Tartaglia

Imagine attack by PhO- and RCO2-. The negative charge is more effectively delocalised across the two oxygens in the carboxylic acid, than the -ve charge in PhO- which is less effectively delocalised within the ring. In this case the larger effective negative charge on the phenoxide will be the most important factor in determining nucleophilicity

You really need the terms "soft" and "hard" when determining what is the best nucleophile. Good nucleophiles are usually ( though not always) "soft". This means the electrons are in a diffuse orbital where the charge is spread out. Bad nucleophiles are usually "hard" where the orbital is tightly bound to the atom and thus can not "reach" easily into the electrophilic centre. Since the OH oxygen has two sp3 lone pairs, which are more diffuse than the two sp2 orbitals on the carbonyl oxygen, I would expect the OH oxygen to be more nucleophilic. The answer does depend to some extent on the electrophilic centre as some centres like hard nucleophiles.

Step I is a disguised aldol reaction Step II is a ketal formation via dehydration. The ketone effectively acts as a protecting group Step III is nucleophilic attack of (CH3)2S+-CH2- at the aldehyde, followed by ring closure to eliminate (CH3)2S and form the epoxide Step IV is nucleophilic attack of amine on the less hindered epoxide carbon, resulting in ring opening Step V is removal of the protecting ketone via acid hydrolysis

In general, the three main branches of chemistry are Inorganic, Organic and Physical. There are obviously many smaller ones. I'm just surprised "physical chemistry" falls the wrong side of the line and does not have its own forum

What is more surprising is that there is no physical chemistry forum

I think steric hinderance will be a problem with many reactions. I also think the answer is to use a method that utilises more electrophilic reagent to react with the amine and that may be difficult for secondary and tertiary alkyl groups

If you use DMSO I would be tempted to use it dry or dryish. I would probably use something more volatile for ease of separation (eg CH2Cl2, diethylether).

I would have thought that any slightly polar aprotic solvent would do

Even though 1 amino-adamantane is a primary amine the adamantyl group is so bulky it is more likely to act as a base rather than a nucleophile with secondary bromides like 2-bromopropane. You could try better leaving groups like iodide or tosylate triflate, but this may still encourage elimination. Doing the reaction the other way around is fraught with problems. The adamantyl tertiary carbon can not adopt a planar configuration due to ring strain and so this makes Sn1 type mechanism much harder. (Sn2 is out as it can't turn inside out)

Are you merely asking how to convert a primary amine into a secondary amine here? Or is there some missing information regarding the adamantyl group?

Sorry Eigenvalue, I didn't see your last post until now. As far as I am aware there are two forms of Co2CO8 both of which have high symmetry and no dipole. (One has bridging carbonyls, one doesn't). Therefore any non polar solvent will probably do. I would probably use 40/60 pet ether

Probably, though I'm far from certain. Amides can react with Wittig reagents in the usual fashion. There is an example on page 853 of "Advanced Organic Chemistry" (3rd edition) by Jerry March, but I suspect the rate is significantly slower

Wittig's definitely react with ketones. The reason that DMF probably didn't react is there is a contributing resonance form of the type Me2N+=C(O-)H, which renders it less susceptible to nucleophilic attack

The green is ferrous chloride (FeCl2) the orange ferric chloride (FeCl3)

sulphonation is reversible so this is not a particularly unusual reaction

It is easier to draw than explain,but there should be four points of inflexion, two near vertical and two near horizontal. Assuming we are adding the strong base to the acid the pattern should go from low to high pH in the following manner 1 A near horizontal point of inflexion (slightly upward sloping) at pKa1 2 A near vertical point of inflexion at first neutralisation point 3 A near horizontal point of inflexion (slightly upward sloping) at pKa2 4 A near vertical point of inflexion at 2nd neutralisation point 1 and 3 are called buffering regions

As long as you don't go too far into the UV I would think this unlikely

The pi back donation to the CO ligands in Co2(CO)8 is pretty weak and so the CO ligands can be knocked off easily. There may be higher oligomers produced eg Co4(CO)12, but as long as oxygen isn't present you should get Co metal and CO gas. If oxygen is present there will probably be some Co(III) oxide produced

AuBr4- ought to be fairly easily synthesised either as YT2095 says from RbBr and AuBr3. There may also be other ways. This is because AuX4- is d8 square planar and these compounds are extremely labile as regards substitution. Thus displacement of Cl- ligands from AuCl4- would probably work in the presence of excess RbBr. The trans directing nature of Cl- and Br- should be similar. Some of the original work on substitution of d8 square planar complexes was carried out on such gold complexes, (1940's and 1950's) which seems a little surprising to me as d8 square planar Pd(II), Pt(II) Rh(I) and Ir(I) have been more thoroughly studied

I think the first statement implies a proton is added with each electron. But I'm not sure exactly what the second implies about protons. Also it would depend which orbital were next in the second statement (and possibly the first too)

I would probably protect the -NH2 group with an acetyl group. Sulphonation I would probably carry out with ClSO3H (in xs) giving p-CH3C(O)NHC6H4SO2Cl. Treat with NH3 and hydrolyse to give sulphanilamide

Funnily enough I very nearly put H2O2 as I had suggested it earlier. There was just enough doubt in my mind as to whether the Cu2+ might catalytically decompose the H2O2 before the insoluble CuCl was oxidised There are ofcourse many other alternatives

Here is an easier approach Divide sample in two Add I- to the first solution and titrate released I2 with S2O32- --> Cu2+ Stir 2nd sample with nitric acid. This should convert all copper ions to Cu2+. Raise PH with NaOH and decant supernatent leaving insoluble precipitate of Cu(OH)2 free from Cl-. Dissolve Cu(OH)2 in dilute H2SO4 add excess NaI and titrate produced iodine with S2O32- --> Cu+ and Cu2+

More commonly known as the nernst equation Are you truly telling me you would have difficulty finding something that oxidises Cu+ to Cu2+, without oxidising Cl-?

Get your standard emf book out. I haven't got one infront of me but I suspect a little H2O2 would do and the excess would be destroyed by the Cu2+ ion. Stir overnight There are plenty of other possibilities