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Topic: Help me at these problems please :)  (Read 2798 times)

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Offline Malekythhexane

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Help me at these problems please :)
« on: November 25, 2011, 10:59:23 AM »
Fiend the hydrocarbon A knowing that:
- it has NE=1;
-it can react whit Br2 in CCl4(which means that it has double bound) and whit the weak alkaline solution KMnO4(idk what it means ,if it say weak maybe whit h2o+Kmn04)
-in hydrogenations(means +H2) whit Ni it makes an  saturated compound whit the formula C5H12;
-1 moll of hydrocarbon A consumes 0,5 L solution K2CR207 on concentration 2M ,mixed whit H2SO4(idk what it means)

I am thinking that if it has NE=1 it is a alkenand if the hydrocarbon +h2=a compound whit c5h12 formula then the hydrocarbon is c5h12  and it has double bound
So if i am right the compound has to have the formula c5h12 ,it has double bound  and NE=1

Offline Malekythhexane

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Re: Help me at these problems please :)
« Reply #1 on: November 25, 2011, 11:24:27 AM »
i think its 3 methyl 1 butene because it has double bound ,it has NE=1 which means its an alken and +H2 it makes c5h12

Offline Dan

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Re: Help me at these problems please :)
« Reply #2 on: November 26, 2011, 03:18:21 PM »
I think your answer is wrong.

Think about the reaction with acidified K2Cr2O7.

How many moles of K2Cr2O7 react with 1 mole of the alkene? What does this tell you about the oxidation reaction - remember, K2Cr2O7 is a 6 electron oxidant.
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Offline Malekythhexane

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Re: Help me at these problems please :)
« Reply #3 on: November 27, 2011, 04:33:33 PM »
I think your answer is wrong.

Think about the reaction with acidified K2Cr2O7.

How many moles of K2Cr2O7 react with 1 mole of the alkene? What does this tell you about the oxidation reaction - remember, K2Cr2O7 is a 6 electron oxidant.
can you explain more? i am sure that the compound i wrote is good until that point

Offline Dan

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Re: Help me at these problems please :)
« Reply #4 on: November 27, 2011, 05:41:11 PM »
Again:

Quote from: Dan
How many moles of K2Cr2O7 react with 1 mole of the alkene?

If you demonstrate effort, you will get all the help you need.
My research: Google Scholar and Researchgate

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