Nucleophilic Substitution & Elimination Chemistry Beauchamp 1 y:\files\classes\315\315 Handouts\315 Fall 2013\2b 315 SN and E & chem catalog, answers.doc Problem 1 - How can you tell whether the S N 2 reaction occurs with front side attack, backside attack or front and backside attack? Use the two molecules to explain you answer. Use curved arrow formalism to show electron movement for how the reaction actually works. Problem 2 - Why might C 3 and C 4 rings react so slowly in S N 2 reactions? (Hint-think about bond angles in the transition state versus bond angles in the starting ring structure.) Problem 3 - Why might C 6 rings react slower in S N 2 reactions? What are the possible conformations from which a reaction is expected? Trace the path of approach for backside attack across the cyclohexane ring to see what positions block this approach. Which chair conformation would have the leaving group in a more reactive position (axial or equatorial)? Is this part of the difficulty (which conformation is preferred and present in higher concentration)?
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y:\files\classes\315\315 Handouts\315 Fall 2013\2b 315 SN and E & chem catalog, answers.doc
Problem 1 - How can you tell whether the SN2 reaction occurs with front side attack, backside attack or front and backside attack? Use the two molecules to explain you answer. Use curved arrow formalism to show electron movement for how the reaction actually works.
Problem 2 - Why might C3 and C4 rings react so slowly in SN2 reactions? (Hint-think about bond angles in the transition state versus bond angles in the starting ring structure.)
Problem 3 - Why might C6 rings react slower in SN2 reactions? What are the possible conformations from which a reaction is expected? Trace the path of approach for backside attack across the cyclohexane ring to see what positions block this approach. Which chair conformation would have the leaving group in a more reactive position (axial or equatorial)? Is this part of the difficulty (which conformation is preferred and present in higher concentration)?
y:\files\classes\315\315 Handouts\315 Fall 2013\2b 315 SN and E & chem catalog, answers.doc
Problem 4 - In each of the following pairs of nucleophiles one is a much better nucleophile than its closely related partner. Propose a possible explanation.
Problem 5 – Write out the expected SN2 product for each possible combination (4x8=32 possibilities).
AlD
D
D
D Li
Nu =H
O
H3CO
O
O NC
HS
H3CS
NN
N1
3 4 6 7
OH
A
B
C
D
2 58
OH O O
O
SH SC
N
NN
ND
O O
O
SH S C
N
NN
ND
OH O O
O
SH S C N N
N
N
D
OH O O
O
SH SC
NN N N D
Problem 6 – Using R-Br compounds from page 2 and reagents from page 3 to propose starting materials to make each of the following compounds. One example is provided. TM-1 is an E2 product (see page 15), all the others are SN2 products.
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Problem 10 – It is hard to tell where the hydride was introduced since there are usually so many other hydrogens in organic molecules. Where could have X have been in the reactant molecule? There are no obvious clues. Which position(s) for X would likely be more reactive with the hydride reagent? Could we tell where X was if we used LiAlD4?
Li
Al H
H
HH
+X
Where was "X"?
H2C
CH2
H2C
CH3
X
X
X
X
X could be on any sp3 carbon atom.If we used LiAlD4 there would be a 'D' where 'X' was.
SN2
Problem 11 - How many total hydrogen atoms are on C carbons in the given RX compound? How many different types of hydrogen atoms are on C carbons (a little tricky)? How many different products are possible? Hint - Be careful of the simple CH2. The two hydrogen atoms appear equivalent, but E/Z (cis/trans) possibilities are often present. (See below, problem 7, for relative expected amounts of the E2 products.)
C
C
CH3
CC1
H3C
CH2H3C
HHH
H
H
C
C
CH3
CC1
H3C
CH2H3C
HH
H
H
H C
C
CH3
CC1
H3C
CH2H3C
HH
H
H
H C
C
CH3
CC1
H3C
CH2H3C
HHH
H
H
E and Z possible here depending on stereochemistry of C and C1.
