18 Isomerism and stereochemistry

Transcription

1 s manual for Burrows et.al. hemistry Third edition 8 Isomerism and stereochemistry s to worked examples WE 8. Structural isomers (on p. 88 in hemistry ) For the following four compounds, A D, identify which are chain isomers, which are position isomers and which are functional group isomers. (Ph stands for phenyl, 6 5.) Ph Ph Ph Ph A B D Determine the molecular formulae of these molecules:. To be structural isomers, these molecules must have the same molecular formulae.. To be position isomers, these molecules must have the same functional groups but at different positions.. To be functional group isomers, these molecules must have different functional groups.. Molecules A, and D are structural isomers as they have the same molecular formulae ( 9 0 ). Molecule B is not a structural isomer since it has a different molecular formula ( 0 ).. Molecules A and are positional isomers as they both contain the same functional groups [a carboxylic acid (- ) group and a phenyl (-Ph) group] but at different positions. xford University Press, 07. All rights reserved. i g h e r E d u c a t i o n

2 s manual for Burrows et.al. hemistry Third edition. Molecule D is a functional group isomer of both molecules A and B, as it contains different functional groups (an alcohol and an aldehyde in D versus a carboxylic acid group in both A and B).. A, and D are structural isomers all have the molecular formula, 9 0. A and are position isomers (the Ph group is at different positions). and D (and A and D) are functional group isomers ( is a carboxylic acid, D is an -hydroxyaldehyde). WE 8. Using Newman projections (on p. 85 in hemistry ) Use Newman projections to draw the eclipsed and staggered conformations of chloroethane. Explain which conformation you would expect to have the higher energy.. Draw out the skeletal structure of chloroethane.. Draw out the staggered Newman projection of chloroethane; this is where all the - bonds and -l bond are staggered.. Draw out the eclipsing Newman projection of chloroethane; this is where all the - bonds and the -l bond are eclipsed.. Determine which Newman projection is more stable.. The condensed structure of chloroethane is: l chloroethane. The staggered Newman projection is where the three distal - bonds (on the back carbon atom) are staggered (by 0 degrees) with the three proximal bonds (two - bonds and the one -l bond) on the front carbon atom. [Note: this distal carbon atom is not visible in this Newman projection.] xford University Press, 07. All rights reserved. i g h e r E d u c a t i o n

3 s manual for Burrows et.al. hemistry Third edition. The eclipsing Newman projection is where the three distal - bonds (on the back carbon atom) are eclipsed with the three proximal bonds (two - bonds and the one -l bond) on the front carbon atom. [Note: this distal carbon atom is not visible in this Newman projection.] steric strain between eclipsing bonds l chloroethane l staggered conformation l eclipsed conformation = proximal carbon = distal carbon. The staggered projection is more stable than the eclipsed projection as there is significantly less steric congestion between the distal and proximal bonds. Staggered projections are in energy minima and eclipsing projections are in energy maxima. Stable conformers are generally those which have their large groups staggered and positioned furthest away from each other. The eclipsed conformation is higher in energy than the staggered conformation because there is steric strain between the chlorine atom and a neighbouring hydrogen atom, and two further -, - eclipsions. l l steric strain staggered conformation eclipsed conformation xford University Press, 07. All rights reserved. i g h e r E d u c a t i o n

4 s manual for Burrows et.al. hemistry Third edition WE 8. E/Z nomenclature (on p. 86 in hemistry ) Draw a skeletal structure for (Z)--ethoxybut--en--ol.. Draw out the longest continuous carbon chain as a zig-zag structure and remember to number the carbon atoms.. Draw out its double bond at its designated position.. A (Z)-isomer is where both high priority substituents are on the same side of the double bond, and an (E)-isomer is where the high priority substituents are on opposite sides of the double bond. The priorities are determined using the ahn- Ingold-Prelog rules (See p. 89 in hemistry ).. Identify the substituents from their suffix, and place them along this carbon chain at their designated positions. 5. heck that this structure is correct!. The longest number of continuous carbon atoms is four [(Z)--ethoxybut--en--ol]. Draw out this carbon chain in a zig-zag arrangement.. The suffix of this molecule is -ol [(Z)--ethoxybut--en--ol]; it is therefore an alcohol. This molecule also contains a double bond [(Z)--ethoxybut--en--ol], and this is positioned at carbon ( ).. The double bond has (Z)-stereochemistry.. There is an ethoxy (-Et) group at carbon-. Et suffix = but--en--ol prefix -ethoxy -ethoxybut--en--ol now consider its stereochemistry Et on () higher priority than on () higher priority than both high priority groups on same side = (Z) (Z)--ethoxybut--en--ol xford University Press, 07. All rights reserved. i g h e r E d u c a t i o n

