Chem211 Main Group Chemistry Study Notes.docx

Chem211 Main Group Chemistry Study Notes Molecular Symmetry Point group – set of symmetry elements Coordination complex – an atom or ion and a surrounding array of bound molecules or anions  2 coordinate complexes – linear, bent  3 –Trigonal planar, T-shaped, Trigonal pyramidal  4 – Tetrahedral, Disphenoidal (interhalogen compounds eg [BrF ] , 4 Square Planar  5 – Trigonal bipyramidal, square-based pyramidal, Pentagonal planar  6 – Octahedral  7 and 8 – Pentagonal bipyramidal (I7 ), Square antiprismatic8[IF ] Symmetry Element – a geometrical entity such as a line, plane or point about which a symmetry operation can be performed. Symmetry Operation – an operation performed on an object, which leaves it in a configuration that is indistinguishable from, and superimposable on, the original configuration Molecular Symmetry Principal axis – axis with the largent C axis Symmetry element Symbol Symmetry operation Identity E Leave the molecule alone Proper axis Cn Rotate the molecule by 360/n degrees around the axis Horizontal plane sn Reflect the molecule through the plane which is perpendicular to the major axis Vertical plane sv Reflect the molecule through a plane which contains the major axis Dihedral plane sd Reflect the molecule through a plane which bisects two C 2xes Improper axis Sn Rotate the molecule by 360/n degrees around the improper axis and then reflect the molecule through the plane perpendicular to the improper axis Invert the molecule through the inversion Inversion center or i center of symmetry center The Identity, E  Possessed by all molecules  Operation is to not do anything Proper axis of symmetry, C n  360/n rotation to produce equivalent config of molecule  n is the order of the axis  Axis of highest order in a molecule is the principal axis o Often the z-axis Plane of symmetry, σ  Plane through which a reflection of all parts of the molecule produces a configuration which is the same as original  Three types o Vertical (σ v  Contain principal axis of symmetry o Horizontal (σ ) h  Perpendicular to the principal axis of symmetry o Dihedral (σ ) d  Contains the principal axis of symmetry and which bisects two adjacent 2-fold rotation axes Improper axis of symmetry, S n  Rotation through 360 /n about an axis (C ), followed by reflection through n a plane perpendicular to the axis (σ h  S 4one twice is a C 2 Inversion Centre or Centre of Symmetry, i  Centre of molecule that if reflection of all of the parts of the molecule produces a configuration indistinguishable from the original molecule  If the molecular orbital has an inversion centre it is a g = gerade and if it doesn’t it is a u = ungerade  No i about a tetrahedral carbon centre  i = C2x σ h S 2 Successive Symmetry Operations  used to get the molecule exactly as it started and not just indistinguishable While a single symmetry operation will leave the molecule indistinguishable from the original molecule, we can carry out a succession of operations which will leave the molecule exactly as it started. 3 S access is C e.g. ammonia (NH ). 3 3 = C3x C x3C = 3 x sigma Categories of Point Groups Point group - molecule has a set of symmetry operations that describes the molecule's overall symmetry Nonrotational Groups: C ,1C s C i Single-Axis Groups: C n C nvC ,nh ,2n ∞v Dihedral Groups: D ,nD nvD ,nh ∞h Cubic Groups: T , O , I d h h The symbol associated with the point group is called the Schönflies notation, which indicates the type of reflection symmetry and the order of the principal axis of symmetry. Non-rotational Groups C1 E CHFClBr • low order • only E or E+ 1 Cs E, σ h FBClBr other symmetry CI E, i HClBrC-CHClBr Single axis groups Cn E, C n H 2 2 • only one rotational Cnv E, C n nσ v H 2, 9-BBN Cnh E, C n σ h S n Boric acid [B(OH) ]3 S2n E, S 2n 1,3,5,7 - tetrafluoracyclooctatetrane C HCl ∞v E, C ∞ ∞ σ v Dihedral groups D n E, C n nC 2 CoN 6 • has nC a2es perp. to D nd E, C n nC ,2 S8(crown sulfur) principal