E and Z possible here depending on which proton is lost.
C1 has only 1 H, could be R or S stereochemistryC2 has two H, they are different, lead to E/Z stereoisomersC3 has three H and they are all equivalent, no E/Z possible
X
XR =
Problem 12 - Reconsider the elimination products expected in problem 11 and identify the ones that you now expect to be the major and minor products. How would an absolute configuration of Cα as R, compare to Cα as S? What about C1 as R versus S?
y:\files\classes\315\315 Handouts\315 Fall 2013\2b 315 SN and E & chem catalog, answers.doc
Problem 13 -Provide an explanation for any unexpected deviations from our general rule for alkene stabilities above. A more negative potential energy is more stable.
Hof = -18.7 kcal/mole
Hof = -22.7 kcal/mole
Hof = -19.97 kcal/mole
Our rules predict that stability is trans-di > cis-di > monosubstituted alkene. However, the data show that the cis alkene is less stable than the monosubstituted alkene. This is due to the large size of the t-butyl group crowding the medium methyl group. This would probably cause twisting of the pi bond and weaken pi overlap.
7 C(gr) + 7 H2(g) = zero reference energy
moststable
leaststable
Steric crowding by t-butyl group changes order of stability. Expected this
alkene to be the least stable, but it's not.
Hf
HfHf
Problem 14 – Order the stabilities of the alkynes below (1 = most stable). Provide a possible explanation.
H C C H R C C H R C C R
"R" = a simple alkyl group
Our rules predict that more substitution at pi bonds should make them more stable by extra inductive donating effect (compared to H) at more electronegative sp hybrid orbitals, so disubstituted > monosubstituted > unsubstituted.
Problem 15 - Propose an explanation for the following table of data. Write out the expected products and state by which mechanism they formed. Nu: -- /B: -- = CH3CO2
-- (a weak base, but good nucleophile).
H2CH3C Br
HCH3C Br
CH3
HCCH Br
CH3
CH3C Br
CH3
CH3
H3C
H3C
percent substitution
percent elimination
100 %
100 %
11 %
0 %
0 %
0 %
89 %
100 %
O
O
O
O
O
O
O
O
our rules
SN2 > E2
SN2 > E2
SN2 > E2(wrong
prediction)
only E2
Acetate is a pretty stable anion due to resonance stabilization on two oxygen atoms. We predict that it will favor SN2 reaction at methyl, primary and secondary RX electrophiles and E2 at tertiary RX. Everything looks as expected here except example 3, which is secondary, but C is tertiary so there is extra steric hindrance at both C and C which appears to push it to E2 > SN2. We will still follow our simplistic rules in this course, but here is an example that shows our rules are a bit too simplistic.
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Problem 16 – One of the following reactions produces over 90% SN2 product and one of them produces about 85% E2 product in contrast to our general rules (ambiguity is organic chemistry’s middle name). Match these results with the correct reaction and explain why they are different.
ClO
pKa = 16
OpKa = 19
H
ClH
SN2 > E2
E2 > SN2
OH Cl
ClO
H
Sterically larger t-butoxide forces E2 reactions even at primary RX.
85% E2
90% SN2
Problem 17 - A stronger base (as measured by a higher pKa of its conjugate acid) tends to produce more relative amounts of E2 compared to SN2, relative to a second (weaker) base/nucleophile. Greater substitution at C and C also increases the proportion of E2 product, because the greater steric hindrance slows down the competing SN2 reaction. Use this information to make predictions about which set of conditions in each part would produce relatively more elimination product. Briefly, explain your reasoning. Write out all expected elimination products. Are there any examples below where one reaction (SN2 or E2) would completely dominate? a.
I
Cl
...versus...I
Cl
more E2 because of extra steric hindrance at C position.
b.
Cl
...versus...
ClO
O O
pKa(RCO2H) = 5 pKa(ROH) = 16
more E2 because base is less stable (higher pKa of conjugate acid, more basic)
c.
Cl
...versus...
Cl
more E2 because of extra steric hindrance at C position.
d.
N C...versus...
N C
ClCl
more E2 because of extra steric hindrance at C position, only E2 at tertiary RX center.
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e.
OR
Br
...versus...OR
Br
more E2 because of extra steric hindrance at C position, which is quaternary, so no SN2, only E2.
f.
Cl
...versus...
N CCl
pKa(RCCH) = 25 pKa(NCH) = 9
more E2 because base is less stable (higher pKa of conjugate acid, more basic)
Problem 18 - (2R,3S)-2-bromo-3-deuteriobutane when reacted with potassium ethoxide produces cis-2-butene having deuterium and trans-2-butene not having deuterium. The diastereomer (2R,3R)-2-bromo-3-deuteriobutane under the same conditions produces cis-2-butene not having deuterium and trans-2-butene having deuterium present. Explain these observations by drawing the correct 3D structures, rotating to the proper conformation for elimination and showing an arrow pushing mechanism leading to the observed products. (Protium = H and deuterium = D; H and D are isotopes. Their chemistries are similar, but we can tell them apart.)
y:\files\classes\315\315 Handouts\315 Fall 2013\2b 315 SN and E & chem catalog, answers.doc
Problem 19 - Draw a Newman projection of the two possible conformations leading to E2 reaction. Show how the orientation of the substituents about the newly formed alkene compares.
B
C C
R3
R4
R2
R1
H
X
staggeredanti E2reaction
reactant conformation 1
rotate about center bond
eclipsedsyn E2reaction C
HC
R3
R4
X
reactant conformation 2
alkene stereochemistryis different
Newmanprojections
B
< 1% > 99%
H
R1R2
X
R4 R3
H
R1
R2
X
R4 R3
H
R1
R2
R3
R1R4
R2
R3
R2R4
R1
alkene stereochemistryis different
diasteromers
B
B
Problem 20 – What are the possible products of the following reactions? What is the major product(s) and what is the minor product(s)? There are 55 possible combinations.
y:\files\classes\315\315 Handouts\315 Fall 2013\2b 315 SN and E & chem catalog, answers.doc
X
X
X
CD
E
NuNu
SN2 > E2 products(all except t-butoxide)
no R/S
E2 productsno E/Z
Nu
SN2 > E2 products(all except hydroxide, alkoxide,
t-butoxide, acetylide)
Nu
E2 products(enantiomers)
R S
Nu
only E2 products (3 ways, tri > gem di)
primary RX secondary RX
tertiary RX
Problem 21 - Explain the differences in stability among the following carbocations (hydride affinities = a larger number means a less stable carbocation = more energy released when combined with hydride).
OCH2 N
CH2
CO CH3
CH
H
H
CR
H
H
CR
H
R
CR
R
R
R groups inductively donate electrons to electron deficient centers likecarbocations (and free radicals). The more R groups the more stable. The carbocation that releases the most energy when combined with hydride was the least stable starting energy.
H2C
HC
CH2 H2C
HC
CH2
Primary carbocations are 270 kcal/mole, but this primary allylic carbocation is only 256, so it is 14 kcal/mole more stable than expectedbecause of resonance sharing by adjacent pi bond.
CH2 CH2CH2CH2
Primary carbocations are 270 kcal/mole, but this primary benzylic carbocation is only 239, so it is 31 kcal/mole more stable than expectedbecause of resonance sharing by adjacent pi bonds.
