Archive for March 2014
In the lab this week, CHM456 (AS202) students were reporting “strange” ranges for their melting point determinations. e.g. 60-150oC.
Most of the pure compounds we use should have melting point ranges within one, two or three degrees only. Hence I think it’s fair to say students don’t really understand what it is they need to see in order to get the melting point range.
I must say, that I’m struggling to understand how this could be. I’d would have thought it quite easy to understand what the process of melting looks like.
Nevertheless, I do understand that what is obvious to an experienced person can be very non-obvious to someone with no experience.
Unfortunately, it’s not feasible or possible to show 20+ students what it looks like during a lab session.
But this site does a good job. It shows a sequence of frames (animated gif) showing what you should see in the melting process.
Another source showing it is here: https://www.youtube.com/watch?v=9aQio1KQKrs (1min 37seconds in)
Yet another one here:
(melting begins about 3mins 18 seconds in)
And here too:
(about 6min 40s into the video)
Please do watch ALL of the videos. They are short.
This one is longer, but still worth watching I feel.
https://www.youtube.com/watch?v=zMjl9MzpySU (starts about 20min 20s in)
Actually this highlights an important point. Studying has NEVER EVER been EASIER. The amount of resources and information available is simply stunning. You just have to put the time in and do the reading or even easier, watch the videos.
Sketch the following molecules and deduce whether the following molecules are polar or non-polar.
1) chloromethane, CH3Cl
2) Propanone, CH3COCH3
3) tetrachloromethane, CCl4
4) phosphine, PH3
5) Cyclohexane, (CH2)6
6) Ethylamine, CH3CH2NH2
7) Carbon monoxide, CO
8) Iodine, I2
9) propene, H2CCHCH3
10) phosphorous pentachloride, PCl5
11) Sulfur trioxide, SO3
12) Methane, CH4
– Answers will be posted later in the comments.
Lets just look a little bit more about C before we discuss the hybridization of other atoms.
Remember, hybridization is important because the hybridization of the atom is related to the molecular geometry(i.e. shape) around that atom, and molecular shape has a very important influence on chemical reactivity and its physical properties.
The hybridization can be deduced by a simple formula:
Because the number of pi bonds are complementary to the number of sigma bonds, you can of course also express that formula in terms of pi bonds.
Take propene, H2C=CHCH3 as an example. According to the pi based formula, first carbon (the blue coloured one) in propene, is sp(3 -1) = sp2
This pi bond based formula can also be used to determine hybridization in other atoms such as N and O etc.
Consider ammonia, NH3.
The number of pi bonds is zero, so, by the pi based formula, N’s hybridization is is sp(3 -0) = sp3
In the molecule pyridine, as shown below, the N has 2 single bonds and one pi bond.
so the N in pyridine is sp(3 -1) = sp2 hybridized.
As for oxygen, it is a similar story. In water, O has two single bonds and zero pi bonds, so the hybridization of O in H2O is sp(3 -0) = sp3 hybridized.
In the group of molecules called esters, there are two oxygen atoms (primarily)responsible for making the molecule an ester. Applying the pi based hybridization formula for each oxygen in the ester….
we get sp(3 -1) = sp2 hybridization for the oxygen coloured in green and sp(3 -0) = sp3 hybridization for the oxygen coloured in red.
You may have realised that the N based compounds and the O based compounds above have LONE PAIRS on them. Remember a lone pair are two valence electrons in a single orbital that are NOT used in bonding). Those lone pairs have the same hybridized as we calculated for the atom.
Finally, you may wonder why we didn’t use the original sigma (σ) bond formula for the N and O species. This is because the number of bonds those atoms form by sharing one electron with a different atom, varies. N has 3 single unpaired electrons and O has two, i.e., the valence of N is three and the valency of O is two. Hence the number of σ bonds formed for N and O compared to C is difference so the σ based formula needs to be adjusted each time an atoms with a different valency is used. But the pi (π) based formula doesn’t need to be adjusted, hence it can can be applied more widely 🙂
When atoms share electrons to bond, the first bond that forms between them is a sigma bond. So single bonds are sigma (symbol = σ) bonds.
Sometimes, further bonding can occur between the same two atoms. If this happens, they can then form a pi (symbol = π) bond. The σ and the π bond are collectively known as a double bond. To form this π bond, two electrons (in addition the two e- already in the sigma bond) will need to be shared.
It is possible for even more electrons to be shared between the same two atoms. If this happens a second pi bond will form, and all these bonds are known collectively as a triple bond.
