nucleophilic substitution & elimination chemistry beauchamp...

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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 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)?



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.

NN

N

?

Br

problem solution

NaN3N3

2-azidopropane

TM = target molecule

SN2

(buy)

SN2

TM-1 TM-2TM-3

OR

Br

E2

Br

NaO

RBr

OK

(make)

SN2

(make)

Na

(make)



N

O

O

O

TM-4TM-5

TM-6 O

N

O

O

Br

O

O

Br(no structure)

(make)

SN2 SN2

(make)

O

CH3C

O

TM-7 TM-8

Br

N

H3COH Br

H3CBr

SN2

SN1

P

PhPh

Ph

P

PhPh

Ph

Br

TM-9

C

N

Br

SN2

Na

(buy)

SN2

(buy)

SPh Ph

TM-10

TM-11

Br

OSPh Ph

Br

OBr

SN2

(buy)

Li

(make)SN2

TM-12 TM-13 TM-14

HO

SROH

Br

SH

BrS

R

Br

Na

(buy)SN2

(buy)SN2

Na Na

SN2

(make)

TM-15

O

O

TM-16

S

S

H

O

O Br Br

S

S

H

Li

SN2

(make)

Li

SN2

(make)

D

TM-17

AlD4

Br

Li

(buy)

SN2

Problem 7 – Show the acyl substitution mechanism for each functional group with hydroxide as the nucleophile.

Na

O HC

R

O

ClCR

O

Cl

OH

CR

O

OH

Cl

CR

O

OClH

acid/baseacylsubstitution

CR

O

OC

O

R CO

O

R

OH

O HC

O

R

O

C

O

R CR

O

OH

acylsubstitution



Na

O HC

R

O

OCR

O

O

OH

CR

O

OH

O

CR

O

OOH

acid/base

R

R R

R

acylsubstitution

Na

O HC

R

O

NCR

O

N

OH

CR

O

OH

N

CR

O

ONH

acid/base

R

R R

R

R

R

RRacyl

substitution

Problem 8 – Write an arrow pushing mechanism for each of the following reactions.

O

H

O

H

H

Na

K

H

pKa

ROH 16-19

H-H 37

pKa = 17

pKa = 19

O

O

H H

(gas)

H H

(gas)

Keq = 10-17

10-37= 10+20

Keq = 10-19

10-37= 10+18

acid/base

acid/base

Problem 9 – Write an arrow pushing mechanism for the following reaction.

Li

Al D

D

DD

C

Cl

CH3

H

+

Na

B H

H

HH

C

Cl

DCH3

+

a.

b.

C

D

HCH3

C

H

CH3D

Cl

Cl

R

S

S

S

The center inverted, but the priorities changed, so still S.

SN2

SN2



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.

No E and Z is possible here.

stability or alkenes = tetrasub > trisub > trans-disub > cis-disub = gem disub > monosub.

C

C

CH3

CC1

H3C

CH2H3C

HHH

H

H

H

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?

(3S,4R)-3-bromo-3,4-dimethylheptane

C

C

H

Br

C3

HH3C

H

C1

H3CCH2CH3

H

H

HC

C

H

C3

HH3C

H

C1

H3CCH2CH3

HH

C

C

H

C3

HH3C

C1

H3CCH2CH3

H

H

H

C

C

C3

HH3C

H

C1

H3CCH2CH3

H

H

H

C

C

C3

H3CH

H

C1

H3CCH2CH3

H

H

H

O R

E alkenetetrasub = most stable, #1

No Z or E alkenegem disub = less stable, #4

E alkenetrisub = middle stable, #2

etc.

Z alkenetrisub = middle stable, #3



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



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.



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.



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.)

CH3

Br H

D

CH3

H

O

Na

(2R,3S)

DCH3

Br H

H

CH3

D

O

Na

(2R,3R)

H

Br

CH3

H

H3C

DO C C

D

H3C

H

CH3

D

cis-2-butene(D present)

One 3D example.

trans-2-butene(no D present)

cis-2-butene(no D present)

trans-2-butene(D present)

Br

H3C H

D

H

H3C

EtO

H3CH3C DH

Br

H3C H

CH3

D

H

EtO

HH3C CH3H

Z alkene, with D E alkene, without D Br

HDC H

H

H

H

EtO

HDHC HH

1-alkene, no E/ZH3C

H3C

(2R,3S)

H

Br

CH3

H

D

H3CO C C

H3C

D

H

CH3

One 3D example.

(2R,3R)

Z alkene (with D) E alkene (with D)

(2R,3S) (2R,3S) (2R,3S)



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.

