chap5a

Views:
 
Category: Education
     
 

Presentation Description

No description available.

Comments

By: radharajasekarram (111 month(s) ago)

This presentation is very informative.

Presentation Transcript

Chapter 5 Stereoisomerism and Chirality: 

Chapter 5 Stereoisomerism and Chirality

Donald Cram: 

Donald Cram Source: Michigan State University, Department of Chemistry http://www.chemistry.msu.edu/Portraits/PortraitsHH_collection.shtml

Chapter 5: 

Chapter 5 Skip p. 388-389 Read p. 475-477 for interest Don’t memorize names of sugars (p. 460-465) or amino acids (p. 472-474)

Chapter 5 Problems: 

Chapter 5 Problems 1 a, b, c, e, g, m 2 4 a, b, c, d, f, g, i 5 a, b, e 11, 12 13 b, c, d 14 - 18 20 a, b, d 23 - 28 30 32 - 36 38 40, 41 43 45 48 52 59

Sect 5.1: Symmetry and Chirality: 

Sect 5.1: Symmetry and Chirality If a molecule is superimposable on its mirror image, it is achiral (non-chiral). If a molecule is not superimposable on its mirror image, it is chiral. Enantiomers are isomers that are non-superimposable mirror-images

Slide6: 

Sect 5.2: Enantiomers Four different atoms are attached to a chiral carbon atom. Mirror images are non-superimposable. rotate this molecule is chiral note that the fluorine and bromine have been interchanged in the enantiomer

Slide7: 

Stereocenters The previous molecule has a stereocenter, and is chiral. A stereocenter is an atom, or a group of atoms, that can potentially cause a molecule to be chiral. stereocenters - can give rise to chirality

Slide8: 

A carbon atom with three identical groups is achiral! There is a symmetry plane in the plane of the paper. An achiral molecule

Slide9: 

plane of symmetry side view edge view

Slide10: 

Another achiral molecule These two structures are superimposable! This molecule has a plane of symmetry in the plane of the paper.

Slide11: 

* Sect 5.3: identification of stereocenters * Look for carbon atoms with four different groups!

Slide12: 

* * *

Slide13: 

Finding stereocenters plane ( indicates no stereoisomers )

Slide14: 

* * * * * * Finding stereocenters The cis isomer (two methyl groups) - achiral! Thetrans isomers has two stereocenters!

Slide15: 

Finding stereocenters * * * * * * * *

Sect. 5.4 Properties of Enantiomers: 

Sect. 5.4 Properties of Enantiomers Enantiomers interact differently with polarized light. Enantiomers have equal magnitude, but opposite signs of rotation Most other properties are identical. Odor may be different!!

Sect. 5.5 Polarimetry: 

Sect. 5.5 Polarimetry dissolve sample in a solvent and put sample into polarimeter obtain sign of rotation value depends on concentration and path length.

Optical Activity: 

Optical Activity

Slide19: 

Dextrorotatory Rotates the plane of plane-polarized light to the right. (+)- d- Levorotatory Rotates the plane of plane-polarized light to the left. (-)- l- TYPES OF OPTICAL ACTIVITY new older new older

Slide20: 

[a]D = a cl Specific Rotation [a]D a = observed rotation c = concentration ( g/mL ) l = length of cell ( dm ) D = yellow light from sodium lamp t = temperature ( Celsius ) t Specific rotation calculated in this way is a physical property of an optically active substance. This equation corrects for differences in cell length and concentration. You always get the same [a]D t value of

Slide21: 

ENANTIOMERS HAVE EQUAL VALUES BUT OPPOSITE SIGNS OF ROTATIONS W C X Z Y W C X Y Z (+)-numbero (-)-numbero dextrorotatory levorotatory Enantiomers The numbers are the same, but have opposite signs. All other physical properties are IDENTICAL.

Slide22: 

Racemic mixture an equimolar (50/50) mixture of enantiomers [a]D = 0o the effect of each molecule is cancelled out by its enantiomer

Sect. 5.6 Configuration: 

Sect. 5.6 Configuration Arrangement in space of atoms or groups around the stereocenter of the molecule Enantiomers have different configurations.

