EXPERIMENT: An SN2 Reaction-Preparation of 1-bromobutane DATA TABLE: 1-butanol N

Question

EXPERIMENT: An SN2 Reaction-Preparation of 1-bromobutane DATA TABLE: 1-butanol NaBr Molar mass (g/mol) as on a /mo. 0.81 g/mL Density or 18.0 M N/A Concentration Mass or Volume 2.13 mL 2.63mL a al Used (g or mL) Moles Equivalents Equivalents: Set the reagent present in the smallest number of moles equal to 1.0 equivalent (eq). Then for each of the other reagents. the #eq will be the ratio of its amoles divided by the smoles of the minimum reagent. The ratio of equivalents is equal to the molar ratio. Show the calculation of the moles and equivalents of all the reagents used: Show calculation of the starred quantities in the large space below:

Explanation / Answer

CH3CH2CH2CH2OH + NaBr ---(H2SO4 )--> CH3CH2CH2CH2Br

First we calculate the number of moles:

NaBr = 2.919 g / molar mass; 102.8938 g/mol

= 0.0284 moles NaBr

Or 0.0284 mole Br-

Volume of 1 butanol = 2.63 ml

Density = 0.81 g/ ml

Density = mass / volume

Therefore

Mass = density * volume

= 0.81 g/ ml*2.63 ml

= 2.1303 g

Number of moles

= 2.1303 g / 74.12 g/ mol

= 0.0287 moles butanol-1

Here Br- is limiting agent , due to following reasons:

Now calculate the moles of 1 Bromobutane as follows:

0.0284 mole Br- * 1 mole 1 Bromobutane / 1 mole Br

0.0284 mole 1 Bromobutane

Amount of 1 Bromobutane= number of moles * molar mass

= 0.0284 mole 1 Bromobutane*137.02 g·mol1

= 3.888 g this is the theoretical yield

Actual yield is always less than theoretical yield here actual yield is greater means other impurity is present of compound is wet.

% yield = actual yield / theoretical yield *100

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