Product obtained by reaction with resorcinol

Transcription

1 33 rd International hemistry lympiad Preparatory Problems 3 ( Z ) 3-( 4-methoxyphenyl )-2-pentenedioic acid d. Two products are possible when compound A reacts with bromine [1] [2] Structures 1 and 2 are enantiomers. e. 2 S R R S f. 2 B Product obtained by reaction with phenol 2 Product obtained by reaction with resorcinol g. In the formation of compound A from anisole, the attack takes place at the p- position of the 3 group. owever, when compound B is formed from phenol, the attack takes place at the o-position of the group. Steric Mumbai, India, July

2 33 rd International hemistry lympiad Preparatory Problems hindrance of 3 group favours the attack at the para position. Steric hindrance of the group is comparatively less. Thus, the attack is possible at the ortho or para positions. owever, addition at ortho position is favoured as it leads to cyclization of the intermediate acid to stable B. h. enol has only one group on the phenyl ring whereas resorcinol has two groups on the phenyl ring at the m-positions. ence, position 4 is considerably more activated (i.e, electron rich) in the case of resorcinol. enol Resorcinol Therefore, under identical reaction conditions, the yield of compound is much higher than that of B. 17. rganic spectroscopy and structure determination a. The given Molecular formula is Therefore, the possible structures are: 3 I 3 3 II III 3 2 IV 3 V VI The NMR spectrum of compound A shows a single peak which indicates that all the protons in A are equivalent. This holds true only for structure I. The IUPA name of this compound is 2-propanone Mumbai, India, July 2001

3 33 rd International hemistry lympiad Preparatory Problems The NMR spectrum of compound B shows four sets of peaks, which indicate the presence of four non-equivalent protons. This holds true for structures III and IV. owever, for structure IV, no singlet peak (see peak at δ = 3) will be observed. So, compound B must have structure III. The IUPA name is 1- methoxyethene. b. b c a 3 Three doublets of doublets centred at 6.5 ppm, 3.9 ppm, 3.5 ppm are seen in the spectrum. The assignments in the spectrum are a : 6.5 ppm b : 3.5 ppm c : 3.9 ppm Due to the presence of electron donating 3, the trans proton b has higher electron density and thus more shielded than c. Thus, b appears upfield as compared to c. There is also a singlet line at δ=3. This corresponds to the in 3. c. oupling constants a : 12, 16 z J ( a, b ) = 12 z J ( a, c ) = 16 z b : 8, 12 z J ( a, b ) = 12 z J ( b, c ) = 8 z c : 8, 16 z J ( b, c ) = 8 z J ( c, a ) = 16 z Note: J = (difference in two lines in ppm) x (Instrument frequency) Geminal coupling < cis-vicinal coupling < trans-vicinal coupling Mumbai, India, July

4 33 rd International hemistry lympiad Preparatory Problems d. Peak positions in z (for 400 Mz instrument) Peak positions in z (for 600 Mz instrument) e. ompound A will react with malonic acid in the following manner Addition ompound A Malonic Acid Meldrum's acid ( ) The structure of Meldrum s acid is consistent with the 1 -NMR and IR data. The peak in the IR spectrum at cm -1 is because of the = stretching. The presence of peaks only between 0 7 δ in the 1 -NMR spectrum indicates that the compound doesn t have any acidic group like or. If compound B reacts, the only possibility is that it will add across the double bond giving a product with molecular formula equal to This molecular formula does not match with the one stated in the problem Mumbai, India, July 2001

5 33 rd International hemistry lympiad Preparatory Problems 3 ompound B 2 Malonic Acid 3 3 f. The increased acidity is due to active 2 group of Meldrum's acid flanked by two groups. The carbanion formed at 2 will be stabilised by these groups, which are coplanar Meldrum's acid ( ) g. The condensation product of Meldrum's acid with an aromatic aldehyde has the structure Ar 18. Polymer synthesis a. Ag cat 2 2 1/ Ethylene P Q b. 2 2 P Q R Mumbai, India, July

6 33 rd International hemistry lympiad Preparatory Problems Sl 2 l l R S l l alc. K S T c NBS peroxide l 4 2 KN Et/ N / 2 / or i) Na / 2 ii) 2 N / / 2 2 p-xylene 3 dimethyl benzene-1,4-bis(acetate) d. Three signals (three singlets for - 3, 2 and aromatic protons) e. Structure of polymer n f. Polymer I) K / 2 / II) / Polymer LiAl g. With Glycerol (being a triol), cross-links between the polymer chains involving Mumbai, India, July 2001

