Chapter 17: Aldehydes and Ketones Ch 17 contents
Aldehyde and Ketone Answers
 
Qu1: Increasing the number of alkyl substituents decreases the electrophilicity of the carbonyl C due to: 
(i) electronic effects (alkyl groups are weak electron donors) and (ii) steric effects (inhibit Nu approach) 

H2C=O  >  CH3CHO  >  (CH3)2C=O
Qu 2: Note the very similar nature of the reactions that the aldehyde and ketone undergo:
(a) LiAlH4 is a hydride reducing agent converting C=O to H-C-OH.
(b) PhMgBr is a Grignard reagent.  The Ph adds to the C=O to give Ph-C-OH.
(c) Alcohols react to give acetals, in this case since it is a diol, we get cyclic acetals.
(d) The arylhydrazine reacts to give the arylhydrazone via addition-elimination.
(e)  The ylid undergoes a Wittig reaction generating the alkene.
(f)  Hydroxylamine reacts to give the oxime via addition-elimination.
Qu 3: First step is to recognise what the products are as that defines the nucleophilic atom.
(i) Hydroxyl amine, NH2OH , reacts through the N to give an oxime, R2C=N-OH.  N is more nucleophilic (better electron donor) than O because the higher electronegativity of O makes it more difficult to donate its' electrons. 

(ii) Semicarbazide, NH2NHCONH2 reacts to give a semicarbazone.  This reaction occurs through the terminal amine type N rather than either of the amide type N which are less nucleophilic due to the involvement of the electrons in resonance. 
 


Qu 4: Here is a scheme collecting possible syntheses together (based on the more important reactions) 

(a)  Ethanal : we need to lose a C... ozonolysis of an alkene formed by elimination of the alkyl halide would allow that and give the aldehyde.
(b)  Propanone : oxidation of the secondary alcohol to get the ketone requires that we eliminate then hydrate the alkene in Markovnikov fashion. 
(c)  Propanal : aldehydes can be obtained by oxidation with the Cr reagents PDC (pyridinium dichromate) or PCC (pyridinium chlorochromate) in methylene chloride (aq. conditions result in over oxidation to the carboxylic acid). The primary alcohol is obtained by substitution.
(d) 2-Butanone : now we need to gain a C. We can do this with Grignard chemistry by adding MeMgBr to the previous product, propanal, followed by oxidation of the secondary alcohol. 
(e)  Pentanal : this time we must add 2 C. Again use a Grignard, but this time add ethylene oxide to the Grignard of the original halide. Finally, controlled oxidation (see (c)).
Qu 5: In both cases we need to add a C atom so we need nucleophilic C systems  which react with cyclohexanone.
Addition of a Grignard reagent gives the tertiary alcohol, but this can only be eliminated to give the more highly substituted endocyclic alkene.  The Wittig reaction is ideal for creating the exocyclic alkene since the Wittig reaction specifically transforms a C=O to C=C at the same location.