5 s manual for Burrows et.al. hemistry Third edition 5. heck that the structure is correct by re-naming it! This structure is (Z)--ethoxybut--en--ol. higher priority higher priority lower priority lower priority (Z)--ethoxybut--en--ol WE 8. Enantiomers and enantiomeric excess (on p. 8 in hemistry ) What is the observed specific optical rotation of a mixture containing 5% ( )-nicotine and 75% (+)-nicotine?. Work out which is the major enantiomer.. Work out the % optical purity of this sample.. Work out its relative composition (divide by 00 to give a ratio out of ).. Multiply the relative composition by its known specific rotation (for the major enantiomer). This will ensure that you have the correct sign. (+)-Nicotine has a specific rotation of The major enantiomer is (+)-nicotine.. The % optical purity of this enriched (+)-nicotine sample is 75% - 5% = 50%.. The relative composition of this sample is = 0.5. The specific rotation of this optically enriched sample of (+)-nicotine is = +8. enantiomeric excess, ee (%) = 75% 5% = 50% xford University Press, 07. All rights reserved. i g h e r E d u c a t i o n

6 s manual for Burrows et.al. hemistry Third edition 50 = observed [ ] D [ D of pure enantiomer x 00 = observed [ ] D +66 x x 66 observed [ ] D = 00 = +8 WE 8.5 R/S nomenclature (on p. 86 in hemistry ) Draw a hashed wedged line structure of butan--ol.. Draw out the longest continuous carbon chain as a zig-zag structure and remember to number the carbon atoms.. Identify any substituents from their suffix, and place them along this carbon chain at their designated positions.. Draw out the stereochemistry. This can be worked out using the ahn-ingold-prelog rules (See p. 89 in hemistry ).. heck that this structure is correct! 5. Draw out the hashed-wedged line structure of this molecule.. The longest number of continuous carbon atoms is four [butan--ol]. Draw out this carbon chain in a zig-zag arrangement.. The suffix of this molecule is -ol [butan--ol]; it is therefore an alcohol.. The chiral centre at carbon-- has stereochemistry. A configuration is where the three highest priority groups (, and ) on a particular conformation can be rotated clockwise ( ), whilst the lowest priority group,, is at the rear of this conformer. xford University Press, 07. All rights reserved. i g h e r E d u c a t i o n

7 s manual for Burrows et.al. hemistry Third edition suffix = butan--ol butan--ol now consider the group priority IP group priorities = (-) = (- ) = (- ) = (-) now consider its stereochemistry now redraw in a "hashed-wedged" structure Et Et butan--ol. heck that the structure is correct by re-naming it! This structure is (Z)--ethoxybut--en--ol. 5. The hashed-wedged line structure of this molecule is: butan--ol Draw in hashed-wedged notation and then rank the groups ( ) directly attached to the chiral centre and apply the IP rule: * first draw butan--ol and identify the chiral centre butan--ol WE 8.6 Enantiomers and diastereomers (on p. 85 in hemistry ) Ephedrine is a naturally occurring compound that is used to treat asthma. xford University Press, 07. All rights reserved. i g h e r E d u c a t i o n

8 s manual for Burrows et.al. hemistry Third edition (a) Assign a configuration to each of the chiral centres in ephedrine. N ephedrine. Identify the two chiral centres in ephedrine, and labelled them.. Redraw this molecule in its condensed form.. Work out the configuration for both chiral centres. It is easier to consider one centre at time. The (R/S)-stereochemistry is assigned using the ahn-ingold-prelog rules (See p. 89 in hemistry ).. heck that your assignments are correct.. The longest continuous carbon chain (of propan--ol) has been numbered (, and ) for ease. There are two chiral centres at carbons- and -.. The condensed form of this molecule is: condensed form N. The configurations at carbons- and - are and, respectively. Group priorities at carbon- N assign configuration (N ) IP group priorities = (-) = (-(N ) ) = (--(=); Ph) = (-) clockwise rotation = xford University Press, 07. All rights reserved. i g h e r E d u c a t i o n