HH
H
OCH2H
NCH2H
H
CO CH3
Primary carbocations are 270 kcal/mole, but the primary carbocation next to oxygen is only 248, so it is 22 kcal/mole more stable than expected even though oxygen's inductive effect works in the opposite direction. When next to nitrogen it is more stable than expected by 52 kcal/mole because less electronegative nitrogen shares better than more electronegative oxygen. The last example, lone pair next to oxygen is 40 kcal/mole more stable than expected. For all of them it is because of resonance sharing by adjacent lone pair of electrons. In each case this helps fill the octet of the carbocation and adds an extra bond.
y:\files\classes\315\315 Handouts\315 Fall 2013\2b 315 SN and E & chem catalog, answers.doc
Problem 22 - Why are vinyl carbocations so difficult to form? (Hint – What is their hybridization?) How does an empty sp2 orbital (phenyl carbocation) or empty sp orbital (ethynyl carbocation) compare to a typical sp2 carbocation carbon with an empty 2p orbital? (An empty 2p orbital is also present on the vinyl carbocation, but the carbon hybridization is sp.)
C
H
H
C H CH C
sp hybridized carbon is the most electronegative carbon because it is 50% s character and makes it more difficult (higher energy) to form. The empty orbital is 2p (the least electronegative). The second example is also sp carbon, but the orbital is sp, which makes it much more difficult to take electrons from. The phenyl example cannot change its shape because of the ring and that makes the empty orbital sp2, which is also more electronegative than 2p and harder to form. The difficulty to form can be seen from the energy released when each carbocation reacts with the hydride donor. The more energy released, the less stable the starting point.
287 386
300
2p
sp3
sp2
sp
2s
electronegativityof orbitals
most
least
hold on electrons
least
most
sp orbitalvery electronegative
2p orbital, but on sp carbon
sp2 orbitalvery electronegative
Problem 23 – The bond energy depends on charge effects in the anions too. Can you explain the differences in bond energies below? (Hint: Where is the charge more delocalized?) We won’t emphasize these differences.
C
CH3
H3C
CH3
X
X = gas phase bond energies
Cl +157Br +149I +140
C
CH3
H3C
CH3
X
The difference in energy cost is due to the different anions since the carbocation is the same in every instance. The larger and more delocalized the electrons the more stable is the anion and easier to form, just like we saw in our acid/base topic.
H
Problem 24 – Draw in all of the mechanistic steps in an SN1 reaction of 2R-bromobutane with a. water, b. methanol and c. ethanoic acid. Add in necessary details (3D stereochemistry, curved arrows, lone pairs, formal charge). What are the final products?
Br
HO
H
H
H H HH
H
HO
H
ab
c
H H HH
H
HOH
H
Adds from both faces, makes R and S chiral center. (SN1)
y:\files\classes\315\315 Handouts\315 Fall 2013\2b 315 SN and E & chem catalog, answers.doc
c H3CH2C: ethyl migrates
CH3C
H
C
H
CH3
CH2
2o carbocationlooks OK
H3CSN1 and E1 are possible here H O
R
SN1 R and S possible
E and Z possibleE1
no E and Z possibleE1
trisubst. > monosubst.
Problem 26 – Write out your own mechanism for all reasonable products from the given R-X compound in water (2-halo-3-methylbutane).
C
CH3
H3C
H
C
X
CH3 HO
H
H
C
CH3
H3C
H
C CH3
Hrearrangement
2o to 3o R+C
CH3
H3C C CH3
H
H
SN1 E1 SN1 E1
C
CH3
H3C C CH3
H
C
CH3
H3C C CH2
H
C
CH3
H3C
H
C CH2
H
HO
H
SN1
H
a
b
c
C
CH3
H3C
H
C CH2
H
HO
H H H
O
H
C
CH3
H3C
H
C CH2
H
HO
H R and S
C
CH3
H3C C CH3
H
H
C
CH3
H2C C CH3
H
HOH
Also, from second carbocation.
E1
H
Problem 27 - Consider all possible rearrangements from ionization of the following RX reactants. Which are reasonable? What are the possible SN1 and E1 products from the reasonable carbocation possibilities?
