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Organic Chem Class > Exam 4 > Flashcards

Exam 4 Flashcards

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1
Q

alcohols will react with carbocations to give __________
- the carbocation can either come from protonation of an alkene or from “ionization” (loss of a good leaving group) of an alkyl halide or similar starting material

A

ethers

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2
Q

Reactive positively charged 3-membered ring intermediates such as halonium ions & mercurinium ions can also be attacked by _________ at the most substituted carbon (Markovnikov) to give “halo ethers” (for halonium ions) and ethers ( if we use Hg(OAc)2 and then add NaBH4).

A

alcohols
- note that alcohol is the solvent here

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3
Q

True or False:
carbocations and 3 membered ring intermediates] are both extremely reactive electrophiles

A

True

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4
Q

In order for a leaving group to leave, it has to be a fairly _____________

A

weak base

  • hydroxyl (HO-) and alkoxy (RO -) groups are both fairly strong bases, and therefore poor leaving groups
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5
Q

Hydroxyl groups (HO-) are poor leaving groups because they’re strong bases. However, if we protonate them by adding _________, then we get an oxonium ion (R-OH2+). The leaving group is now H2O – a weak base & a great leaving group

A

acid
- can now participate in Sn1 & E1 reactions as leaving group step

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6
Q

One of the keys to the reactions of alcohols is that the Conjugate Acid is a better ____________ and the Conjugate Base is a better _____________

A

leaving group
nucleophile

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7
Q

Hydroxyl groups in R–OH are poor nucleophiles because they’re neutral and the electron pair is held tightly to the oxygen. However, if we remove a proton by adding a __________ we then get an alkoxide ion (RO-) which has much higher electron density, and is a much better nucleophile (as well as being a strong base)

A

base
- can now participate in Sn2 as nucleophile & E2 as strong base

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8
Q

Alcohols are mild __________
Alcohols are also ___________

A

*acids
- typical aliphatic (i.e. “alkyl”) alcohols such as ethanol, isopropanol, and t-butanol have a pKa of about 16-18, making them slightly more acidic than water (pKa of 15.7).
*weak bases
- they can react with strong acids to give oxonium ions which have a pKa of about -2

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9
Q

Alcohols that are in conjugation with a pi bond or aromatic ring will be (___MORE or LESS___) acidic since the conjugate base is resonance-stabilized.

A

more
- resonance-stabilized & nearby electron-withdrawing groups will stabilize the negative charge of the conjugate base through inductive effects

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10
Q

The conjugate base of an alcohol is called an_____________

A

alkoxide

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11
Q

The conjugate acid of an alcohol is called an ____________

A

oxonium ion

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12
Q

True or False:
In acid-base reactions, the equilibrium will favor the direction where a stronger acid & stronger base produces a weaker acid and a weaker base

A

True
- look at Acid & Conjugate Acid (the weaker the higher the pKa thus the more favorable to form)

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13
Q

True or False:
The stronger the acid, the weaker the conjugate base.
The weaker the acid, the stronger the conjugate base.

A

True

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14
Q

Alcohols where the conjugate base is resonance stabilized will be (___MORE or LESS___) acidic

A

more
- any factor which stabilizes the conjugate base will increase acidity (resonance affects & inductive affects)

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15
Q

In Williamson Ether Synthesis, an alkyl halide (or sulfonate, such as a tosylate or mesylate) undergoes _____________ by an alkoxide to give an ether

A

SN2
- best results are obtained with primary alkyl halides or methyl halides. Tertiary alkyl halides give elimination instead of ethers. Secondary alkyl halides will give a mixture of elimination and substitution
- the alkoxide RO– can be those of methyl, primary, secondary, or tertiary alcohols
- the reaction is often run with a mixture of the alkoxide & its parent alcohol (e.g. NaOEt/EtOH or CH3ONa/CH3OH)

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16
Q

When an alkoxide and alkyl halide are present on the same molecule, an ________________ reaction may result to give a new ring. This works best for 5- and 6-membered rings

A

intramolecular

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17
Q

One of the simplest and most versatile ways for making an _________ is the SN2 reaction between an alkoxide (RO–, the conjugate base of an alcohol) and an alkyl halide

A

ether

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18
Q

Since it’s not always an option to use the alcohol as solvent, another option is to generate the alkoxide by using a strong base that will irreversibly deprotonate the alcohol. A popular choice is the _____________________ , which is the conjugate base of hydrogen gas (pKa about 35) and is also a poor nucleophile. Sodium hydride (NaH) or potassium hydride (KH) can be added to the ‘starting alcohol’ to generate the alkoxide. The hydrogen gas byproduct then bubbles out of solution into the atmosphere.

