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Video lessons › Chemistry · Unit 17 · Organic Compounds Containing Oxygen · Chapter 6 · Carboxylic Acids and Their Derivatives

IIT-JEE Questions (1981–1995): Acids

8-minute video lesson in English, narrated with a computer voice.

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HVZ reaction past questions from IIT-JEE 1983 and 1995, solved step by step: why ethanoic acid is halogenated by chlorine and red phosphorus but methanoic acid is not (no alpha-hydrogen), and how propanoic acid gives 2-bromopropanoic acid and then propenoic (acrylic) acid with excess alcoholic KOH, with the substitution or elimination trap.

Chapters

  1. 0:00Introduction
  2. 0:51The HVZ reaction and the α-hydrogen
  3. 2:32After HVZ: aqueous or alcoholic alkali?
  4. 3:42Example 1: why methanoic acid gives no HVZ reaction
  5. 5:09Example 2: propanoic acid to acrylic acid, HVZ then alc KOH
  6. 7:11Quick revision: HVZ, alpha-H, aqueous vs alcoholic alkali

Transcript

0:00Introduction

Hello everyone! Let us start with a small mystery from the IITJEE paper of nineteen eighty three. Take ethanoic acid, the acid in vinegar, and treat it with chlorine and a little red phosphorus. One of its hydrogens is replaced by chlorine. Now take methanoic acid, the simplest carboxylic acid of all, and do exactly the same thing. This time, no hydrogen is replaced. Both are carboxylic acids, and both have a hydrogen joined to carbon. So why does one react, and the other not? Interesting, right? Today we solve this one, and then a harder question from nineteen ninety five, built on the same reaction. The alpha hydrogen idea behind both still appears in JEE questions, so let's begin.

0:51The HVZ reaction and the α-hydrogen

So, first, let us name the carbons. On screen, COOH is the carboxyl group, and each dash is a bond. The carbon joined directly to the carboxyl group is the alpha carbon, shown as C with a small Greek letter alpha written low beside it. The next carbon along the chain is the beta carbon, with the Greek letter beta. Hydrogens on the alpha carbon are called alpha hydrogens, okay? And the map on the right is the course's summary of carboxylic acid reactions; we will read it in a moment. Now look at that map on the right. The second box from the bottom on the right reads X two, red P, and HVZ in brackets. HVZ is short for the Hell-Volhard-Zelinsky reaction. X two is a halogen, the small two written low meaning two atoms, and we add only a little red phosphorus, the capital P on screen. Right? Now, a carboxylic acid has hydrogens in different places: one on the oxygen of the carboxyl group, and others on its carbons. So which hydrogen does the halogen replace? An alpha hydrogen, yes, exactly! X two is chlorine, C l two, or bromine, B r two, and one halogen atom takes the place of one alpha hydrogen, the alpha H on screen. Clear? So the product is an alpha halo acid, just as that box on the map says. Halo means a halogen atom now sits on the alpha carbon. With me? And this gives us the condition: the acid must have an alpha hydrogen. No alpha hydrogen, no HVZ reaction. Keep this rule ready, because our opening mystery depends on it. Got it?

2:32After HVZ: aqueous or alcoholic alkali?

Now, what can an alpha halo acid do next? Suppose we add an alkali. The alkali also reacts with the COOH group and turns it into its salt. So at the end, we add acid to get the free COOH back. Okay? Now the halogen on the alpha carbon. The alkali can be dissolved in water, or in alcohol. Does that change the product? Yes, it changes the product completely. With aqueous alkali, O H replaces X, the halogen, which is a substitution, and we get an alpha hydroxy acid. Right? But with alcoholic KOH, that is potassium hydroxide dissolved in alcohol, H X is removed instead: X leaves the alpha carbon, and a hydrogen leaves the beta carbon. This is elimination, not substitution, and it needs a hydrogen on the beta carbon. With me? So a double bond forms between the beta and alpha carbons, right next to the carboxyl group; the equals sign on screen is that double bond. Aqueous gives substitution, alcoholic gives elimination. Remember this pair, because Example two is built on it. Clear?

3:42Example 1: why methanoic acid gives no HVZ reaction

Time for Example one, our opening mystery, from IITJEE nineteen eighty three. Ethanoic acid can be halogenated with chlorine and red phosphorus, but methanoic acid cannot, and the question asks us why. Pause the video and try it first. Done? Okay, let's check. Ethanoic acid is C H three COOH, as on screen. The carbon of the C H three group is joined directly to the carboxyl group, so it is the alpha carbon, and it carries three alpha hydrogens. Right? So chlorine, with red phosphorus, both written above the arrow, replaces one of them. We get C l C H two COOH, chloroethanoic acid. Got it? Now, methanoic acid, HCOOH, also has a hydrogen joined to carbon. So, is that hydrogen an alpha hydrogen? No, it is not, and that is exactly the trap in this question. Look where that hydrogen sits: on the carboxyl carbon itself. An alpha hydrogen must sit on the next carbon along, and methanoic acid has no next carbon at all. Okay? So methanoic acid has no alpha carbon, so no alpha hydrogen, and the HVZ reaction cannot happen. So the answer is that methanoic acid has no alpha hydrogen. And our opening mystery is solved! With me?

5:09Example 2: propanoic acid to acrylic acid, HVZ then alc KOH

Now Example two, a harder one, from IITJEE nineteen ninety five. Propanoic acid is treated with phosphorus and bromine to give compound A, okay? Then compound A, with excess alcoholic KOH followed by acid, gives compound B, and we must identify both. Pause the video and try it first. Done? Okay, let's check. Propanoic acid is C H three C H two COOH, on the left of the arrow. Its alpha carbon is the C H two, right next to the carboxyl group. Bromine with phosphorus, B r two and P written above the arrow, replaces one alpha hydrogen, so we get C H three CHB r COOH, where B r is bromine. Right? So compound A is two bromopropanoic acid. The carboxyl carbon is carbon one, so the alpha carbon is carbon two, the C dash two on screen. Clear? Now the second step. KOH brings O H, so the O H simply replaces the bromine, and B is two hydroxypropanoic acid, right? That is the trap, so be careful! Read the condition again: alcoholic KOH, not aqueous. With alcoholic KOH, H B r is eliminated: the bromine leaves the alpha carbon, and a hydrogen leaves the beta carbon, the C H three. Got it? And why do we need excess KOH? Because part of it is used up turning the COOH group into its salt. Then H plus, the dilute acid, with the small raised plus for its charge, gives the COOH back at the end. Okay? So a double bond forms between the alpha and beta carbons, the equals sign on screen, and we get C H two, double bond, CHCOOH, on the right of the arrow. Compound B is propenoic acid, also called acrylic acid. See the two steps: HVZ puts bromine on the alpha carbon, and alcoholic KOH pulls it off along with a beta hydrogen. With me?

7:11Quick revision: HVZ, alpha-H, aqueous vs alcoholic alkali

So, quick revision. In the HVZ reaction, a halogen with a little red phosphorus replaces an alpha hydrogen, so the acid must have one. Okay? Methanoic acid has no alpha carbon at all, so it gives no HVZ reaction. Right? And an alpha halo acid gives substitution with aqueous alkali, but with alcoholic KOH, it loses H X, which is elimination. Now go through the other past questions on this page, and try each one before you read its solution. All the best!

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