C
CH3
H3C
CH3
CH
X
CH3CH2
C
H2C
CH3
CHX
CH3
H2C
CH2
CHC X
CH3H3Ca. b. c.
initial R+SN1 = 2E1 = 1
subsequent R+SN1 = 1E1 = 2
initial R+SN1 = 2E1 = 1
subsequent R+SN1 = 6E1 = 8
initial R+SN1 = 2E1 = 1
subsequent R+SN1 = 6E1 = 8
total = 6 products total = 17 products total = 17 products
y:\files\classes\315\315 Handouts\315 Fall 2013\2b 315 SN and E & chem catalog, answers.doc
CH3
X
H
CH3
CH3
H
CH3
2o carbocation
CH3
H
CH3
same as b above
CH2
C
H2C
C
CH3
CH3H
same as c' above
3o carbocation
d e
e dc.
addnucleophile
losebeta H
addnucleophile
losebeta H
CH3
H
CH3
CH3
OH
H
CH3
*
3o carbocation
R and SSN1
E1
Similar to above.
CH2
C
H2C
CH
CH3
CH3
addnucleophile
losebeta H
3o carbocation
same as c above
* = chiral center
around 17 possible products
*
d. What would happen to the complexity of the above problems with a small change of an ethyl for a methyl? Use the key of “b” and “c” as a guide. This problem is a lot more messy than those above, (which is the point of asking it). There are too many possibilities to consider listing every answer.
y:\files\classes\315\315 Handouts\315 Fall 2013\2b 315 SN and E & chem catalog, answers.doc
Problem 28 – Lanosterol is the first steroid skeletal structure in your body on its way to cholesterol and other steroids in your body. It is formed in a spectacular enantiospecific cyclization of protonated squalene oxide. The initially formed 3o carbocation rearranges 4 times before it undergoes an E1 reaction to form lanosterol. Add in the arrows and formal charge to show the rearrangements and the final E1 reaction.
R
CH3
H
H
CH3
CH3
H
CH3
HOH
R
CH3
H
CH3
CH3
CH3
HOH
H
lanosterol precursor
lanosterol
B H
R
O protonatedsqualene
acidcatalysis
rearrangement1
R
CH3H
CH3
CH3
H
CH3
HOH
H
rearrangement2
R
CH3
CH3
H
CH3
HOH
H
H
rearrangement4
CH3
R
CH3
H
CH3
HOH
H
HCH3
CH3
R
CH3
CH3
CH3
H
CH3
HOH
H
H
rearrangement3
E1 reaction
19 more steps
H3CH
CH3
CH3
HO
H
H H
H
cholesterol
otherbody
steroids
H
squalene
Requires 5 arrows to show the reaction.
B
Problem 29 - Propose a synthesis of monodeuterated cyclohexane from cyclohexanol.
y:\files\classes\315\315 Handouts\315 Fall 2013\2b 315 SN and E & chem catalog, answers.doc
Problem 8 (29b) – Write a detailed arrow-pushing mechanism for each of the following transformations.
S
O
O
Cl N
H
S
O
O
N
H
H
N
a.
b.
O
Cl
O
NN
H
HNR3
H
NH
Cl
NR3H
Cl
S
O
OCl
N
H
HS
O
OCl
N
H
NH
Cl
ON
H
H
Cl
ON
H
toluenesulfonyl chloride
N-ethyltoluenesulfonamide
ethanoyl chloride(acetyl chloride)
N-ethylethanamide(N-ethylacetamide)
Problem 30 – We can now make the following molecules. Propose a synthesis for each. (Tosylates formed from alcohols and tosyl chloride/pyridine via acyl substitution reaction, convert “OH” from poor leaving group into a very good leaving group, similar to iodide).
OTs
OTsH3C OTs
OTs
OTs
OTs
OTsOTs
OTs
1 2 34 5
6
7 8 9
NaBr(SN2)
no rearrangement
Br
All of these tosylates can be made from alcohols + tosyl chloride/pyridine.
