Yes you'll find that here:

You'll find this in the below video:

Yay! We love hearing you got it right 🎉

Recall that with a chair conformation the bulky substituents are most stable in the equatorial position. So all major conformations should have the bulky groups in the equatorial position, if they do not then they are NOT representing the conformational equilibrium. You'll find this concept explained in the below video with even more examples.

Hi Medha, at this time we don't have printable slides unfortunately it's due to copyright issues. Some past students tried to leak a lot of visuals. 

But we do have some downloads for the practice problems and exams. 

Under each Study with Us (Green thumbnails), Practice problems (blue thumbnails) and Practice exams (orange thumbnails) you'll find each pdf for that specific video. 

Yes this is shown in the below video: 

This isn't tertiary it's secondary as stated and labelled at 5:59 , since yes it's connected to two other carbons. 

What Ari was referring to was that Elimination reactions, both E1 and E2, favor tertiary and secondary alkyl halides. The wedged doesn't do anything here it's just showing stereochemistry which doesn't matter here since elimination just changes the product to an alkene. The stereochemistry (wedged or dashed) matters when we are doing substitution reactions since we will either have inversion (SN2)  or both inversion and retention (racemization, SN1). 

The end product is yes, it is also written on the slide in the black text for reference. 

Ashley Wampler Yes this is still an acid-catayzed tautomerization.

At 0:59 it shows the mechanism then at 4:41 it shows another example of the mechanism.