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.

metals are good conductor of electricity. it has free electrons on their surface. according to electron pool theory, each atom present in a metal crystal losses all of its valence electron and as a result a pool of free is form. these pool of electrons are free to move and   conduct electricity easily   while making metal as good conductor.

the correct answer is a. the ability of electrons to flow throughout the metal.

if it's not useful, feel free to report for mistake in answer.