- Aqueous KOH favors nucleophilic substitution (SN1/SN2) reactions, where OH− acts as a nucleophile, forming alcohols.
- Alcoholic KOH, due to the presence of alkoxide ions (RO−), is a stronger base and favors the elimination (E2) reaction, leading to alkene formation.
Explain with the help of chemical reactions: (I) Acetone is treated with semicarbazide.

Draw a rough sketch for the curve $y = 2 + |x + 1|$. Using integration, find the area of the region bounded by the curve $y = 2 + |x + 1|$, $x = -4$, $x = 3$, and $y = 0$.