A

hydride ion H–

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19
Q

In the transition state of the SN2 there is a _______-coordinate geometry about the carbon with partial bonds to the nucleophile and to the leaving group.

A

five

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20
Q

With any ether there are two potential sites where a new C-O bond could be formed, which gives you _________ alkyl halide + alkoxide combinations to choose from.

A

two combinations
- just make sure the alkyl halide is methyl/primary since its SN2

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21
Q

Ethers can be made from the reaction of alkenes with ________ in the presence of alcohols

A

acid
- useful when making an ether through SN2 reaction would not work due to elimination

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22
Q

If we dissolve an tertiary alkyl halide in an alcohol solvent, eventually the leaving group will leave, forming the carbocation – which is then trapped by the alcohol solvent. Deprotonation of alcohol occurs, and we’re left with our _______

A

ether
- classic Sn1 reaction

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23
Q

If we start with an alkene in an alcohol solvent, and treat with a strong acid (H2SO4 or TsOH) the carbocation will be generated, which is then trapped by the alcohol solvent then deprotonates and we are left with our _________

A

ether

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24
Q

Expect _______________ if we form a secondary carbocation adjacent to a tertiary or quaternary carbon, expect a hydride or alkyl shift (respectively) that will result in a more stable carbocation.

A

rearrangements

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25
There is a way we can form ethers from alkenes in a way that doesn’t involve a carbocation intermediate and it is by __________________
oxymercuration-demercuration - involves dissolving the starting alkene in an alcohol solvent and adding a source of mercury(II) like Hg(OAc)2 . A “mercurinium” ion is formed, which is then attacked at the most substituted position by one of the molecules of alcohol solvent. After removal of a proton, we’re left with the product of “oxymercuration”. The mercury can then be removed by treatment with sodium borohydride (NaBH4).
26
Not a very common method but... Symmetrical ethers can be made from the acid-catalyzed dehydration of ___________ alcohols
primary - it’s limited to symmetrical ethers. If we try to make unsymmetrical ethers using this process, we will end up with mixtures that will need to be separated, giving us low yields of each individual component - temperature has to be carefully optimized, because there are lots of side reactions possible. For example the optimal temperature for the formation of diethyl ether is about 130-140 degrees C. Once the temperature gets to 150 degrees and above, elimination starts to compete, leading to the formation of ethylene gas
27
Synthesis Of Symmetrical Ethers via Acid-Catalyzed Dehydration of Alcohols 3 key steps ____________ ____________ ____________
First : 1 equivalent of alcohol is protonated to its conjugate acid – which has the good leaving group, OH2 (water, a weak base) Second : 2nd equivalent of the alcohol can now perform nucleophilic attack at carbon (SN2), leading to displacement of OH2 (water) and formation of a new C-O bond. This is an SN2 reaction Third : deprotonation of the product by another equivalent of solvent (or other weak base), resulting in our ether product.
28
Ethers do not undergo very many reactions. One key reaction of ethers is that they can undergo cleavage to _____________ in the presence of strong acids, such as HI, or strong Lewis acids such as boron tribromide (BBr3)
alcohols - these reactions involve protonation of the Ether oxygen, followed by either an SN1 or SN2 reaction pathway, depending on structure - ethers of primary alcohols tend to undergo cleavage via SN2 - ethers of tertiary alcohols tend to undergo cleavage via SN1) - the only significant reaction of ethers you need to know…. is how to break them!!!!!!
29
The first step in acidic cleavage of Ethers is __________ of oxygen
Protonation - the usual strong acid of choice is usually hydroiodic acid (HI). Not only is it powerful (pKa of –10), as we’ll see the iodide counter-ion plays a role as well.
30
For methyl and primary Ethers, the second step of Ether cleavage is __________
