Chem S-20ab Purple Book Solutions Week 2 - Page 1 of 27 R and S Nomenclature

1. Circle all the stereocenters in the following molecules. For each stereocenter, assign an R or S designation according to the standard rules.

Cl H R H

R

H R

OH S H S H3C S

S O R S HO N

2. Draw an unambiguous structural representation of the following molecules:

(S)-4,5-dimethyl-(Z)-2-hexene (R)-3,5,6-trimethyl-(E)-3-heptene

H H

3. Provide the best possible systematic name for the following compound:

(S, E)-5-isopropyl-2,3,7-trimethyl-4-octene Chem S-20ab Purple Book Solutions Week 2 - Page 2 of 27 Chirality

1. For each of the following species, indicate if it is chiral or achiral. For those molecules that are chiral, circle all stereocenters, and draw the structure of the enantiomer.

H H

F F H C CH 3 Cl Cl 3

chiral

H CH3 H3C H

chiral

H CH3

achiral

H Br Br H

H Br Br H chiral

H Br

Br H

achiral ("meso")

H3C CH3 H3C CH3

H3C CH3 H3C CH3 chiral Chem S-20ab Purple Book Solutions Week 2 - Page 3 of 27 Enantiomers and Diastereomers

1. For each of the following pairs of stereoisomers, indicate if they are enantiomers, diastereomers, or identical compounds.

diastereomers

identical

enantiomers

identical

CH3 CH3

Br Br

CH Br 3 diastereomers H3C H3C

Br CH3 Chem S-20ab Purple Book Solutions Week 2 - Page 4 of 27 Identifying Isomers

1. For each of the following pairs of isomers, indicate whether they are Identical, Enantiomers, Diastereomers, or None of the above by circling the appropriate term for each.

Identical Enantiomers Diastereomers None

HN HN Me H Identical Enantiomers Diastereomers None

H Me

Identical Enantiomers Diastereomers None HN HN H H Me Me

Identical Enantiomers Diastereomers None

H Identical Enantiomers Diastereomers None H

Identical Enantiomers Diastereomers None Chem S-20ab Purple Book Solutions Week 2 - Page 5 of 27 Stereochemical Concepts

1. Which of the following are true? Give counterexamples for those that are false.

(a) In some cases, constitutional isomers are chiral. T

(b) In every case, a pair of enantiomers have a mirror-image relationship. T

(c) Mirror-image molecules are in all cases enantiomers. F identical

CH3 CH3 (d) If a compound has an enantiomer, it must be chiral. T

H (e) Every chiral compound has a diastereomer. F has no diastereomer F H C 3 Cl

(f) If a compound has a diastereomer it must be chiral. F and are diastereomers

H Br (g) Any molecule containing an asymmetric carbon must be chiral. F achiral ("meso")

Br H (h) Any molecule with a stereocenter must have a stereoisomer. T

(i) Some diastereomers have a mirror image relationship. F; no diastereomers have a mirror image relationship.

(j) Some chiral compounds are optically inactive. F; all PURE chiral compounds are optically active

(k) All chiral molecules have no plane of symmetry. T

H Br

(l) If a compound has a stereocenter, it has an enantiomer. F achiral ("meso")

Br H

(m) A meso compound will necessarily have at least two diastereomers. T Chem S-20ab Purple Book Solutions Week 2 - Page 6 of 27 Working with Stereoisomers

1. Consider the species 2,3-butanediol, whose structural formula is:

HO OH 2,3-butanediol

H3C CH3 a) There are three distinct, pure, configurational stereoisomers of 2,3-butanediol, all of which have the same structural formula above. Using any unambiguous notation, draw the three different stereoisomers in the boxes below. (The boxes have been arbitrarily labeledA,B,andC.) ABC

HO OH HO OH HO OH

H3C CH3 H3C CH3 H3C CH3

b) By circling the appropriate terms below, indicate whether each of the three species you drew above is chiral or achiral.

A: chiral achiral B: chiral achiral C: chiral achiral

c) Two of the above species have the same melting point (19°C). The other species has a different melting point (34°C). Circle the letter of the species which has a melting point of 34°C:

ABC

What characteristic of the other two stereoisomers causes them to share the same melting point (19°C)?

B and C are enantiomers, and enantiomers always have identical physical properties; for instance, they will have the same melting point. Chem S-20ab Purple Book Solutions Week 2 - Page 7 of 27 Principles of Reaction Stereochemistry

1. Each of the following statements is false. Discuss why the statement is false. Then make the statement true by changing only the underlined word or phrase.

