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kakado
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Chemical compound

Post by kakado »

I have a question about C4H8Cl2

A compound with the molecular formula C4H9Cl reacts with chlorine ( ultraviolet light is used as the catalysator). I would like to know wich structure the compound (C4H9Cl) has if it can react into exactly three different dichloride compoundes?

Is it 1-chlorobutane, 2-chlorobutane, 1-chloro-2-methylpropane or 2-chloro-2-methylpropane? I think it´s 2-chloro-2-methylpropane, but I´m not sure.

thanks for your help!

Also-

I have a though question about ethanol. The ethanol in a 750 cm3 wine bottle has oxidated into acetic acid. How much NaHCO3 do you need to neutralize the acid. The alcohol concentration in the wine bottle is 20% and the density of ethanol is 0,789 grams/cm3.

I can probably solve the question myself, but I need some help with the reaction. In some weird way the neutralization doesn´t seem to work if you write the reaction like this:

CH3COOH + NaHCO3 = H2O + NaCO3 + CO2

I can´t find the right molaramounts of the chemicals either!

And-

Can octane and ethanol have the same amount of carbondioxide discharge when incinirated?

The "energiamount" of octane is 5470 000 Joule/molar
and ethanol is 1367 000 Joule/molar

(Combined Threads-Lou)
Last edited by kakado on Sun Apr 15, 2007 1:48 am, edited 1 time in total.
clarke Slemon
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Post by clarke Slemon »

First of all there is an error in the question. You are not reacting with chloride but with chlorine. Neither a saturated chloride like C4H9Cl or chloride anion interact with light.

I will assume that your teacher is not including consideration of chirality 9optical isomers). If you are told that there are exactly three different compounds formed you need to realize that this means there are three different types of hydrogens which can be attacked. 1-chlorobutane would have 4 different types of hydrogens. it would give 1,1-dichlorobutane, 1,2-dichlorobutane, 1,3-dichlorobutane and 1,4-dichlorobutane so that would not be correct. Get the idea? Now you do the problem. Good luck!
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