Sect. 5.7 Specification of Configuration: Cahn-Ingold-Prelog rules : 

Sect. 5.7 Specification of Configuration: Cahn-Ingold-Prelog rules

Sir Christopher Ingold: 

Sir Christopher Ingold Source: Michigan State University, Department of Chemistry http://www.chemistry.msu.edu/Portraits/PortraitsHH_collection.shtml

Specification of Configuration: 

Specification of Configuration SEQUENCE RULE 1: priority depends on the atomic numbers of the 4 atoms attached to the stereocenter; atom with higher atomic number receives the higher priority. If two atoms are isotopes of the same element, the heavier isotope is assigned the higher priority.

Slide27: 

R S

Specification of Configuration: 

Specification of Configuration SEQUENCE RULE 2: If the relative priority of two groups cannot be decided by Rule 1, it shall be determined by a similar comparison of the next atoms in the groups (and so on, if necessary), working outward in ranks from the stereocenter.

Specification of Configuration: 

Specification of Configuration SEQUENCE RULE 3: A doubly-bonded atom A is treated as if there were two C-A single bonds. Priorities in the expanded representations are assigned on the basis of Rule 2.

Remember! : 

Remember! The atoms shown in parentheses (the duplicate representations) do not exist! They are written only for purposes of assigning priorities.

More...: 

More... A triply-bonded atom A is treated as if there were three C-A bonds, as in:

Corollary of Rule 3: 

Corollary of Rule 3 If no other distinction can be made, a real atom outranks a “fictional” atom. NOTE CAREFULLY: This exception is used only as a last resort! You will rarely see this happen.

Sect. 5.8: Compounds with more than one stereocenter: 3-Chloro-2-butanol: 

Sect. 5.8: Compounds with more than one stereocenter: 3-Chloro-2-butanol S R R S R R S S

Slide34: 

3- Chloro-2-butanol: Fischer formulas (Sect. 5.9) pair of enantiomers-1 pair of enantiomers-2

Slide35: 

How Many Stereoisomers Are Possible? maximum number of stereoisomers = 2n, where n = number of stereocenters sometimes fewer than this number will exist For the previous example; two stereocenters = 4

Slide36: 

* * 22 = 4 stereoisomers * * * 23 = 8 stereoisomers Determining the number of possible stereoisomers

Slide37: 

meso enantiomers diastereomers S R S S R R 2,3-Dichlorobutane

2,3-Dichlorobutane: 

2,3-Dichlorobutane meso

Slide39: 

2,3-Dichlorobutane: Fischer formulas (Sect. 5.9) Meso! “Pair” becomes one! pair of enantiomers

Slide40: 

(as a minor component) ALSO FOUND (+)-tartaric acid (-)-tartaric acid Tartaric Acid (from fermentation of wine): There are three stereo-isomers [a]D = 0 meso -tartaric acid

Sect 5.9: Fischer Formulas: 

Sect 5.9: Fischer Formulas

Slide42: 

H O H O H H H C H 2 O H C H O O H EVOLUTION OF THE FISCHER PROJECTION Substituents will stick out toward you like prongs Fischer Projection Main chain bends away from you “Sawhorse” Projection Orient the main chain vertically with the most oxidized group at the top.

Slide43: 

Mirror images (enantiomers) are created by switching substituents to the other side. OPERATIONS WITH FISCHER PROJECTIONS All stereocenters must be switched to get an enantiomer (the mirror inverts them all). diastereomers enantiomers If you switch only one of the stereocenters, but not both, you get a diastereomer.

Slide44: 

Rotation by 180o in the plane of the paper does not change the molecule. No other rotation is allowed. 180o Legal operations with Fischer formulas .