Two molecules which are diastereomers should be expected to have the same melting point.

enantiomers

Two molecules which are enantiomers would exhibit different rates of reactivity with an achiral reagent.

diastereomers

An achiral starting material can react with an achiral reagent to give a single chiral product.

single achiral product

or

racemic mixture of products

Two molecules which are enantiomers would exhibit identical reaction rates when reacting with a chiral reagent (such as a biological enzyme).

different

When an achiral starting material is treated with an achiral reagent to yield a pair of diastereomers,both products must always be produced in equal amounts.

enantiomers Chem S-20ab Purple Book Solutions Week 2 - Page 8 of 27 Mechanism and Stereochemistry

1. a) In the box below, draw the product of the following reaction. Your product must match the chemical formula provided. Indicate stereochemistry clearly, if appropriate.

O

OH Br2

Br (+/-)

Chemical Formula: C6H11BrO

b) Provide a complete curved arrow mechanism for the transformation indicated above. Explain the stereochemical outcome, if any.

Br Br Br OH OH

Bromonium ion can form OH attacks from the opposite above or below the plane, leading to the plane, leading to a diastereoselectivity (although, racemic mixture of since only one sterocenter is bromonium ion formed in this reaction, only one intermediates. This in pair of enantiomers is possible. turn leads to a racemic solv mixture of products. H O O

Br Br

c) Explain the regioselectivity of this reaction; in other words, why is the particular isomer you drew in (a) the one that's formed.

The intermediate bromonium ion is unsymmetrical: there is much more partial positive charge on the tertiary carbon. This is similar to the relative stabilization conferred by tertiary vs. secondary carbocations. The nucleophile (ROH) then reacts by breaking the weaker of the two C–Br bonds. Chem S-20ab Purple Book Solutions Week 2 - Page 9 of 27 Stereochemistry of Alkene Additions

1. Fill in each box with the organic product(s) of the indicated transformation. Be sure to indicate the correct stereochemistry. (If the product is formed as a racemic mixture, please indicate this!)

Br Br +

Br Br2 Br (racemic mixture) Br Br ** Note: The chair forms are shown just + to give you extra practice converting planar drawings to chairs. Br Br

Br Br +

OH Br2,H2O OH (racemic mixture) OH Br +

Br OH

OH OH +

1. BH3,THF – (racemic mixture) 2. H2O2,OH H

OH + OH H

(achiral)

H2 ,Pt/C Chem S-20ab Purple Book Solutions Week 2 - Page 10 of 27 Stereochemistry of Alkene Additions

1. Fill in each box with the starting materials thatwouldleadtotheindicatedproduct.Besuretoindicatethe correct stereochemistry.

H Br Br2

Br H

HO H 1) BH3,THF

- 2)H2O2,OH Me H (+/-)

H D2, Pt/C D D (+/-)

Br O HO Me 2 Me Et Et Br H (+/-) Chem S-20ab Purple Book Solutions Week 2 - Page 11 of 27 Cyclohexanes 1. Look at a chair. Notice the axial and equatorial substituents. Pay close attention to the parallel lines!

A chair has 3 sets of parallel lines, plus vertical lines for axial groups. Once that is drawn in, every subsequent line must be parallel to a line that has already ben drawn! Try drawing t-butyl groups in the equatorial positions. 2. Draw each of the following dimethyl cyclohexanes (on the planar ring), then identify whether the substituents would be either: "Axial and Equatorial" or "Both Equatorial" 1,2-cis 1,3-cis 1,4-cis

Ax. and Eq. Both Eq. Ax. and Eq.

1,2-trans 1,3-trans 1,4-trans

Both Eq. Ax. and Eq. Both Eq. 3. Draw each of the following in the most stable chair conformation:

Et Chem S-20ab Purple Book Solutions Week 2 - Page 12 of 27 Chair Conformations

1. Consider the following molecule: CH3

Cl

H3C CH3 a) This molecule is expected to have two relatively stable "chair" conformations. Draw clear representations of the two different conformations in the boxes below. (The boxes have been arbitrarily labeled A and B.) In addition, please be sure to draw the isopropyl substituent in its best conformation in each case. AB

H3C

H3C H H Cl

CH3

H3C H Cl H3C H

CH3

b) Which of the two chair conformations is more stable? (circle one of the statements below)

1) A is more stable than B. 2) Both conformations are equally stable. 3) B is more stable than A. Chem S-20ab Purple Book Solutions Week 2 - Page 13 of 27 Bicyclic Compounds and Bredt's Rule

1. Draw each of the following bicyclic compounds in a good "perspective" drawing.

CH3 CH3 trans-decalin: two chairs

CH3 CH3

CH3

norbornane skeleton:

CH H3C 3 CH3

2. The molecule trans-cyclooctene is known to exist. (It is chiral, by the way). Why is the analagous molecule trans-cyclohexene unstable? Far too strained to have a trans-alkene in a six-membered ring.

