purplebook-week2-ans
TRANSCRIPT
CH3H
CH3H
H
H3CCl
F
Br
Br
H
H
Br
H
H
Br
CH3
H3C CH3
H3C
H3C H
Chirality
H
CH3ClF
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.
chiral
chiral
Br
Br
H
H
achiral
chiral
achiral ("meso")
chiral
H3C
CH3H3C
CH3
Week 2 47
H
H
HCl
O
NHO
H3C
OHH
2. Draw an unambiguous structural representation of the following molecules:
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.
(S)-4,5-dimethyl-(Z)-2-hexene
H
R
R
R
SS
S
R
S
S
(R)-3,5,6-trimethyl-(E)-3-heptene
H
3. Provide the best possible systematic name for the following compound:
(S)-(E)-5-isopropyl-2,3,7-trimethyl-4-octene
Week 2 48
CH3 CH3
H3CCH3
Br
Br
H3CBr
Br
CH3
Enantiomers and Diastereomers
1. For each of the following pairs of stereoisomers, indicate if they are enantiomers, diastereomers, or identical compounds.
diastereomers
diastereomers
identical
enantiomers
identical
Week 2 49
Identifying Isomers
H
H
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
Identical Enantiomers Diastereomers None
Identical Enantiomers Diastereomers None
Identical Enantiomers Diastereomers None
HN
H
Me
HN
Me
H
HN
Me
H
HN
Me
H
Identical Enantiomers Diastereomers None
Identical Enantiomers Diastereomers None
Week 2 50
CH3 CH3
1. Which of the following are true? Give counterexamples for those that are false.
(a) In some cases, constitutional isomers are chiral.
(b) In every case, a pair of enantiomers have a mirror-image relationship.
(c) Mirror-image molecules are in all cases enantiomers.
(d) If a compound has an enantiomer, it must be chiral.
(e) Every chiral compound has a diastereomer.
(f) If a compound has a diastereomer it must be chiral.
(g) Any molecule containing an asymmetric carbon must be chiral.
(h) Any molecule with a stereocenter must have a stereoisomer.
(i) Some diastereomers have a mirror image relationship.
(j) Some chiral compounds are optically inactive.
(k) All chiral molecules have no plane of symmetry.
(l) If a compound has a stereocenter, it has an enantiomer.
(m) A meso compound will necessarily have at least two diastereomers.
H
H3CCl
F
Stereochemical Concepts
Br
H
H
Br
identical
has no diastereomer
and
are diastereomers
achiral ("meso")
Br
H
H
Br
F; no diastereomers have a mirror image relationship.
F; all PURE chiral compounds are optically active
achiral ("meso")
T
T
F
T
F
F
F
T
T
F
T
Week 2 51
2,3-butanediol
H3C
HO OH
CH3
Working with Stereoisomers
1. Consider the species 2,3-butanediol, whose structural formula is:
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 labeled A, B, and C.)
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:
A B C
What characteristic of the other two stereoisomers causes them to share the same melting point (19°C)?
H3C
HO OH
CH3 H3C
HO OH
CH3 H3C
HO OH
CH3
B and C are enantiomers, and enantiomers always have identical physical properties;for instance, they will have the same melting point.
A B C
Week 2 52
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.
Two molecules which are enantiomers would exhibit different rates of reactivity with an achiral reagent.
An achiral starting material can react with an achiral reagent to give a single chiral product.
Two molecules which are enantiomers would exhibit identical reaction rates when reacting with a chiral reagent (such as a biological enzyme).
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
diastereomers
single achiral product
or
racemic mixture of products
different
enantiomers
Week 2 53
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.
OH Br2
O
Br
Chemical Formula: C6H11BrO
b) Provide a complete curved arrow mechanism for the transformation indicated above. Explain the stereochemical outcome, if any.
OH
Br Br
OH
Br
O
Br
Hsolv
O
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.
Bromonium ion can form above or below the plane.
OH attacks from the opposite plane, either from above or below, leading to a racemic mixture.