SN2 - If an excess (2 equiv or more) of HI is present, that alcohol product can be converted into an alkyl iodide product through 2 subsequent steps (protonation / SN2)
31
For tertiary Ethers, the second step of Ether cleavage is __________
SN1 - again, if excess HI is present then that alcohol product will be converted into an alkyl halide product
32
For secondary Ethers, the second step of Ether cleavage is _____________
mixture of SN2 & SN1 - can proceed through both pathways
33
True or False: Intramolecular cleavage of cyclic Ethers are fundamentally no different than those that form and break in intermolecular processes
True
34
Epoxides (oxiranes) are cyclic ethers that have unusually high reactivity due to ___________
ring strain
35
True or False: The 3-membered ring of epoxides can be opened under both acidic and basic conditions
True
36
______________ can be synthesized from alkenes via epoxidation with a peroxyacid like m-CPBA, or from halohydrins via treatment with base.
Epoxides - key point about epoxides is that they are not typical ethers, and thus deserve their own discussion
37
The interior bond angles of epoxides are about _________ Contrast that with the “ideal” bond angle of 109.5° for tetrahedral carbon.
60° - ring strain
38
Treating an epoxide with aqueous acid [H3O+] will open an epoxide to provide a ___________
1,2 -diol - vicinal diol or glycol
39
Unlike the vast majority of ethers, epoxides can also be cleaved with a ________
base - treatment of an epoxide such as ethylene oxide with sodium hydroxide in water similarly leads to formation of a vicinal diol
40
There’s 2 important ways to make epoxides from alkenes, one “direct” and one “indirect” 1. 2.
1. Reaction of Alkenes with a Peroxyacid 2. Treating Halohydrins with Base
41
A second "indirect" way to make epoxides is via a two-step process. Starting with an alkene, if one adds a halogen (Br2 or Cl2) and water as solvent, we make a species known as a _______________. Then treatment with strong base (such as NaH or NaOH) leads to deprotonation of the OH to give O- , which then displaces the adjacent halide via SN2 reaction to provide the resulting epoxide.
halohydrin - SN2 reactions proceed via a backside attack, leading to inversion of configuration. If the halohydrin is “locked” in position (as part of a ring, for example) and the alkoxide [O- ] cannot approach the backside of the C-Br bond, then the SN2 cannot occur and therefore an epoxide will not be formed (so OH & X must be trans)
42
Epoxides undergo reaction with nucleophiles under _________ conditions at the more substituted end of the epoxide
acidic - the epoxide oxygen is protonated first, making it a better leaving group, then the nucleophile tends to attack the more substituted carbon, which breaks the weakest C-O bond - think more positive on more substituted area thus it gets attacked - :Nuc & OH are trans to each other
43
Epoxides undergo reaction with nucleophiles under _________ conditions at the least substituted end of the epoxide
basic - similar to an SN2 attack at less substituted C - :Nuc & OH are trans to each other
44
Treating alcohols with HCl, HBr, or HI (“HX”) results in the formation of _____________
alkyl halides - Primary alcohols tend to proceed through an SN2 mechanism - Tertiary alcohols tend to proceed through an SN1 mechanism - Watch out for rearrangements where a secondary carbocation may be formed
45
Anytime a reaction proceeds through a carbocation intermediate, we need to be on the lookout to see if it is ___________ to a carbon which can generate a more stable carbocation through a shift of a C-H or C-C bond.
adjacent
46
A Good Rule of Thumb for Secondary Alcohols with HX is to assume _________
SN1 - treatment of secondary alcohols with HX leads to a mixture of products from SN1 and SN2 pathways
47
Another example of a favorable rearrangement is when a secondary carbocation is adjacent to an “allylic” or “benzylic” hydrogen. Rearrangement results in a secondary carbon which is more stabilized by ____________
resonance
48
Only alkyl alcohols (alcohols on _______ hybridized carbons) will undergo SN1 and SN2 reactions
sp3 hydridized
49
The OH group can be converted into a much better leaving group through conversion to a ______________ group such as p-toluenesulfonyl (“tosyl”, abbreviated Ts) or methanesulfonyl (“mesyl”, abbreviated Ms)
sulfonate - installing these groups does NOT affect the stereochemistry of the alcohol - very good leaving group
50
Alcohols can be converted to sulfonates by treating them with _______________