H H = H H H H

trans-cyclooctene trans-cyclooctene trans-cyclohexene???? (planar representation) (perspective representation)

3. Draw each of the following bicyclic alkenes in a good "perspective" representation. Only one of these three compounds actually exists. Which one, and why?

In these two, the alkenes can't be planar. Highly strained. (Try to model!) BAD! BAD! Ok!

Or, notice the ** Bredt's Rule: trans-alkenes Can't have sp2 carbon at in small rings. a bridgehead. (No alkenes, no carbocations!) Chem S-20ab Purple Book Solutions Week 2 - Page 14 of 27 More Practice With Reaction Stereochemistry

1. Consider compounds A and B shownontheright: a) Is A chiral? (circle) Yes No b) Would a solution of B in a polarimetry cell at room temperature Yes No AB rotate plane-polarized light? (circle) c) What is the stereochemical relationship of A and B? Diastereomers

Draw the products of the following transformations. Be sure to include all unique stereoisomers that will be produced in each reaction. Br Br

Br2 Br d) A Br CH2Cl2 diastereomers

achiral achiral

OH H2O e) B + H3O

chiral diastereomers

H2 f) A Pd/C achiral achiral

diastereomers

H2 g) B Pd/C same compound achiral

h) Label each of the products above as chiral or achiral as appropriate. (You do not need to worry about labeling anything meso.) i) For all products of the same formula in d-g above, identify pairwise stereochemical relationships. Chem S-20ab Purple Book Solutions Week 2 - Page 15 of 27

Introduction to SN2, E2, SN1, and E1 Mechanisms 1. Provide complete curved-arrow mechanisms for the following transformations.

Br NaOH OH

:OH

Br KOtBu

H :OtBu

H2O Br OH

:OH2 :OH2 H O

H

conc. H2SO4

OH

H :sol H OH2 Chem S-20ab Purple Book Solutions Week 2 - Page 16 of 27 Mechanisms and Stereochemistry

1. Predict the product.

– H CH3 CH3S H CH3

I SCH3

2. Provide a curved-arrow mechanism, and explain the observed stereochemistry.

Br S :OH :OH– S HO O SH H O O + H :O With SN2, "retention" is ALWAYS "double inversion" O :OH Br O O

O SH O S: :O SH

O 3. Explain the difference in reactivity.

Br but Br KOtBu KOtBu

no reaction!

Br axial Br equatorial Br can't get antiperiplanar perfect for Br antiperiplanar with any H, so can't E2 elimination do E2 elimination. H :B 4. Explain the difference in reactivity.

conc. conc. but no reaction! H SO 2 4 H2SO4 HO OH This tertiary carbocation E1 elim. OK, with cannot be planar, secondary carbocation and the product violates which can be planar + Bredt's Rule (as required) + Chem S-20ab Purple Book Solutions Week 2 - Page 17 of 27 Choosing a Reaction Pathway 1 Consider the following conditions:

for SN2 reactions: need good steric interaction (Me, 1° is good, 2° OK, 3° bad)

for E2 reactions: need strong attacking base (at least as strong as OH–)

for SN1/E1 reactions: need stable carbocation (3° is good, 2° is OK, 1° is bad) All things being equal, choose the reaction higher up on the above list. For each of the following, predict the major product, and indicate the type of mechanism.

