Question

The acid strength of active methylene group in (I) $CH_3COCH_2COOC_2H_5$ (II) $CH_3COCH_2COCH_3$ (III) $C_2H_5OOCCH_2COOC_2H_5$ decreases as

• A. I > II > III
• B. III > I >II
• C. III > II > I
• D. II > I > III

Answer 1

Answer: (D)

II > I > III

Answer 2

Acid strength in active methylene compounds can be decided by two factors (i) Presence of electron withdrawing group (ii) Stability of enolate anion obtained after removal of $H ^{+}$. Higher the electron withdrawing ability of substitutents attached to electron withdrawing groups higher will be acidic strength of methylene group. Electron withdrawing ability of $CH _{3} CO -$ is greater than increases acidic strength of active methyl e ne compound to greater extent than $CH_3CH_2O-\overset{\underset{||}{O}}{C}-$ Number of $CH_3 - \overset{\underset{||}{O}}{C}-$ groups Number of $-\overset{\underset{||}{O}}{C}- OCH_2CH_3$ groups On the other hand stability of enolate anion obtained after the removal of $H ^{+}$ can be explained as keto group stabilises enolate anion to more extent than ester as ketone group stabilise enolate anion by resonance through one side only while ester stabilises by both side of keto group which can be shown as less tautomerism will be seen due lesser availability of $C ^{1}$ atom for tautomerism (as at this $C$ -atom $> C = O$ bond is in conjugation with the lone of neighbouring oxygen attached). Further, in as both the C-atoms of $> C = O$ bond have similar situations hence, tautomerism decreases further. That's why correct order of acidity can be arranged as II $>I>$ III and correct choice is (c).