RO TsCl / pyridine
RO
Ts
Tosylate is a very good leaving group (SN/E chemistry is possible). Tosylates can be used when a usual 'OH' to 'Br' transformation would lead to rearrangement.
y:\files\classes\315\315 Handouts\315 Fall 2013\2b 315 SN and E & chem catalog, answers.doc
Problem 31– Look back at the table of R-Br structures on page 2. Include stereoisomers together. Be able to list any relevant structures under each criteria below.
1. Isomers that can react fastest in SN2 reactions This would include all primary RBr, except when a neopentyl center is at a C position.
2. Isomers that give E2 reaction but not SN2 with sodium methoxide This would include all tertiary RBr and secondary neopentyl, when there is a C-H.
3. Isomers that react fastest in SN1 reactions This would include all tertiary RBr.
4. Isomers that can react by all four mechanisms, SN2, E2, SN1 and E1 (What are the necessary conditions?) This would include secondary RBr. SN1/E1 conditions would use weak nucleophile/bases (neutral) and SN2/E2 conditions would use strong nucleophile/bases (usually anions). When a neopentyl center is at a C position the SN2 reaction will not work.
5. Isomers that might rearrange to more stable carbocation in reactions with methanol. This would include secondary RBr structures, especially if a tertiary carbocation can form.
6. Isomers that are completely unreactive with methoxide/methanol This would include primary neopentyl centers (unreactive by all four mechanisms).
7. Isomers that are completely unreactive with methanol, alone. This would include all primary RBr compounds.
y:\files\classes\315\315 Handouts\315 Fall 2013\2b 315 SN and E & chem catalog, answers.doc
Problem 32 – Predict possible products of a. water and b. ethanoate (acetate) with structures 2, 10 and 13. Only consider rearrangements to more stable carbocations where appropriate.
y:\files\classes\315\315 Handouts\315 Fall 2013\2b 315 SN and E & chem catalog, answers.doc
Only axial 'X' conformations are shown, equatorial 'X' is too slow reacting in SN2/E2 reactions. Axial or equatorial RX can react via SN1/E1.
X
CH3
1 = 3o RX
H
H
H
H
X
H
CH3
H
H
H
X
H
H
CH3
H
H
X
H
H
H
H
CH3
2 = 2o RX 3 = 2o RX 4 = 2o RX
Q1. SN1/E1 both conformationsQ2. No SN2 because 3o RXQ3. E2 is OK at both C-HQ4. SN1 = 1, E1 = 1, SN2 = 0, E2 = 1Q5. NAQ6. NAQ7. achiralQ8. achiral
Q1. SN1/E1 both conformationsQ2. No SN2 because anti CH3
Q3. E2 is OK at at rear C-HQ4. SN1 = 2, E1 = 2, rearrangement likely to 3o, SN2 = 0, E2 = 1Q5. enantiomer with 3, diastereomer with 4 and 5Q6. diastereomer products are possibleQ7. chiralQ8. chiral
Q1. SN1/E1 both conformationsQ2. No SN2 because anti CH3
Q3. E2 is OK at at front C-HQ4. SN1 = 2, E1 = 2, rearrangement likely to 3o, SN2 = 0, E2 = 1Q5. enantiomer with 2, diastereomer with 4 and 5Q6. diastereomer products are possibleQ7. chiralQ8. chiral
Q1. SN1/E1 both conformationsQ2. SN2 is possibleQ3. E2 is OK at at both C-HQ4. SN1 = 2, E1 = 2, rearrangement likely to 3o, SN2 = 1, E2 = 2Q5. enantiomer with 5, diastereomer with 2 and 3Q6. diastereomer products are possibleQ7. chiralQ8. chiral
X
H
5 = 2o RX
H
H
