Question

I need to calculate the theoretical yield and percent yield for 1-bromobutane (Synthesis of 1-Bromobutane via SN2) using this attached data but I am having trouble figuring it out.

Mass NaBr: 6.985 g

Water: 7.50 mL

1-butanol: 5.00 mL

Note: Look up the density of 1-butanol

Mass of product container: 28.112 g

Mass of product container & 1-bromobutane: 35.432 g

Flame turned from blue to green when wire placed in flame

Note: look up what the Beilstein test is and what it tells us

Answer 1

Synthesis of 1-Stormobetare SH :- via H₂O 1-Butanol + Ba 1- Bromobutane Giza:- Mass of Na Ba : 6.985g Volume of Water = 7.50 ml Volume of 1- butanol = 5.00 ml Mass of Product container m = 28.112g Mass of product container f o r i-bromobutane cm = 35.4329 ! Solution :- Hoze dl-bromobutane - mam, = 25.4.82 - 28.112) 9 Mass of I- bromobutano? = 7.329 (Actual yield) / 329 Moss o i- beetand : Density x volume Density of 1-butanol = 810 kg 1m3 = 0.81 g/m2 Mass of 1-butanol = 0.819/opt x 5.00ml Mood -4.259 Molar mass of 1- butand = 74.121 g/mol Molax mass of NaBr = 102.9 and Molar mass of 1- baomobuitare = 137.02 g/mol

Calculate theoretical yield: Moles of 1-butanol & mass / mdar mass = 4:058/74.1 glmal 1. butand = 0 . 0546 mal Moles of Nabe = 6.9868/10219 glad NaBr 0.0679 and 1 Mole of 1- but and reacts with indle of Na Bar to give Imole of 1- bromabutane Limiting geactant is I- but and .: 0.0546 mol of I- butand give 0.0546 mol of 1- bromobutane : Mass of 1- bromabutane - Moles & Holod mare Theoretical yield Calculation Peacent : 0.0546 ml x 137.02 g/mnd = 7.48 9 of percent yeolady wield - ( Actual yield ) |x 100 % =(7.329 x 100%. 7.489 : Peacent yield - 97.86% = 98%

Beilstein best Beilstein test is a flame test used to detect the presence & halogens in organic compounds. A green flame indicate the presence of halogen compounds Cuo + Substance - Cux Where x-halgen

The theoretical yield of 1-bromobutane = 7.48g

The percent yield of 1-bromobutane = 98%