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:

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. 

The video below walks you through all of this and more, Melissa lists what you should look for when ranking acids and determining the strongest acid and base. I highly recommend watching this lesson first then watching the Study with Us video for extra practice and common exam questions. 

At timestamp 12:37 you'll see the intermediates are the ones that canceled out. This goes back to Chem 2 Kinetics where the intermediates are the ones that cancel out when finding the overall balanced reaction. Intermediates commonly formed in free radical halogenation are in the initiation step ( the 2 halogen radicals, 2 Cl• or 2 Br•), in the propagation step it's the alkyl radical formed (like methyl radical, CH₃•) and the regenerated halogen radical (Cl• or Br•). There are none in the termination step. 

In this video you'll also find intermediates shown at 29:35, this is specifically referring to energy diagrams and identifying the intermediate from the energy diagram. 

If you were looking for other nonradical reactions with examples of intermediates, we have a video that shows the intermediates for SN1 and E1 reactions it starts at timestamp 12:08 in the below video:

Lesson: Drawing Energy Diagrams for SN2, SN1, E2 and E1 Reactions

Does this answer your question?

Covering all the different types of Ochem 1 reactions will happen, it just typically takes the entire semester. The way you can cover Ochem 1 reactions is by watching each video for the specific type of reaction. 

You'll find Nucleophilic Substitution and Elimination Reactions (SN1, SN2, E1 and E2)  in the following section, each one is a different type of reaction used in Ochem 1.  https://chemmunity.com/categories/category-HNTCdf5Mrws  I recommend watching the first 5 videos in this category.

To understand organometallic reactions watch the first two videos in the Organometallic compounds category: https://chemmunity.com/categories/category-yIVgr3ARaxQ

Yes you can watch the Free Radical Halogenation video this goes into a type of reaction and talks in detail about reactivity, thermodynamics, kinetics and other key terms you'll need to understand. However if you haven't covered radicals yet or Free radical halogenation then I don't recommend watching this yet. 

Hi Riley Johnson, our Chemmunity AI alerted us that you would benefit from watching this video. At timestamp 33:00 you'll find your exact question explained. Also, if you haven't done so already I recommend watching the entire video to fully understand this concept. 

Yes that's correct, a benzylic radical is highly stable. You'll see this again typically in the second semester of Organic Chemistry. This type of reaction that focuses on benzylic radicals is known as free radical allylic bromination or NBS (timestamp 4:15 of the below video).

Chemmunity Team 

3. DMP (Dess-Martin Periodinane)

Reagents: Uses Dess-Martin periodinane, which is an iodine-based oxidizing agent.

Conditions: Very mild and efficient, often used at room temperature.

Products:

Primary alcohols → Aldehydes.

Secondary alcohols → Ketones.

Selectivity: Like the others, it’s selective and stops at the aldehyde stage for primary alcohols. It does not oxidize further to carboxylic acids.

Advantages: DMP is known for being user-friendly and producing fewer byproducts compared to chromium-based reagents like PCC.