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Organic acid-base mechanisms

How to use curved arrows to draw mechanisms for organic acid-base reactions.

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  • blobby green style avatar for user Samantha Nordin
    In the very start how did you know that the H on the OH was going to be the acid when the other molecule also has an OH bond? I am just confused since they look very similar to me, how did you identify them?
    (16 votes)
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    • mr pink red style avatar for user Anthony Scaletti
      One way to look at it is you want to minimize charge. If you take the H from the hydroxide ion on the right, you have an O atom with a -2 formal charge and the O atom of the molecule on the left will have a +1 formal charge.

      On the other hand, if we take the H from the left, we get H2O with no formal charges, and an acetate ion on the left with a -1 formal charge.
      (13 votes)
  • piceratops tree style avatar for user Adam Davisson
    I am wondering how you can just tell that something is an BL acid or a BL Base. For instance in the video he just says, this is going to act as the acid and the other like the base. What am I looking for to tell me which is an acid and which is a base? How can you so easily distinguish one from the other?
    (13 votes)
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  • starky ultimate style avatar for user Noah Hubbell
    Why does the sodium move to the conjugate base along with with the electrons in the acetic acid reaction? Why couldn't the base donate one of the lone pairs instead?
    (6 votes)
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  • spunky sam blue style avatar for user Learner
    At the end of the video you show acetone reacting with the hydronium ion. Why does the resultant conjugate acid have a C=OH (double bond between the C and O). This would mean the O has 3 bonds. Why do the electrons in the pi orbital not form a second lone pair on the oxygen? Thank you
    (5 votes)
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  • hopper cool style avatar for user Preetodeep Dev
    At , what does the crosshair-like symbol below the reversible double-headed arrow symbol indicate? It isn't present in the previous examples.
    (3 votes)
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  • blobby green style avatar for user Brian Erwin
    But how do we start? How do we identify the acid and base before we begin drawing arrows? Maybe re-record this, because I see multiple people asking this question.
    (2 votes)
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    • leaf red style avatar for user Richard
      So taking the first reaction as an example, most people new to organic chemistry wouldn't recognize the molecule on the left potentially just by its structural formula. It's acetic acid and instead they might know it from its chemical formula C2H4O2. As someone becomes more familiar with organic functional groups it should be obvious that it's a carboxylic acid and should therefore act as the acid in the reaction.

      The right molecule however should be recognizable to any chemistry student as sodium hydroxide which is obviously a base. So if the right molecule is the base, then the left molecule (whatever it is) should be the acid.

      With the second reaction, again the left molecule would look unfamiliar to students new to organic chemistry, but the right molecule should be immediately identified as hydronium which we know is an acid. So if the right molecule is an acid, then the left molecule is a base.

      Hope that helps.
      (3 votes)
  • blobby green style avatar for user Noor Shujaat
    aoa. i did not find any lecture regarding the increasing and decreasing order of acidity and basicity of aromatic acids and bases with substituent either electron donating group or electron with drawing (attached on ortho or meta or para position) group can you please help me regrading that?
    (1 vote)
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    • spunky sam blue style avatar for user Ernest Zinck
      Electron withdrawing groups increase acidity.
      Electron donating groups decrease acidity.
      Almost all groups increase acidity — electron withdrawing via resonance and electron donating by inductive withdrawal of electrons.
      No good correlation due to steric effects.
      (5 votes)
  • blobby green style avatar for user alexandra duran
    where did water come from on 1 minute and 10 seconds part of the video
    (1 vote)
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  • leaf green style avatar for user Nalin Rajput
    in the example that was given at , why isn't the hydrogen attached to the alpha carbon removed by NaOH? Because that conjugate base that is obtained by removing H from the alpha carbon is also resonance stabilised right?
    (1 vote)
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    • female robot grace style avatar for user tyersome
      Carbon is much less electronegative than oxygen and so isn't stable holding a negative charge.

      Another way of saying this is that the methyl hydrogens are very poor acids, while the carboxylic acid hydrogen is a good acid.

      I would guess that if you removed a hydrogen from the methyl group and there wasn't another source of hydrogen§ you would get a double bond between the two carbons, but I'm pretty sure that would still be a very unstable structure and would react with something ...‡

      §Note: You would then be dealing with ethanoate rather than ethanoic acid and would have to do the reaction in an aprotic solvent.