9 s manual for Burrows et.al. hemistry Third edition Group priorities at carbon- N IP group priorities = N (-N ) = (-()Ph) = (- ) = (-) assign configuration then rearrange into a more user-friendly conformer Ph() N ()Ph N. heck that the structure is correct by naming it! This structure is (R,S)--(methylamino)--phenylpropan--ol (ephedrine). (R) - N (R,S)--(methylamino)--phenylpropan--ol ephedrine hiral centre on the left-hand side N anticlockwise rotation = xford University Press, 07. All rights reserved. i g h e r E d u c a t i o n

10 s manual for Burrows et.al. hemistry Third edition hiral centre on the right-hand side Rotate so that the group of lowest priority is pointing away from you The configurations are: N N N (b) Draw the structure of a diastereoisomer of ephedrine.. Draw out a non-superimposable non-mirror images of (R,S)-ephedrine.. Two non-superimposable non-mirror images (diastereoisomers) of ephedrine can be drawn. These are the (R,R)- and (S,S)-diastereoisomers. The remaining stereoisomer (S,R)- is the enantiomer of (R,S)-ephedrine. N N (R,R)-diastereoisomer (S,S)-diastereoisomer xford University Press, 07. All rights reserved. i g h e r E d u c a t i o n

11 s manual for Burrows et.al. hemistry Third edition s diastereomers of ephedrine or N N xford University Press, 07. All rights reserved. i g h e r E d u c a t i o n

12 s manual for Burrows et.al. hemistry Third edition s to end of chapter questions (on p. 858 in hemistry ). The synclinal conformation of,-dichloroethane is.8 kj mol higher in energy than the anti-periplanar conformation. The two energy barriers for rotation about the bond in,-dichloroethane are.5 kj mol and 8.9 kj mol higher than the energy of the anti-periplanar conformation (a) Sketch a graph of energy versus angle of rotation about the bond (dihedral angle) for,-dichloroethane.. Draw out the most stable Newman projection of,-dichloroethane.. Rotate clockwise the proximal (front) carbon atom of this projection (from part ) by 60 o and consider its relative energy.. Repeat this 60 o clockwise rotation until a full revolution (80 o ) has been achieved.. Sketch the change in energy versus the relative angle of rotation. For ease, set the most stable projection has having 80 o ; this is its dihedral angle. 5. Label each anti-periplanar, anticlinal, syn-periplanar and synclinal projections. l l l l syn-periplanar l l l syn-periplanar l anticlinal anticlinal energy.5 kj mol 8.9 kj mol.8 kj mol l l l l synclinal l l anti-periplanar synclinal angle of rotation (dihedral angle) ( ) xford University Press, 07. All rights reserved. i g h e r E d u c a t i o n

13 s manual for Burrows et.al. hemistry Third edition (b) What conformation of,-dichloroethane has the highest energy? Eclipsed conformers are higher in energy than staggered conformers. Draw out the three possible eclipsed projections of,-dichloroethane, and consider which projection has the greatest eclipsing interactions. f the three eclipsing interactions, there are two identical synclinal projections and a single syn-periplanar projection. These synclinal projections contain two -l/- and one -/- eclipsing interactions. The remaining syn-periplanar projection contains two -/- and one -l/-l eclipsing interactions; this -l/-l eclipsing interaction is by far the most unfavoured. Therefore, the syn-periplanar conformer of,-dichloroethane has the highest energy. Eclipsing interactions for,-dichloroethane eclipsed interactions; for, = l l l l for,l = l l for l,l = synclinal syn-periplanar synclinal increasing steric strain The syn-periplanar conformation of,-dichloroethane has the highest energy.. The following questions relate to meso-butane-,-diol. (a) Draw sawhorse and Newman projections of the syn-periplanar conformation of meso-butane-,-diol. xford University Press, 07. All rights reserved. i g h e r E d u c a t i o n

14 s manual for Burrows et.al. hemistry Third edition. Draw the condensed structure of butane-,-diol and remember to label its chiral centres.. It will be easier to first draw out the Sawhorse projection of meso-butane-,-diol as this is an eclipsing projection. It is worthy of note, for a molecule to be meso-, it must have at least two chiral centres and contain either a plane or point of symmetry. If you can orient your conformer so there is an internal mirror plane, this will automatically give the required syn-periplanar conformer.. Draw out the Newman projection of meso-butane-,-diol. It is worth remembering that a syn-periplanar conformer is an eclipsing conformer, where the same groups on both the proximal and distal carbon atoms (of its Newman projection) are in a synand eclipsing arrangement.. The condensed structure of butane-,-diol and the chiral centres are shown below: condensed structure of butane-,-diol = chiral centre. The sawhorse projection of meso-butane-,-diol is shown below. If drawn correctly, all three different groups (, and ) will be syn-periplanar. sawhorse projection of meso-butane-,-diol syn-periplanar conformer mirror plane of symmetry. The syn-periplanar Newman projection can be drawn by looking down the central ()-() bond. xford University Press, 07. All rights reserved. i g h e r E d u c a t i o n

