[tmo]

"Add 3ml of 0.1M tert.ButylChloride to a 25ml flask. in another flask add 0.3ml of 0.1M NaOH and 6.7ml of distilled water, then mix the contents of the two flasks."

whats the concentration of NaOH?
and the conc of tert.butyl chloride?

and what happens if we repeat that step but before mixing them, we add 10ml of 7:3 water-acetone mixture to the flask containing the NaOH ? what will the concs be then?
Thanks

[Hunter2]

1. Calculate the total volume of all solutions.
2. Calculate the mole of the Butylchloride
3. Calculate the mole of the NaOH
4. Calculate the chemicals after reaction
5. Recalculate the new concentrations.

[tmo]

so its conc would be :
for NaOH = (0.3 x 0.1)/10?

and for ButCl = (3 x 0.1)/10?
for the initial concs?

[Hunter2]

Use correct units!!

What happens if the Butylchloride and NaOH  react together?

[tmo]

total volume becomes 10ml and the 10^-3 will cancel out as top and bottom both have 10^-3
is that correct?

[Hunter2]

The numbers are correct.

So what happend if the acetone mixture is used, and why?

[tmo]

with acetone there will be a Sn2 reaction as acetone is aprotic.

[Hunter2]

Acetone is solvent so it is easier to substitute the Chloride to the alcohol.

[discodermolide]

This is a tertiary butyl chloride, is elimination of isobutyelene not more likely?

[Hunter2]

Maybe not

http://en.wikipedia.org/wiki/Tert-Butyl_chloride

[discodermolide]

These are basic conditions so an elimination of HCl is perhaps more likely? Your link pointed to the carbocation formed under acidic conditions.

[Hunter2]

I mean this part:

When tert-butyl chloride is dissolved in water, a polar and protic solvent, the bulky chloride substituent is carried away by it, and isolated from the aliphatic chain, causing an heterolytic rupture of the compound, giving rise to a carbocation which eventually becomes a tertiary alcohol after a water molecule reacts with it, releasing hydrochloric acid as the final product. If a different, stronger nucleophilic agent is present at the moment of reaction, reaction product may not be an alcohol, but a tertiary carbon with the nucleophile as a substituent.

[discodermolide]

Sorry I mis-understood.

[easyorganic]

This is a standard SN1 reaction scheme used in undergraduate labs.  The reaction rate is independent of the NaOH or water.  The acetone/water is merely to change the initial concentration of the t-butyl chloride so the student will see a rate dependence on t-butylchloride concentration as you'd expect for 3-halide substitution.  The hydroxide ion is there to keep pH above 7 until you hit 10% conversion, and the when pH swings below 7 you see some color change (by product of this reaction is HCl).  Do you see that the NaOH is 10% of your t-butylCl?  You must be using some type of indicator (BromoPhenol Blue maybe).


What will the pH in the solution be when the reaction is at 10% conversion?

[tmo]

Im still really confused about the concentration calculations
I do understand that the conc of NaOh does not effect the rate as the rate determining step is the first step where naoh isnt involved
And i understand when there is more acetone there will be less solvation of butylchloride hence rate will decrease?
And how do i calculate conc of butylchloride?