Question

Two isomeric compounds A and B of formula C6H10 were isolated and subjected to spectroscopicanalysis. in the C¹³ NMR A shows 93.0, 62.5, 30.9, 27.2 ppm. In the DEPT 135 two lines appear at 62.5 and 30.9 ppm with non inverted. In the IR A shows 3311 (moderate sharp), 2973 (strong), 2108 (weak)cm^-1. In the C¹³ NMR B shows 81.0, 14.5 and 12.6 ppm.n the DEPT 135 two lines appear at (*)14.5 and 12.6 (with 12.6 inverted) The IR of B shows no peak above 3000 cm^{-1 .} Prepare interpretation table for each Isomer. Give the structures for A and B. *****************Please show all calculations and give an explaination of how you pull out the atoms for the "carbon type".

The table should be set up something like this.... 5 across    4 down (here are all the ans. please show How (show all work) Thanks so Much.

Carbon     I   Chemical shift     I    Chemical Shift                                I Carbon Type C, CH, CH2, CH3            I          Carbon Environment

                                                       predicted by ChemWindows              as detertmined by DEPT-135

                                                                                                                         or otherwise

      a.                    27.2                            26.5                                                 C - no H                                           (CH3)3C-(sheilding group)                             

      b.                      30.9                            27.8                                                 CH3 or CH                                       CH3C(CH3)2C(many a B effects)

      c.                    62.52                           66.8                                                 CH3 or CH                                        -C(triple Bond)CH

      d.                     93.0                             92.6                                                  C - no H                                            (CH3)3C-C(triple bond)C

Answer 1

Molecular formula for A and B = C6H10

Degree of unsaturation : 2

For Compound A

Dept-135 : showed 2 signals both non-inverted, so it can be either CH3 or CH at 62.5 and 30.9 ppm

13C NMR : showed 4 signals, and incorporating Dept-135 data, we can say, signal at 27.2 and 93.0 ppm are for quaternary carbons.

IR : showed 2973 cm-1 for aliphatic C-H stretching ; 3311 cm-1 is for alkyne C-H stretch ; 2108 cm-1 for alkyne C-C stretching.

A possible structure would be thus shown below.

For Compound B

Dept-135 : showed 2 signals 14.5 ppm and 12.6 ppm (inverted). So 12.6 is due to CH2 carbon. 14.5 ppm could be CH3 or CH carbon.

13C NMR : showed 4 signals, and incorporating Dept-135 data, we can say, signal at 81.0 ppm is for quaternary carbon.

IR : showed no peak above 3000 cm-1, so alkyne group is absent.

A possible structure for compound B is given below.

Compound A and B shown below.