sulfonyl chlorides - tosyl chloride - mesyl chloride
51
Since acidifying -OH produces -OH2 a good leaving group that follows a pathway that could possibly result in C+ rearrangement and altered stereocenters, is there any way in changing -OH into a good leaving group without this occuring>
- tosylates - mesylates - A third option, less commonly seen in introductory courses, is the “triflate” (Tf) group, which replaces the three hydrogens on methanesulfonate with three fluorines. The conjugate acid, trifluoromethanesulfonic acid, is among the more acidic species known (pKa of -13!) and for this reason, triflate is a really “hot” leaving group when attached to alkyl groups: it likes to leave of its own accord! It’s more commonly used on aromatic alcohols and other species that don’t form carbocations as easily
52
Alcohols can be converted into alkyl halides with PBr3 or SOCl2: Inversion of configuration occurs but it is much more mild and predictable than using HBr or HCl to convert alcohols to alkyl halides since it avoids the possibility of _________________________
carbocation rearrangements
53
What is a key advantage of using PBr3 and SOCl2 over strong acids like HCl or HBr for converting alcohols to alkyl halides?
PBr3 and SOCl2 do not cause rearrangements
54
- What is the "activation" step in the reaction of an alcohol with PBr3? - What is the "activation" step in the reaction of an alcohol with SOCl2?
- The alcohol forms a bond with phosphorus, displacing a bromide ion and creating a good leaving group. - The alcohol attacks sulfur, displacing a chloride ion.
55
What is a key limitation of using tosylates or mesylates as leaving groups?
They cannot be used to make Grignard reagents - Grignard reagents typically made from Alkyl halides
56
What is the byproduct of the reaction of SOCl2 with alcohols?
Sulfur dioxide (SO2) and HCl.
57
Why is the removal of SO2 gas from the reaction vessel important in the reaction of SOCl2 with alcohols?
It makes the reaction irreversible and drives it to completion
58
What type of reaction mechanism is involved in the substitution step with PBr3 and SOCl2?
SN2 mechanism
59
If a strong acid such as H2SO4 or p-TsOH is used, the most likely result is _____________, since the conjugate bases of these acids (HSO4 – and TsO– ) are poor nucleophiles and unlikely to add to the carbocation.
elimination
60
What is the overall outcome of dehydrating an alcohol?
Formation of an alkene (removal of water).
61
The best conditions for promoting _______________ of alcohols is to use strong acid with a poorly nucleophilic conjugate base like H2SO4 (sulfuric acid), H3PO4 (phosphoric acid), and TsOH (p-toluenesulfonic acid) with heat
Elimination - HX acids result in substitution reaction
62
What is the typical outcome of treating tertiary alcohols with strong acids like H2SO4?
Elimination reaction to form an alkene - E1 mechanism (Elimination, Unimolecular).
63
What is Zaitsev's rule in the context of elimination reactions?
The most substituted alkene (the one with the most alkyl groups attached to the double bond) is the major product
64
What is a key concern in elimination reactions of secondary alcohols?
Carbocation rearrangements can occur.
65
What type of reaction can occur with primary alcohols and H2SO4?
Elimination to form an alkene (via E2 mechanism) and some ether formation.
66
What is a ring expansion?
A type of carbocation rearrangement where a strained ring expands to form a less strained ring.
67
Phosphorus oxychloride (POCl3) is a useful reagent for cleanly performing ______________ reactions on alcohols to form alkenes
elimination
68
What are the advantages of using POCl3 for alcohol elimination compared to strong acids?
One-step process (E2 mechanism) to give the more substituted double bond No carbocation rearrangements Milder reaction conditions
69
What is the role of the base pyridine in POCl3-mediated elimination?
Remove a H, leading to the formation of the alkene.
70
What is the first step in the reaction of an alcohol with POCl3?
Formation of a chlorophosphate ester, which is a good leaving group.
71
Does POCl3-mediated elimination work with primary, secondary, and tertiary alcohols?
Yes, it's effective for all types of alcohols.
72
What are the two main drawbacks of the two-step alcohol-to-alkene conversion (via alkyl halide)?
It requires two separate reactions, making it less efficient.
73
Why is the chlorophosphate ester a good leaving group?
The strong oxygen-phosphorus bond stabilizes the negative charge that develops on the oxygen upon leaving.