AlD

D

D

D

BD

D

D

D LiNa

Nu

Nu = HO

RO C

OH3C

H3CCH3

O

O NC C

C

HH

SR

S

NN

N

1 2 3 4 5 6

7

X

A

OROR

8

OR

9 O

enolates S

S10

S

Ph

Ph

CH2

sulfurylids

11

N

O

Oimidate

OH OR

O

O

SH C

NC

CH

N3 D

O

S

S

S

Ph

Ph

CH2

N

O

O

hydroxide alkoxide t-butoxide carboxylate hydrogenthiolate cyanide acetylidealkylthiolate

azide borodeuteride aluminumdeuteride dithiane anion

primary RX

12

1 1 2 3 4 5 6

7 8 9 10 1112

Nu

X

B

S

Ph

Ph

CH2

N

O

O

(E2 > SN2)3 possible

(E2 > SN2)3 possible O

O

SH C

N

(E2 > SN2)3 possible

N3D

O

S S

secondary RX 12 3 4 5 6

7 8 9 10 1112



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.

methyl = 315 primary = 270 secondary = 249 tertiary = 232 allyl = 256

benzyl = 239

= 248 = 218 = 230



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)

HO

H

H H HH

H

HOH

a

b and c

from c from b

a

2E alkene 2E alkene 1 alkene

(SN1)R/S



Br

H3CO

H

H

H H HH

H

HO

H

ab

c

H H HH

H

HOH3C

H

Adds from both faces, makes R and S chiral center. (SN1)

H3CO

H

H H HH

H

HOH3C

(SN1)R/S

a

b and c

from c from b

a

similar steps and similar products (ether instead ofan alcohol) 2E alkene 2E alkene

1 alkene

H3CC

O

OBr

H

H H HH

H

ab

c

similar steps and similar products (ester instead ofan alcohol or ether),plus some minor alkene products.

H H H HH

H

HOC

H3C

O

(SN1)R/S

Problem 25 – What are the likely SN1 and E1 products of the initial carbocation and the rearranged carbocations from “a”, “b” and “c”?

a H: hydride migrates

C

CH2

H3C C

H

CH3

H

3o carbocationlooks very good

SN1 and E1 are possible here

H3C

OR

SN1 (achiral)E and Z possible no E and Z possible

E1 E1

trisubst. > gem disubst.

H3C: methyl migrates

b

C

H

C

CH3

CH3

H

2o carbocationlooks OK

CH2

H3CSN1 and E1 are possible here

H OR

SN1 R and S possible

no E and Z possibleE1

E and Z possibleE1

trans disubst. < trisubst.



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





total = 6 products total = 17 products total = 17 products



CH2

C

H2C

CH3

CHX

CH3

b.

CH2

C

H2C

CH3

CH

CH

H H

c

b

a

b

a

a

c

b

a

CH2

C

H2C

CH

CH3

CH3

CH3

CH3

H

CH2

C

H2C

CH3

CH2

CH

good = 2o to 3o c'

CH2

C

H2C

C

CH3

CH3H

OK, 3o to 3o

these two are different

CH3

CH3

ionization

rearrange

good = 2o to 3o

poor = 2o to 1o

b'

these two are equivalent

OK, 3o to 3o

addnucleophile

losebeta H

addnucleophile

losebeta H

H

This rearrangement is not going to happen.

addnucleophile

losebeta H

H

Assume water is the solvent. Possible SN1/E1 products are:

CH2

C

H2C

CH3

CHX

CH3

b.

CH2

C

H2C

CH3

CH

CH

H H

c

bb

cb

CH2

C

H2C

CH

CH3

CH3

CH3

CH3

H

good = 2o to 3oc'

CH2

C

H2C

C

CH3

CH3H

OK, 3o to

ionization

rearrange

good = 2o to 3o

addnucleophile

losebeta H

addnucleophile

losebeta H

addnucleophile

losebeta H

addnucleophile

losebeta H

Possible products

OH

OH

OHOH

E1SN1 E1 E1 E1

*

* = chiral center

*

*

*

R and S R and S

SN1(RR, SS, RS SR)

E1 E1

E1SN1 SN1

R and S

around 17 possible products

from initial R+ from subsequent R+

3o



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


*

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.

d.

rearrange

addnucleophile

losebeta H

addnucleophile

losebeta H

addnucleophile

losebeta H

Possible products

OH

OH

OHOH

E1 SN1 E1E1

E1

*

* = chiral center

*

*

*

R and S

SN1(RR, SS, RS SR)

E1 E1

E1

SN1 SN1

R and S


addnucleophile

losebeta H

addnucleophile

losebeta H

E1 E1

**

R and S

R and S R and S

*

*

SN1(RR, SS, RS SR)

OH

E and Z E and Z

E and Z

**

R and S R and S



H2CCH3

CHX

CH3

H2CCH3

CH3

X

similar



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.

H

OH

HBrH

Br

1. Mg2. D2O

H

DSN1



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.

OH

HBr

major product

OH Br

major product

Br1. TsCl/py2. NaBr

H

rearrangement norearrangement



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.



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.

C

H3C

H3C

H3C

H

H

CCH3

CH3

H3CH

H

severe stericrepulsion

19,999

Keqt-butyl substituent locksin chair conformationwith equatorial t-butyl

X

X

X

X

1213

Cyclohexane structureshave two chair conformationspossible.

