posted by Kellie .
I am still not understanding this one at all.
Draw the structure of the reaction product of 2-methyl-1pentene with H2 in the presence of platinum catalyst. This is the exact question it does not say cis or trans.
I didn't follow DrBob222's reply, but I thought I had missed a previous dialogue.
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There are no geometric isomers as both R groups (CH3 and CH2CH2CH3) are on the same C atom, C#2, (i.e. not on opposite sides of the double bond) and on the other C atom, C#1, both atoms are H.
When this compound is reacted with hydrogen (H2) the hydrogen molecule will add across the double bond so that C#1 and C#2 get one H each.
If you want to post your answer I can check it.
Where do I put the platinum catalyst?
The reaction is carried out in a solvent that is inert to the reaction but dissolves the hydrocarbon. 2-methy-1-pentene is a liquid at room temperature (boiling point 60.5°C)and soluble in similar hydrocarbons and some polar solvents like ethanol (bp 78.5°C). The choice of solvent will depend on the scale on which you want to carry out the reaction. On a large scale it could be carried out neat, by suspending the Pt catalyst in the starting material. More usually it would be with a solvent as the reaction is easier to control. The choice of solvent should be to allow easy separation from the product, which incidently has almost the same bp as the starting material. The procedure is to suspend the Pt catalyst in the solvent (say ethanol) in a flask with a stirer. The flask is connected to a source of H2 and a means of measuring the H2 used. Purge the system of air by replacing it with H2. Add the starting material in a small amount of solvent and note the amount of H2 taken up. Once the calculated amount of H2 has been taken up the mixture is filtered to remove the catalyst and distilled to recover the product.
Just spotted the structure. No this is not the stucture for 2-methyl-1-pentene. I drew it in the previous answer.
When H2 adds across a double bond the H atoms attach themselves to opposite ends of the double bond, and the double bond disappears. For example:
(CH3)2C=C(CH3)2 + H2 => (CH3)2HC-CH(CH3)2
It is usual to show the catalyst (Pt)above the arrow, but not easy to do here.
I think I got it.
DrRuss--You are right, of course, and I realized my error in even talking about cis and trans isomers in this problem last night as I was trying to go to sleep. As I recall, that was an afterthought as I was about to push the submit button. It would have been better to let the afterthought go. However, again as I recall, the remainder of my advice of adding a H atom to each C carrying the double bond and removing the double bond was ok. By the way, I'm glad to see you show up here on this forum. I hope you continue to check the board for I am not an organic chemist and I have been trying to read Morrison and Boyd as a way to help on those organic questions.
Nothing wrong with reading Morrison and Boyd, an excellent text. I am still a fan of Finar, both volumes of course, but I like reading Stuart Warren's and John McMurray's books as well.