Carbon
•Organic chemistry: transformations (reactivity) and creation (synthesis) of organic structures (C) ▪Carbon -1s2 2s2 2p2 = Valence of 4 and can form 4 covalent bonds -C-C and C-H bonds are strong and unreactive = provides solid scaffolds (backbone) -Versatile - C can bond to itself, metals and heteroatoms (atom that is not H or C) -Geometrically flexible – capable of catenation: forms chains, rings and multiple bonds ▪Nitrogen -forms 3 covalent bonds and has 1 lone pair -forms polar bonds to carbon -but lone pair is available for reaction ▪Oxygen -forms 2 covalent bonds and has 2 lone pairs -forms more polar bonds to carbon -lone pairs are less available for reaction – it holds the lone pairs more closely to itself ▪Fluorine -forms 1 covalent bond and has 3 lone pairs -forms extremely polar bonds to carbon -lone pairs are unavailable for reaction (high electronegativity) Elements involved in organic molecules: H, B, C, N, O, F, Si, O S, Cl, Br, I Representations of Molecular Structure •Stick representations of a molecule 1. Lines represent bonds 2. C is basis of structure (chains and rings of carbon atoms) - Carbon atoms not shown - assumed to be at intersections and ends of lines - C – H bonds omitted (hydrogen count assumed from valence) - All heteroatoms shown (+ H’s bound to heteroatoms) 3. Valence indicates the number of covalent bonds that an element will typically make 4. Geometry depends on hybridisation - Indicate bond angles (~120° or 180°) – sp 180°, sp2 120°, sp3 90° 5. In condensed formulae, atoms bonded to a C atom are listed after it 6. Formal charge = Actual number of bonds formed – Typical number of bonds formed (valence)
•Functional groups in a molecule
•Geometry and hybridization state (Carbon) -The valence electrons of a Carbon atom sit in 1s2 2s2 2p2 orbitals that are different in energy. It has 2 x 2s electrons + 2 x 2p electrons are available to form 4 covalent bonds. These orbitals mix (hybridise) to form new orbitals. 3
① sp
hybridisation – Alkanes If the 2s orbital mixes with three 2p orbitals 1 x s + 3 x p = 4 x sp3 = 4 identical orbitals = 4 σ orbitals 4 σ bonds = 4 single bonds
∴
2
② sp
hybridisation – Alkenes If the 2s orbital mixes with two 2p orbitals 1 x s + 2 x p = 3 x sp2 (+ 1 x p unhybridised) = 3 identical orbitals + 1 different orbital = 3 σ orbitals + 1 π orbital 1 σ bond + 1 π bond = 1 double bond
∴
③ sp hybridisation – Alkynes, Allene
If the 2s orbital mixes with one 2p orbital 1 x s + 1 x p = 2 x sp (+ 2 x p unhybridised) = 2 x identical orbitals + 2 different orbitals = 2 σ orbitals + 2 π orbitals 1 σ bond + 2 π bonds = 1 triple bond
∴
-π bond is weaker (more unstable higher energy form) than σ bond as the σ orbitals overlap better than the π orbitals which can only overlap from the sides.
•IUPAC names and structures 1. Find and name the longest continuous carbon chain -Maximum number of multiple bonds must be included – indicated by number of the first carbon - If there is more than one double bond, the suffix is expanded to include a prefix that indicates the number of double bonds present (-adiene, -atriene, etc) – adiyne, -anediol -C branches are indicated by methyl, ethyl, propyl with number -If chains of equal length are competing for selection as the parent chain, then the choice goes in series to: a) the chain which has the greatest number of side chains b) the chain whose substituents have the lowest numbers c) the chain having the greatest number of carbon atoms in the smaller side chain d) the chain having the least branched side chains 2. Identify and name substituents (groups attached to the chain) 3. Number the carbon chain that gives the substituents the lowest numbers -Double and triple bonds have priority over alkyl and halo substituents -When comparing a series of numbers, the series that is the "lowest" is the one which contains the lowest number at the occasion of the first difference -If two or more side chains are in equivalent positions, assign the lowest number to the one which will come first in the name 4. Designate the location by an appropriate number and name -If the same substituent occurs more than once, the location of each point on which the substituent occurs and the number of times the substituent group occurs is indicated by a prefix (di, tri, tetra, etc.). 5. Assemble the name, listing groups in alphabetical order using the full name -prefixes di, tri, tetra etc are excluded and iso-, cyclo- are included when alphabetizing •Properties of alkanes -Molecules are held together by dispersion forces which increase with increasing size. MP and BP increase with size of the alkane (but branching lowers the MP and BP as it forces the chains away from each other a bit) -Chemically inert -Good substrates for radical reactions -Large amounts of energy liberated in combustion -React with oxygen (combustion) - used as fuels -short-chain alkanes burn efficiently (complete) -bigger alkanes don’t burn as well (incomplete) •Distinguish between conformational and configurational isomers of alkanes Conformational isomers include single bonds which allow the spinning movement around the bond axis. Configurational isomers include double bonds. A double bond contains σ bond and π bond. π bonds result from porbital overlap and are directional. Rotation around the axis requires breaking this π bond.
