Solution - Career Point

2. An organic compound (A) C 7 H 15 Cl on treatment withalcoholic KOH gives a hydrocarbon (B) C 7 H 14 . (B)on treatment with O 3 and subsequent hydrolysis givesacetone and butyraldehyde. What are (A) and (B) ?Explain the reactions.Sol. In numerical, following data are given :C7HCl ⎯⎯⎯15(A)Alkyl halide∆Alc.KOH ⎯∆→C7H 14(B)AlkeneO⎯⎯→3C7H14O3Ozonide⎯H 2⎯O ⎯⎯/ Zn → CH 3 COCH 3 + CH 3 CH 2 CH 2 CHOThe alkene contains seven carbon atoms. Theposition of C = C double bond can be locatedas follows :HCH 3–2[O]C = O + O = C.CH 2 CH 2 CH 3CH 3CH 3 – C = CH.CH 2 CH 2 CH 3CH 3Thus, alkene (B) is 2-methyl hexene-2The ozonolysis reaction is as follows :CH 3 – C = CH.CH 2 CH 2 CH 3CH 3(B)O OCH 3 – C CH – CH 2 CH 2 CH 3OCH 3H 2O/Zn∆O 3CH 3 – C = O + O = C – CH 2 CH 2 CH 3CH 3 HSince alkene (B) is produced by the removal of onemol of HCl from alkyl halide (A) and thus (A) can beeither (I) or (II).Cl(I) CH 3 – C – CH 2 CH 2 CH 2 CH 3or(II)CH 3CH 3 – CH – CH – CH 2 CH 2 CH 3CH 3 ClThe dehydro halogenation reaction by(I) or (II) yeildsClKOHCH 3 – C – CH 2 CH 2 CH 2 CH 3alc.CH 3(I). CH 3 – C = CH.CH 2 CH 2 CH 3CH 3Main product (Saytzeff's rule)KOHCH 3 – CH – CH – CH 2 CH 2 CH 3Alc.CH 3 Cl(II) CH 3 – C = CH.CH 2 CH 2 CH 3Thus, both (I) and (II) give main product as 2-methylhexene-2, hence (A) is either (I) or (II).Cl(A) CH 3 – C – CH 2 CH 2 CH 2 CH 3or(B)CH 32-chloro-2-methyl hexaneCH 3 – CH – CH – CH 2 CH 2 CH 3CH 3Cl3-chloro-2-methyl hexaneCH 3 – C = CH.CH 2 CH 2 CH 3CH 32-methyl hexene-23. An unsaturated hydrocarbon (A), C 6 H 10 readily gives(B) on treatment with NaNH 2 in liquid NH 3 . When(B) is allowed to react with 1-chloro propane, acompound (C) is obtained. On partial hydrogenationin the presence of Lindlar catalyst (C) gives (D),C 9 H 18 . On ozonolysis (D) gives 2, 2-dimethylpropanal and butanal. Give structures of (A), (B), (C)and (D) with proper reasoning.Sol. The structure of compound (D) can be obtained byjoining the products of ozonolysis.CH 3–2[O]CH 3 – C – CH = O + O = CH.CH 2 CH 2 CH 3ButanalCH 32,2-dimethyl propanal CH 3CH 3 – C – CH = CH.CH 2 CH 2 CH 3CH 32,2-dimethyl heptene-3 (D)Ozonolysis equation of (D) is :CH 3(I) O 3CH 3 – C – CH = CHCH 2 CH 2 CH 3(II) H 2O/ZnCH 3(D)CH 3CH 3 – C – CHO + CH 3 CH 2 CH 2 CHOCH 3Alkene (D) is obtained by the partial hydrogenationof (C), thus (C) contains a – C≡C – triple bond at C 3 .CH 3CH 3 – C – C ≡ C – CH 2 CH 2 CH 3CH 3(C)CH 3H 2Lindlar catalystCH 3 – C – CH = CHCH 2 CH 2 CH 3CH 3Main product (Saytzeff's rule)CH 3(D)XtraEdge for IIT-JEE 35 APRIL 2010