Slide45: 

The molecule does not have an enantiomer because it is not chiral. It had a plane of symmetry, rendering it a meso molecule. This molecule is meso! There isn’t an enantiomer plane of symmetry

Slide46: 

H C H 2 O H CHO O H To determine the configuration (R/S). Make two “switches”. Place the priority 4 group in one of the vertical positions. H O H OHC C H 2 O H 1 2 3 4 1 2 3 4 R alternatively: H C H 2 O H CHO O H 1 2 3 4 HOCH2 CHO O H 2 1 4 3 H R Both H’s are in back = same result R

Slide47: 

Draw all of the stereoisomers!

Slide48: 

Any more?? NO!

Slide49: 

There are only four stereoisomers!

Sect 5.10: cyclic compounds: 

Sect 5.10: cyclic compounds

Slide51: 

plane ( indicates no stereoisomers ) 20 = 1 21 = 2 21 = 2 stereoisomers (max) :

Slide52: 

* * * * * * plane ( but only if cis ) 22 = 4 22 = 4 22 = 4 stereoisomers (max) :

Slide53: 

cis trans diastereomers 1-Bromo-2-chlorocyclopropane note that the cis/trans isomers are also diastereomers R S S R R R S S

Slide54: 

Br B r B r Br Br B r B r Br mirror image identical meso diastereomers 1,2-Dibromocyclopropane cis trans

Slide55: 

1-Bromo-2-chlorocyclohexane diastereomers cis trans cyclohexanes may be analyzed using planar rings

Slide56: 

1,2-dichlorocyclohexane cis trans meso diastereomers mirror image identical

Slide57: 

Interchanges: See problem 59 CH3 Br F H CH3 F Br H enantiomer ONE STEREOCENTER 1..3..5…etc interchanges = enantiomer 2..4..6...etc interchanges = original compound Odd: Even:

Slide58: 

CH3 Br F H CH3 H F Br PROBLEM: ARE THESE IDENTICAL OR ARE THEY ENANTIOMERS? You can use interchanges to answer this question! Next slide.

Slide60: 

CH3 Br F H CH3 H F Br Back to the problem: are these identical or are they enantiomers? They are enantiomers! Three switches = enantiomers

Slide61: 

MORE THAN ONE STEREOCENTER To form an enantiomer you must interchange all of the stereocenters. enantiomer two stereocenters two interchanges

Slide62: 

If a compound has more than one stereocenter, and you only interchange one of them, you willI form a distereomer. two stereocenters one interchange diastereomer

Sect 5.11: Stereocenters other than carbon: 

Sect 5.11: Stereocenters other than carbon

Sect 5.12: Other types of chirality: 

Sect 5.12: Other types of chirality

Sect 5.13: Resolution (Skip): 

Sect 5.13: Resolution (Skip)

Sect 5.14-5.16: carbohydrates: 

Sect 5.14-5.16: carbohydrates

Emil Fischer: 

Emil Fischer Source: Michigan State University, Department of Chemistry http://www.chemistry.msu.edu/Portraits/PortraitsHH_collection.shtml

Glyceraldehyde (an aldotriose): 

Glyceraldehyde (an aldotriose)

Slide69: 

Notice that the D enantiomer has the R configuration in the R/S sytem.

The Aldotetroses: 

The Aldotetroses Notice that the D enantiomers are (-)! Notice that the L Enantiomers are (+)!

Slide71: 

There is no relationship between D or L and the sign of rotation in a polarimeter! D can either be (+) or (-) L can either be (+) or (-)

Fig. 5-27: the aldopentoses: 

Fig. 5-27: the aldopentoses

Slide73: 

There are four natural pentose sugars. They have the D configuration. OH points to right! D OH points to right! D OH points to right D OH points to right D

Fig 5-28 “natural” D-aldohexoses: 

Fig 5-28 “natural” D-aldohexoses

The “unnatural” L-aldohexoses : 

The “unnatural” L-aldohexoses

Sect 5.17: natural products and amino acids: 

Sect 5.17: natural products and amino acids

Slide77: 

Amino acids belong to the L-series. Can You assign the R or S configuration to the following Fischer structure of Alanine? enantiomer Same! The original structure is, therefore, (R) (S)

Slide78: 

These compounds are enantiomers! They have identical physical properties, except for odor and rotation of light in a polarimeter!

authorStream Live Help