CH3
H
X
H
H
H
H
H
X
H
H
H
H
H
X
H
H
H
H
H
6 = 2o RX 7 = 2o RX 8 = 2o RX
H3C
H
H3C
H CH3
H
Q1. SN1/E1 both conformationsQ2. SN2 is possibleQ3. E2 is OK at at both C-HQ4. SN1 = 2, E1 = 2, rearrangement possible, SN2 = 1, E2 = 2Q5. enantiomer with 7, diastereomer with 8 and 9Q6. diastereomer products are possibleQ7. chiralQ8. chiral
Q1. SN1/E1 both conformationsQ2. SN2 is possibleQ3. E2 is OK at at both C-HQ4. SN1 = 2, E1 = 2, rearrangement possible, SN2 = 1, E2 = 2Q5. enantiomer with 6, diastereomer with 8 and 9Q6. diastereomer products are possibleQ7. chiralQ8. chiral
Q1. SN1/E1 both conformationsQ2. SN2 is possibleQ3. E2 is OK at at both C-HQ4. SN1 = 2, E1 = 2, rearrangement possible, SN2 = 1, E2 = 2Q5. enantiomer with 9, diastereomer with 6 and 7Q6. diastereomer products are possibleQ7. chiralQ8. chiral
Q1. SN1/E1 both conformationsQ2. SN2 is possibleQ3. E2 is OK at at both C-HQ4. SN1 = 2, E1 = 2, rearrangement likely to 3o, SN2 = 1, E2 = 2Q5. enantiomer with 4, diastereomer with 2 and 3Q6. diastereomer products are possibleQ7. chiralQ8. chiral
X
H
9 = 2o RX
H
H
H
H
X
H
H
H
H
H
10 = 2o RX 11 = 2o RX 12 = 2o RX
H
CH3
Q1. SN1/E1 both conformationsQ2. SN2 is possibleQ3. E2 is OK at at both C-HQ4. SN1 = 2, E1 = 2, rearrangement possible, SN2 = 1, E2 = 2Q5. achiral, diastereomer with 11Q6. diastereomer products are possibleQ7. achiralQ8. achiral
Q1. SN1/E1 above conformation onlyQ2. SN2 is possibleQ3. E2 is OK at at both C-HQ4. SN1 = 2, E1 = 2, rearrangement possible, SN2 = 1, E2 = 2Q5. achiral, diastereomer with 13Q6. diastereomer products are possibleQ7. achiralQ8. achiral
Q1. SN1/E1 both conformationsQ2. SN2 is possibleQ3. E2 is OK at at both C-HQ4. SN1 = 2, E1 = 2, rearrangement possible, SN2 = 1, E2 = 2Q5. enantiomer with 8, diastereomer with 6 and 7Q6. diastereomer products are possibleQ7. chiralQ8. chiral
H
CH3
X
H
H
H
H
HH3C
H
Q1. SN1/E1 both conformationsQ2. SN2 is possibleQ3. E2 is OK at at both C-HQ4. SN1 = 2, E1 = 2, rearrangement possible, SN2 = 1, E2 = 2Q5. achiral, diastereomer with 10Q6. diastereomer products are possibleQ7. achiralQ8. achiral
y:\files\classes\315\315 Handouts\315 Fall 2013\2b 315 SN and E & chem catalog, answers.doc
13 = 2o RX 14 = 2o RX 15 = 2o RX
Q1. SN1/E1 above conformation onlyQ2. SN2 is possibleQ3. E2 is OK at at both C-HQ4. SN1 = 2, E1 = 2, rearrangement possible, SN2 = 1, E2 = 2Q5. achiral, diastereomer with 12Q6. diastereomer products are possibleQ7. achiralQ8. achiral
X
H
H
H
H
HC
HH3C
H3C
CH3
H
H
X
H
C
HH3C
H3C
CH3
H
H
H
H
H
X
C
HH3C
H3C
CH3
H
H
Q1. SN1/E1 above conformation onlyQ2. SN2 is possibleQ3. E2 is OK at at both C-HQ4. SN1 = 2, E1 = 2, rearrangement possible, SN2 = 1, E2 = 2Q5. achiral, diastereomer with 15Q6. diastereomer products are possibleQ7. achiralQ8. achiral
Q1. SN1/E1 above conformation onlyQ2. SN2 is possibleQ3. E2 is OK at at both C-HQ4. SN1 = 2, E1 = 2, rearrangement possible, SN2 = 1, E2 = 2Q5. achiral, diastereomer with 14Q6. diastereomer products are possibleQ7. achiralQ8. achiral
Problem 33 – What are the oxidation states of each carbon atom below? All of the atoms in these examples have a formal charge of zero.