S 2n,nσ v axis D nh E, C n nC ,2 BF3, naphthalene(C H10 10 σ h nσ v D ∞h E, i, C , CO 2 ∞ ∞σ Sv, ∞ ∞C , σ 2 h Cubic groups Td E, C 3 C 2 S 4 CCl4 • 7 groups but 3 (tetrahedron) common σ d O h E, C 3 C 2 C 4 SF6 • high order (Octahedron) (T d24, i, S4, S6, σd, σ h O h48, -2 I=120) Ih E, C 3 C 2 C 5 [B12 12 (icosahedron i, S , S , σ and 10 6 dodecahedron)  In D point group C 2s grouped with sigma v  S2is the same thing as i When determining point group you need to check  High symmetry (cubic) T ,dO ,hI h  linear C∞, D∞h  very low symmetry C ,1C ,sC, i 4,6,8…  All linear molecules have a C principal axis and an infinite number of σ ∞ v planes of symmetry o A centrosymmetric (a molecule with a centre of inversion point group) linear molecule also has a center of inversion σ hnd a σ h plane, and is in a D point group ∞h Cubic shapes Tetrahedron – touches corners of cube Octahedron – touches faces of the cube Icosahedron – has a flat edge that touches cube Cubic shapes touches corners touches faces of touches cubedge that of cube the cube tetrah(dron octah(e)dron icosch)edron Assigning moleculesThe polyhedra associated with the cubic point group fit symmetrically inside a cube. 1. look to see if the molecule seems very symmetric or unsymmetric a. if it is probably belongs to either: i. low symmetry: C , C , 1 or sineai C ∞v or D ∞h ii. high symmetry: T , O ordI h h 2. For all other molecules, find the rotation axis with the highest n value, the principal C axns of the molecule Housecroft and Sharpe h always has priority over v if you have both Example BF 3 For three boxes trick you need to check  High symmetry (cubic) T , O , Id h h  linear C ,∞D ∞h  very low symmetry C , C , C1 S s i 4,6,8… if it doesn’t have these three then do the following h always takes priority over v if you have both Step 2: Step 1: Step 3: • D if nC C n of • h if σ exists 2 n h • C if not principal • v or d vf d /σ exist (v for C, d for D) axis • blank if no σ exists What symmetry elemeWhat symmetry elements does this molecule have? What is its point gWhat is its point group? Molecular Symmetry – Character Tables  Used to sort out properties of the molecule that depend on molecular symmetry  http://chemwiki.ucdavis.edu/Physical_Chemistry/Symmetry/Character_Ta bles_for_Symmetry_Groups Muliken symbols singly degenerate state which is symmetric with respect to rotation about the principal axis, singly degenerate state which is antisymmetric with respect to rotation about the principal axis, doubly degenerate, triply degenerate, ' = symmetric with respect to a horizontal symmetry plane , " = antisymmetric with respect to a horizontal symmetry plane . also the various symmetry labels have various subscripts:  Subscript 1- symmetric with respect to the Cn principle axis, if no perpendicular axis, then it is with respect to σv  Subscript 2-asymmetric with respect to the Cn principle axis, if no perpendicular axis, then it is with respect to σv  Subscript g- symmetric with respect to the inverse  Subscript u- asymmetric with respect to the inverse Bond vibrations in water Bond Vibrations Symmetrical Asymmetrical Bend Stretch Stretch  Symmetrical stretch (in the case of H2O A ) ke1ps the molecule in the proper form allowing for the same structure, if the molecule is unchanged when the symmetry operations are test then it is +1  Asymmetrical stretch (in the case of H2O B ) ch1nges the structure of the molecule and causes some of the symmetry options to change, when they are changed the molecule is revered and the character becomes –1  The symmetric and asymmetric vibration modes match the two irreducible representations with the symmetry labels A and B 1 1 o Infrared (IR) selection rules  Must change the dipole moment of the molecule  Has to have a linear term to show up in the IR spectrum Raman selection rules  Must change the polarizability of the molecule o Polarizability is the measure of the change in a molecule's electron distribution in response to an applied electric field, which can also be induced by electric interactions with solvents or ionic reagents.  