Chem 351 Fall 2012 MT Mechanism

Part 7: 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.

mechanism

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. Both systems are C-H bonds. Aldehydes have a pKa of about 17 due to the acidity of the hydrogens in the group adjacent to the C=O. Removal of such a proton gives a carbanion as the conjugate base that is stablised by a resonance interaction that delocalises the -ve charge on to the electronegative oxygen in the carbonyl group (i.e. an enolate). Alkanes have a pKa > 50 since the deprotonation of the C-H bond gives an unstabilised carbanion.

resonance

c. In the analogous reaction, ethanal would need to be deprotonated and reacted with 1-bromobutane to give hexanal. C2 + C4 = C6.

hexanal symthesis

Common errors:

a.

could not draw the structure of ethanal, methyl bromide, propanal, diisopropyl amine correctly
could not draw lithium diisopropylamide, LiN(CH(CH₃)₂)₂ and / or diisopropyl amine HN(CH(CH₃)₂)₂correctly. The "tough" one here was the "amide", but if you drew the amine and then reversed the simple acid / base reaction and removed a proton, it would have "told" you the amide ion.
not removing the correct H (i.e. most acidic H is from the methyl group adjacent to the carbonyl group, it is not the aldehyde H in the CHO group)
drawing structures that break valence rules
incorrect formal charges (extra or missing)
incorrect, vague, backwards and incomplete curly arrows. Arrows are required to account for all the bonding changes.
showing extra steps not described in the question (so in effect answering a different question)
using single headed arrows (radical) despite the question saying to use double headed arrows
having removed the wrong H from the aldehyde, did not realise that the reaction would give propan-2-one rather than propanal (or just ignored the issue which should have told you that you had a problem)

b. Did not correctly show or describe the resonance stabilisation of the enolate.

c. Did not count carbon atoms correctly and therefore suggested 1-bromopropane (C3) or 1-bromopentane (C5)