Question

1a) consider the mixed aldol condensation reaction of 1-methylcyclopentane-1-carbaldehyde (shown below) and 3,3-dimethyl-2-butanone (shown below) in the presence of sodium hydroxide. provide the expected major mixed aldol condensation product for this reaction (can exist as e or z isomer so you can draw either)

Answer 1

Aldehydes and ketones having α-hydrogen atoms, undergoes aldol condensation, in present of base (NaOH). 

The initial product formed during this reaction is β-hydroxy alcohol, which then undergoes dehydration to form α,β-unsaturated aldehyde or ketone.

In present case, 3,3-dimethyl-2-butanone  has 2α-hyrogen atom, while methylcyclopentane-1-carbaldehyde has 1α-hydrogen atom. So the major product formed during cross aldol condensation reaction of these reactants is:
 5-hydroxy-4,4-dimethly-1-(2-methylcyclopentyl)pent-1-en-3-one. 

The complete reaction product formed is shown below. 

Answer 2

Explanation:

Aldol condensation refers to a reaction between an enolate ion or enol and a carbonyl compound to produce beta hydroxy aldehyde or beta hydroxy ketone. The initial step in a crossed aldol condensation is ketone deprotonates or enolizable aldehyde.  

An enolizable ketone or aldehyde signifies that it should exhibit one or more alpha hydrogens. In the given case, only 3,3-dimethyl-2-butanone possess alpha hydrogen. Thus, it will get deprotonated by the base (hydroxide ion) in the initial step. In the succeeding step, it will be supplemented to the carbonyl carbon of the aldehyde. And in the ultimate step alkoxide gets protonated by water.