If the molecule has a quadratic term it will show up in the raman spectrum  2 bands one for symmetric and one for asymmetric Point group representations Representation – set of characters which forms a row of the character table and are given Mulliken symbols (eg A or1a ) 1 Irreducible Representations – the simplest possible representations (as found in the character tables)  product of two rows of a character table must also be a row in the table. For example B x 1 = A2(thi2k like C x C 2 E) 2  the product of any two columnss of a character must also be a column in the table Reducible Representation – arise from the sum of two irreducible representations For example the two 1s orbitals on the hydrogens in an H2O molecule h = 1s + 1 B 1s A h 2 1s –B1s A To get the vibrational mode rotations for the overall molecule we add the symmetric and asymmetric rows of the table  A 1nd B w2ll be seen in the IR and Raman spectrums as they have a linear and quadratic term To get this table you have to follow several steps: 1. Structure (figure out using VSEPR) 2. Symmetry elements 3. Point group (determine point group where molecule belongs) 4. Test operations on vibrators (each O-H is one) Stereochemistry Chirality Chirality – a molecule that cannot be superimposed upon its mirror image Achiral – structures that can be superimposed on their mirror images Enantiomers - Two non-superimposable objects which are mirror images of each other (used to describe pair of molecules not individuals) To test for non-superimposability: 1. Turn the mirror image of the molecule so that the two substituents are aligned 2. Move the mirror image of the molecule to the original. If it doesn’t match it is non-superimposable or chiral  Chiral molecules can have different functions and properties even with the same formula o Milk is made up of many chiral molecules Any molecule that does not have an S axis of rotation is a chiral molecule  Also does not have a plane of symmetry or an inversion centre Chiral point groups: C , C , C , …, and D , D , … 1 2 3 2 3  Sometimes a molecule can be chiral on a very short timescale but achiral on a long timescale, this is due to the molecule constantly moving. Eg H2O 2 Chiral Centres  Chiral centre is loosely referred to as an atom that has four different groups bonded to it  For N centres - if it's three groups and a lone pair it is achiral as it moves frequently. However if there are four different groups attached it can be chiral  N centre don’t have inversion centres, so if there are 4 different substituents, it would be a chiral centre  Molecules are said to be achiral when it consists of rapidly equilibrating enantiomeric conformations that cannot be separated on any reasonable Isomers – constitutiotimescaleereoisomers Racemic mixture – has equal amounts of left and right-handed enantiomers of a chiral molecule Isomers are compounds that contain the same atoms bonded together in different ways. Isomers – constitutional vs stereoisomers how many stretches is  Compounds that contain same atoms bonded in difbasically asking how many bonds. Constitutional isomers: Different connectivity between atoms.ectivity between atoms have very different properties Br H 2 C OH vibrations 3 T2 modes 1 CH OH H3C CH bonds in IR 1 or 0 (has 3 H 3 C vibrations but one) mode H 2 Br because the vibrations have the same energy 2-Bromo-propan-1-ol 1-Bromo-propan-1-ol Stereoisomers: Same connectivity between atoms. Stereoisomers – Same connectivity between atoms (different cis/trans isomers) different cis/trans isomersHDoeBr't Br H have to be chiral C OH C OH H3C C H3C C H2 H2 two enantiomers of 2-Bromo-propan-1-ol properties Br H 2 C OH vibrations 3 T2 modes 1 CH OH H 3 CH bonds in IR 1 or 0 (has 3 H 3 C vibrations but one) mode H 2 Br because the vibrations have the same energy 2-Bromo-propan-1-ol 1-Bromo-propan-1-ol Stereoisomers: Same connectivity