Lecture 1 Introduction Monday, 25 July 2016 9:02 AM
Organic
containing a C-C bond and/or C-H bond Formed in living systems
Why carbon? Can concatenate to itself: forming chains, rings, multiple bonds Mid-range electronegativity o Bonds with itself o Bonds with metals, not so much o Bonds with heteroatoms C-C and C-H are strong and unreactive Remember: Carbon forms 4 bonds (ALWAYS) If bond is not drawn, assume C-H bond
Lecture 2 Hybridization Tuesday, 26 July 2016 12:44 PM
Hybridisation Carbon has 4 valence electrons. 2 x 2s electrons and 2 x 2p electrons. But hydrogen bonds to them all equally Sigma bond A bond along the intermolecular axis Strong Can rotate Pi bond A bond around the intermolecular axis Weaker than sigma bonds Cannot rotate, break by twisting Count the number of things attached 2 things attached
sp
3 things attached
sp2
4 things attached
sp3
Lecture 5 Mass Spectrometry
Wednesday, 3 August 2016 11:07 AM
Steps to determine structure 1. Isolate and purify a. Chemical means Acid-base separation b. Physical means Precipitation, crystalisation Distillation Chromatography 2. Determine elements present and ratio between them (empirical formula) a. Burn it in an oxygen-rich atmosphere -- promote complete burning b. Products are easily measured (except for oxygen) 3. Determine molecular formula a. Measure the mass of the compound. Mass spectrometry. 4. Identify functional groups 5. Identify structural formula Mass spectrometry uses the motor effect: moving charge through a magnetic field experiences a force Shoot the sample through an electron beam (to ionize) and then through a magnetic field Heavier atoms will curve less, lighter atoms will curve more
If we make the detector plate large enough and sensitive enough, we can separate molecules which very close mass like isotopes.
ONLY IONS ARE DETECTED IN MASS SPEC When we ionise a molecule, there is a chance we break the ionic bonds You may see daughter ions -- broken bits of the original molecular ion Only the charged daughter ions will be detected Daughter ions can also fragment -- forming secondary daughter ions o For this reason, we DO NOT go too far toward zero Look at the maximum mass and work backwards, finding the differences Base peak
The highest peak. We assign this 100% and everything else is relative to this.
Back to isotopes Isotopes happen in different ratios If you see two peaks in a 1:1 ratio -- there must be a bromine in it somewhere If you see two peaks in a 3:1 ratio -- there must be a chlorine in it somewhere Recall, you can know which elements in your molecule by burning it You can also determine which elements using a high resolution mass spec
Lecture 6 Structure determination Thursday, 4 August 2016 8:46 AM
IR spectroscopy visualises bond vibrations within a molecule, so reveals the presence (or absence) of particular groups with the presence (or absence) of key signals in the IR spectrum. UV/Vis spectroscopy visualises electronic transitions within a molecule, so reveals the presence (or absence) of conjugation. (Conjugation lowers the energy gap between the highest occupied molecular orbital (HOMO) and the lowest unoccupied molecular orbital (LUMO), which is the gap across which an electron is excited by UV/Visible light.) Mass spectrometry measures the mass:charge ratio (m/z ratio) of individual ions, including the molecular ion (i.e. the ionised form of the molecule). This effectively allows us to 'weigh' individual molecules, giving molecular weight information from which we can deduce a molecular formula. 1H NMR (nuclear magnetic resonance) spectroscopy shows the connectivity (bonding) and shape (stereochemistry) within a molecule. Each different type of hydrogen atom gives rise to a different signal in the spectrum, and the position and shape of this signal depend on the exact chemical environment around that hydrogen atom (i.e. the other atoms and bonds nearby).
Lecture 14 Elimination
Friday, 26 August 2016 9:07 AM
Zaitsev's Rule: Where there is a choice, the most substituted alkene forms. Look at the ends of the double bond Count the number of carbons on each side This is because hyper conjugation stabilities the product Applies to E1 reactions, where can carbocation intermediate forms
Dehydrohalogenation Halogens are a good leaving group You only need to try and get rid of the H Use a base E2, one step process But wait Dehydrohalogenation does NOT obey Zaitsev's rule There is no carbocation intermediate Hoffman's Rule: Where there is a choice, the less substituted alkene forms For E2 reactions The attacking molecule will latch onto the most accessible carbon Substituted carbons are 'hidden' within and difficult to get to Contradictory to Zaitsev's rule? This is because no carbocation forms in E2 reactions.