Consider the bromination of 1-pentanol by HBr to form 1-bromopentane. CH3(CH2)3CH2OH + HBr ---> CH3(CH2)3CH2Br +...

Consider the bromination of 1-pentanol by HBr to form 1-bromopentane.

CH3(CH2)3CH2OH + HBr ---> CH3(CH2)3CH2Br + H2O

1 pentanol mol. weight 88 g/mol

hydrobromic acid 80.9 g/mol

1-bromopentane 151 g/mol

nitric acid 63 g/mol

a) If you had 44 g of 1-pentanol, 34 g of HBr and 1.7 g of nitric acid, what would be the limiting reagent in this reaction?

b) What is the theoretical yield of 1-bromopentane in (a) ?

c) If you obtained 16.0 g of 1-bromopentane, what is the % yield for the reaction?

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Expert Solution

queen_honey_blossom answered 1 year ago

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