Acid chloride formation | Carboxylic acids and derivatives | Organic chemistry | Khan Academy

[@lblake11]

Good details and explanation was clear and concise. Only problem I face with Ochem in college is solving more complicated reactions and compounds. I think I should actually practice problems from the book more often.

[@vedant8002]

Are you alive?

[@MrSmudger687]

@TheCarnagekidd6 - yes you can but thionyl chloride is preferable as the by products of the reaction (HCl and SO2) are both gaseous.

[@Darkstar88Gaming]

good job!! explains alot! you da man, man!!!!

[@vedantsinha7374]

thank you !

[@MrTitchfield]

quite effectiive and explanation is clear

[@ncamaa]

Isn't it much more likely that in the 3rd state, the Cl(-) will attack the H of the O(+) instead of the Carbonyl?

[@niloofarnoormohammadi9898]

Perfect👍

[@asisable]

Thank you so much for sharing...❤️🙏🏾💯

[@paysonbiker]

I'm confused; my o-chem prof 'actived' the oxygen from the carbonyl through resonance, and then used the subsequent, negatively charged oxygen as the nucleophilic site to attack the electrophilic sulfur site. The rest is basically the same and I end up with the correct product. Are both ways possible? Or is my professor wrong?

[@quinnpuffer7901]

I've seen it done both ways

[@TheCarnagekidd6]

We can use dry PCl5 to change acetic acid to acetyl chloride as well?

[@Acumen06]

yup, the explanation was clear but hopefully next time, aside from discussing the whole mechanism, the narrator could also discuss the state/phase of the beginning and end product, and the effect of the reaction, if it is endothermic, exothermic or toxic... i have a question sir, is it possible for this reaction to also form acetic anhydride? thanks

[@callycap3]

The mechanism is exactly the same. One of the Chlorine atoms from PCl5 acts first as a leaving group, then as a nucleophile, just in the same way as in this video.

[@Ara2061]

thank u.....thank u ....thank u ... *-*

[@MrSmudger687]

@TheCarnagekidd6 - No problem

[@thomaswilliams1963]

Great vid...but good god the repeating is annoying...

[@younismohamed6251]

Why is oxalyl chloride a better reagent for the formation of acid chloride than thionyl sulfide?

[@OsamaShehzad1995]

whoever is watching this reaction, please do NOT follow it. This mechanism is completely WRONG according to fundamentals of organic chemistry. In the first step, lone pairs on Oxygen from (-OH) will NEVER attack in the presence of oxygen from carbonyl. but why? because if oxygen from carbonyl (C=O) attacks the chloride, the resulting intermediate can be resonance stabilized which cannot happen in case of attack from OH. Resonance stability shows that C=O oxygen remains more nucleophilic at all times

[@sebastiancioban1476]

Osama, can you help me with somenthing? I'm trying to find the radicals of dicarboxylic acids, like oxalic, succinic etc.., I know that there is 2 radicals (OC-CO and HOOC-CO), one of them is oxalyl for oxalic, but I don't know how to name the divalent RADICAL

[@lizzyadeola675]

Why

[@jdjdjdndn9369]

Hi ✋ ... can I use HCl with carboxylic acid , if no why cannot use ?!

[@MuhammadRafiMaulaAzmi]

nicee

[@55studebaker]

Your initial net rxn is shown as irreversible, but everything depicting the mechanism to the end is shown as in equilibrium. Don't get it.

[@TheCarnagekidd6]

@MrSmudger687 Thank you.

[@zambrocca]

radical reaction??? are you sure?

[@4SticksOfGum]

Does anyone know how you would form an acid chloride using PCl5?

[@pranatipramanik123]

RCOOH+PCl5 ->RCOCl

[@jhyland87]

You repeat yourself quite a bit... repeat yourself quite a bit. :-P Great video tho! Question - Why am i having such a difficult time finding acetyl chloride (in the states)? I can't find it on Amazon or eBay. It doesn't appear to be on the list of chemicals the DEA restricts/watches... So i don't get why it's seemingly difficult to source.

[@feistymind4915]

(y)

[@thehaunted508]

third

[@acmilanshevachels]

fourth

[@xcolonel]

second

[@wookiemaster73]

first

[@user-zu6ko2wx5e]

Organic Chemistry sucks...

[@vedantsinha7374]

thank you !