Hybridisation and Aromaticity PDF

Hybridisation and Aromaticity Course Medical Chemistry Created time @October 3, 2024 3:09 PM Last edited time @December 18, 2024 4:12 PM Nucleus takes up most of the atoms mass but most of its volume is taken up...

Hybridisation and Aromaticity Course Medical Chemistry Created time @October 3, 2024 3:09 PM Last edited time @December 18, 2024 4:12 PM Nucleus takes up most of the atoms mass but most of its volume is taken up by the atomic shell around the nucleus. Electrons surrounding the nucleus exists in discrete spatial limits - atomic orbitals. Atomic orbitals describe the probability of finding an electron within a given space. Different orbitals represent energy levels that can be adopted by electrons. The further away an orbital / electron is from an nucleus, the higher the energy level. Orbitals of similar energy levels are grouped together in shells. The two shapes of orbitals we will be dealing with now are S orbitals and P orbitals. Hybridisation and Aromaticity 1  S- ORBITALS Always spherical Each orbital can hold two electrons. P- ORBITALS Dumbell shaped The greater the number of nodes the greater the energy of the orbital In each shell (after the first), there are 3 orbitals which have the same shape but pointing in different directions. They are mutually perpendicular to each other. (Px. Py, Pz) Each of these p-orbitals in the shell are exactly the same energy level. Orbitals of this type are called degenerate orbitals. Atoms can also contain d and f orbitals but we dont do them in this course. ELECTRONIC CONFIGURATION Hybridisation and Aromaticity 2  Rules for electronic configuration 1. Aufbau Principle: Orbitals are filled so that those of lowest energy are filled first. So we fill up 1s before 2s and then 2p. 2. Pauli’s exclusion principle: A maximum of two electrons maybe placed in each orbital but only when the spins of the two orbitals are paired. 3. Hund’s rule: When we have equal energy (degenerate orbitals) such as three p orbitals, we add one electron to eachwith their spins unpaired until each of the degenerate orbitals contain one electron. Then we begin adding a second to each degenerate orbitals until they are paired. so its : Note: Valence shell is the most outermost shell. Arrangements of atoms and bonds give every molecule a specific 3-d shape (a key feature of the “mode of action” of most drugs) Hybridisation and Aromaticity 3  BOND LENGTH AND BOND STRENGTH Bond formation releases energy Bond breaking requires energy There is an optimum distance between atoms forming a bond. Therefore strength and length are characteristic for each bond. Length and strength both depend on the type of orbitals that have overlapped to give the bond. eg: SP3 HYBRIDISATION Model molecule : methane How can the four CH bonds on the methane molecule be that same? Hybridisation and Aromaticity 4  In the eletronic configuration of carbon it appears only two electrons are in the valence shell and so carbon has only two electrons to share “covalently”.Only unpaired electrons take part in covalent bonding. However carbon in its excited state: This electronic configuration accounts for the tetravalent carbon but doesnt offer a satisfactory model for the carbon of methane because 3 of the bond will result from overlap with a p orbital and the other will result from overlap with an s orbital. this will result in bonds of differing lengths and strengths. But methane bonds are identical. A satisfactory model of methanes structure, based on quantum mechanics can be obtained through an approach called orbital hybridisation. Hybridisation and Aromaticity 5  SP3 HYBRIDISATION Orbital hybridisation is the combining of individual wave functions for s and p to obtain wave functions for new orbitals. The new orbitals have in varying proportions, the properties of the original orbitals taken seperately. These new orbotals are called Hybrid atomic orbitals. LIKELY EXAM QUESTION : Using electronic configuration, explain how carbon is able to form 4 identiocal bonds to seperate atoms in methane. Answer: The 2s orbital and three 2p orbitals all mix together to give 4 identical hybrid orbitals and these 4 orbitals give a tetrahedral geometry to the carbon atom and give it a bond angle of 109.5 degrees to the bonds. Hybridisation and Aromaticity 6  SP2 HYBRIDISATON Model molecule : Ethene How do we get a double bond in a molecule and how are the