15 s manual for Burrows et.al. hemistry Third edition Sawhorse projection of meso-butane-,-diol Newman projection of meso-butane-,-diol syn-periplanar conformer both groups are syn-periplanar both atoms are syn-periplanar view from here both groups are syn-periplanar Sawhorse projection Newman projection (b) Assign or configuration to each of the chiral centres in meso-butane-,-diol. Assign the configurations of this molecule using the ahn-ingold-prelog rules (See p. 89 in hemistry ). For ease, use a conformer where the lowest priority groups on both chiral centres are facing away from you. Rotate the sawhorse projection in (a), by 80 o using a vertical axis, gives the more userfriendly sawhorse projection for determining these configurations. rotate 80 o around a vertical axis xford University Press, 07. All rights reserved. i g h e r E d u c a t i o n

16 s manual for Burrows et.al. hemistry Third edition Assigning the configuration for the chiral centre on the left-hand side of this molecule using the ahn-ingold-prelog rules: IP group priorities = (-) = (-() ) = (- ) = (-) assign configuration () anti-clockwise rotation = Assigning the configuration for the chiral centre on the right-hand side of this molecule using the ahn-ingold-prelog rules: IP group priorities = (-) = (-() ) = (- ) = (-) assign configuration () The overall configurations for meso-butane-,-diol are given below: anti-clockwise rotation = xford University Press, 07. All rights reserved. i g h e r E d u c a t i o n

17 s manual for Burrows et.al. hemistry Third edition Rotate molecule so that the lowest priority group () is facing away from you chiral centre on the left-hand side () Rotate molecule so that the lowest priority group () is facing away from you chiral centre on the right-hand side () (c) Draw a Fischer projection of meso-butane-,-diol and use this to explain why this is a meso-compound. For a molecule to be meso-, it must contain two or more chiral centres and contain either a plane or point of symmetry. A Fischer projection is simply a D representation of a D sawhorse projection. Butane-,-diol contains two chiral centres. xford University Press, 07. All rights reserved. i g h e r E d u c a t i o n

18 s manual for Burrows et.al. hemistry Third edition The sawhorse projection in part a can be converted into a Fischer projection by simply drawing its vertical projection as shown below. As this Fischer projection has a plane of symmetry (or a horizontal mirror plane) passing through the middle of its ()-() bond, and contains two chiral centres, it is meso-. In addition, meso-butane-,-diol is not chiral as its mirror image is superimposable. meso-butane, diol view from here convert into a Fischer projection Sawhorse projection Fischer projection mirror plane of symmetry This is a meso-compound because it contains two chiral centres and has an internal plane of symmetry (and so is achiral). internal plane of symmetry (d) Draw a Fischer projection of a diastereoisomer of meso-butane-,-diol. Draw out the two non-superimposable non-mirror images of meso-butane-,-diol. xford University Press, 07. All rights reserved. i g h e r E d u c a t i o n

19 s manual for Burrows et.al. hemistry Third edition Two non-superimposable non-mirror images (diastereoisomers) of meso-butane-,-diol can be drawn. These are the (S,S)- and (R,R)-diastereoisomers. The remaining projection (S,R)- is identical to its original (R,S)- by 80 o rotation. Fischer projections of butane-,-diol (R,S)- (S,R)- (S,S)- (R,R)- meso- A diastereoisomer of meso-butane-,-diol has a different configuration at one of the chiral centres. a different configuration to meso-butane-,-diol at this position. The citric acid cycle (Krebs cycle) is a series of reactions in the body involved in the oxidation of fats, proteins, and carbohydrates to form carbon dioxide and water (p.955 in hemistry ). ne step in the citric acid cycle is the addition of water to fumaric acid to make malic acid. (a) Does the alkene double bond in fumaric acid have the (E)- or (Z)-configuration?. Draw out the condensed structure of fumaric acid to include all -atoms.. For molecules containing a single stereoisomeric double (=) bond. Its (Z)-isomer is where both high priority substituents are on the same side of the double bond, whereas, its (E)-isomer is when these high priority substituents are on opposite sides xford University Press, 07. All rights reserved. i g h e r E d u c a t i o n