– 1. CH3I+OH CH3OH excellent SN2

horrible f or SN2, but no other reactive pathway 2. CH3I+ – exists O OCH3

OH 3. Br +OH– good SN2

Br – CN excellent S 2 4. +CN N

Br nucelophile much too bulky f or SN2, so 5. + E2 prevails O–

6. Br +OH– when nucleophile is also a strong base, secondary substrates will pref er E2 to SN2

excellent nucleophile that is not basic, so S 2 – N 7. Br +CN CN will dominate

tertiary substrate cannot undergo S 2 reaction, so – N 8. +OH E2 is the only pathway available Br

SN1 and E1 always compete, but SN1will + 9. +H2O always dominate in such situations Br OH major minor

S 1 and E1 always compete, but S 1will + N N 10. +CH3OH always dominate in such situations Br OCH3 major minor

alcohols dehydrate in mineral acids only via E1 11. +H2SO4 OH Chem S-20ab Purple Book Solutions Week 2 - Page 18 of 27 Choosing a Reaction Pathway 2

1. Which product (or) products would you expect to obtain from each of the following reactions? In each case identify the mechanism by which each product is formed (SN1, SN2, E1, or E2) and predict the relative amount of each (i.e. would the product be the only product, major product, minor product, etc?). a) NaOMe + Br OCH3

SN2major E2 minor

t b) KO Bu Br E2 only product tBuOH

c) NaOEt (CH3)3CCl E2 only product

d) EtOH +

Br OEt E1 minor (transor E favored S 1major N over cis or Z) ***Not always true in E2 elimination, where the anti-periplanar conf ormation is required!!!***

t e) KO Bu + + tBuOH Br E2 major, E2 minor, Zaitsev product E2 minor, Hofmann product Zaitsev product (KOtBu is *both Zaitsev products are bulky enough possible here since there are to give some two protons available f or Hofmann product) deprotonation, each giving rise to a different alkene geometry Chem S-20ab Purple Book Solutions Week 2 - Page 19 of 27 Substitution and Elimination: Synthesis 1. Each of the following products can be synthesized in one step from an alkyl halide. Show the starting material that could serve as an immediate precursor to the indicated target, and fill in the reactions conditions required to obtain the desired product. Be sure to consider stereochemistry where relevant!

H3C H 1. NaCN X CN

*SN2, so remember to account f or inversion! (where X = Cl, Br, I)

KOH, NaOH, NaOR, t X 1. or KO Bu etc.

(where X = Cl, Br, I) *Since no reaction pathway is possible except for E2, so any strong base is okay.

X

1. H2O OH

(where X = Cl, Br, I) *Stereochem is unimportant here since this is a tertiary center, making only an SN1 mechanism operative. Chem S-20ab Purple Book Solutions Week 2 - Page 20 of 27 Carbenes and Cyclopropanes

1. Chloroform can react with strong bases to yield dichlorocarbene, :CCl2. Provide a curved-arrow mechanism for this transformation: Cl Cl t CH KO Bu Cl C Cl Cl

t :O Bu Cl

C Cl Cl

This is an alpha-elimination.

2. Dichlorocarbene can react with alkenes to yield cyclopropanes. For instance: H Cl Cl + C Cl Cl H This reaction involves two significant donor-acceptor orbital interactions between the two reactants. Identify the donor and acceptor orbitals in each interaction. • Include both a verbal description and a picture of each orbital.

Cl Donor:Interacts With Acceptor: Cl Carbene Alkene * nC C-C

Cl

Donor:Interacts With Acceptor: Cl Alkene Carbene C-C 2pC Chem S-20ab Purple Book Solutions Week 2 - Page 21 of 27 Putting It Together: Products

1. Fill in each box with the major organic product of the indicated transformation. Be sure to indicate stereochemistry when relevant.

Br NaOH

CH3 CH3

The Zaitsez product will dominate.

H+ O HO OH

Mechanism? Intramolecular SN1

1. Zn CH2I2 2.

Form carbenoid, then use it to make cyclopropane

1. Mg, Et2O

2. H2O Br

Form Grignard, then protonate it-get an alkane! Chem S-20ab Purple Book Solutions Week 2 - Page 22 of 27 More Putting It Together: Products

1. Fill in each box with the single major organic product of the indicated transformation. (Any chiral starting materials are provided as single, pure enantiomers.) Be sure to give the stereochemistry of the product if it is relevant!

H3C Br

a) H3C OH HBr Br CH3

Racemic mixture (from SN1reaction)

H3C CH3 H3C CH3 NaN3 b)

I I N3 I (those are iodine substituents)

CH CH 3 KOtBu 3 c)

Cl Note: The Zaitsev product is not possible here, since there is no H anti to the leaving group on the more substituted carbon.

H

t 1. KO Bu H3C H d) (+/-) 2. IZnCH2I Cl Note: While you will get some of the Hofmann elimination product, Zaitsev will still be the most prominent. Chem S-20ab Purple Book Solutions Week 2 - Page 23 of 27 Making Alcohols into Leaving Groups

1. The OH from an alcohol is not a good leaving group. Why?

Good leaving groups are weak bases, but OH- is a strong base!