74
What is an oxidation reaction in organic chemistry?
An oxidation reaction forms a C-O bond and breaks a C-H bond on the same carbon.
75
What are "weak" oxidants?
Oxidants that oxidize primary alcohols to aldehydes & secondary alcohols to ketones.
76
What are "strong" oxidants?
Oxidants that oxidize primary alcohols to carboxylic acids & secondary alcohols to ketones.
77
Can tertiary alcohols be oxidized?
No, tertiary alcohols do not undergo oxidation with the reagents discussed.
78
Give three examples of "weak" oxidants.
PCC (pyridinium chlorochromate), DMP (Dess-Martin Periodinane), Collins Reagent (CrO3/pyridine) & Swern oxidation [(COCl)2, DMSO, NEt3]
79
Give two examples of "strong" oxidants.
KMnO4 (potassium permanganate), H2CrO4 (chromic acid).
80
What is the oxidation product of a primary alcohol?
An aldehyde (with weak oxidants) or a carboxylic acid (with strong oxidants).
81
What is the oxidation product of a secondary alcohol?
A ketone
82
What are the two categories of strong oxidants?
1. potassium permanganate (KMnO4) 2. Cr(VI) species - K2Cr2O7, Na2Cr2O7, Na2CrO4, K2CrO4, CrO3/H3O+, Jones reagent
83
Ladder
84
Why don't ketones oxidize further?
There's no C-H bond that can be broken to form a new C-O pi bond.
85
What happens to the oxidant in an oxidation reaction?
It is reduced
86
What happens to the substrate in an oxidation reaction?
It is oxidized
87
What is the oxidation state of carbon in CH4?
-4
88
What is the oxidation state of carbon in CO2?
+4
89
How does a C-H bond affect the oxidation state of carbon?
Decreases it by 1
90
How do C-C bonds affect the oxidation state of carbon?
They do not affect it.
91
92
Give an example of "strong" reductant
LiAlH4 (LAH)
93
Give an example of "weak" reductant
NaBH4 - only aldehyde to primary alcohol
94
What is a free radical?
A free radical is a highly reactive chemical species containing an unpaired electron.
95
What is homolytic bond cleavage?
Homolytic bond cleavage is the breaking of a bond in which each bonding partner receives one electron from the bond.
96
What is the reaction of methane with Cl2 in the presence of light called?
Free radical substitution - Chlorination of Methane
97
Why is the chlorination of methane with Cl2 not dependent on solvent polarity?
Because the reaction proceeds through neutral intermediates, not charged ones. - same reaction rate if placed in polar solvent or nonpolar solvent - no charged intermediates (-)
98
What is the driving force for the high reactivity of free radicals?
The strong drive to form a full octet of electrons.
99
What are two factors that stabilize free radicals?
1. Increasing the number of alkyl groups on the carbon bearing the free radical (more substitution) 2. Delocalization (resonance) through overlap with adjacent pi bonds.
100
What is the geometry of free radicals?
A shallow pyramid. - essentially trigonal planar
101
What are three factors that influence the stability of free radicals?
Hybridization, electronegativity, and polarizability.
102
How does electron density affect the stability of free radicals?
Free radicals are destabilized by the removal of electron density.
103
How does the s-character of the orbital containing the free radical affect its stability?
Radical stability decreases with increasing s-character.
104
How does electronegativity affect the stability of a free radical?
Radical stability decreases with increasing electronegativity of the atom
105
How does polarizability affect the stability of a free radical?
Radical stability decreases as polarizability is decreased.
106
Why are larger atoms (e.g., iodine) better at stabilizing free radicals than smaller atoms (e.g., fluorine)?
Larger atoms are more polarizable, allowing the electron-deficient orbital to be spread out over a greater volume.
107
What is the relationship between reactivity and instability in free radicals?
In free radicals, reactivity is synonymous with instability.
108
What are two main factors that destabilize free radicals?
1. Bringing the half-filled orbital closer to the nucleus (greater s-character, higher electronegativity). 2. Restricting the delocalization of the free radical (decreasing polarizability).
109
What is the relationship between BDE and radical stability?
Low BDE reflects the formation of stable free radicals.
110
Besides free radical reactions, where else are BDEs important?
Understanding fragmentation patterns in mass spectroscopy.
111
Does bond breaking require energy input or release energy?
