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.

Common stumbling blocks for some students were :

• could not draw the esters
  
• could not draw the amide or ammonia
  
• incorrect formal charges
• incorrect or vague curly arrows
• removed the wrong proton, H+  from the ester
• poor understanding of drawing resonance structures obeying the basic rules of valence.
  
• poor rationalisation of the acidity in b(ii)
  

(a)

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.

In this problem, if you didn't know what sodium amide was, then you should have been able to work it out from the knowing what ammonia was and that amide forms ammonia after an acid - base reaction (i.e. simple proton transfer) and given that it tells you that amide is a base in the question....

If you were not sure which H+ to remove, then there are 2 clues (a) only one set give a carbanion that is resonance stabilised and (b) by looking at the location of the new methyl group in the product.

(b)
(i) The significant resonance structures can be drawn by delocalising the charge on the C to the O atom by using the -ve lone pair to form a new C=C and breaking the C=O. The two major  contributors are the most significant. Note the question asked for the carbanion.

(ii) Acidity : The methyl propanoate is more acidic than propane as indicated by the lower pKa (25 versus 50+ respectively).  The major factor is the resonance stabilisation of the conjugate base of the ester (as shown above) where the negative charge on C can be delocalised to the more favourable (i.e. more stable) location on the electronegative O atom. Propane, only has a simple, unstabilised carbanion.

(iii) If you wanted to prepare methyl 2-methylhexanoate from methyl propanoate using this type of reaction, you should react the carbanion of the propanoate with 1-iodobutane (aka butyl iodide) = CH₃CH₂CH₂CH₂I instead of the methyl iodide (as stated in the question the ester undergoes alkylation when reacted with a base and then an alkylating agent) that was used in the first part of the question (count C atoms... in the question 4 + 1 gives 5.... need 8 have 4.... use C4 alkylating agent).  A simple comparision of the overall reaction in part (a) reveals the answer: