Part 6: MECHANISM
The following diagram shows the solution to the mechanistic question. Note that all the information applies to a single reaction sequence that has been completely described verbally. There is no need for extra reagents or extra steps etc. The curly arrows are drawn specifically to match the text in the question. The biggest problem students have is making sure they understand the language of chemistry. Most students have trouble because they can't draw the structures from the IUPAC names (that means they don't know their nomenclature well enough). Read the words carefully, and then make the curly arrows tell that same story. There is NO need for extra steps. Remember curly arrows go from electron rich to poor and to balance the formal charges at each step - errors on formal charges were common.
a.
(3 marks for each step)
If you struggled with this part of the question, first draw the compounds whose names were provided, then think about the types of reactions (e.g. acid / base) and try to fill in the structures in the gaps, then finally add the required curly arrows to account for all the bonding changes.
b. Hydroxide (HO-) or an alkoxide (RO-) would be the best choices since their pKa's (about 15) mean that they are certainly strong enough to deprotonate a carboxylic acid (pKa = 5). (1 mark)
c. Since we are making another benzoate ester, we still use benzoic acid but now with 2-iodopropane (aka isopropyl iodide) or 2-bromopropane (aka. isopropyl bromide): (1.5, 0.5 marks per item)
d. This time we need to make an ether, so use phenol with ethyl iodide but use a stronger base (since the phenol is less acidic): (1.5, 0.5 marks per item)
e. A phenol is less acidic than a carboxylic acid (pKa's 10 and 5 respectively). We can explain this by looking at the conjugate bases of each system. In a carboxylic acid, the -ve charge sits on an electronegative oxygen atom and there is resonance that delocalises the -ve charge to a second electronegative oxygen atom. In a phenol, the -ve charge sits on an electronegative oxygen atom and there is resonance that delocalises the -ve charge to 3 of the carbon atoms in the benzene ring. Therefore the key difference is that the conjugate base of the carboxylic acid charge delocalisation is to a second electronegative atom which gives better charge stabilisation. (3 marks)
Common errors:
a.
b. Poor knowledge of acids and bases e.g. suggesting things like Cl- or HSO4- which are very weak bases.
c & d. Did not use part a as the basis of working out the answer. Omitting one or more components, incorrect base selected.
e. Did not know the correct structure for a phenol (e.g. confusing it with a phenyl group). Poor knowledge of acidity. Rationale based on pKa alone (which effectively said less acidic because it is less acidic). Incomplete rationale.