Were you referring to the example at timestamp 15:17 ? For this we were showing all the possible fragments which included the CH₃⁺ , notice that the fourth fragment shows that variation of the longer carbon chain as the radical and the methyl being the carbocation now. This is really just to show all the possible fragments and values that you'll see in the mass spectrum data. 

If a CH₃ group breaks off, and the CH₃ keeps the charge, that means the other fragment became the neutral radical. That may seem strange since CH₃⁺ is not a very stable carbocation.Fragmentation doesn’t always favor the most hyperconjugated carbocation.

Sometimes, a methyl shift or cleavage happens simply because it’s energetically favorable or statistically more likely, even if it forms a less stable carbocation.

So even if the longer chain could form a more stable carbocation, the fragmentation mechanism may have favored the methyl group carrying the charge.

Great Question! I recommend watching these videos, they will help you understand how to approach these types of questions. The first video has multiple practice problems, the last question is on answering questions with IR, H NMR and C NMR and Mass Spectrometry. 

Study With Us: Identifying Unknown Compounds Using IR and NMR Spectroscopy and Mass Spectrometry 

Practice Problem: Determining the Identity of a Compound (C8H14) Using IR and HNMR

Practice Problem: Finding the Structure of a Molecule Using IR, NMR, and MS

The below video has a worksheet:

Study With Us: Ranking Acidity, Using pKa, and Drawing Arrows in Acid-Base Reactions

As does this other practice exam:

Practice Exam: Structure, Bonding, Acids and Bases

We do have more practice problems here and we'll be adding worksheets shortly to these:

Practice Problem: Identifying the Most Acidic Proton in a Molecule

Practice Problem: Which of the Following Is the Strongest Acid?

Practice Problem: Which of the Following Is the Strongest Base?

Practice Problem: Listing Alkynes and Alkenes in Order of Increasing Acid Strength

Chemmunity Team 

"Sharpless reagents" usually refers to the set of reagents used in Sharpless asymmetric reactions, invented by K. Barry Sharpless. The two most famous ones are:

Used for: Epoxidizing allylic alcohols (alkenes with an -OH on the adjacent carbon)

Reagents:

Titanium(IV) isopropoxide (Ti(OiPr)₄)

(–)- or (+)-diethyl tartrate (DET) → gives the stereoselectivity!

tert-Butyl hydroperoxide (TBHP) → the oxidant

Product: Chiral epoxide with high enantioselectivity→ Which enantiomer you get depends on whether you use (+)-DET or (–)-DET

Used for: Converting alkenes to vicinal diols (2 adjacent OH groups), enantioselectively

Reagents:

OsO₄ (osmium tetroxide, catalytic)

A chiral ligand like (DHQ)₂PHAL or (DHQD)₂PHAL

NMO (or another co-oxidant)

Product: A chiral diol with predictable stereochemistry→ Again, you choose the enantiomer using the right ligand

At this time we do not have any videos on this topic but I'll let the team know. For now here's information on these topics:

Oxone is the brand name for a triple salt:

2KHSO₅ · KHSO₄ · K₂SO₄

The key component here is potassium peroxymonosulfate (KHSO₅), a powerful oxidizing agent.

What is it used for?

Epoxidation of alkenes

Oxidation of sulfides to sulfoxides/sulfones

Oxidation of alcohols (in some conditions)

Oxidative cleavage of alkenes to carbonyl compounds (less common)

Happy to hear you liked the video :) 

A common mistake students make during Hofmann elimination, aside from not choosing the correct β-carbon, is forgetting to form the quaternary ammonium salt before attempting the elimination step.

Here’s what happens:

Students sometimes jump straight into thinking it’s like an E2 reaction and try to eliminate directly from a regular amine (like a tertiary or secondary amine).

But Hofmann elimination requires converting the amine into a quaternary ammonium salt (usually using excess methyl iodide), followed by treatment with a strong base like silver oxide and water. Without this step, the elimination won't happen.

Other related mistakes include:

Using the wrong base — like hydroxide or an E2-style base instead of silver oxide and water.

Assuming Zaitsev product will form — Hofmann elimination is unique because it usually gives the least substituted (Hofmann) alkene, due to steric hindrance, not the most substituted like in E2 reactions.

Will do! We have another practice problem on this too!

Here's an example of the worked out mechanism:

Thank you for bringing this to our attention, we will fix that. So sorry about that! 

Gustav Hestholm Yay! We are so happy to hear the video helped and that you're learning a lot. If there is any video you need help finding feel free to let us know.