Archive for September 2012
I’ve added a few points and asled a questions on the CHM556 [Organic Lab Practicals] oage. Links: https://intechemistry.wordpress.com/2012/06/22/chm556-organic-lab-practicals/
I changed my mind and now want a photostat of ALL th pages you had in your jotters relating to this experiment.
Please check out the page. Thanks
On the CHM456 [Organic] Page, the “Topic 2.0 Introduction to Organic Reactions: Acids and Bases” sub-page, there are links to the mechanism videos. Just click here: https://intechemistry.wordpress.com/2012/09/10/chm456-topic-2-0-introduction-to-organic-reactions-acids-and-bases/
In class, I asked you to imagine what happens in the last step of the aldol reaction, where a base had removed an alpha hydrogen on a beta-hydroxy carbonyl. Here’s the answer.
https://intechemistry.files.wordpress.com/2012/09/aldol-condensation-step-only.gif (you can get the larger image file using the link)
The first picture shows a lone pair of electrons on the hydroxde base ( OH- ) attacking the alpha-hydrogen. The attack is shown with the bottom arrow. p.s. let me call the the alpha-hydrogen the H(alpha) atom. The attack begins to cause the formation of a dative covalent bond on the O in the hydroxide to the H(alpha). The electrons in the C-H(alpha) bond must therefore break away from the H(alpha), as hydrogen cannot have two bonds. The electrons go onto the more electronegative atom which is carbon. The second upper arrow in the first molecule shows this.
As C was neutral and gained electron density, it now becomes negatively charged. You can use the ‘formal charge’ calculation to show that it has a -1 charge now.
Look at the top molecule in the box. We have a lone pair of e- on the alpha-carbon, then a sigma bond to the left and then a pi bond. We should be able to recognise this sequence as being able to undergo resonance (i.e de localisation of charge via electron movements.)
Follow the red arrows: The e- pair on the C atom move up towards the delta positive carbonyl carbon (the one in the C=O bond) because they are attracted towards that charge. However the carbonyl carbon already has 4 bonds so if a new electron pair are going to make a bond with the carbonyl-carbon then one bond on that carbonyl-carbon has to break.
The bond that’s going to break is the pi bond. Why? because it’s the weakest bond. Where to the pi electrons go? They will go onto the O atom. Why? because O is more electronegative than C, it can accommodate electrons more easily (this is often said that oxygen ‘likes’ electrons), so the electron pair in the pi bond moves onto the O atom. The O has gained charge forming an O- atom form being a carbonyl oxygen. Note this O- atom is still attached to the C atom by a sigma bond.
The electron pair on the alpha-C atom formed a second bond to the carbonyl-carbon (there was already a sigma bond between them) hence this electron pair forms a pi bond between the alpha-C and the carbonyl C
The resonance structure shown by the re-distribution of electrons in the red arrows is shown as the middle-bottom structure (the bottom structure that’s enclosed in the box). Of course these electrons can do the reverse of what they have just done (i.e. the C=C pi bond breaks now, and the e- go back onto the alpha-C and the O- reforms the pi-bond and this brings us back to the upper resonance structure again.
Using our ‘average of all the resonance diagrams’ method, we can get a picture of what the molecule actually looks like. It looks like this:
The reason for this happening is of course to SPREAD CHARGE AROUND THE MOLECULE, which gives GREATER STABILITY (by lowering electron repulsion). But the molecule doesn’t stay like this. It can do another step which lowers the energy even more. It can lose the OH group in the beta-position. This is shown using the blue arrows and produces our final product.
The electron pair on the alpha-carbon (more accurately the delocalised electron density around the carbonyl oxygen to the alpha-carbon) can move to the other side, in-between the alpha and beta carbons. This would form a pi bond in-between the alpha and beta carbons (shown in blue on the final product). But it would cause the beta carbon to have 5 bonds. C cannot have 5 bonds. So as the alpha-beta bond is forming, the carbon(beta)-O bond breaks. The electrons go onto the most electronegative element – i.e. the O and the C(beta)-O sigma bond breaks, forming an OH- ion. The O is negatively charged as it gained electron density. The beta-carbon atoms gained electron density by the formation of the C(alpha)-C(beta) pi bond, but then the beta-carbon lost electron density again when the C(beta)-O bond broke, so the charge on the beta-carbon doesn’t change.
WE are left with an alpha-beta unsaturated carbonyl. This molecule is stabilised as the C-O pi bond can undergo conjugation with the newly formed C=C bond, once again facilitating the delocalization of electrons in that molecule. The molecule is also not charged any more. In general organic compounds ‘prefer’ no overall ionic charge. The OH- used in the beginning is regenerated by the loss of OH- from the carbonyl, showing OH- is a catalyst. The overall loss of a H2O molecule from the hydrocarbon means this reaction is a condensation (or dehydration) reaction. The loss of the water leads to an increase in entropy as now two molecules exist causing more ‘disorder’.
Note: Formal charge calculation: formal charge = Valency of atom – (# of e- on that atoms in bonds)/2 – # of unbonded e- on the atom. The negative carbon atom has valency of 4. 6e- were used in 3 covalent bonds. There was one lone pair on the C atom. So formal charge is 4 – (6/2) – 2 = 4 – 3 -2 = -1
Here you go.
Try those questions. I’ll give answers later.
For Amer (A Level 10G5)
This page holds links to all sub-pages relating to CHM456. sub-pages will eventually contain (some) notes and points for discussion as well as host Q & A.
0.5) ** NEW 8-Nov-2012 ** Powerpoint slides for this course [password protected] .
2) Lesson Plan
2.5) Nomenclature: http://www.acdlabs.com/iupac/nomenclature/
Powrpoints of the
Tutorial Questions on Topic 1: chm456-tutorail-sheet-1-word2003
Answers to Tutorial questions1: answers-to-chm456-tutorail-sheet-1-word2003-old
Tutorial questions #2a: CHM456 Tutorial #2a
Answers to Tutorial questions #2a: CHM456 Tutorial #2a – answers
Tutorial Questions 2b: CHM456 Tutorial #2b
Answers to Tutorial Questions 2b: Answers to CHM456 Tutorial #2b
A) Syllabus: Syllabus-CHM456-Students-copy
The syllabus is a very useful document which lists (albeit in general terms) what topics you must know for the exam. You can use it to help with effective revision strategies.
B) Lab practicals: CHM456 LABORATORY-Students-copy
Note: You must read the relevant practical in advance of attending the lab session. Indeed, the lab manual states the following:
Laboratory Note Book/Jotter
Do not come into a laboratory unprepared. Read and understand the experiments ahead of time; no more reading should be done during lab session. If you do not do this, you will be unable to plan and use your time efficiently. There are several lab techniques that you need to read on your own for each lab session for example distillation, extraction, crystallization etc. They are self-explanatory and if you are unclear about something always consult the lecturer. Students must write/draw a brief outline/flowchart of the procedure of the experiment before coming to the lab. Your note book will be checked by your lecturer for assessment.
In addition, you often have to do some work/calculations/data look-up BEFORE you enter the lab. You MUST make sure you do this.
And of course, the usual warnings of safety in the lab apply.