Exam 4 Flashcards
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
ethers
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).
alcohols
- note that alcohol is the solvent here
True or False:
carbocations and 3 membered ring intermediates] are both extremely reactive electrophiles
True
In order for a leaving group to leave, it has to be a fairly _____________
weak base
- hydroxyl (HO-) and alkoxy (RO -) groups are both fairly strong bases, and therefore poor leaving groups
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
acid
- can now participate in Sn1 & E1 reactions as leaving group step
One of the keys to the reactions of alcohols is that the Conjugate Acid is a better ____________ and the Conjugate Base is a better _____________
leaving group
nucleophile
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)
base
- can now participate in Sn2 as nucleophile & E2 as strong base
Alcohols are mild __________
Alcohols are also ___________
*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
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.
more
- resonance-stabilized & nearby electron-withdrawing groups will stabilize the negative charge of the conjugate base through inductive effects
The conjugate base of an alcohol is called an_____________
alkoxide
The conjugate acid of an alcohol is called an ____________
oxonium ion
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
True
- look at Acid & Conjugate Acid (the weaker the higher the pKa thus the more favorable to form)
True or False:
The stronger the acid, the weaker the conjugate base.
The weaker the acid, the stronger the conjugate base.
True
Alcohols where the conjugate base is resonance stabilized will be (___MORE or LESS___) acidic
more
- any factor which stabilizes the conjugate base will increase acidity (resonance affects & inductive affects)
In Williamson Ether Synthesis, an alkyl halide (or sulfonate, such as a tosylate or mesylate) undergoes _____________ by an alkoxide to give an ether
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)
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
intramolecular
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
ether
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.
hydride ion H–
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.
five
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.
two combinations
- just make sure the alkyl halide is methyl/primary since its SN2
Ethers can be made from the reaction of alkenes with ________ in the presence of alcohols
acid
- useful when making an ether through SN2 reaction would not work due to elimination
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 _______
ether
- classic Sn1 reaction
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 _________
ether
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.
rearrangements
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).
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
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.
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!!!!!!
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.
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)
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
For secondary Ethers, the second step of Ether cleavage is _____________
mixture of SN2 & SN1
- can proceed through both pathways
True or False:
Intramolecular cleavage of cyclic Ethers are fundamentally no different than those that form and break in intermolecular processes
True
Epoxides (oxiranes) are cyclic ethers that have unusually high reactivity due to ___________
ring strain
True or False:
The 3-membered ring of epoxides can be opened under both acidic and basic conditions
True
______________ 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
The interior bond angles of epoxides are about _________
Contrast that with the “ideal” bond angle of 109.5° for tetrahedral carbon.
60°
- ring strain
Treating an epoxide with aqueous acid [H3O+] will open an epoxide to provide a ___________
1,2 -diol
- vicinal diol or glycol
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
There’s 2 important ways to make epoxides from alkenes, one “direct” and one “indirect”
1.
2.
- Reaction of Alkenes with a Peroxyacid
- Treating Halohydrins with Base
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)
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
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
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
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
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
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
Only alkyl alcohols (alcohols on _______ hybridized carbons) will undergo SN1 and SN2 reactions
sp3 hydridized
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
Alcohols can be converted to sulfonates by treating them with _______________
sulfonyl chlorides
- tosyl chloride
- mesyl chloride
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
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
Ladder
