*Please write the identity of the base you will use in the table. 2-bromo-2-methylbutane HON 1-But KotBu 1-Prop Base: Molar mass (g/mol) 151.04 g/mol 54.106 74.12112.21 60.1 g/mol Density or concentration 1.18 g/mL 2.04 0.81 0.902 0.803 Mass or volume used (g or mL) 1.3mL 12.5 12.5 12.5 12.5 mL Moles 0.00976 0.5 0.137 0.10 0.167 Equivalents** 1.0 1:50 14 1: **Equivalents: The reagent present in the smallest molar amount is set to 1.0 equivalent (eq). For each other reagent, #eq equals the ratio of its #moles divided by the #moles of the minimum reagent. The ratio of equivalents is equal to the molar ratio. 2. Present a calculation of the number of moles of BOTH of the reactants: 2-bromo-2-methylbutane] [KOH DEM/V - 204ginal= m/ 12.5 mx Dm=25.5g n= 25.5%/56.106gimo n=0.45 Density= m/v 1.18 glmx = m / 1.3 m m= 1.534 g les (n)= SS 1-Butanol Dam/v 0.81 gimm n= 10.1250 n = 1.5348/(51.04 g/mo armassn= 0.00976 KOLBU D=m/v 0.902 glmm/12.5 myl m11.2759 12.5 74.12 g/mol =P m = 10.125 g Pin = 0.137 mol 1-Propanol D=m/v = 0.803 g/m2 = m/12.5mx n = 11.2759/112.21 g/mol +10.10 mol/ m= 10.03159 n= 10.0375 \n = 0.167mol 60.18 Imo
*Please write the identity of the base you will use in the table. 2-bromo-2-methylbutane HON 1-But KotBu 1-Prop Base: Molar mass (g/mol) 151.04 g/mol 54.106 74.12112.21 60.1 g/mol Density or concentration 1.18 g/mL 2.04 0.81 0.902 0.803 Mass or volume used (g or mL) 1.3mL 12.5 12.5 12.5 12.5 mL Moles 0.00976 0.5 0.137 0.10 0.167 Equivalents** 1.0 1:50 14 1: **Equivalents: The reagent present in the smallest molar amount is set to 1.0 equivalent (eq). For each other reagent, #eq equals the ratio of its #moles divided by the #moles of the minimum reagent. The ratio of equivalents is equal to the molar ratio. 2. Present a calculation of the number of moles of BOTH of the reactants: 2-bromo-2-methylbutane] [KOH DEM/V - 204ginal= m/ 12.5 mx Dm=25.5g n= 25.5%/56.106gimo n=0.45 Density= m/v 1.18 glmx = m / 1.3 m m= 1.534 g les (n)= SS 1-Butanol Dam/v 0.81 gimm n= 10.1250 n = 1.5348/(51.04 g/mo armassn= 0.00976 KOLBU D=m/v 0.902 glmm/12.5 myl m11.2759 12.5 74.12 g/mol =P m = 10.125 g Pin = 0.137 mol 1-Propanol D=m/v = 0.803 g/m2 = m/12.5mx n = 11.2759/112.21 g/mol +10.10 mol/ m= 10.03159 n= 10.0375 \n = 0.167mol 60.18 Imo
Organic Chemistry: A Guided Inquiry
2nd Edition
ISBN:9780618974122
Author:Andrei Straumanis
Publisher:Andrei Straumanis
Chapter12: Chirality
Section: Chapter Questions
Problem 10CTQ
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Question
sample weight:
KOH: 1g
KOtBu 1g
1-propanol: 1g
1-butanol 0.742g
calculate the thoeretical and percent yields of methylbutenes
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