Isomer
Constitutional Isomer
Stereoisomer
•Different sequence of bonding/ nature -Number of constitutional isomers increase rapidly with number of C -Physical and chemical properties differ particularly when different functional groups are present
•Different arrangement of same groups in space -same nature and sequence of bonding
Conformational Isomer •Differ by rotation about a single bond and can be interconverted
Configurational Isomer •Interconverted only by breaking and remaking bonds
Enantiomers
Diastereoisomer
•Non-superimposable mirror images
•Not mirror images
-Identify the stereogenic centre -Assign priority to each group directly attached to the centre in order of atomic number -If identical, work along the chain -Multiple bonds are expanded to the same number of single bonds
→Assign priority to each group on each carbon of C=C -The higher the atomic number of the atom attached, the higher the priority -If identical atoms attached, work along the chain until the first point of difference, then go by atomic number
R (clockwise)
S (anti-clockwise)
-Orientate the molecule with lowest priority group at the back and trace 1,2,3
-Orientate the molecule with lowest priority group at the back and trace 1,2,3
Z (cis) •If high priority groups are on the same side of C=C
E (trans) •If high priority groups are on the opposite sides of C=C
▪Differences in energy -Energy is the highest (disfavoured) during the eclipsed conformation due to steric interaction (electron repulsion causing instability) and hyperconjugation. -Energy is the lowest during the staggered conformation (when the substituents are furthest away from each other) -Large groups at the two ends of the bond near each other = high energy •A chiral molecule often contains a stereogenic centre (stereocentre or chiral centre). A stereogenic centre is a (carbon) atom with four different groups attached to it. If a molecule contains a stereocentre, its mirror image is not superimposable. In these molecules, there is no plane of symmetry. A pair of these molecules would be enantiomers. (N.B. Centre can be attached to one ring – the two halves of the ring are considered as two different groups) Almost all the properties of a pair of enantiomers are identical except for the way they interact with: -other chiral molecules (two enantiomers behave identically when reacting with an achiral reagent) -plane polarized light Hence, enantiomers can be distinguished using plane polarized light. Plane polarized light consists of waves oscillating in only one plane (polarizing filter filters out oscillations in all other planes that are perpendicular to the direction of the light wave)
When plane polarized light is passed through a solution of one pure enantiomer, the plane of polarization is rotated. The enantiomer is said to be optically active. The amount of rotation (α) is characteristic of the enantiomer. The enantiomers will rotate the light by same amount but in opposite directions: One enantiomer causes clockwise rotation (+) and the other causes anti-clockwise rotation (-). It is not possible to predict with enantiomer is (+) and which (-) without performing the experiment. Racemate/ Racemic mixture has 1:1 mixture of the two enantiomers. This occurs when an atom can bond in two opposite directions to form a chiral molecule (SN1 reactions). It gives an overall rotation of zero.
-Absolute configuration of a stereogenic centre = exact three dimensional arrangement of the groups -The absolute configuration does not correlate directly to the direction of optical rotation. Enantiomers rotate plane polarised light in opposite directions and have opposite absolute configurations but direction of rotation is not directly related to configuration For example,
•For n stereogenic centres, there are 2n possible isomers For a molecule with two stereogenic centres: there are 4 possible isomers (RR, SS, RS, SR).