between atoms. different cis/trans isomers.HoesBrt have to be chiral Br H C OH C OH H 3 C H 3 C H2 H2 two enantiomers of 2-Bromo-propan-1-ol Octahedral stereoisomers  Octahedral stereoisomers  Fac and mer require only 3 of the same molecule in the arrangements only need 3 molecules in the fac or mert to be Isomhttp://www.ch.kcl.ac.uk/kclchem/staff/mno/booknet/chapter2/c2fig.htm Configurational vs. conformational isomers ConfiguraConfiguIsomers – configurational vs conformationalat cannot behanged without breaking bonds somewhere in the molecule.where in the molecule. Configurational HsomBrs are differBrt Holecules that cannot be interchanged without breaking bonds somewhere in the molecule. C OH C OH enatiomers H3C H BC H3C H C C H2 OH H 2 enatiomers H3C C H C C C OH two confiH2rations of 2-H2omo-propan-1-ol Conformational isomers of a molecule are the same molecule, and Conformatcan be interconverted by simply rotating around bonds in thele, and interconverted by simply rotating around bonds in the molecule, rather than molecule, rather than breaking them. breaking them. H Br H C HH Br CH 3 H Br 3 Br CH3 C C OHOH CC OHOH C C OH OH H3CH C C C Br CC H H C C 3 H2H2 H2 2 H2 H2 Representing 3-D molecules in 2-D projectionsations of one enantiomer of 2-bromo-propan-1-ol-ol Representing 3-D molecules filled in towards and vertically drawn towards you are away from  filled in wedges - towards you (3-D hash-wedge)  horizontally drawn lines are towards and vertically drawn are away from you (2-D Fischer projection) towards you are away from Newman projections Staggered ethane Eclipsed ethane Newman projections Newman Projections Staggered ethane Eclipsed ethane StaggeredStaggered ethane – Eclipsed ethane Newman projections you can see the behindtre so molecules even though this is an eclipsed ethane Staggered ethane Eclipsed ethane Eclipsed ethane- http://www.ch.kcl.ac.uk/kclchem/staff/mno/booknet/chapter1/c1fig.htm yodrawn slightly off centre so moyou can see the behind thmolecules even thoughane this is an eclipsed ethane Newman projections http://www.ch.kcl.ac.uk/kclchem/staff/mno/booknet/chapter1/c1fig.htm http://www.ch.kcl.ac.uk/kdrawn slightly off centre sopter1/c1fig.htm you can see the behind molecules even though this is an eclipsed ethane http://www.ch.kcl.ac.uk/kclchem/staff/mno/booknet/chapter1/c1fig.htm Racemic mixtures http://www.ch.kcl.ac.uk/kclchem/staff/mno/booknet/chapter1/c1fig.htm If the starting materials of a reaction are achiral, and the products are chiral, they are formed as a racemic mixture of two enantiomers. Racemic mixtures The products are a racemic mixture of non-superimposable mirror images of Racemic mixtures O CN- HO CN Racemic mixtures H3C C H3C C H2 H H2 H each other and thus chiral enantiomers HO CN the aldehyde functional group lies in a plane. CN- H3C C The cyanide ion (also achiral) can attack the C H plane of the carbonyl group. Do the two of the H2 reactions give the same product? O HO CN NC OH Two products are H3C C formed in equal C CH C H amounts H 3 C 3 H2 C H H C NC OH H H 2 2 CN- H 3 C C H H2 Cahn-Ingold-Prelog priority rules The two products are a racemic mixture of non-superimposable  only for chiral molecules mirror images of each other and thus are chiral enantiomers  figure out priorities of the groups attached to carbon following rules  if two of the four groups are the same it is not chiral Assign a priority number to each substituent at the chiral center. 1. Higher atomic number wins. If the same, then go to step 2. 2. List atoms attached to the first atom, in order of decreasing atomic number. At the first difference, higher atomic number wins. 3. If still the same, then repeat step #2 with the next layer out. Eg.  