Solution - Career Point

All the above three acids on heating with soda-limeyields only benzene.COOH COOH, ,COOHCOOHCOOHCOOHNaOH + CaO∆+ 2CO 2Of the three acids, one which on heating gives ananhydride, is o-isomer.COOHCOOH∆–H 2 OCOCOOne acid which on nitration gives a mono nitrocompound is p-dicarboxylic acid.COOHCOOHCOOHHNO3NO 2∆; H 2SO 4COOHOne acid which on nitration gives three mono nitrocompounds will be the m-isomer.COOHCOOHHNO 3H 2SO 4COOHNO 2COOHOCOOHNO 2COOHCOOHNO 2COOH5. Two moles of an anhydrous ester (A) are condensedin presence of sodium ethoxide to give a β-keto ester(B) and ethanol. On heating in an acidic solutioncompound (B) gives ethanol and a β-keto acid (C).(C) on decarboxylation gives (D) of molecularformula C 3 H 6 O. Compound (D) reacts with sodamideto give a sodium salt (E), which on heating with CH 3 Igives (F), C 4 H 8 O, which reacts with phenyl hydrazinebut not with Fehling reagent. (F) on heating with I 2and NaOH gives yellow precipitate of CHI 3 andsodium propionate. Compound (D) also givesiodoform, but sodium salt of acetic acid. The sodiumsalt of acetic acid on acidification gives acetic acidwhich on heating with C 2 H 5 OH in presence of conc.H 2 SO 4 gives the original ester (A). What are (A) to(F) ?Sol. (i) Acetic acid on heating with C 2 H 5 OH givesoriginal compound (A).CH 3 COOH + C 2 H 5 OH⎯ H ⎯ 2 SO ⎯4∆→CH3COOC2H5(A)+ H 2 O(ii) CH 3 COOC 2 H 5 (A) on heating with C 2 H 5 ONaundergoes Claisen condensation to give (B), which isaceto acetic ester.CH 3 CO OC 2 H 5 + H CH 2 COOC 2 H 5(A)C 2H 5ONaReflux+ C 2 H 5 OH + CH 3 COCH 2 COOC 2 H 5(B)(iii) (B) on heating in acidic solution gives (C) andethyl alcohol.CH + HOH ⎯→3COCH2COOC2H5(B)2(C)H⎯ +CH 3COCHCOOH + C 2 H 5 OH(iv) (C) on decarboxylation gives acetone (D).CH 3COCHCOOH2(C)∆⎯ ⎯→−CO 2CH 3COCH 3(D)(v) (D) reacts with NaNH 2 to form sodium salt (E),which on heating with CH 3 I gives butanone (F).CH 3COCH 3 + NaNH 2(D)∆⎯ ⎯→−NH 3CH3COCH2Na(E)CH 3⎯ I ⎯ ⎯–NaI → CH 3COCH2CH3(F)(vi) CH 3COCH2CH3+ 3I 2 + 4NaOH ⎯→(vii)(F)⎯ ∆CHI 3 + CH 3 CH 2 COONa + 3NaI + 3H 2 OCH 3COCH 3 + 3I 2 + 4NaOH ⎯⎯→∆(D)CHI 3 + CH 3 COONa + 3NaI + 3H 2 OCH 3 COONa ⎯ HCl ⎯→ CH 3 COOH + NaClThus, (A) CH 3 COOC 2 H 5(B) CH 3 COCH 2 COOC 2 H 5(C) CH 3 COCH 2 COOH(D) CH 3 COCH 3(E) CH 3 COCH 2 Na(F) CH 3 COCH 2 CH 3XtraEdge for IIT-JEE 37 APRIL 2010