20 s manual for Burrows et.al. hemistry Third edition of the double bond. These priorities are determined using the ahn-ingold-prelog rules (see p. 89 in hemistry ).. The condensed structure of fumaric acid is shown below: condensed structure of fumaric acid. For this double bond, assign the priority of its substituents using the ahn-ingold- Prelog rules (See p. 89 in hemistry ); the carbon atom () of the carboxylic acid group has higher priority than the -atom. This alkene has (E)-configuration as both high priority substituents are on opposite sides of this double bond. assigning the configuration of fumaric acid = high priority = low priority (E)- has a higher atomic number than has a higher atomic number than (E)-configuration (b) Maleic acid is a configurational isomer of fumaric acid. Draw the structure of maleic acid. xford University Press, 07. All rights reserved. i g h e r E d u c a t i o n

21 s manual for Burrows et.al. hemistry Third edition The only configuration in fumaric acid is its (E)-double (=) bond. As maleic acid is a configurational isomer, it must be the same condensed structure as fumaric acid (butenedioic acid) but a different alkene configuration. Fumaric acid is the (E)-isomer of butenedioic acid. Maleic acid is the (Z)-isomer of butenedioic acid. condensed structure and configuration of maleic acid = high priority = low priority (Z)- (Z)-configuration (c) Explain why the conversion of fumaric acid to ( )-malic acid is an example of an enantioselective reaction. fumaric acid Enzyme (fumarase) (-)-malic acid For a reaction to be enantioselective, an unequal mixture of enantiomers must be formed. A common example involves the conversion of an achiral starting material to an enantiomerically enriched product. This reaction involves the selective formation of one enantiomer (-)-malic acid; its mirror image, (+)-malic acid, is not preferred. As there are two potential products {(-)- and (+)-malic acids} which can be formed; this reaction is selective as only i g h e r E d u c a t i o n xford University Press, 07. All rights reserved.

22 s manual for Burrows et.al. hemistry Third edition one of these two products are formed. This reaction is not specific. A specific reaction is where only one product can be formed. This is an enantioselective reaction because ( )-malic acid is formed in preference to (+)- malic acid. (d) Assign the or configuration to the chiral centre in ( )-malic acid. Assign the configuration of this molecule using the ahn-ingold-prelog rules (See p. 89 in hemistry ). For ease, use a conformer in which the lowest priority group on the chiral centre is facing away from you. Rotate the projection in (c), by 80 o using a vertical axis, gives the more user-friendly sawhorse projection for determining this configuration. rotate 80 o around a vertical axis (-)-malic acid Assigning the configuration for this chiral centre using the ahn-ingold-prelog rules: xford University Press, 07. All rights reserved. i g h e r E d u c a t i o n

23 s manual for Burrows et.al. hemistry Third edition assign configuration IP group priorities = (-) = (-(=)) = (- ) = (-) anti-clockwise rotation = (-)-malic acid The configuration for ( )-malic acid is as shown below: (-)-malic acid Rotate molecule so that the lowest priority group () is facing away from you xford University Press, 07. All rights reserved. i g h e r E d u c a t i o n

24 s manual for Burrows et.al. hemistry Third edition. Tetrodotoxin is a potent poison that is found in the puffer fish. It is estimated to be more than times deadlier than cyanide ions. The structure of tetrodotoxin contains a group of interconnected 6-membered rings. Which of the bonds on the cyclohexane ring of tetrodotoxin are in an axial position and which are in an equatorial position? N N N tetrodotoxin. Pick out the cyclohexane ring.. ighlight the equatorial and axial positions on this cyclohexane ring. This has been illustrated below for cyclohexane. a a a e a a a e e e e e e e e e a e a a cyclohexane a a a axial hydrogen ( a ) atoms e equatorial hydrogen ( e ) atoms. f the four potential six-membered rings, there is only one cyclohexane ring. The remaining three rings are heterocyclic. xford University Press, 07. All rights reserved. i g h e r E d u c a t i o n

