Solution - Career Point

UNDERSTANDINGOrganic Chemistry1. An organic compound (A), C 4 H 9 Cl, on reacting withaqueous KOH gives (B) and on reaction withalcoholic KOH gives (C) which is also formed onpassing vapours of (B) over heated copper. Thecompound (C) readily decolourise bromine water.Ozonolysis of (C) gives two compounds (D) and (E).Compound (D) reacts with NH 2 OH to give (F) andthe compound (E) reacts with NaOH to give analcohol (G) and sodium salt (H) of an acid. (D) canalso be prepared from propyne on treatment withwater in presence of Hg ++ and H 2 SO 4 . Identify (A) to(H) with proper reasoning.Sol.C 4 H 9 Cl(A)(Alkyl halide)Alc. KOH∆;– HClC 4 H 8(C)(Alkene)Aq. KOHCuC 4 H 9 OH∆; –KCl (B) ∆; –H 2 O(Alcohol)We know that p-alcohol on heating with Cu givesaldehyde while s-alcohol under similar conditionsgives ketone. Thus, (B) is a t-alcohol because it, onheating with Cu gives an alkene (C). Since at-alcohol is obtained by the hydrolysis of a t-alkylhalide, hence (A) is t-butyl chloride.Thus, (A) isClOH||CH3− C − CH3and (B), is CH3− C − CH3||CH3CH3The alkene (C) on ozonolysis gives (D) and (E),hence (C) is not symmetrical alkene. In thesecompound (E) gives Cannizaro's reaction withNaOH. So, (E) is an aldehyde which does not containα - H atom. Hence it is HCHO. Compound (D) canalso be prepared by the hydration of propyne in thepresence of acidic solution and Hg ++CH 3 – C ≡ CH + H 2 OHg ++H + CH 3 – C = CH 2OHCH 3 – C – CH 3O(D)Hence (D) is acetone and (E) is formaldehyde.Therefore, alkene (C) is 2-methyl propene.CH 3 – C = CH 2CH 3(C)(D) reacts with hydroxyl amine (NH 2 OH) to formoxime (F).CH 3–HC = O + H 2 NOH 2 O CH 3C = NOHCH 3CH 3(D)(F)Thus, (B) is CHReactions :ClCH 3 – C – CH 3CH 3(A)3OH|− C − CH|CH33and (A) is CHOHAq. KOHCH 3 – C – CH 3∆; –KClCH 3(B)CH 3 – C = CH 2 + H 2 OAlc. KOH/∆–KCl; –H 2 OCH 3 – C = CH 2(I) O 3(II) H 2O/ZnCH 3CH 3CH 3(C)C = O + H 2 NOH(D)CH 3CH 3CH 3∆–H 2O(C)3CH 3 – C = CH 2CH 3(C)Cl|− C − CH|CH3Cu/300ºC–H2OOC = O + H –C – H(D) (E)CH 3CH 3C = NOH(F)2HCHO + NaOH → CH 3 OH + HCOONa(E) (G) (H)OCH 3 – C ≡ CH + H 2 O Hg++ CH 3 – C – CH 3H +(D)3XtraEdge for IIT-JEE 34 APRIL 2010