Question

Name Alkoxide base heptene isomers cacluate from the formula / structure or look up in the table on molar mass (g/mol) 2-bromoheptane cacluate from the formula / structure or look up in the table on page 77 density = ? g/ml from PC table (0.22 mL = ?g cacluate from the formula / structure or look up in the table on page 77 page 77 other you calculate 7 .17 you calculate C target amount target mmoles (1 mmole = 0.001 mole) maximum mmoles of product based on each reactant limiting reagent theoretical yield (you calculate you calculate mark one as the limiting reactant you calculate actual amount used fill this in after measuring out chemicals in lab fill this in after measuring out chemicals in lab fill this in after measuring out chemicals in lab fill this in after measuring out chemicals in lab actual mmols updated theoretical yield (mass & mmols) fill this in after measuring out chemicals in lab

the alloxide base will be Na ethoxide. please fill out the circled items and show work. Thx in advance!

Answer 1

thanol) PT I is sonicated for in mixture was stirred a 2-bromoheptane MW: 179. 10g/mol density = 1.142g ImL .: 0.22mL *g density = mass & mass=density x volume volume X 1.142x 0.22 >=0.2512492 target immoles No. of mmoles mass x 103 Molecular mass = 0.25/24 103 174.10 y = 11.4028 mmoles ! sodium ethoxide MW = 68.05g/mol. CH₂ CH-CH₂ - CH₂ Natt CH CH = CH(CH2) - CH₂ cis & trans % for damol of a-bromoheptane, we need emol of Nalet. fox 4026mmol is needed Hledzsmmol X 68.05g/mol = 2,0454.htis round 10.4889 is rared

maximum immoles of product based on 2-become the plane = 1.4028 mmoles Maximum immoles of product based on Nadet 2.7. lit immoles. with Imol of BH. 1 mol product is formed - 1 mol of Nadet. " Nalet is available in excess. - 2-bromoheptane is the limiting reagent. € 1.4028 mmoles of heftene isomess will be formed which gives theoretical jield = 1.4028 x 16 x98.19 +0.134449