Question

Any extra detail in the solution would be greatly appreciated!!

3) (5 marks) A certain molecule has the gross structure shown below, and NMR data following (all coupling constants given in hertz). Using the template below, show the correct stereochemistry/conformation for the sugar. Justify your answer by indicating clearly all coupling constants on the figure? Draw a splitting diagram to scale, for peaks 4) & 5). Меон -оме HO 1) 1.33 (3H, d, J=6) 2) 1.61 (1H, broad s, exchanges with D20) 3) 1.87 (1H, ddd, J=14, 3, 3.5) 4) 2.21 (1H, ddd, J=14, 3, 1.5) 5) 2.87 (1H, dd, J=10, 3) 6) 3.40 (3H, S) 7) 3.47 (3H, s) 8) 3.99 (1H, dq, J=10, 6) 9) 4.24 (1H, ddd, J=3,3, 3.5) 10) 4.79 (1H, dd, J-3.5, 1.5).

Answer 1

4.24(14,ddd) 1.87C14. ddd 5=14,3,3.5 H2) I 1,33638,,5=6H2) 3.40 310 Date Date Ans> given structure 1.61 LH 6.39 (14,24,1335,15) lodis IH pe . 66133-739 ਹੈ| 3-47 311,5) men Wd 2-87- C .de 399 (141,9, 5210,6 H2 IX Correct given steriochemical con Foymautional_ot. sugge molecule.. ARO L 930_ 1 - 2.21(1H, ddd, 5=14,3,.5 Hz) splitting_digram for_signal ☺- 2:21T1Hidad, J=14,3,1.5 H2) ome bazkedome group_is_axiw_due to_an_Anomeric effect _90+ stable. - M mome Somle J=14.42-tax coupling J = 3H2-Cis coupling J=1.542 - 54 coupling MO 2_Assigning on the all coupling constait valdes_for_each_et -proton. H-3-:-3_ 1_Intensity_xatio

splitting diagram for signal 2.87 (1H, dd IF10, 3 H2) J=10112, trans coupling J =3 H2, cis coupling 2 : 1- Intensity ratio
Summery : Here in given Structure marked methoxy qroup is Axial due to an Anomeric Effect get stable due to bonding intraction of oxygen non bonding electron to methoxy non bonding electron pair.

And The assigning an Coupling constant values are done by Using J3 coupling and Longe Range coupling in some proton. Intensity ratio can be measured by useing Pascal's Triangle.