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Like watching paint dry, beautiful

Thank you

bad job

Was the goal to find concentration OH and H30? you did know the base constant = 1.8 x 10^-5. So, what then? i didn´t follow

Hey girl, just wanted to say that I love your videos so much! They have been so helpful in organic chemistry II and I honestly plan on getting an A on this Friday's exam because of these videos. I will update you soon! Have a great night.

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but what about quiz, misleading title

Can you provide a reference for this mechanism?

Thanks for this! My teacher took a whole class period to explain this, and it still didn't make sense. You clarified in 32 second :)

cannot hear anything.

Sometimes it's difficult to make out what you're trying to show with your arrows, like the one where the N is being deprotonated in the third step.

Maybe you should consider using an iPhone earbud/microphone. It would make it easier to hear your narration instead of the background noise (mostly that loud teacher).

If you are using Explain Everything, you should be able to zoom in and out on the screen. It seems that you weren't aware of that feature. If you're using Doceri, you do not have unlimited screen real estate. I can't tell from what's written here which app you were using.

Whoa! You said that you know the rate law is first order because the graph was linear. You didn't mention what was graphed. The question on the handout said that [C2H5OH] vs time was graphed. That graph being linear suggests the reaction is zero order, not first order. So you used the wrong equation to determine the concentrations.

It seems like you considered 150 s to be two sig figs. I considered it as 3 sig figs when I wrote the key. Unfortunately the way the problem was written, the number of sig figs is ambiguous.

It's a little tough to hear your voice over all the background voices. Maybe you could use one of those microphone headsets.

Maybe it would be better if you wrote some parts ahead of time before you start recording. I know it's difficult to write and talk at the same time.

It would be nice if we could see something while we hear your voice. It's a little disconcerting for the screen to stay black for the first half minute and then go black again.

The words are not legible on the handout in the early part of your video, unfortunately.

Make sure you always show the negative charge on the base!

It would be nice if you would explain the mechanism as you draw it.

The first two steps of your mechanism are perfect. But after your first alkylation, the negative charge is now on the bromide rather than the alkyne. In order to carry out a second alkylation step, it is necessary to react the alkyne with another molecule of base.

Your product is perfect. Your depiction of the transition state suffers because it's awfully hard to depict partial bonds and stereochemistry with the same set of lines. I encourage you consider drawing the structure so that the partial bonds can be right in the plane of the page. That way you can show the partial bonds with dotted lines. And you can show relative positions (stereochemistry) with positioning on the page.

You guys should speak up! I can barely hear you above the voices of others in the room.

Methanol (your solvent) is not acidic, so you will not have an appreciable amount of methoxide (CH3O-). It's less acidic than water, which has 1E-07 M hydroxide for a neutral solution. Instead of using methoxide as the nucleophile, you should be using methanol as the nucleophile. That means the structure formed after the carbocation will have a positive charge on the oxygen, which has three bonds. It also means you'll need an additional step, to deprotonate that oxygen and give a neutral product.

The stereochemistry on your carbocation intermediate is not shown appropriately. The carbon with the positive charge is planar, so the methyl group should no longer be pointed back.

Too bad that woman in the background talks so loudly.....

I'm not sure why your video started with the swirly doodle before you got to your naming example. Nice job showing that you understand the naming rules.

Great job demonstrating your understanding of the nomenclature rules. It can be a little tricky when you try to draw the t-butyl substituent when it's not attached to anything. In your example you added another carbon, and then you had 2,2-dimethylpropane rather than a t-butyl group. We don't have a nice way to show things as groups unless we attach them to other bigger molecules.

That approach certainly helps me spot my own mistakes. ;)

I like how you numbered the carbons to keep track of them.

You should be referring to Michael donor and Michael acceptor rather than electron donor and electron acceptor.

Nice explanation of the mechanism for your Michael reaction.

What other reducing agents could you use to reduce the nitro group to an amino group?

The arrow you have when water is leaving is backwards. The oxygen takes the electrons from the C-O bond when water leaves.

I don't hear any narration during your video, even with my volume turned all the way up. You should speak up a bit when you describe your reaction!

I love the use of the laser pointer to bring the viewer's attention to the changes in the reaction that you are describing.

Nice mechanism. It was cool to see your starting material converted to the enolate ion that reacted to form the ring. Remember when you draw mechanisms for tests, you need to show both the enolate and the neutral starting material as separate and distinct structures.

You should include a second arrow for the ring-closing nucleophilic attack, showing that one of the two carbon-oxygen bonds in the carbonyl group breaks when the nucleophile attacks.

Some of those arrows have the pen setting so broad that it's tough to see the bonds after the arrows get added.

I was worried we wouldn't get to see the diacid product formed when permanganate oxidized the two methyl groups, but you showed it!

It seems that you missed the alkyl groups on the acid you show before the decarboxylation.

I think you should have called this video the acetoacetic ester synthesis rather than acetic ester synthesis. Hopefully the handout didn't have that mistake on it.....

You confused me by calling the final step of the synthesis your first step. Otherwise everything looks great!

In your first step, you talk about making a ketone. But your structure shows an aldehyde, without the methyl attached to the carbonyl. The reaction should work okay whether the carbonyl has a hydrogen or methyl group attached. At the end, you snuck that methyl group back in there, changing the substituent to an ethyl group. Be careful with counting your carbons!

Nice work with the laser pointer in Explain Everything!

Nice use of the laser pointer in the app! Too bad that teacher talks so loud while you're recording your videos. In the final step you talked about the decarboxylation. But you didn't mention the ester hydrolysis that you also showed.