1. Which conformation is reactive?

2. Is SN2 possible? Requires an open approach

at C and C .

3. Is E2 possible? Requires anti C -H.

4. How many possible products are there?

5. What is the relationship among the starting

structures?

6. What is the relationship among the products?

7. Are any of the starting structures chiral?

8. Are any of the product structures chiral?

Cyclohexane Examples

1

X X

XX

X X

X

X XX X2

3

4

5

6

7

8

9

10

11

chair 1 chair 2

X = ClBrI

OTs

14 15

Mechanism predictions with:

OR H

R

O

OH

OH HSome examples

OH Na OR Na R

O

O

Na

O K

weaknucleophiles

strong nucleophile/bases + other anions shown above



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 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


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

X

H

H

H

H

HC

HH3C

H3C

CH3



13 = 2o RX 14 = 2o RX 15 = 2o RX


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



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 = _____?-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 = +2

oxidation state of Cr = +4Carbon oxidation occurs in this step.



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

acid/base



Problem 36 – We can now make the following molecules. Propose a synthesis for each from our starting materials.

C

O

H HC

O

H3C H

C

O

H3C CH3

C

O

CH2

H

C

O

CH3

H3C

O

C

O

CH HH3C

CH3

C

O

H C

O

CH2

H

C

O

H OH

C

O

H OCH3

C

O

H3C OH

C

O

CH2

OHH3C

C

O

CH OHH3C

CH3

C

O

OH C

O

CH2

OH

C

O

H3C O

C

O

CH2

OH3C C

O

CH OH3C

CH3

CH3CH3

CH3 C

O

OC

O

CH2

O

CH3

CH3

aldehydes

ketones

C

O

CH

HH2C C

O

CH

OHH2C C

O

CH

OH2C CH3

O

C

O

C HH2C

CH3

C

O

C OHH2C

CH3

C

O

C OH2C

CH3

CH3

conjugated aldehydes, carboxylic acids and esters

C

O

H3C Cl

C

O

CH2

ClH3C

C

O

CH ClH3C

CH3

C

O

ClC

O

CH2

Cl

C

O

H NH

C

O

H3C NH

C

O

CH2

NH

H3C

C

O

CH NH

H3C

CH3

C

O

NH C

O

CH2

NH

(not stable)

CH3 CH3 CH3

CH3CH3

CH3

anhydrides

C

O

H3C OC

O

HC

O

H3C OC

O

CH3

C

O

H3C OC

O

CH2

CH3C

O

H3C OC

O

CHCH3

CH3

C

O

H3C OC

O

carboxylic acids

acid chlorides

amides

esters

nitriles

H3CC

NC

NC

N CN

CN

CN

CN

We haven't learned how to make this epoxide yet.We will when we cover alkenes.

O

OH

1. LDA2. workup

PCC



H3CH Br2

h H3CBr NaOH

SN2H3C

OH CrO3py.

H2CO CrO3

H2OCH

O

OH

1. NaOH2. CH3Br CH

O

O CH3

C

O

H3C OC

O

H

C

O

H3C Cl

H3CNH2

CH

O

N CH3

H

NaN3SN2

H3CN3

1. LiAlH42. workup

H3CNH2

more reactive C=O

NaCNSN2

H3CC

N

Ph-S-Ph

H3CS

Ph

Ph

n-BuLi

O

alkane bromoalkane

nitrile

1o amine

alcohol aldehydecarboxylic

acidester

acid chloride

anhydride

epoxide

1o amide

We can't make the 1C acid chloride, but we can make the 1C anhydride, which does similar reactions

O

H

O

H

O

OH

H Br2h

Br NaOHSN2

OH CrO3py. O CrO3

H2O

O

OH

1. NaOH2. CH3Br

O

O CH3

O

O

O

O

Cl

H3CNH2

O

N CH3

H

NaN3SN2

N3

1. LiAlH42. workup

NH2

NaCNSN2

CN

H

SOCl2

t-butoxide (E2)

n-BuLi

Ph-S-Ph

S

Ph

Ph O

2 Br2h

1. NaNR22. Br

1. NaNR22. CH3Br 1. NaNR2

2. workup (zipper)

excess

H2SO4/

Many of the syntheses of other target molecules are similar. A few that go farther are shown. Not all are shown.

Br2h

HBrROOR

h BrNaOH

SN2 OHCrO3py.

CrO3H2O

O

OH

Br2h

HBrROOR

h BrNaOH

SN2 OHCrO3py.

CrO3H2O

Br

Br

t-butoxide(E2)

t-butoxide(E2)

O

O

1. LDA2. workup

OHCrO3py.

O

H

CrO3H2O

O

OH

OHCrO3py.

O

H

CrO3H2O

O

OH

1. LDA2. workup

1. NaOH2. CH3Br

O

O

O

O

etc.

nucleophilic substitution & elimination chemistry beauchamp...

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