Week 2 54
Br
Br Br
Br
OH
Br HO
Br
OH
OH
H H
H
H
Stereochemistry of Alkene Additions
2. H2O2, OH–
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
Br2, H2O
Br
1. BH3
Br
H2 / Pt
+
(racemic mixture)
+
OH
(racemic mixture)
Br
+
OH
(racemic mixture)
Br
(achiral)
H
** Note: I have shown theserings as planar hexagons toemphasize the syn- or anti-relationships. I have also drawnthe proper chair forms below.
H
+
OH
+
OH
+
Br2
Week 2 55
Cyclohexanes
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
1,2-trans 1,3-trans 1,4-trans
3. Draw each of the following in the most stable chair conformation:
1. Look at a chair. Notice the axial and equatorial substituents. Pay close attention to the parallel lines!
Ax. and Eq. Ax. and Eq.
Ax. and Eq.
Both Eq.
Both Eq. Both Eq.
Notice the confomation of thatethyl group: avoid syn-pentane!
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.
Week 2 56
1. Consider the following molecule:
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.
b) Which of the two chair conformations is more stable? (circle one of the statements below)
Chair Conformations
CH3
H3C CH3
1) A is more stable than B.
2) Both conformations are equally stable.
3) B is more stable than A.
Cl
Cl
CH3
H
H3C
H3C
H
H
H
H3C
H3C
Cl
CH3
A B
Week 2 57
CH3
CH3
CH3
CH3
H
H
Bicyclic Compounds and Bredt's Rule
H
H
1. Draw each of the following bicyclic compounds in a good "perspective" drawing.
H
H
2. The molecule trans-cyclooctene is known to exist. (It is chiral, by the way). Why is the analagous molecule trans-cyclohexene unstable?
=
trans-cyclooctene(planar representation)
CH3
CH3
trans-cyclooctene(perspective representation)
H3C CH3
trans-cyclohexene????
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?
trans-decalin: two chairs
norbornane skeleton:
Far too strained to have a trans-alkene in a six-membered ring.
BAD! BAD!Ok!
In these two,the alkenescan't be planar.Highly strained.(Try to model!)
Or, notice thetrans-alkenesin small rings.
** Bredt's Rule:
Can't have sp2 carbon at
a bridgehead. (No alkenes,
no carbocations!)
Week 2 58
1. Consider compounds A and B shown on the right:
Chairs and Stereochemistry
A B
a) Is A chiral? (circle) Yes No
b) Would a solution of B in a polarimetry cell at room temperature rotate plane-polarized light? (circle)
Yes No
c) What is the stereochemical relationship of A and B? Diastereomers
Draw the products of the following transformations in their lowest energy conformations. Be sure to include all unique stereoisomers that will be produced in each reaction.
d) A
Br2
CH2Cl2
e) B
H2O
H3O+
f) A
H2
Pd/C
g) B
H2
Pd/C
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.
CH3
H3C
CH2Br
Br
CH3
H3C
Br
CH2Br
achiral achiral
diastereomers
CH3
CH3 OH
CH3 only one compound and one lowest energy conformer.
chiral
CH3
H3C CH3
CH3
H3C
CH3achiral achiral
diastereomers
CH3
H3C
CH3achiral
same compound
diastereomers
Week 2 59
Br OH
Br
Br OH
OH
An Introduction to SN2, E2, SN1, and E1 Mechanisms
H
1. Provide complete curved-arrow mechanisms for the following transformations.
conc. H2SO4
OH
H
+ Br–
tBuO:–+ Br–
+ tBuOH
+
OH
H
+
:Solvent
+
H
+
:Solvent
OH–
KOtBu
H2O
HO:–
:OH2
H+
Week 2 60
I
CH3H
Br
O
O
SHHO
O
S
Br Br
OHHO
Mechanisms and Stereochemistry
SCH3
CH3H
H
1. Predict the product.
3. Explain the difference in reactivity.
Br
–:O
O
SH
2. Provide a curved-arrow mechanism, and explain the observed stereochemistry.
no reaction!
SH
O
O
but
S:–
O
O
conc.
4. Explain the difference in reactivity.
conc.no reaction!
–:O
O
S
but
Br
H
HO—H
axial Brperfect forantiperiplanarE2 elimination
With SN2, "retention" is ALWAYS "double inversion"
Br
equatorial Brcan't get antiperiplanarwith any H, so can'tdo E2 elimination.
