at the same time, in a single step.
82
E2 Mechanism
83
Order of reactivity for E2
84
When may Zaitsev's Rule not apply?
85
Effect of the base E1 vs E2
The nature of the base is the single most important factor in determining whether an elimination will go by the El or E2 mechanism. If a strong base is present, the rate of the bimolecular reaction will be greater than the rate of ion- ization, and the E2 reaction will predominate (perhaps accompanied by the SN2). If no strong base is present, then a good solvent makes a unimolecular ionization likely. Subsequent loss of a proton to a weak base (such as the solvent) leads to elim- ination. Under these conditions, the El reaction usually predominates, usually accom- panied by the SN1.
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Effect of the Solvent E1 vs E2
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Effect of the substrate E1 vs E2
88
Nomenclature of Alkenes
- ending is changed from ane to ene - The chain is numbered starting from the end closest to the double bond, and the double bond is given the lower number of its two double-bonded carbon atoms. - two double bonds, diene. three double bonds, triene. four double bonds, tetraene. - try to have as many double bounds in the main chain as possible
89
E-Z Nomenclature
Assign first and second priorities on each side of the double bond. If the two first priority items are on the same side of the double bond, it's Z. If the two first priority items are on opposite sides of the double bond, its E.
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Stability of Alkenes
91
Which solvents proceed by E2/ SN2?
SN2: SH- RS- CN- I- E2: -OH RO- -NH2 R-C=C- (trip bond), bulky bases
92
What is the reagant used in allylic bromination?
NBS and light
93
What reagant is necessary to add an Iodide group to an alkane?
KaI and acetone
94
H2SO4 Reagant and heat results in
E1 mechanism. Also will help eliminate an OH group with a double bond
95
transdiaxial
96
What does E2 elimination on cyclohexanes require?
97
Most stable arrangement in an E2 involving single bond (rotation)
98
NO SN2 on what?
SP2!!!
99
H20 and heat result in...
substiution of OH from Br
100
How do you create an SH group from a Br group?
Nash!
101
2n rule?
The 2n rule suggests we should look for a maximum of 2n stereoisomers. We may not always find 2n isomers, especially when two of the asymmetric carbon atoms have identical substituents.

Organic Chemistry (8 decks)

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