One way of making it a good leaving group is to protonate it. This often creates problems with carbocations, rearrangements, etc. It can be used in limited cases:

2. Provide curved-arrow mechanisms for the following transformations:

HClOH HCl Cl OH Cl HCl HCl

OH2 :Cl OH2 :Cl

The best ways of turning alcohols into leaving groups (at least for SN2 reactions) are to use either SOCl2 or to form a sulfonate ester (most often a "tosylate"):

3. Provide the intermediates and products in the following synthetic sequences. Watch stereochemistry!

OH Cl CN SOCl2, pyridine NaCN

OH OTs CN TsCl, pyridine NaCN Chem S-20ab Purple Book Solutions Week 2 - Page 24 of 27 Making Alcohols into Leaving Groups: Mechanisms

1. Provide a complete curved-arrow mechanism for each of the following transformations. Be sure to pay attention to stereochemistry! R OH O2S Cl OTs TsCl, pyridine

O

TsCl = Cl S CH3 =ClSO2R O

pyr: O O O O Pyridine is: H S S O R O R N

O

OH S Cl Cl Cl SOCl2, pyridine

pyr: +SO2 (g) O O +Cl– H S S O Cl O Cl

SN2 displacement inverts stereochemistry Cl:–

Br

OH P Br Br Br PBr3

Br

H P + HPOBr2 O Br

SN2 displacement inverts stereochemistry Br:– Chem S-20ab Purple Book Solutions Week 2 - Page 25 of 27 Putting It Together: Reagents 1. Each of the following transformations can be carried out in one or two steps. Fill in the reagents required for each step. If a second step is not needed, please put an "X" through the second box.

N N OH SOCl2 OR TsCl, pyridine N 1.

NaN3 2.

1. conc. H2SO4 OH

H ,Pd/C 2. 2

OH PBr OR conc. HBr Br 1. 3

2.

OH conc. H SO OH 1. 2 4

Br

Br2,H2O 2. (plus enantiomer)

OH conc. H SO 1. 2 4

H2,Pd/C 2. Chem S-20ab Purple Book Solutions Week 2 - Page 26 of 27 Ether Syntheses 1

1. Devise efficient syntheses for each of the following types of ether.

Primary—Primary example: O Choose either SN2 route (typical Williamson synthesis):

+ CH I O:– 3

Primary—Secondary example: O

Use the secondary alkoxide as the nucleophile, and a primary halide in an SN2 process:

+ O:– Br

Secondary—Secondary: tough! Either way gives a strong probability of elimination! Show SN2andSN1 example: O

+ + HO OH HO with acid, use large excess – + with acid, of the alcohol. O: Br likely to give some E1 (Can you show mechanism?) likely to give E2

Tertiary—anything

Use a tertiary alcohol or an alkene, a trace of acid, and lots of the other alcohol, in an SN1-type route: example: O

+ + HO OH HO with acid, use large excess with acid, of the alcohol. use large excess of (Can you show the primary alcohol mechanism?) Chem S-20ab Purple Book Solutions Week 2 - Page 27 of 27 Ether Syntheses 2

1. a) The following ethers can be synthesized by an SN2 reaction from two different combinations of an alkoxide and an alkyl halide. For each molecule, show the combinations of alkyl halide and alkoxide that could combinetoformbond1 and the combination that could combine to form bond 2 and circle the pair that will react to provide a higher yield of the ether.

SN2 Starting Materials: SN2 Starting Materials: 1 2 X O 1 O Me 2 + +

O X

SN2 Starting Materials: SN2 Starting Materials:

XCH3 Me Me OCH3 1 2 + Me + CH O Me CH H3C 3 H3C 3

O CH3 1 2 X CH3

b) Consider using an SN1 reaction to form the same ethers. For each molecule, show the combination of alkyl halide and alcohol that could combine to form bond 1 and the combination that will combine to form bond 2 and circle the pair that will react to provide a higher yield of the ether.

SN1 Starting Materials: SN1 Starting Materials: 1 2 X OH 1 O Me 2 + +

OH X

SN1 Starting Materials: SN1 Starting Materials:

XCH3 Me Me HO CH3 1 2 + Me + CH O Me CH H3C 3 H3C 3

HO CH3 1 2 X CH3