E1 elim. OK, withsecondary carbocationwhich can be planar(as required)
+
This tertiary carbocationcannot be planar,and the product violatesBredt's Rule +
CH3S–
:OH–
KOtBu KOtBu
H2SO4 H2SO4
:OH–
:OH–
:B
Week 2 61
+ OH–
Choosing a Reaction Pathway 1
Consider the following conditions:
for SN2 reactions:
for E2 reactions:
O–
need good steric interaction (Me, 1° is good, 2° OK, 3° bad)
Br
Br
Br
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.
3.
O–
2.
Br
Br
4.
5.
6.
7.
Br8.
Br9.
Br
OH11.
CH3I + OH–
CH3I +
+ OH–
+ CN–
10.
+
+ CN–
+ OH–
+ H2O
+ CH3OH
+ H2SO4
OCH3
rule 1
rule 2
rule 3
sterics poor, but no possibility for E2 or SN1/E1
sterics are only OK, but base is not so strong, so SN2
tertiary carbocation stable, mainly SN1
OH
CN
CN
OH
OCH3
excellent SN2
excellent SN2
sterics make SN2 bad, so goes E2
excellent SN2
sterics terrible; E2 all the way
With secondary carbon, get some SN2, but the main
product is likely to be the alkene from E2 elimination.
tertiary carbocation stable, mainly SN1
tertiary carbocation stable, conc. sulfuric favors E1
CH3OH
Week 2 62
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)
Br + CH3ODMSO
25° COCH3
+
SN2: major E2
b)
Br + (CH3)3CO25° C
E2
(CH3)3COH
c)+(CH3)3CCl
25° C
E2
(CH3)3COHCH3O
d)+
50° C
EtOHEtOH
Br OEt
SN1
+
E1
+
E1: minor
e)+
25° C
(CH3)3COH(CH3)3CO
Br
E2: major(saytzeff)
+
E2: minor
+
E2: minor
Week 2 63
Substitution and Elimination: Synthesis
CN
H3C H
OH
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!
1.
I
1.
H3C H
Br
1.Br
KOH or anystrong base (E2)
(could also be E1)
H2O (SN1)
(SN2, so inversion!)
NaCN
Week 2 64
C
Cl
Cl
H
H
Cl
Cl
Carbenes and Cyclopropanes
1. Chloroform can react with strong bases to yield dichlorocarbene, :CCl2. Provide a curved-arrow
mechanism for this transformation:
2. Dichlorocarbene can react with alkenes to yield cyclopropanes. For instance:
+
This reaction involves two significant donor-acceptor orbital interactions between the two reactants. Identify the donor and acceptor orbitals in each interaction.
CH
Cl
Cl
Cl C
Cl
Cl
• Include both a verbal description and a picture of each orbital.
Donor: Acceptor:
Donor: Acceptor:
Interacts With
C
Cl
Cl
Cl—
Carbene lone pair
(sp2-hybridized)ClCl
Interacts With
p* orbital of alkene
ClCl
Carbene vacant p-orbitalp orbital of alkene
This is the (perhaps surprising) geometry requiredfor all the orbital overlap between the carbene and thealkene orbitals!
KOtBu
tBuO–
Week 2 65
CH3
Br
Br
Putting It Together: Products
OHHO
1. Fill in each box with the major organic product of the indicated transformation. Be sure to indicate stereochemistry when relevant.
CH3
1. Mg
2. H2O
O
often prefer more subst. alkene
Mechanism? Intramolecular SN1
Form carbenoid, then use it to make cyclopropane
Form Grignard, then protonate it-get an alkane!
1. Zn
2.
NaOH
H+
CH2I2
Week 2 66
More Putting It Together: Products
I I
CH3
Cl
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!
a)
b)
c)
d)
CH3
H3C CH3NaN3
N3 I
H3C CH3
(those areiodine substituents)
OHH3C
BrH3C
CH3Br
Racemic mixture (from SN1 reaction)
HBr
KOtBu
1. NaOEt
2. IZnCH2I H3C
H
H
Cl
Week 2 67