WK 1 Sapling #10

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Earl Garrovillo 2L
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Joined: Wed Sep 30, 2020 9:55 pm

WK 1 Sapling #10

Postby Earl Garrovillo 2L » Thu Oct 08, 2020 4:37 pm

I know like a million people have posted about this question so I apologize if this is a bit repetitive (or if it's a stupid question lol)

#10 asks:
"Consider the nucleophilic addition reaction of 2-butanone with excess propyl magnesium bromide, made in situ by reacting 1-bromopropane with metallic magnesium, to make 3-methyl-3-hexanol.

2-butanone (d= 0.81g/mL) + 1-bromopropane (d=1.35 g/mL) + Mg -------> 3-methyl-3-hexanol (d= 0.82g/mol)

A reaction was performed in which 0.55mL of 2-butanone was reacted with an excess of propyl magnesium bromide to make 0.49g of 3-methyl-3-hexanol. Calculate the theoretical yield and percent yield for this reaction."


I was able to get the correct answer but I'm not exactly sure why my answer was correct. I think I understand the calculations behind my answer. I converted the given volume of 2-butanone into moles and then converted that into grams of 3-methyl-3-hexanol. But I did that under the assumption that 1 mole of 2-butanone and excess of other reactants yields 1 mole of 3-methyl-3-hexanol and I'm not entirely sure why that was true.
Basically, can someone explain how to balance the chemical equation given in the problem? Or if the given chemical equation is already balanced, why is it already balanced? Where did the bromine and magnesium in the reactants go? The structure of 3-methyl-3-hexanol didn't look like it contained either element.

Rylee Mangan 1K
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Re: WK 1 Sapling #10

Postby Rylee Mangan 1K » Thu Oct 08, 2020 4:50 pm

I completed this question under the assumption that the chemical equation was already balanced. So if someone could explain why it is/how to make sure that would be great.

I found the moles of 2-butanone by finding the mass first - (volume x density = mass) and I got 0.365g. Then I converted this to moles by diving the grams by the molar mass (.365g / 72.11 g/mol) to get .0051 moles. So basically we can assume 3-methyl-3-hexanone also is .0051 moles? And solve from there?

Lucy Wang 2J
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Re: WK 1 Sapling #10

Postby Lucy Wang 2J » Thu Oct 08, 2020 5:46 pm

Yeah thats what I did and it worked for me! Since it is a 1:1 ration, I just used the number moles of 2-butanone and multiplied it by the molar mass of 3-methyl 3-hexanol to get the theoretical yield of the product in grams.

IanWheeler3F
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Re: WK 1 Sapling #10

Postby IanWheeler3F » Fri Oct 09, 2020 4:07 pm

I think the bromine and magnesium are spectators? Like they don't play a meaningful part in the reaction and just end up as ions in solution and don't precipitate out (make a solid). And also this looks like an organic chem reaction so I don't think any of us understand exactly why it is the way it is, hopefully someday though.

Jerry_T
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Re: WK 1 Sapling #10

Postby Jerry_T » Fri Oct 09, 2020 11:47 pm

How did you guys obtain the molar mass? I don't remember how to interpret skeleton drawings of organic molecules.

Giselle_zamora_1L
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Re: WK 1 Sapling #10

Postby Giselle_zamora_1L » Fri Oct 09, 2020 11:59 pm

I also had questions about balancing that equation / assuming it was balanced. I googled 3‑methyl‑3‑hexanol, which I think just confused me more. But assuming it is balanced how were you guys able to tell? This might be repetitive sorry but I'm just trying to understand.

Allan Nguyen 2G
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Re: WK 1 Sapling #10

Postby Allan Nguyen 2G » Sat Oct 10, 2020 12:02 am

I went to a workshop and someone asked this question, and the UA ended up saying that the molar mass should have been given because there would have been no way for us to find it.

Kandyce Lance 3E
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Re: WK 1 Sapling #10

Postby Kandyce Lance 3E » Sun Oct 11, 2020 10:26 pm

Allan Nguyen 2E wrote:I went to a workshop and someone asked this question, and the UA ended up saying that the molar mass should have been given because there would have been no way for us to find it.

I ended up pressing the give up button on that one. I was able to find the theoretical yield but couldn't for the life of me find the percentage yield (which is arguable the easier part of the question). Either way,, I only missed 5 points by pressing give up but I had already tried 13 attempts so.....


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