[NISHANT]

I have concluded these points that one must look at before deciding what type of a reaction it is going to be. Are all these correct?
1.If alkyl halide is primary and the base is C2H5ONa, sustitution is preferred.
2.If the alkyl halide is secondary or tertiary and the base is C2H5ONa,elimination is preffered{Acc. to sytzeff}
3.Alkyl halide may be primary,sec. or ter. and base is C4H9ONa,elemination is preffered{Acc. to hoffmann}.
4. If the base is C2H5ONa but the groups:
-F
-CF3
-NR4{+}
are there then elimination is preffered acc. to hoffmann

[Donaldson Tan]

1. substitution is the most probable outcome

2. for elimination vs substitution, also don't forget the effect of solvent on the reaction pathway.

[NISHANT]

but in case of NR4+ it is hoff. elimination which is E2.THEN HOW SUBSTITUTION?

[Donaldson Tan]

the solvent molecules form solvent-solute bonds with the nucleophile of interest. if the solvent is of heavy molecular weight, the nucleophile becomes relatively bulkier and less mobile, thus its ability to act as a substitute decreases. Instead, the residual charges on the nucleophile maybe able to abstract hydrogen or halogen, bringing about elimination.

[NISHANT]

WHY IS ALWAYS HOFFMANN PREFFERED FOR
-F
-NR4+_
-CF3

[Demotivator]

Those groups listed are electronegative and favor the development of negative charge. Also, F bonds are strong and do not weaken as fast as the C-H bond.

As a result, the transition state for those groups have carbanion character instead of the alkene character in saytzeff elimination with other groups.
Carbanion transition states are more stable with less substituted carbons  resulting in hoffman elimination.