Question

Synthesis of 1-Bromobutane: An SN2 Reaction Pre-Laboratory Exercises

PRE-LABORATORY EXERCISES NAME: Cynthesis of 1-Bromobutane: An SN 2 Reaction Butylbromide was prepared by refluxing in a 100-ml round bottomed flask a solution containing 13:35 Br 15 ml of water, 10 mL of n-butyl alcohol. 15 ml of concentrated H.SO. and a few boiling stories, 1. 2. R." d 1.275 laborales Nel CH3CH2CH2CH2OH + NaBr H2SO4 Superletalyst CHCH2CH2CHBr + NaHSO + HO (eq. 16) 1-butanol bp 118 C 1-bromobutane d 0.810 bp 101.6°C MW 74.12 MW 137.03 1. Determine the limiting reactant in this synthesis. b and y to mess DV Dereito (D)= m = 0.810.10nd - 0.13 moles of Br 8.169 d. 11 moles of we buty! - mol=m/MW. -8.109/74.12 3.3 -0./2423 mol 107.8914 met of No Br = 0.10928 molot butanol 2. Calculate the theoretical yield of 1-bromobutane. amount rontorived it rennent to completion I wol 33 mol 133 met BB Butual 3.30 mal Butano! m=nol. Mw=0.10928 md of butonet. 137.03= 14 974 - 15 ed bronebutane 3. If 0.055 mole of the bromide was recovered at the end of the reaction, what is the % yield? child actually 100% on theoreticalia 103

Answer 1

Calculation for conducting the reaction:

Mole Ratio S.No Name of chemicals n-Butyl alcohol NaBr Water Con. H2SO4 Quantity 8.1 g (or) 10 ml 13.3 g 10 ml 21 g (or) 11.5 ml Density 0.81 g/cm3 Solid 1 g/cm3 1.83 g/cm3 M.Wt 74.12 102.89 18 98.07 Mole 0.1092 0.1292 1 volume 0.2141

Reaction Scheme:

H2SO4 + NaBr – H20 H Br → H3C H3c O n-Butyl alcohol bp 118 °C d = 0.81 M.Wt = 74.12 1-Bromobutane bp 101.6 °C d = 1.27 M.Wt = 137.03

Mechanism SN2 Reaction:

Hoc mont 0-5-OH hs H3c7V OH + - + 0-5-OH H3C O=o=O n-Butyl alcohol Sulfuric acid + Na Bri Sodium Bromide Br H2O + H2C7 1-Bromobutane

Yield:

Theoretical yield of 1-Bromobutane = 14.97g (or) 11.79 ml

Side Products:

O= "+10-5-OH 5-OH - OH + H3C O=y=0 H30 V n-Butyl alcohol Sulfuric acid Na Br Sodium Bromide H20 + H3C V Br 1-Brom obutane CH3 H3C Br 2 Bromobutane Another side product (Because of carbocation more stabilized molecule) CH2 H3C 1-Butene One of the side product (Elimination of H 20 molecule)

Demonstrate existence of side product:

CH3 Нас - CH2 НАС Нас- Br Br 1-Butene 1-Bromobutane bp 101.6 °C bp = -6.47 °C 2 Bromobutane bp = 91 °C

Before collection the 1-bromobutane first 1-butene gas will comes out (butene gas dipped in dilute bromine water, then the bromine water will get decolorize). Followed by 2-bromo butane will comes at boiling point of 91 which shows clear compound of 2-bromobutane.