The C is counted as being attached to 3 carbons  C,C,C vs C,C,C same so repeat step 2 for next layer. C,C,H (attached to 2 carbons and a H as it is triple bonded to carbon) vs H,H,H Determining R and S configurations 1. Assign a priority number to each substituent at the chiral center 2. Point the lowest priority substituents away from you 3. Mentally move from substituent priority 123 R = clockwise (rectus) S = counterclockwise (sinister) E/Z double bond isomers E/Z double bond isomers 1-bromo-1-chloro-1-propene CH 3 Br CH 3 Cl  If both the higher-priority substituents are on the same side, the C C C C arrangement is Z; if on opposite sideH, the Clrangement is H Br 1. Assign priority for the substituents on the left C the 2 substituents on the left C. E/Z double bond isomers 2. Assign priority for the subs2. Assign priority for the 2 substituents on the right C. 3. Assign Z or Eloro-1-propene 3. Assign Z or E. CH 3 Br CH 3 Cl C C C C H Cl H Br 1. Assign priority for the 2 substituents on the left C.  Enantiomers interact with other chiral molecules differently 2. Assign priritInteract with polarized light differently 3. Assign Z or E. Handling enantiomers  Make pure enantiomers by using chiral reagents or chiral catalysts.  Separate racemic mixtures of enantiomers, by employing the fact that different enantiomers interact with other chiral molecules in different ways.  Determine how much of each enantiomer is present (enantiomeric excess or optical purity) using their rotations of plane-polarized light. Polarized light Polarimetry  Helps tell us how much of an enantiomer we have by how they react with polarized light. Polarimetry o As it passes through the sample the sample will rotate the light based on the amount of the enantiomer o Depending on which form we have the rotation will be clockwise (R) or counterclockwise (S) helps to tell how much of an entanimor we light. As it passes through the sample the sample will rotate the light based on the amount of the enantiomer the rotation will be clockwise ( R ) or counterclockwise (S)  a racemic mixture will not rotate the light. http://www.ch.kcl.ac.uk/kclchem/staff/mno/booknet/chapter3/c3fig.htm  if there is an excess the light will be between the maximum for a molecule and the light's original position more to the right(clockwise) means more of one enantiomer and more to the left (counterclockwise) means the other  clockwise rotation is positive  counterclockwise rotation in negative specific rotation is usually determined at 20 C using 589 nm (‘D” line) of sodium Specific rotations Example: A sample of 65.0 mg of R-mandelic acid was O 3 C OH dissolved in 2.00 cm of ethanol. In a polarimeter cell of 1.00 dm (10 cm), an optical C rotation of -5.05° was observed at 20 °C using OH 589 nm light. H (R)-mandelic acid [a]D= a/c l magnitude of rotation is = (-5.05)/((0.0650/2.00) x 1.000) related to the size of the main carbonts attached to the = -155 (units are usually not used)  Magnitude of rotation is related to the size of the substituents attached to the main carbon  In molecules with a lone pair, the lone pair can matter depending on theity molein the four groups attached to the chiral carbon center + and – enantiomers  Just because you have an R enantiomer doesn't mean that it is + or –  R and S is a way to name a compound while the + and - is how it reacts with light (plain polarized light)  When the net rotation of plane-polarized light is positive it is a positive rotation (+)  When the net rotation of plane-polarized light is negative it is a negative rotation (-) D and L enantiomers  If the enantiomer of your compound can be transformed into D- glyceraldehyde, then your enantiomer is D.  