What is the coupling constant of the alkene signals of the NMR of your product? (The two alkene signal are doublets at 7.28 and 7.14 ppm.) How does this information tell you that you isolated the trans and not cis product?

Explanation:

I need more information to answer this question and a better figure.

1. What is the frequency of the NMR machine?

Possible solution:

1. coupling constant Jab (in ppm) is given by

\nu_{a} - \nu_{b} = 4J_{ab}

2. Jab (in ppm) * Frequency of machine in (MHz)/106 is Jab in Hz

3. for cis vicinal Hydrogen Jab = 6-14 Hz

4. for trans vicinal Hydrogens Jab = 11-18 Hz

Now, considering 2 doublets are centered at 7.14 and 7.28 ppm , it gives

7.14 -7.28 = 4 Jab  

thus, Jab = 0.07 ppm

Now if we consider a 100 MHz machine,

Jab = 7 Hz , thus indicating cis product

but if machine is 300 MHz

then Jab = 21 Hz , thus indicating a trans product.

But, most probably I feel it is a trans product. I hope it helps.

percent abundance for cu-63 can be shown as:

63 x + 65 (1-x) = 62.9296

65 - 62.9296 = 2 x

x = 1.035

thus, percent abundance for cu-63 is 1.035

percent abundance for cu-65 can be shown as:

63 x + 65 (1-x) = 64.9278

65 - 64.9278 = 2 x

x = 0.0361

thus, percent abundance for cu-63 is 0.0361

the mass spectrum of cu-63 showing the intense peak as, this isotope is found in nature more abundantly around 92.38 % in comparison to cu-65 which is found 62.58 % naturally