1) A&B and C&D are enantiomers (non-superimposable mirror images) -Enantiomers have opposite configuration at all stereocentres 2) A&C and A&D and B&C and B&D are diastereoisomers (not mirror images) -Diasteroisomers have opposite configuration at some stereocentres •Meso compounds = molecules which have 2 stereogenic centres and an internal plane of symmetry →one stereogenic centre is the mirror image of another and they are superimposable -Meso form can be expected when the same groups are attached to adjacent stereocentres. The single bond can be rotated to allow an internal plane of symmetry. -Meso form and either of the enantiomers are diastereoisomers. -Meso form is not optically active
There are only three isomers: pair of enantiomers + meso form
•Fischer projections = 2D representations of 3D molecules -Carbon atom 1 is at the top -Horizontal lines stick out of page (a flat cat looking towards me) -Vertical lines stick into page
•Functional groups in a molecule
•Geometry and hybridization state (Carbon) -The valence electrons of a Carbon atom sit in 1s2 2s2 2p2 orbitals that are different in energy. It has 2 x 2s electrons + 2 x 2p electrons are available to form 4 covalent bonds. These orbitals mix (hybridise) to form new orbitals. 3
① sp
hybridisation – Alkanes If the 2s orbital mixes with three 2p orbitals 1 x s + 3 x p = 4 x sp3 = 4 identical orbitals = 4 σ orbitals 4 σ bonds = 4 single bonds
∴
2
② sp
hybridisation – Alkenes If the 2s orbital mixes with two 2p orbitals 1 x s + 2 x p = 3 x sp2 (+ 1 x p unhybridised) = 3 identical orbitals + 1 different orbital = 3 σ orbitals + 1 π orbital 1 σ bond + 1 π bond = 1 double bond
∴
③ sp hybridisation – Alkynes, Allene
If the 2s orbital mixes with one 2p orbital 1 x s + 1 x p = 2 x sp (+ 2 x p unhybridised) = 2 x identical orbitals + 2 different orbitals = 2 σ orbitals + 2 π orbitals 1 σ bond + 2 π bonds = 1 triple bond
∴
-π bond is weaker (more unstable higher energy form) than σ bond as the σ orbitals overlap better than the π orbitals which can only overlap from the sides.
•IUPAC names and structures 1. Find and name the longest continuous carbon chain -Maximum number of multiple bonds must be included – indicated by number of the first carbon - If there is more than one double bond, the suffix is expanded to include a prefix that indicates the number of double bonds present (-adiene, -atriene, etc) – adiyne, -anediol -C branches are indicated by methyl, ethyl, propyl with number -If chains of equal length are competing for selection as the parent chain, then the choice goes in series to: a) the chain which has the greatest number of side chains b) the chain whose substituents have the lowest numbers c) the chain having the greatest number of carbon atoms in the smaller side chain d) the chain having the least branched side chains 2. Identify and name substituents (groups attached to the chain) 3. Number the carbon chain that gives the substituents the lowest numbers -Double and triple bonds have priority over alkyl and halo substituents -When comparing a series of numbers, the series that is the "lowest" is the one which contains the lowest number at the occasion of the first difference -If two or more side chains are in equivalent positions, assign the lowest number to the one which will come first in the name 4. Designate the location by an appropriate number and name -If the same substituent occurs more than once, the location of each point on which the substituent occurs and the number of times the substituent group occurs is indicated by a prefix (di, tri, tetra, etc.). 5. Assemble the name, listing groups in alphabetical order using the full name -prefixes di, tri, tetra etc are excluded and iso-, cyclo- are included when alphabetizing •Properties of alkanes -Molecules are held together by dispersion forces which increase with increasing size. MP and BP increase with size of the alkane (but branching lowers the MP and BP as it forces the chains away from each other a bit) -Chemically inert -Good substrates for radical reactions -Large amounts of energy liberated in combustion -React with oxygen (combustion) - used as fuels -short-chain alkanes burn efficiently (complete) -bigger alkanes don’t burn as well (incomplete) •Distinguish between conformational and configurational isomers of alkanes Conformational isomers include single bonds which allow the spinning movement around the bond axis. Configurational isomers include double bonds. A double bond contains σ bond and π bond. π bonds result from porbital overlap and are directional. Rotation around the axis requires breaking this π bond.