      ‡ADDENDUM: Finally looked this up — I believe the molecule you would get is known as ethenone (aka ketene) and is in fact very reactive.
      See: https://en.wikipedia.org/wiki/Ethenone
      (2 votes)
  • male robot donald style avatar for user samarthingalagavi
    Is it the protons that is donated or the election in Acid and Base reaction?
    (1 vote)
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Video transcript

- [Voiceover] Drawing acid-base reactions is really an important skill when you're doing organic chemistry mechanisms. So let's look at an acid-base reaction. On the left, acetic acid is gonna function as our Bronsted-Lowry acid. It's gonna be a proton donor. On the right we have sodium hydroxide. And hydroxide is going to accept a proton. It is going to be a Bronsted-Lowry base. So when you're drawing the mechanism, you used curved arrows to show the flow of electrons. And these two electrons on hydroxide, let's say, are the two that are going to grab the acidic proton on acetic acid. So there is my curved arrow. Only the proton, so these two electrons are left behind on this oxygen. So let's draw the products, the products, I should say, of this acid-based reaction. So on the left we would have our carbon double bonded to our oxygen. And then now this oxygen we have three lone pairs of electrons around it which gives this oxygen a negative one formal charge. We'd also have a sodium cation here, so we could think about that. Write an ionic bond. And then what do you get if you add an H plus to OH minus? You would get H2O or water. So let me go ahead and draw water in here. And I'll put in my lone pairs of electrons. Let's follow our electrons along so the two electrons, right, this lone pair right here on the hydroxide anion picked up this proton. So let's say those two electrons in magenta are these two electrons, and this was the proton that they picked up. And then we also need to follow the electrons, the electrons in here, I'll make them blue. So these electrons in blue come off onto the oxygen. So let's say those electrons in blue are right here, which gives the oxygen a negative one formal charge. So this is an acid-based reaction. And we can even identify conjugate acid-based pairs here. So on the left, right, on the left this was acetic acid. This was our Bronsted-Lowry acid. What is the conjugate base to acetic acid? Well, that would be over here, right. Just take away a proton, and this would be the conjugate base. Let me identify this as being the conjugate base. This is the acetate anion, right. So this is our conjugate base. For hydroxide, hydroxide on the left side functioned as a base, right. So the conjugate acid must be on the right side. So if you add a proton to OH minus, you get H2O. So water is the conjugate, oops. I'm writing conjugate base here, but it's really the conjugate acid, right. So we've identified our conjugate acid-base pairs. All right, the biggest mistake that I see when students are drawing acid-base mechanisms is they mess up their curved arrows. So the biggest mistake I see, and I'll do this in red so it'll remind you not to do it, is they show this proton right here moving to the hydroxide anion. And that is incorrect, all right. That's a very common mistake, because curved arrows show the movement of electrons, right. And that's not, this is not what's happening here. These two electrons up here in magenta are the ones that are taking that acetic proton. So this is incorrect. Don't draw your acid-base mechanisms like this. Let's do one more acid-base mechanism for some extra practice here. So on the left we have acetone and on the right we have the hydronium ion, H3O plus. So the hydronium ion is gonna function as our Bronsted-Lowry acid. It's going to donate a proton to acetone, which is going to be our Bronsted-Lowry base. Remember when you're drawing an acid-base mechanism, your curved arrows show the movement of electrons. So if acetone functions as our base, a lone pair of electrons on this oxygen could take this proton right here and leave these electrons behind on this oxygen. So let's show the results of our acid-base mechanism. So on the left, right, the lone pair on the left of the oxygen didn't do anything. The lone pair on the right of the oxygen picked up a proton, formed a bond, and so we get this with a plus one formal charge on the oxygen. We'd also form water here, so H2O, let me draw that in and show our lone pairs of electrons. And let's follow our electrons again. So the electrons in magenta right here on the oxygen picked up this proton forming this bond, right, so this bond right here, the two electrons in magenta. And then the electrons in blue here move off on to the oxygen to add another lone pair of electrons onto that oxygen, giving us water. So identifying our conjugate acid-base pairs again, on the left hydronium H3O plus is functioning as our Bronsted-Lowry acid, right. And you take a proton away from that, and you're left with a conjugate base. So on the right would be water, which is our conjugate base. And on the left, acetone is functioning as a Bronsted-Lowry base. So on the right, this right here must be the conjugate acid. So this is the conjugate acid on the right. We've identified our conjugate acid-base pairs. And we've shown the movement of electrons using curved arrows. So practice your acid-base mechanisms because they really are extremely important. And you have to be able to do them fairly quickly when you're writing an organic chemistry mechanism.