A sigma bond is formed when the valance orbitals of one atom, directly overlap, head-on, with the valence orbitals of a different atom
When carbon, oxygen and nitrogen bond covalently to other atoms, they hybridize. Note: some atoms do not hybridize, but for this course we focus strongly on C,N,O which do hybridize.
You may notice the atoms, C,N and O have for their valence shell, the n=2 shell, involving the 2s orbital, the 2p(x) orbital, 2p(y) orbital and the 2p(z) orbital. Remember an orbital can only hold 2 e- and when this happens, they must have opposite spin!
Hybridization (mixing) of an atoms internal orbitals allows better overlap between atoms, hence stronger bonds form, hence more energy is released when they form, hence the resulting species is more stable than it would be if no hybridization takes place.
- If an s orbital hybridizes (mixes) with a p orbital on the same atom, you get an sp orbital.
- If an s orbital hybridizes (mixes) with two lots of p orbital on the same atom, you get an sp2 orbitals.
- If an s orbital hybridizes (mixes) with three lots of p orbital on the same atom, you get an sp3 orbitals.
The diagram below is to try and help illustrate that. It shows a 2p(x) orbital of one atom (e.g. a carbon atom) bonding/overlapping with an s orbital of another atom (e.g. a hydrogen atom).
Summary: Sigma bonds (single bonds) form by head on overlap of the atomic orbitals of different atoms
What about pi bonds?
Pi bonds are formed from NON-hybridized p orbitals overlapping in a sideways manner.
The electron density in the atoms is a pi bond is formed by the sideways overlap of non-hybridized orbitals. This is illustrated below (note the preceding sigma bond between the C atoms is not shown for the sake of clarity, but in reality, it is there!)
A very useful statement:
The shortest distance between two points is a straight line. The e- in a sigma bond essentially lie in a straight line (with a bit of bulging) between the attracting nuclei. So the e- are at the shortest distance that’s possible between to two attracting nuclei, hence they are attracted (i.e. bonded) as strongly as they can be, and so are ‘tied’ up, and not so easily available for reaction.
The electrons in pi bond are NOT directly in between the attracting nuclei, so the distance of the attracting nuclei to the electrons in a pi bond is greater than it is for sigma bond.
HENCE, the e- in a pi bond are attracted less strongly than a corresponding sigma bond, so the e- in the pi bond are MORE reactive than the electrons in the sigma bond.
The result of this is that pi bonds are more reactive than sigma bonds, so pi bonds will react first. This helps you identify some reactive sites in molecules, and so you can begin to ‘feel’ how a chemical can react.
Skeletal diagrams show C-C bonds and bonds to heteroatoms (e.g. C,N,O,F,Cl,Br,I). Bonds between C and H are usually left off – especially if they are not involved in a discussion or a reaction.
Get familiar with using skeletal sketches of molecules. They will be used on the exam paper.
Dear students, please students to fill in the entrance survey in i-learn
portal before the third week of semester!
I will give you a hard copy (printed copy) of the labs and an instruction manual next week (Monday, 10th March). I will probably do a lab briefing IN the lab next week (10th-14th March). I’ll let you know in class again on Monday.
P.S. AS 229
Re: The timetable clash. I brought it to the attention of the timetable committee and they have registered each one of you for the lab. All you have to do is split yourselves into two EVENLY SIZED groups, one group (group “x”) doing organic chem on Tuesday morning, the other doing (‘y’) doing analytical. Then elsewhere in the week (thursday I think), ‘x’ will enter the analytical chem while group ‘y’ go for organic.
This was necessary because your group is quite large. Not enough lab spaces.
Hello there AS229-2A (46? students) and AS246-3A (14 students)
Welcome to CHM412 – Organic Chemistry For Technology
I hope you will use this blog to it’s full potential. Sadly, last semesters students hardly used it at all.
Clicking https://intechemistry.wordpress.com/2013/09/09/chm-412-sept-dec-2013/ will take you to the ‘home page’ of CHM412. Feel free to ask any question in the relevant, clickable, sub-pages listed there. I do ask that if you want to ask a question, then please state your current understanding of the issue you want to ask about.
This semester, because of past abuses, I will not give take-home tests, and may give a number of ‘spot tests’ instead. So you must attend every class or have a PROPER reason for not attending.
The code has not changed since last semester, hence the material here is just as relevant to you as it was for last semesters students.
Please read ALL that this blog has in relation to CHM412 and download all the necessary documentation.
Best wishes and I look forward to your succeeding in this code.