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Percent yield of alkyl halide (Show all calculations for limiting reactant, theoretical yield and percent yield.)...

Percent yield of alkyl halide (Show all calculations for limiting reactant, theoretical yield and percent yield.)

Given:

R-CH2OH +NaBr +H2SO4---------> R-CH2BR +H2O+ NaHSO4

- started with - unkown alcohol: 3-methyl-1-butanol (18.996 g) started with,

-NaBr ( 10.509g) used

-H2SO4 (25 mL of 8.7M)

-My alkyl halide is 1-bromo-3- methylbutane

-H2O

-NaHSO4

Note! I ended with 1.746g of the alkyl halide is 1-bromo-3- methylbutane

Solutions

Expert Solution

Given Reaction :

R-CH2OH +NaBr +H2SO4---------> R-CH2BR +H2O+ NaHSO4

alcohol (Reactant ) : 3 - methyl - I - butanol ( 18.996 g )

NaBr : 10.509 g

H2So4 : 25 mL of 8.7 M

Product alkylhalide : I - bromo - 3 - methylbutane ( 90 + 1.746 g )

Write a balanced equation for this rreaction :

We know mass of the rreactants calculate first moles of the reactants.

(i) moles of 3 - methyl - I - butanol : weight = 18.996 g

molecular weight = 88.1482 g / mol

moles = weight / molecular weight =

next , find the limiting reagent km stoichiometric ratio & mole ratio method

Actual molar ratio = moles of alcohol / moles of NaBr = 0.2155 / 0.1021 = 2.11 mol alcohol / 1 mol NaBr

queen_honey_blossom answered 2 years ago
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