1. Calculate the moles of 3-methyl-1-butanol and acetic acid from the starting volumes. Given that we are adding 5.0 mL of 3-methyl-1-butanol, 7.0 mL of glacial acetic acid all intoto a 50 mL round bottom flask. 1a. If we are reacting 3-methyl- 1-butanol (AKA isopentyl alcohol or isoamyl alcohol) with an excess of anhydrous acetic acid( acetic acid )to form isopentyl acetate also calculate the theoretical yield of isopentyl acetate in grams

1. Calculate the moles of 3-methyl-1-butanol and acetic acid from the starting volumes. Given that we are adding 5.0 mL of 3-methyl-1-butanol, 7.0 mL of glacial acetic acid all intoto a 50 mL round bottom flask. 1a. If we are reacting 3-methyl- 1-butanol (AKA isopentyl alcohol or isoamyl alcohol) with an excess of anhydrous acetic acid( acetic acid )to form isopentyl acetate also calculate the theoretical yield of isopentyl acetate in grams

Organic Chemistry: A Guided Inquiry

2nd Edition

ISBN:9780618974122

Author:Andrei Straumanis

Publisher:Andrei Straumanis

Chapter4: Polar Bonds, Polar Reactions

Section: Chapter Questions

Problem 7E

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Recall that the formula for ΔHrxn = [(-Ccal)(moles of limiting reactant)/(ΔT)]If the actual concentration of the sodium hydroxide used in the calibration part of the experiment was lower than the stated value, will the value of Ccal increase, decrease, or remain the same? Please explain. How about if the actual concentration of the sodium hydroxide used in the calibration part of the experiment was lower than the stated value. However, the NaOH solution used in the determination of ΔHrxn was of the correct concentration. Will the magnitue of ΔHrxn increase, decrease, or remain the same? Please explain.
please help- so which one is better part b has a percent yield of 76.8 percent and its melting point is what was expected also in part A but percent yield is 71.67 percent. this would be a sucess exeriment and pure? please explain Part A NaoH= 2.5g = 0.0625 mol Benzaldehyde= 2.65 g =0.0250 mol Acetone 0.725 g = 0.0125 mol Theoretical yield=0.0125 mol Actual yield= 2.1g Percent yield=71.67% Expected melting point= 110.5-112 degree Celsius Melting point observered= 112 degree Celsius Part B acetophenone 0.5 g, 4.2 mmol Para-methylbenzaldehyde 0.5 g, 4.2 mmol NaOH=0.18g = 4.5 mol Theroretical yield=0.93g =0.0519 mol Percent yield=76.8% Actual yield= 0.715 g Melting point observered= 93 degree Celsius Expected melting point =93-94.5 degree Celsius
1.Each of the student statements below is wrong. You will need to explain why they are not correct. (a) All my compound dissolved right away in the small amount of solvent I added at room temperature, and I didn’t need to heat it at all. This means I’m going to get lots of pure compound out. (b) I did a recrystallization of naphthalene and my percent recovery was very high (99%), so I must have pure product. (c) When you’ve dissolved all your compound in hot solvent, and you’re in a hurry, it’s ok to just place it straight into the ice bath.
Please answer the question at the bottom, Thanks. LAB NOTES FOR PREPARATION OF BUTYL MAGNESIUM BROMIDEAND ITS SUBSEQUENT CONVERSION TO AN ALCOHOL To the reaction flask were added 2.5967 gram of magnesium turnings, about 20 mlof diglyme, and a solution of 13.5942 gram of butyl bromide in about 25 mldiglyme.The procedure as per the lab manual was initiated. After the required heating timewas completed the reaction mixture was allowed to cool to room temperature.Then a solution of 6.1123 gram of acetone in about 15 ml of diglyme was addeddropwise as per the lab manual.The rest of the lab manual procedure was completed, ultimately affording 2.5532gram of an oily, clear and slightly yellow liquid, presumed to be 2-methyl-2-hexanol. Boiling point as measured by distillation was 138 – 141 deg C.About 0.8 ml of the presumed 2-methyl-2-hexanol was placed in an NMR tube towhich was added 2 drops of TMS. The tube was capped, then inverted and righted20 times to thoroughly mix the product and…
Please answer the question at the bottom, Thanks. LAB NOTES FOR PREPARATION OF BUTYL MAGNESIUM BROMIDEAND ITS SUBSEQUENT CONVERSION TO AN ALCOHOL To the reaction flask were added 2.5967 gram of magnesium turnings, about 20 mlof diglyme, and a solution of 13.5942 gram of butyl bromide in about 25 mldiglyme.The procedure as per the lab manual was initiated. After the required heating timewas completed the reaction mixture was allowed to cool to room temperature.Then a solution of 6.1123 gram of acetone in about 15 ml of diglyme was addeddropwise as per the lab manual.The rest of the lab manual procedure was completed, ultimately affording 2.5532gram of an oily, clear and slightly yellow liquid, presumed to be 2-methyl-2-hexanol. Boiling point as measured by distillation was 138 – 141 deg C.About 0.8 ml of the presumed 2-methyl-2-hexanol was placed in an NMR tube towhich was added 2 drops of TMS. The tube was capped, then inverted and righted20 times to thoroughly mix the product and…
Give the wet chemical tests (#1-6) that work for ALDEHYDES only.
3. Suppose you have a mixture of saturated aqueous sodium chloride (brine) and your 2-chloro-2-methylbutane product in a separatory funnel. Use densities to predict which phase will be the top layer in the funnel. a. 2-chloro-2-methylbutane (organic phase) b. saturated sodium chloride (aqueous phase) c. hard to predict since densities are both very close to 1.18 g/mL d. there would only be one phase since the substances are miscible
Please help with both parts! 41 and 42. An explanation for each one would be super!
Why do you multipy the # of H2S mols by 4, when 4 corresponds to the reactory amount of O2?The logic there is somewhat confusing to me.I understand how/why you found out the mols of the equation but dividing the sample's MM by the equation confuses me.
Each of the student statements below is wrong. Explain why they are not correct. (a) All my compound dissolved right away in the small amount of solvent I added at room temperature, and I didn’t need to heat it at all. This means I’m going to get lots of pure compound out. (b) I did a recrystallization of naphthalene and my percent recovery was very high (99%), so I must have pure product. (c) When you’ve dissolved all your compound in hot solvent, and you’re in a hurry, it’s ok to just place it straight into the ice bath.
How would each of the following errors affect the experimental value of your molarity of acetic? Would your Molarity value be too high, too low, or unaffected? Explain your answer. if you titrated your acid sample to a bright pink rather than faint pink endpoint? if your "acetic acid" burette was still wet inside with deionized water when you filled it with acetic acid? if you recorded the final volume of NaOH as "39.88 mL" instead of "38.88 mL"?
Please help write the final product, reaction equation, table of physical properties and percent yield please for the Isopentyl Acetate lab. My data is 5.0 ml of isopentyl alcohol 7.0 ml of acetic acid 1.0 ml sulfuric acid 10 ml of DI water 5 ml of sodium bicarbonate 5 ml of saturated sodium chloride 125 degrees is the boiling point final product weight is 0.633 g