Tartaglia

I would cut the lithium up into bits using a stanley knife, drop carefully into ethanol (goggles) Add nitric or hydrochoric acid to neutral pH, and evapourate. Recrystallise if necessary.

Benzyl grignard + cyclopentanone, work uo to give tertiary alcohol, treat with sulphuric acid to give alkene, treat with MCBPA to give epoxide

Well you can determine the electrostatic attraction/repulsion part of the lattice energy due to point charges (this involves a constant called the Madelung constant), though not the repulsion due to overlapping electron clouds. The solvation energy is largely a function of the polarising power of the ions, which is determined by charge density. There are also obviously entropy considerations

Organic chemistry is largely slog. If you learn the basics well then you will do very well, if you don't come to terms with the basics you will do badly. It does not suffer the conceptual difficulties that say theoretical chemistry has.

They are going to be very close frankly.

Well aqua regia will certainly work, but I expect it is a "sledge hammer to crack a nut" approach. The NO3- is non coordinating and nitrate salts are often far more soluble than chlorides, so it may be less of a problem than first appears.

I suspect the oxidation is a kinetic problem rather than a thermodynamic problem, but tbh NiCl26H2O is so cheap I would just buy a little bit!

Well, acetyl chloride could be used, but I expect you don't have that either. Also as this is a typical A level prep and your knowledge of chemistry seems limited, the use of acetyl chloride is far more risky than acetic anhydride

The problem with this prep is that the phenolic oxygen lone pair is not very nucleophilic so the usual ester prep from the alcohol and carboxylic acid will not work.

Its a d9 (essentially square planer complex), its crystal field stabilisation energy will be lowish and it will therefore be substitutionally labile. In dilute concentation I would expect it to return to the hexaaqua complex

Treating Co2(CO)8 with OH- in thf gives Co(CO)4- anion which can be protonated to give HCo(CO)4, which I believe is relatively unstable. Don't protonate it in thf though you'll polymerise the solvent.

Co(CO)3H is a 16e transient species, which is probably in eqm with the 18e HCo(CO)4. Cobalt carbonyls are notoriously labile

I would look to see if there are any chiral 3,4 dihydroxy amphetamine like natural products available and then you won't have to resolve but can build the methylene five membered ring in after

By the way, I expect this post will get some attention from the moderators - for my penny's worth, you look too credible to be doing something dodgy!

I should just wash it out with some dilute HCl and then water. Dry the DCM with a basic drying agent after

It is really the solubility of the base (NaOH, but it would be better to use a stronger one) which thf will increase. Thf is a coordinating solvent and will bind to Na+/K+.It is also not an electrophile and so will not undergo substitution reactions which CH2Cl2 can.

I think bearing in mind you must use a stoichiometric amount of base, I would simply use a standard procedure, which presumably you can look up.

A long shot here - you aren't doing this for Mike Turner at Manchester are you?

Thf is quite a good solvent for these sort of things. You only have to get a little bit in solution anyway.

I think the important point here is the need for a stoichiometric amount of base which UC pointed out and I (ashamedly) missed.

Use a stronger base or a solvent in which NaOH is more soluble - thf for instance.

Afterwards how about adding a mild oxidant and washing the carboxylic acid out with base

Factors which affect the change in valency include.

1) Successive ionisation energies

2) Lattice energies

3) strength of covalent ligand to metal bonds

4) whether ligands pi donate, or pi accept

5) crystal field stabilisation energy particularly low spin/high spin changes

6) solvation energies

7) etc etc etc

Actually its not even a full answer as nobody has mentioned "spin exchange energy"

Hermantrude You could always do the appropriate engineering council exams in your spare time.

http://www.cityandguilds.co.uk/documents/ind_engineering_ECUKexams/9107-02_vqhandbook_v3.pdf

i don't think that its -O-S(O)Cl- is the leaving group in case of SOCl2 and same for PCl5 , have a look on mech.

 

http://www.cem.msu.edu/~reusch/VirtualText/Images2/alcolrx1.gif

If you look at your post you'll see I'm right. -O-S(O)Cl- leaves and rearranges giving SO2 and Cl-, likewise -O-PCl4- leaves giving POCl3 and Cl-.

Because OH- is not the leaving group. In the case of conc HCl, H2O will be the leaving group. In the case of SOCl2, -O-S(O)Cl- is the leaving group, in the case of PCl5, -O-PCl4- is etc etc.

It is perfectly possible to become a good chemist without being a good mathematician, but in my experience, it is not the way to put the odds in your favour. You'll be selected out of physical and theoretical cheimstry for a start. Even inorganic chemists (which is what I first trained as) often require good maths. There wasn't a day during my PhD when I didn't use group theory as I did a lot of spectroscopy. The physical organic chemistry can also be pretty mathematical, so immediately you get narrowed down to organic synthetic, some aspects of inorganic and possibly some biochemical type disciplines

In most chemistry courses Schroedinger's equation will be introduced in the 2nd year so if you can't handle second order partial differential equations then you are at a distinct disadvantage.

I would also echo the above post on the importance of group theory which is crucial in spectroscopy and in molecular orbital theory and is imo generally poorly understood by many chemists. (You only have to look at how little symmetry is discussed on this forum.)

I always recommend anyone wishing to study chemistry improve their mathematical skills to well above A level

I do remember a PhD student "accidentally" pressing sodium wire into CHCl3 to dry it when I was postdocing in Germany. When his supervisor found out, they got a robot type thing to drop the choroform bottles out of the lab window after having cleared three or four floors of laboratories. All extremely embarassing for the people involved.

There is a brief description at the bottom of this page - http://chem.ch.huji.ac.il/nmr/whatisnmr/whatisnmr.html

The CH2O protons are diastereomeric and form an AB quartet by a roofing second order effect, these are further split by the adjacent CHMe proton

Vedmecum, The point of the question is the 1e/3e interconversion of the NO ligand. This is a rather common undegraduate organometallic chemistry exam question