Isomer
Constitutional Isomer
Stereoisomer
•Different sequence of bonding/ nature -Number of constitutional isomers increase rapidly with number of C -Physical and chemical properties differ particularly when different functional groups are present
•Different arrangement of same groups in space -same nature and sequence of bonding
Conformational Isomer •Differ by rotation about a single bond and can be interconverted
Configurational Isomer •Interconverted only by breaking and remaking bonds
Enantiomers
Diastereoisomer
•Non-superimposable mirror images
•Not mirror images
-Identify the stereogenic centre -Assign priority to each group directly attached to the centre in order of atomic number -If identical, work along the chain -Multiple bonds are expanded to the same number of single bonds
→Assign priority to each group on each carbon of C=C -The higher the atomic number of the atom attached, the higher the priority -If identical atoms attached, work along the chain until the first point of difference, then go by atomic number
R (clockwise)
S (anti-clockwise)
-Orientate the molecule with lowest priority group at the back and trace 1,2,3
-Orientate the molecule with lowest priority group at the back and trace 1,2,3
Z (cis) •If high priority groups are on the same side of C=C
E (trans) •If high priority groups are on the opposite sides of C=C
▪Differences in energy -Energy is the highest (disfavoured) during the eclipsed conformation due to steric interaction (electron repulsion causing instability) and hyperconjugation. -Energy is the lowest during the staggered conformation (when the substituents are furthest away from each other) -Large groups at the two ends of the bond near each other = high energy •A chiral molecule often contains a stereogenic centre (stereocentre or chiral centre). A stereogenic centre is a (carbon) atom with four different groups attached to it. If a molecule contains a stereocentre, its mirror image is not superimposable. In these molecules, there is no plane of symmetry. A pair of these molecules would be enantiomers. (N.B. Centre can be attached to one ring – the two halves of the ring are considered as two different groups) Almost all the properties of a pair of enantiomers are identical except for the way they interact with: -other chiral molecules (two enantiomers behave identically when reacting with an achiral reagent) -plane polarized light Hence, enantiomers can be distinguished using plane polarized light. Plane polarized light consists of waves oscillating in only one plane (polarizing filter filters out oscillations in all other planes that are perpendicular to the direction of the light wave)
When plane polarized light is passed through a solution of one pure enantiomer, the plane of polarization is rotated. The enantiomer is said to be optically active. The amount of rotation (α) is characteristic of the enantiomer. The enantiomers will rotate the light by same amount but in opposite directions: One enantiomer causes clockwise rotation (+) and the other causes anti-clockwise rotation (-). It is not possible to predict with enantiomer is (+) and which (-) without performing the experiment. Racemate/ Racemic mixture has 1:1 mixture of the two enantiomers. This occurs when an atom can bond in two opposite directions to form a chiral molecule (SN1 reactions). It gives an overall rotation of zero.
-Absolute configuration of a stereogenic centre = exact three dimensional arrangement of the groups -The absolute configuration does not correlate directly to the direction of optical rotation. Enantiomers rotate plane polarised light in opposite directions and have opposite absolute configurations but direction of rotation is not directly related to configuration For example,
•For n stereogenic centres, there are 2n possible isomers For a molecule with two stereogenic centres: there are 4 possible isomers (RR, SS, RS, SR).
1) A&B and C&D are enantiomers (non-superimposable mirror images) -Enantiomers have opposite configuration at all stereocentres 2) A&C and A&D and B&C and B&D are diastereoisomers (not mirror images) -Diasteroisomers have opposite configuration at some stereocentres •Meso compounds = molecules which have 2 stereogenic centres and an internal plane of symmetry →one stereogenic centre is the mirror image of another and they are superimposable -Meso form can be expected when the same groups are attached to adjacent stereocentres. The single bond can be rotated to allow an internal plane of symmetry. -Meso form and either of the enantiomers are diastereoisomers. -Meso form is not optically active
There are only three isomers: pair of enantiomers + meso form
•Fischer projections = 2D representations of 3D molecules -Carbon atom 1 is at the top -Horizontal lines stick out of page (a flat cat looking towards me) -Vertical lines stick into page