If the enantiomer of your compound can be transformed into L- glyceraldehyde, then your enantiomer is L. D and L enantiomers  Amino acids are usually L enantiomers and sugars are usually D H OH this is how the D and L 2 C C nomenclature is applied NOBr H2N C O HNO2 HO relate the molecules but without changing H OH (+)-isoserine H OH the chiral centre H2C C H2C C Br C O HO C O HO HO (+)-glyceric acid Na/Hg HgO H OH H C C H OH 3 H C C C O 2 HO HO C O D-(-)-lactic acid D-(+)-glyceraldehyde Λ or Δ enantiomers labels wwwchem.uwimona.edu.jm/courses/IC10Kiso.html  think of it like a propeller o propeller always moves in the direction of the leading edge o use hand rule to determine direction o always octahedral and chiral L and D enantiomer labels  Λ is for left and Δ is for right left hand rule right hand rule Λ ∆  Fingers are the direction it is rotating and thumb is the way it is sticking out clockwise counterclockwise towards us towards us Meso Compounds  Contain chiral centres but are not chiral think of molecule as a propeller and ask the  Compounds which have chiral centres but which are actually the same question which direction does the propeller move? compound as they have inversion centres (i) or planes of symmetry (σ)  They are R on one side and S on the other X X X X X X trans ! ² Diastereomers  Differ in cis/trans bonds Diastereomers due to a double bond  So they are cis/trans isomers be cis/trans isomers  They can be achiral O H O H O H H C O O C C O C C C C O C H H H O H Cis isomer Trans isomer Maleic Acid Fumaric Acid m.p. 140-142 °C m.p. 299-300 °C same formula different connectivity. acetone and isopropanyl differ in cis and trans Enantiomers and Diastereomers  If R switches to S or vice versa they are diastereomers  If both R and S switch they are enantiomers Enantiomeric excess  what you're looking for is the relative amount of each compound  How much of R and how much S is required. It is important as one can be Enantiomeric excess active in a drug while the other is inactive The enantiomeric exEnantiomeric excess (ee) is given by: [a]D(observed) [R] – [S] = = e.e. [a]D(pure) Example: What would be the enantiomeric excess if a sample of 2-chloro-1- phenylethanol had an observed optical rotation of -35? What would be the enantiomeric excess if a Cl sample of 2-chloro-1-phenylethanol had an CH2 observed optical rotation of -35? ee = (-35/-50) = 0.7 C (therefore enantiomer H OH excess of 70%) (R)-2-chloro-1-phenylethanol [D] = -50 Preparing a Pure Enantiomer 1. Synthesis directly from chiral starting reagents 2. Synthesis using chiral catalysts 3. Chiral chromatography 4. Preparing and separating diastereomeric salts Chiral Catalysis Chiral Catalysis  Catalyst comes together to make a cavity that results in specific reactions and it imposes chirality on to the achiral starting material. Similar to an enzyme in the body With R = phenyl, for example, the reaction produces 92% of the R enantiomer and 8% of the S enantiomer. What ee is that? With R = phenyl, for example, the reaction produces 92% of the R [ ] [ ] enantiomer and 8 % of the S enantiomer. What e.e. is that?icates Silicates [R] - [S] / [R] + [S] = 0.92 - 0.08/ 0.1 = .84 = 84% Polysilicates Po lysi lica te s The [SiO4] (silicate) ion is the basic building block for a large number of silicon oxide (silica) particles used in chromatography Column Chromatography  Column chromatography in chemistry is a method used to purify individual chemical compounds from mixtures of compounds Diastereometric salt formation  Synthesizing a salt that has completely different physical properties from each other and can be separated by solubility Diastereomeric salt formation synthesising a salt that has 2 completely different physical and can be seperated byher solubility.  When separated into the R and S enantiomers there may be parts of the Rtiomer_resolution_via_diaste.htm enantiomer in the S enantiomer sample changing the rotation of the sample. Sugars - Aldoses and Ketoses Sugars – Aldoeses and Ketoses called carbohydrate because  called carbohydrate because people thought that it carbon with waters added to itt a SuSugars - Aldoses and Ketoses • empirical formula C H O  Two types: n 2n n people thought that it was just a • a chain of C atoms with one bearing a carbonyl groucarbon with waters added to it having OH groups. • empiricl fn-2 is nh2nnunber of chiral carbons • a chain of C atoms with one bearing a carbonyl group and the rest  it also contains an aldehyde Two types:H groups. O chiral carbon is in the middle carbons