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)

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.  

the seeing the electronic configuration we can say that the element is fluorine.

fluorine is the most electronegative element in the periodic table.

but by only seeing the configuration we cannot say about its reactivity because we don't know if its configuration after excitation or by addition of electrons.

but if not then, it is the lightest halogen and exists as a highly toxic pale yellow diatomic gas at standard conditions. as the most electronegative element, it is extremely reactive: almost all other elements, including some noble gases, form compounds with fluorine.