OHCC C O C O
HOH
CO O
H3CC
OH H3CC
O
H
H3CC
O
OH
H3C C
H H
C O
HO
HO
H2O
oxidation state of carbon = _____?
oxidation state of carbon = _____?
oxidation state of carbon = _____?
oxidation state of carbon = _____?
oxidation state of carbon = _____?
oxidation state of carbon = _____?
oxidation state of carbon = _____?
oxidation state of carbon = _____?
oxidation state of carbon = _____?
oxidation state of carbon = _____?-4 -2 0 +2
+4
+4+3+1-1-3
H
H
H
H
H
H
H
H
H
H
H H
Problem 34 – What are the oxidation states below on the carbon atom and the chromium atom as the reaction proceeds? Which step does the oxidation/reduction occur? (PCC, B: = pyridine and Jones, B: = water)
C
H
R
R
O
H
Cr
O
O
O C
H
R
R
O
H
Cr
O
O
O
B
C
H
R
R
OCr
O
O
O
oxidation state of C = 0
oxidation state of Cr = +6
BC
R
R
OCr
O
O
O
HB
HB R.D.S. = slow step
E2 reaction
partial negative of oxygen bonds to partial positive of chromium
(step 1)
acid/base (step 2)
(step 3)
oxidation state of C = 0
oxidation state of Cr = +6
oxidation state of C = 0
oxidation state of Cr = +6
oxidation state of C = +2
oxidation state of Cr = +4Carbon oxidation occurs in this step.
y:\files\classes\315\315 Handouts\315 Fall 2013\2b 315 SN and E & chem catalog, answers.doc
Problem 35 – Supply all of the mechanistic details in the sequences below showing 1. the oxidation of a primary alcohol, 2. hydration of the carbonyl group and 3. oxidation of the carbonyl hydrate (Jones conditions).
H3CCH2
CH
OH
H
Cr
O
OOH3C
CH2
CH
OH
H
Cr
O
O
O
H3CCH2
CH
O
H
Cr
O
O
O
H3CCH2
CH
OCr
O
O
O
Jones = CrO3 / H2O / acid primary alcohols oxidize to carboxylic acids
(water hydrates the carbonyl group, which oxidizes a second time )
primary alcohols
aldehydes
HO
HH
OH
H
HO
H
H
H3CCH2
CH
OH
H3CCH2
CH
OH
HO
H
hydration of the aldehyde
Under aqueous conditions, a hydrate of a carbonyl group has two OH groups which allow a second oxidation, if another C-H bond is present. This is only possible if the starting carbonyl group was an aldehyde (true when starting with methyl and primary alcohols).
chromicanhydride
inorganicesternucleophile
addition at Cracid/base
H
O
H
resonance
nucleophile addition at C
E2 reaction(oxidation here)
H3CCH2
C
OH
O
H
H
H
HO
HH3C
CH2
C
O
O
H H
HO
H
H
Cr
O
OO
H
H3CCH2
C
O
O
H H
H Cr
O
O
OHO
H
H3CCH2
C
O
O
H H
Cr
O
O
OH
OH
H
HO
H
H3CCH2
C
O
OH
HO
H
H
Cr
O
O
O
second oxidation of the carbonyl hydrate
carboxylic acid from an aldehyde or a primary alcohol
acid/base
nucleophile addition at Cr
acid/base
inorganicester
E2 reaction(oxidation here)
1
2
3
1. oxidation of 'OH' to C=O2. hydration of C=O group3. oxidation of 'OH' to C=O