The decreasing order of ease of alkaline hydrolysis for the following ester isIIIIIIIV
(A)III > II > IV > I
(B)III > II > I > IV✓ Correct
(C)IV > II > III > I
(D)II > III > I > IV
Answer: (B)
Solution
Base-promoted ester hydrolysis proceeds through hydroxide attack on the carbonyl carbon, so anything making that carbon more electron-poor speeds it up. A ring nitro group withdraws electron density strongly and accelerates the attack most, chlorine does so moderately, an unsubstituted ring is intermediate, and an electron-donating methoxy group pushes density in and slows it most. Ranking by how electron-hungry the carbonyl is left gives the ease of hydrolysis as III>II>I>IV.
JEE Main · 2019 · 10 Jan 2019 · Shift 1 · Single correct option