Aldose: HO H Two types: n-2 O parenthesis and isn OH the middle carbon Alo Ketose HO H Ketose: O n-2 n is total number of  n-3 is the number of chnumber of chiral carbons  It contains a ketone group HO O Ketose: n-3 carbons, while n-2 is OH number of chiral carbons OH HO n-3 OH Sugar Stereoisomers  As more carbons are added more chiral carbons can be present and thus increasing the amount of isomers of the molecule n  2 is the number of isomers where n is the amount of chiral centre o may be less than 8 but 8 is the max amount Drawing all of the stereoisomers 1. Draw the molecule with the carbon skeleton in the zig-zag form 2. Identify the possible chiral centres 3. Identify the diastereomers first, using syn and anti labels to differentiate the diastereomers (based on two atoms in a row) 4. Check for meso compounds (i or σ). 5. Draw the enantiomers of any chiral diastereomer by inverting all of the stereogenic centers  To control the stereochemistry of a polymer we use the polymerization catalysis to get the desired configuration  Polypropylenes can be: o Isotactic - all the substituents are located on the same side of the macromolecular backbone o Syndiotactic - the substituents have alternate positions along the chain Polypropylenes o Atactic - substituents are placed randomly along the chain Isotactic HHC 3 H HC3 HHC 3 HHC 3 H HC3 H HC 3 H H 3 H H3C H H3C Syn diotactic HC 3 H HC 3 H HC3 H Atactic HHC 3 H HC3 H3C H HHC 3H 3 H H3C H Conformational Analysis Newman projections  Eclipsed is not favored because the hydrogens interfere with each other due to molecular orbitals repelling Conformations of Ethane eclipsed is not favaoured interfere with eachother repellinglecular orbitals  Barriers to rotation are mainly the van der Waals’ interactions and Coulombic interactions (Attractions or repulsion between charges as they The barriers to rotation are mainly due to van der Waals interactions become closer to each other) between groups on the carbon atomsween groups on adjacent carbon atoms.  Energy is required to rotate the molecule from one conformation to the next  Eclipsed being a high energy state and staggered being a low energy state gives us an energy graph like: high in enelow in energy  Really cold conditions would cause the conformation toreally cold conditions would cause the conformation to not have enough energy to rotate making it stagnant energyt to rotate making it stagnint Nuclear magnetic resonance spectroscopy (NMR) Nuclear magnetic resonance spectroscopy, most commonly known as NMR spectroscopy, is a research technique that exploits the magnetic properties of certain atomic nuclei to determine physical and chemical properties of atoms or the molecules in which they are contained. Nuclear magnetic resonance (NMR) is a physical phenomenon in which magnetic nuclei in a magnetic field absorb and re-emit electromagnetic radiation. Barriers to rotation  Some molecules are either rotating or moving so slowly that the NMR can’t tell it’s rotating  In some amides the bond is half way between a single and double bond and therefore doesn’t rotate easily  Addition of a methyl group to a ethane causes the eclipse to be even worse as the molecule has several molecules interfering o Since there are more groups more energy is required Conformations of Butane  The staggered that is okay but not the best is called gauche and the staggered that is the best is anti which staggered that is the best is degress apart and in the gauche theys are 180 degrees apart and in the gauchare 60 degrees aparts apart There are three possible conformations of butane that are associated with the 0 van der Waals radii that differ by multiples of 60 : gauche: A staggered conformation for butane in which the methyls are not anti.
