Chemical Forums
Chemistry Forums for Students => Organic Chemistry Forum => Topic started by: farazali on November 23, 2007, 02:08:50 AM
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The problem goes as follows...
"Methylcyclohexane reacts with molecular bromine (Br2) in the presence of light and heat. The major product obtained from this reaction is reacted with sodium ethoxide in ethanol at 55°C. Draw the skeletal structure of the major organic product resulting from the second reaction."
So for the first reaction, I'm assuming that methylcyclohexane with Br2 (and light) will form 1-bromo-1-methylcyclohexane (because there's a tertiary hydrogen). Taking that through the 2nd rxn with yeild 1-ethoxide-1-bromocyclohexane (ethoxide=strong base in protic solvent)
Hopefully I am right, i have no way to check the answer. I tried to search but could not find an answer and want to know if my logic is flawed or not. Also, is there going to be an E1 (minor) rxn after adding sodium ethoxide? or not?
I appreciate anyones help.
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I like the E1 idea.
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Moreover, I suppose two products: exo and endo double bond
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So Ill take it that 1-ethoxide-1-methylcyclohexane is not the right answer. Rather, there are going to be elimination products.
Following through with this logic, I'm hoping to get the following elimination products.
methylenecyclohexane
1-methylcyclohexene
These are the only two elimination products I can come up with. I can't tell which one will be minor or major. I'm assuming that since methylenecyclohexane is more highly substituted, it will win over the bet.
Can someone verify this for me? I'm trying to study for an exam (two weeks from now, no rush), and this question doesn't have an answer in the back of the book. It would be really nice to know if I'm right or not. And if not, where I'm going wrong.
To AWK- I'm not exactly sure what endo and exo double bonds are... would you mind elaborating please?
Thank you guys for your feedback. Much appreciated.
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I'm assuming that since methylenecyclohexane is more highly substituted, it will win over the bet.
Right idea, wrong isomer ;) 1-methylcyclohex-1-ene is trisubstituted, methylenecyclohexane only di, so the former is probably more favoured.
To AWK- I'm not exactly sure what endo and exo double bonds are... would you mind elaborating please?
I'm not AWK, but I've only seen endo / exo refer to bicyclics or higher, as in http://en.wikipedia.org/w/index.php?title=Endo-exo_isomerism&oldid=172136842 (http://en.wikipedia.org/w/index.php?title=Endo-exo_isomerism&oldid=172136842). I think you may be able to extend to exo being "outside" the main ring, so methylenecyclohexane (in this case), and endo more 1-methylcyclohex-1-ene.
S
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oops thats my mistake. But for the most part, I think that your feedback has cleared up major issues with me. I only have one more question, will this be an E2 or an E1 reaction (I'm guessing E2 since the nucleophile is a strong base -- so one concerted step).