25 s manual for Burrows et.al. hemistry Third edition N N N cyclohexane core. n this cyclohexane ring, the axial positions are vertical and the equatorial positions point away (by 9 degrees) from the equator of the ring. N N N axial positions highlighted in red N N N equatorial positions highlighted in blue N N N Axial bonds Equatorial bonds 5. Lamivudine (Epivir) is used for the treatment of IV (Aids) and is an example of a class of drugs called nucleoside reverse transcriptase inhibitors. These inhibitors stop IV xford University Press, 07. All rights reserved. i g h e r E d u c a t i o n

26 s manual for Burrows et.al. hemistry Third edition from infecting cells in the body. nly the ( )-enantiomer of lamivudine is registered for the treatment of IV because it is more active and less toxic than either the (+)-enantiomer or the racemate. In the ( )-enantiomer, the chiral centre at position- has the configuration and the chiral centre at position-5 has the configuration. Draw the structure of ( )-lamivudine using hashed wedged line notation. N N N 5 S (±)-lamivudine. Redraw this molecule in a condensed form.. Arbitrarily assign both chiral centres in (±)-lamivudine using the ahn-ingold- Prelog rules (See p. 89 in hemistry ). It will be easier if you consider one chiral centre at time.. Pick out the correct diastereoisomer (R,5S)-.. Draw out this diastereoisomer using hashed-wedged line notation.. The condensed structure of lamivudine is shown below: condensed structure of lamivudine N N N 5 S. Arbitrarily assign the configurations of this molecule using the ahn-ingold-prelog rules. For ease, use a conformer in which the lowest priority groups on both chiral centres are facing away from you. Assign the configuration for the chiral centre at carbon- using the ahn-ingold- Prelog rules: the chiral centre at carbon- has stereochemistry. A configuration is where the three highest priority groups (, and ) on a particular xford University Press, 07. All rights reserved. i g h e r E d u c a t i o n

27 s manual for Burrows et.al. hemistry Third edition conformation can be rotated clockwise ( ), whilst the lowest priority group,, is at the rear of this conformer. Assigning the configuration at () N N N S IP group priorities = S (-SR) = (-R) = (- ) = (-) assign configuration R RS clockwise rotation = Assign the configuration for the chiral centre at carbon-5 using the ahn-ingold- Prelog rules: the chiral centre at carbon- has stereochemistry. A configuration is where the three highest priority groups (, and ) on a particular conformation can be rotated anticlockwise ( ), whilst the lowest priority group,, is at the rear of this conformer. Assigning the configuration at (5) N N N IP group priorities = (-R) = N (-NR ) = (- S-) = (-) S assign configuration R N SR anti-clockwise rotation =. The required (R,5S)-diastereoisomer is shown below. xford University Press, 07. All rights reserved. i g h e r E d u c a t i o n

28 s manual for Burrows et.al. hemistry Third edition N N N 5 S (R,5S)-lamivudine. The hashed-wedged structure of (R,5S)-lamivudine is shown below. N N N 5 S hashed = wedged = (R,5S)-lamivudine N N N S ( )-lamivudine 6. The following questions relate to ( )-menthol, a naturally occurring compound isolated from peppermint oil. 5 (-)-menthol (a) Draw the preferred chair conformation of ( )-menthol. xford University Press, 07. All rights reserved. i g h e r E d u c a t i o n

29 s manual for Burrows et.al. hemistry Third edition. Draw a chair conformer for cyclohexane, and then draw out the axial and equatorial positions at carbons-, - and -5.. Insert the substituents (at carbons-, - and -5) in (-)-menthol onto this cyclohexane template. [Place the largest (isopropyl) group in the less steric demanding equatorial position.]. The axial and equatorial positions at carbons-, - and -5 for cyclohexane are illustrated below. e e a 5 e a a. Replace the less steric demanding equatorial e atom with the largest isopropyl group at carbon-. As the neighbouring group at carbon- is on the opposite sides of this molecule (anti-stereochemistry between carbons and ), replacing the e atom at carbon- with this group will give the correct stereochemical relationship between these substituents. The remaining group at carbon-5 is on the same face as this group, and therefore will be in an equatorial position. e e a 5 e replace e on () with the isopropyl group e a 5 e replace e on () with the group a a a a a 5 e replace e on (5) with the group a 5 omit atoms a a a a xford University Press, 07. All rights reserved. i g h e r E d u c a t i o n