orbitals involved in this type of bonding giving trigonal geometry to the atoms at 120 degrees bonding angle? Each of the carbons in this molecule is sp2 hybridised Only three hybrid orbitals are formed leaving one very important unchanged p orbital as opposed to sp3 hybridisation which all the p orbitals are hybridised. It is the 3SP2 hybrid orbitals that give the trigonal planar shape to the molecule Hybridisation and Aromaticity 7  The way the sp2 hybridised orbitals come toghether: The sp2 hybridised carbon atoms in a c-c bond overlap. One part of the double bond results from a (head on) overlap of sp2 orbitals this is called a sigma (σ) bond, amd the other part results from a (sideways) overlap of “unhybridised” p orbitals, this is called pi (π) bond. SP HYBRIDISATION Model molecule: Ethyne ( CHCH) Liner bond, 180 degrees Only two hybrid orbitals are formed leaving 2 p orbitals unchanged. NOTE: ALL HYBRIDISED ORBITALS ( EVEN IN SP3 AND SP2) ARE DEGENERATE. Hybridisation and Aromaticity 8  The way sp hybridised orbitals come together: One part of the triple bond results from a (head on ) overlap of sp hybridised orbitals called sigma (σ) bond. The next two bonds are formed from a (side ways) overlap of unhybridised p orbitals called pi (π) bond. Hybridisation and delocalisation also work with atoms other than carbon eg. N, O, S etc. What are electron configuration for nitrogen and oxygen atoms in these compounds? TIP: You can tell the hybridisation state from the type of bond attached to the atom. If it only has single bonds → SP3 Hybridised, Double bonds → sp2 hybridised, Triple bond → sp hybridised. Conjugation If multiple double bonds ion series are seperated by no more than 1 single bonds they are conjugated. So if they are seperated by two more single bonds or not seperated at all, they are not in conjugation. Hybridisation and Aromaticity 9  In a conjugated system π electrons are no longer associated with specific bonds. They are delocalised in overlapping p orbitals over all atoms involved in the conjugated system (electron in the p orbital of multiple bonds are known as π electrons) In conjugation A conjugated pi system can be considered as ONE PI system which is orbital overlapped all way through. Conjugated system Hybridisation and Aromaticity 10  In a conjugated system , P orbital 2 has equal opportunity to overlap with p orbital 3 and 1 equally. There is equal overlap in electron density of each of the p orbitals. Every carbon in a conjugated system is sp2 hybridised. Therefore each carbon has a p orbital allowing for electron interaction across the whole system. Isolated double bonds In isolated system : They dont have the sideways overlap so the conjugated pi system is not extended through this chain because there is an sp hybridised carbon (single bonds) in the way. Delocalisation increases thermodynamic stability (decreases reactivity), (Delocalisation : distribution of electrons across several atoms rather than being confined to an atom or pair of atoms. AROMATICITY Aromatic compounds are commonly represented as a system of alternating conjugated double bonds. (kekule representation) The thermodynamics stability of benzene and its unusual bond lengths are a result of delocalisation of the pi electrons all over 6 carbons. Hybridisation and Aromaticity 11  Some cyclic molecules with conjugated double bonds have been found to be “unusually stable” compared to other conjugated alkene system. This class of stabilsed cyclic molecules are known as arenes or more commonly Aromatic compounds. Aromatic compounds (or arenes) are stabilised cyclic molecules. all aromatics are cyclic but not all cyclic are aromatic. eg. Benzene More stable than the hypothetical 1,3,5 cyclohexatriene by 152 kj/mol C-C bonds lengths are all the same ( 1.39Å) which is half way between a C-C single bond are C-C double bond. A more accurate description of the pi system of an aromatic ring is given below: in conjugation To be aromatic a compound must Be cyclic Fit huckel’s rule - must contain 4n + 2 = number of pi bonds. solving for n , n must be a whole number( where n= 1,2,3,) Hybridisation and Aromaticity 12  Show unusual thermodynamic stability Be completely planar (flat) Exhibit a π system (double bonds) that is conjugated Lone pairs at the side of electrons also conjugate and so pyrrole is conjugated and fits huckels rule and so it is aromatic Hybridisation and Aromaticity 13