Bond breaking requires an input of energy.
112
How does light provide energy for bond cleavage?
Photons collide with molecules and impart energy.
113
What is the net change in the number of free radicals during initiation? What is the net change in the number of free radicals during propagation? What is the net change in the number of free radicals during termination?
Net increase No net change Net decrease
114
How many monochlorinated products can be formed from methane or ethane? How many monochlorinated products can be formed from propane?
Only one Two (1-chloropropane and 2-chloropropane).
115
Why does chlorination of propane not give a 75:25 ratio of 1-chloropropane to 2-chloropropane?
The reaction is not completely random. The stability of the intermediate free radical plays a role in determining the product distribution. - It indicates that not all hydrogen atoms in propane are equally reactive towards substitution.
116
How does the number of possible monochlorinated products increase with the size of the alkane?
The number of possible constitutional isomers increases, leading to a greater variety of monochlorinated products.
117
What is the selectivity in free radical halogenation?
The preference for forming one type of free radical intermediate over another, leading to a non-statistical distribution of products.
118
How do you calculate the selectivity for a particular type of hydrogen in free radical halogenation?
Divide the observed yield of the product formed from that hydrogen by the number of those hydrogens in the starting material. - chlorine is 3.66 times more likely to abstract a secondary hydrogen than a primary hydrogen in propane
119
Why does the stability of the intermediate free radical affect selectivity?
More stable free radicals are formed faster, leading to a preference for products arising from those intermediates.
120
What is the key factor influencing the difference in selectivity between bromination and chlorination?
The difference in activation energies for the C-H bond breaking step. - The greater the difference in activation energies, the larger the selectivity. - The transition state for bromination resembles the products (late transition state), which are farther apart in energy - The transition state for chlorination resembles the reactants (early transition state), which are closer in energy.
121
Radical Halogenation at Tiffany's Analogy (explain selectivity for Br2 & Cl2)
explains the difference in selectivity between chlorine and bromine radicals in free radical halogenation. It compares the C-H bond to a customer, and the chlorine and bromine radicals to two salespeople. The chlorine radical, being unstable and reactive, is like a customer in a hurry with lots of money, readily buying any diamond (C-H bond) regardless of price (bond strength). The bromine radical, being less unstable and more selective, is like a slow and methodical customer who only buys the cheapest diamond (weakest C-H bond). This analogy highlights that chlorine radicals are less selective and react with various C-H bonds, while bromine radicals are more selective and prefer to react with the weakest C-H bonds.
122
Why is NBS preferred over Br2 for allylic bromination?
NBS maintains a low concentration of Br2, preventing unwanted side reactions like dibromination.
123
What is allylic bromination?
The process of replacing a hydrogen atom in the allylic position of an alkene with a bromine atom.
124
What is benzylic bromination?
The process of replacing a hydrogen atom in the benzylic position of an aromatic ring with a bromine atom.
125
What is the benzylic position?
The carbon atom directly attached to an aromatic ring.
126
What is the allylic position?
The carbon atom adjacent to a carbon-carbon double bond.
127
Why are allylic and benzylic C-H bonds weaker than other types of C-H bonds?
The resulting free radicals are stabilized by resonance.
128
With radical reactions, how does resonance contribute to allylic rearrangement?
The allylic radical intermediate has resonance forms, allowing for delocalization of the radical and potential reaction at different sites - formation of a more stable product, often driven by factors like Zaitsev's rule
129
What is the advantage of using free-radical bromination in organic synthesis?
It allows for the installation of a good leaving group (bromine) on an otherwise unreactive molecule.
130

Organic Chem Class (19 decks)

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