30 s manual for Burrows et.al. hemistry Third edition 5 rotate by 80 o in a horizontal axis all three substituents are equatorial (b) Assign or configuration to each of the chiral centres in ( )-menthol. Assign the configurations of this molecule using the ahn-ingold-prelog rules (See p. 89 in hemistry ). For ease, assign each configuration separately and use a conformer in which the lowest priority group on each chiral centre is facing away from you.. Redraw (-)-menthol in a condensed form.. Assign all three chiral centres in (-)-menthol using the ahn-ingold-prelog rules. It will be easier if you consider one chiral centre at time.. The condensed structure of menthol and the three chiral centres are shown below. condensed structure of menthol 5. Assign the configurations of (-)-menthol using the ahn-ingold-prelog rules. For ease, when assigning each chiral centre use a conformer in which the lowest priority group is facing away from you. xford University Press, 07. All rights reserved. i g h e r E d u c a t i o n

31 s manual for Burrows et.al. hemistry Third edition Assign the configuration for the chiral centre at carbon- using the ahn-ingold- Prelog rules: the chiral centre at carbon- has stereochemistry. A configuration is where the three highest priority groups (, and ) on a particular conformation can be rotated clockwise ( ), whilst the lowest priority group,, is at the rear of this conformer. Assigning the configuration at () assign configuration -R R clockwise rotation = IP group priorities = (-) = (-( ) ( -)) = (- -) = (-) Assign the configuration for the chiral centre at carbon- using the ahn-ingold- Prelog rules: the chiral centre at carbon- has stereochemistry. A configuration is where the three highest priority groups (, and ) on a particular conformation can be rotated anti-clockwise ( ), whilst the lowest priority group,, is at the rear of this conformer. xford University Press, 07. All rights reserved. i g h e r E d u c a t i o n

32 s manual for Burrows et.al. hemistry Third edition Assigning the configuration at () assign configuration R -R ( ) IP group priorities = (- R) = (-( ) ) = (- R) = (-) rotate 80 o R- R ( ) anti-clockwise rotation = Assign the configuration for the chiral centre at carbon-5 using the ahn-ingold- Prelog rules: the chiral centre at carbon-5 has stereochemistry. Assigning the configuration at (5) assign configuration R R clockwise rotation = IP group priorities = (- R) = (- R) = (- ) = (-) The three configurations of (-)-menthol are: xford University Press, 07. All rights reserved. i g h e r E d u c a t i o n

33 s manual for Burrows et.al. hemistry Third edition 5 (-)-(R,S,5R)-menthol Rotate molecule so that the lowest priority group () is facing away from you xford University Press, 07. All rights reserved. i g h e r E d u c a t i o n

34 s manual for Burrows et.al. hemistry Third edition (c) Draw the enantiomer of ( )-menthol. Draw out the non-superimposable mirror image of (-)-menthol. Two non-superimposable mirror images (enantiomers) of menthol can be drawn. These are the (-)-(R,S,5R)- and (+)-(S,R,5S)-enantiomers (-)-(R,S,5R)-menthol (+)-(S,R,5S)-menthol mirror plane (+)-menthol (d) Draw a diastereoisomer of ( )-menthol. xford University Press, 07. All rights reserved. i g h e r E d u c a t i o n

35 s manual for Burrows et.al. hemistry Third edition Draw out the non-superimposable non-mirror images of (-)-menthol. Six non-superimposable non-mirror images (diastereoisomers) of (-)-menthol can be drawn. These diastereoisomers, (S,S,5R)-, (R,R,5R)-, (R,S,5S)-, (S,R,5R)-, (S,S,5S)- and (R,R,5S)- are shown below. The first three of these can be drawn by inverting a single chiral centre of (-)-(R,S,5R)-menthol, and the last three can be drawn by inverting two chiral centres of (-)-(R,S,5R)-menthol. Interestingly, changing all three chiral centres of (-)-(R,S,5R)-menthol gives its enantiomer, (+)-(S,R,5S)- menthol. hanging a single chiral centre of (-)-menthol at () gives (S,S,5R)-menthol at () gives (R,R,5R)-menthol at (5) gives (R,S,5S)-menthol hanging two chiral centres of (-)-menthol at () and () gives (S,R,5R)-menthol at () and (5) gives (S,S,5S)-menthol at () and (5) gives (R,R,5S)-menthol xford University Press, 07. All rights reserved. i g h e r E d u c a t i o n

36 s manual for Burrows et.al. hemistry Third edition or or or or s provided by J. Eames (j.eames@hull.ac.uk) xford University Press, 07. All rights reserved. i g h e r E d u c a t i o n