 anti: The largest substituents are as far from each other as possible. (Lowest energy) eclipsed: The groups on the two carbons are oriented as close to each other as possible. (Highest energy) skewed: any conformation that is different than the other three Cycloalkanes  Dihedral angle is the angle in between two planes (in this case the one the 2Similarily, we can look at the combustion of a series of cyclic that is between the 2 methyl groups) alkanes CycloCycloalkanes 60° 90° 108° (CH )2 n + 3n/2 O 2 n H 2 + n CO 2 120° 128.5° 135° DH comb(cyclo(CH 2 n = n x DH rxn ) 2 DH (CH ) = DH (cyclo(CH ) )/n rxn 2 comb 2 n Ring strain is the tendency of a cyclic molecule, such as cyclopropane, to destabilize when its atoms are in non-favorable high-energy spatial orientations. Linear alkanesinimum strain is found at cyclopentane (if rest are planar) Linear alkanes CH 3 + 7/2 O 2 3H2O + 2 CO 2 (CH ) 2 n DH rxn= -1560 kJ/mol CH 3 (DH for burning ethane (n = 0)) + 3n/2 O n H O + n CO 2 2 2 DH comb = n x DHrxnH )2 DH comb(alkane) = -1560 + n x DH rxn ) 2 Cyclopentane  Can be in envelope shape, which allows it to be staggered instead of eclipsed called envelope shape, Cyclopentane which allows it ot be staggered instead of eclipsed Cyclopentane  The flap part of the envelope molecule is constantly moving but we say it’s flat  The molecule would rather be in the equatorial position that the axial position H C p n t u m c o p Enthalpy of combustrather be in equatorial position h than the axial position n 700 e 695 the flap part of the envelope 690 in the molecule is constanly make the big one 685 moving but we say it is flat equatorial 680 o / 675 k p 670 r 665 2g 660 655 650 4 6 8 10 12 14 16 18 ring size n Cyclohexane Drawing Cyclohexane Drawing cyclohexane Drawing Cyclohexane axial equitorial Cyclohexane can interconvert between chair conformations  if the cyclohexane switches from a chair to another chair conformation the Interconversion+ of+ the+ chair+ conforma0ons! hydrogens chang! from axial to equatorial which means the hydrogens ! are equal if they change fairly easily 5 -1 0 Barrier to interconverison is 43 kj/mol with a rate of 2x10 s at 25 C if the cyclohexane switches from a chair to a boat conformation the hydrogens Chair5to5chair+interconversion!! change from axial to equitorial which Easiest way to convert one chair conformation into the other is to go through the means the hydrogens are equal if they change fairly easily boat conformation. twist boat H 3 a 5 2 Ha 4 3 2 2 1 4 1 H b H b 6 Ha 4 5 6 3 6 5 1 chair Hb chair H 4 5 2 5 a H 2 1 H 3 6 1 a b H 4 3 6 b half chair half chair  The equatorial and axial conformations start to matter when we add substituents as they will repel the hydrogens and cause a certain configuration to be less favourable Conformational Analysis of Substituted Cyclohexanes Conformational Analysis of Substituted Cyclohexanes X K X axial conformer equatorial conformer [equatorial conformer] K = ———————————— [axial conformer] For X ≠ H, the equatorial conformer is more stable  A cyclohexane with an OCH3 attached only has a 17% chance of being trans5Disubs0tuted+Cyclohexanes! oxygen reacts with the hydrogen and reduces the van der waal repulsion. ! !  The oxygen is more favourable in the equatorial position but the electronegativity counteracts this  the greater the repulsion between the substituents the less amounts of that configuration are present Trans-Disubstitued Cyclohexanes eachother regardless of theed to conformation they are in With a cis- bond the substituents can be either both axial or both equatorial and in this case either one is favoured 1,4-Disubstituted Cyclohexanes In the 1,4 – Disubstituted Cyclohexanes there is no a,a conformation but there is an e,a conformer which is in equilibrium with an e,e conformer in which the cyclohexane ring is in a twisted boat, which is much more stable H C CH 3 3 CH 3 CH3 CH H 3 H C 3 3 CH 3 H 3 H3C CH 1,4-Disubstituted Cyclohexanes CH 3 3 There is no a,a conformer of 1,4-di-t-butylcyclohexane, but there is an e,a conformer which is in equilibrium with an e,e conformer in which the cyclohexane rinCH3s in a twisted boat, which is much H3C C3 CH3 CH3 more stable 3 C H3C C3 H3C H3C