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  •  » Find MF of unknown alcohol to alkyl bromide by SN2 substitution

#1 2010-10-02 22:33:02

orgopete
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Find MF of unknown alcohol to alkyl bromide by SN2 substitution

I did this lab about preparation of n-butyl bromide by an SN2 substitution reaction.

NaBr + H2SO4 + butanol--> butyl bromide + NaHSO4 +H2O

Then I received this question which I don't know how solve:
When an unknown primary alcohol was subjected to this same experiment, using excess NaBr, a maximum yield of 15.8 g of product was obtained by careful treatment of 10.0 g of alcohol. What is a possible formula for the alcohol?

This is as much a word problem as a chemistry problem. How do you do the calculation? Here goes:

If the alcohol is saturated, then its MF = H(CH2n)OH
Let x = mw of alcohol, x-18 = mw of alkyl portion, x-18+80.9 = mw of bromide
Let m = moles of alcohol, moles of bromide

10/x = m

15.8/(x-18+80.9)= m

10/x = 15.8/(x+62.9)

10x + 629 = 15.8x

629 = 5.8x; x = 108 g/mol (the mw of the alcohol)

H(CH2)nOH = 108 - 18 (-H2O) = 90 = (CH2)n; 90/14 = 6.4; therefore there must be 6 or 7 carbons.

C6H13OH, mw 102.2, C6H13Br, mw 165.1, theo wt bromide = 16.28 g ((10.0 g/102.2 g/mol)*165.1 g/mol = 16.28 g)
C7H15OH, mw 116, C7H15Br, mw 179.1; theo wt bromide = 15.42 g

You could not get 15.8 g of product from 10 g of alcohol if the MF = C7H15OH; therefore it must be C6H13OH.

****************************

Or make a table
(CH2)n    Alcohol (moles)    Theoretical Wt. Bromide (g)
1    0.313    29.656
2    0.217    23.674
3    0.167    20.483
4    0.135    18.500
5    0.114    17.148
6    0.098    16.167
7    0.086    15.422
8    0.077    14.838

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