7 33 rd International hemistry lympiad Preparatory Problems the secondary hydroxyl group will form giving a three-dimensional network polymer is possible. 2 2 Glycerol The polymer is unsuitable for drawing fibers because of its cross-linked, resinlike property. 19. rganic synthesis involving regioselection a. The product obtained in the presence of catalyst SbF 6 is m-bromophenol. From the mass spectra given in the problem, direct bromination of phenol gives o/p bromo derivatives as group present in phenol is o/p- directing. b. SbF 6 ( at ) ( 3 ) 2 S 4 Na 2 enol A 3 Mg / TF / toluene B 3 Mg ompound B may undergo nucleophilic reaction at the carbon bearing bromine. ompound contains a carbanion and hence functions as a Mumbai, India, July

8 33 rd International hemistry lympiad Preparatory Problems nucleophile and will attack an electrophile. Thus, reactivity of B is reversed on its conversion to (umpolung). c. 2 N N 3 yclohexanone D 3 3 Mg N 3 3 N 3 3 D Tramadol d. 3 2 N( 3 ) N( 3 ) N( 3 ) N( 3 ) 2 Tramadol has two asymmetric carbon atoms. It has two pairs of enantiomers. 20. arbon acids a. The molecular formula of the keto ester is Since X and Y are keto esters, they must have the following unitsketo group ester group This accounts for 4 3.The remaining part comprises of 8. Thus, only two types of ester groups are possible, methyl or ethyl Mumbai, India, July 2001

9 33 rd International hemistry lympiad Preparatory Problems For a methyl ester: 3 will be a part of the ester moiety. This leaves 5 to be attached. For an ethyl ester: 2 3 will be a part of the ester group. Therefore 3 unit needs to be accounted for. Therefore, possible structures of the keto esters are: Structure I Structure II Structure III b. Reaction sequence for keto esters * Na 3 Structure I Keto acid ( ) LDA ( 1equiv. ) 3 I / Keto acid ( ) * Na Structure Ill LDA ( 1equiv. ) 3 I / Keto acid ( ) Mumbai, India, July

10 33 rd International hemistry lympiad Preparatory Problems Structure ll NaMe 2 LDA ( 2 equiv. ) MeI LDA ( 1 equvi. ) / MeI / 2 2 / Bl - Keto acid B l - Keto acid ( ) 2 ( ) Structure I gives a keto acid with molecular formula which matches with the formula of the keto acid obtained from Y. Structure I is Y. Structure II gives a neutral compound with molecular formula that matches with the molecular formula of the neutral acid stated for X. Structure II is X. Structure III gives a keto acid with molecular formula that also does not match with any given molecular formula Mumbai, India, July 2001

11 33 rd International hemistry lympiad Preparatory Problems ence the two keto esters are : ompound Y ompound X (Structure I) (Structure II) α-keto ester β-keto ester 3 c. The β-keto ester gives on hydrolysis a β-keto acid. This acid readily undergoes decarboxylation involving a 6-membered transition state, giving a neutral product ( Ketone ) d. i. When 1 equivalent of LDA is used compound X produces a carbanion (monoanion) as shown below. 3 3 LDA ( 1 equiv. ) arbanion ompound X ( monoanion ) ii. Use of 2 equivalents of LDA leads to the formation of a dianion. 3 3 LDA ( 2 equiv. ) ompound X 2 Dianion Mumbai, India, July

12 33 rd International hemistry lympiad Preparatory Problems 21. Amino acids and enzymes a. The protonated amino group has an electron withdrawing effect. This enhances the release of proton from the neighboring, by stabilizing the conjugate base -. This effect is greater when the - is physically closer to N 3. As N 3 group is present on the α-carbon, the effect is greater on α- than on the γ-. So the pka value of α- is lower than that of γ-. b. The ratio of ionized to unionized γ- group is obtained by using enderson-asselbalch equation, p = pk a [ ] log [] The p = 6.3 and pka of γ- group is 4.3. Substituting these values in the above equation we get, [ ] 6.3 = 4.3 log [] [] = = 0.99% c. Glutamic acid has two pka values lower than 7.0 and one pka value higher than 7.0. Thus, the isoelectric point (pi) for glutamic acid will lie between the two acidic pka values. at pi = ( )/ 2 = 3.25 p 6.3 At p = 3.25, net charge on glutamic acid will be zero since this p coincides with pi of glutamic acid. ence, glutamic acid will be stationary at p Mumbai, India, July 2001