[GoGoGadget]

I just had a couple of questions regarding a Grignard reaction we did in my lab a few weeks ago. My professor gave us a mass spectrum of the product we obtained, triphenylmethanol. The highest peak is at 260 which I know is the product. There was a peak at 183 which I know is representative of a fragment ion of C13H9O. In our directions, it says to "draw the structure of the molecular ion or fragment represented by each peak." For the peak at 183 for example, I had a hard time trying to figure out what the actual structure would look like and am wondering if it would be appropriate to leave it as [C13H9O]+ or if there is a way to draw it out. There are other peaks I have to figure out too and just am wondering if there is or isn't a good way to draw out fragments.

My other question, we have to calculate how much water in mL would destroy the phenylmagesium bromide we prepared from reacting magnesium turnings and bromobenzene together. I used .306 g of Mg and 1.32 mL for the bromobenzene and am trying to figure out how to best calculate it for the water. Any help would be great.

[Dan]

There are other peaks I have to figure out too and just am wondering if there is or isn't a good way to draw out fragments.

Start by drawing out tripenylmethanol. Now break bonds, this generates possible fragments. Draw out as many possibilities as you can think of and start matching them up with the peaks. By the way, [C₁₃H₉O]⁺ would be 181, [C₁₃H₁₁O]⁺ would be 183.

My other question, we have to calculate how much water in mL would destroy the phenylmagesium bromide we prepared from reacting magnesium turnings and bromobenzene together. I used .306 g of Mg and 1.32 mL for the bromobenzene and am trying to figure out how to best calculate it for the water. Any help would be great.

Start with a balanced chemical equations for the formation of PhMgBr, and the reaction of PhMgBr with water.
Using your data, calculate the moles of PhMgBr formed, and from that the moles of water required to destroy it.
Finally convert moles of water to mass and finally volume.

[GoGoGadget]

Thanks a ton! That helps a lot.  For a mass peak at 77, the closest I can get is C4H5O which totals to 69. Is it appropriate to add H's to account for the remaining mass or would that be incorrect?

[Dan]

I don't think you're going about this in the best way. I advise that you first forget the numbers, draw triphenylmethanol and by breaking different sigma bonds draw as many fragments as you can. Now calculate the mass of each fragment you came up with and compare it to the peak list you have.

If you do this you'll see that for 77 your suggestion of [C₄H₁₃O]⁺ is not sensible - for a start, there is too much hydrogen in it - it would not be possible to draw a cation with that formula. Second, it does not coincide with any sensible fragment, I would not expect to see any C₄ fragments.

Hint: You worked out that the 183 fragment was. Tripenylmethanol has a relative mass of 260. Note that 260-183=77..