Ethanol and Ethanoic Acid

Ethanol, $\text{C}_2\text{H}_5\text{OH}$, is a colourless liquid that mixes with water in all proportions. Its chemistry in this chapter revolves entirely around the –OH (hydroxyl) group hanging off the ethyl chain.

• With sodium metal. The –OH hydrogen is loosely held enough that reactive metals can displace it, exactly as they do in water:

where: $\text{C}_2\text{H}_5\text{ONa}$ is sodium ethoxide (a salt-like product) and $\text{H}_2\uparrow$ is the hydrogen gas that bubbles off. The parallel to $2\text{Na} + 2\text{H}_2\text{O} \rightarrow 2\text{NaOH} + \text{H}_2\uparrow$ is deliberate — it is the fastest way to remember the reaction.

• Dehydration to ethene. Heating ethanol with excess concentrated $\text{H}_2\text{SO}_4$ at about 443 K pulls out a water molecule, leaving the double bond of ethene:

where: concentrated $\text{H}_2\text{SO}_4$ here acts purely as a dehydrating agent, not as an acid reacting by proton transfer — it soaks up the water that leaves.

• Uses. Solvent for varnishes, perfumes and tinctures (tincture iodine); as a fuel additive; in medicines. Excess or denatured alcohol consumption harms the body.

Ethanoic acid, $\text{CH}_3\text{COOH}$, is a weak acid whose dilute (5–8%) aqueous solution is sold as vinegar. Two reactions dominate the syllabus.

• With carbonates and bicarbonates. Like every acid, it displaces $\text{CO}_2$ from carbonates:

where: $\text{CH}_3\text{COONa}$ is sodium ethanoate (a salt) and the $\text{CO}_2$ turns limewater milky — the standard confirmatory test.

• Esterification. With a little concentrated $\text{H}_2\text{SO}_4$ as catalyst, ethanoic acid combines with an alcohol to form a fragrant ester:

where: $\text{CH}_3\text{COOC}_2\text{H}_5$ is ethyl ethanoate, a sweet-smelling ester used in perfumes and flavourings. The reverse reaction — an ester (or fat/oil) boiled with strong base (NaOH) hydrolysing back to a salt (soap) and an alcohol — is saponification, the basis of soap-making.

The whole chapter hinges on which part of the molecule is reactive and what the reagent does to it — get that pairing right and every equation follows.

Why sodium reacts with ethanol's –OH. The –OH group in ethanol behaves chemically like the –OH in water: the O–H bond is polar enough that an active metal such as sodium can insert itself in place of the hydrogen, releasing that hydrogen as $\text{H}_2$ gas. This is why the equation for ethanol + sodium looks structurally identical to water + sodium, just with $\text{C}_2\text{H}_5\text{ONa}$ instead of $\text{NaOH}$.

Why concentrated $\text{H}_2\text{SO}_4$ dehydrates rather than protonates. In the ethene reaction, concentrated $\text{H}_2\text{SO}_4$ is used in excess and hot specifically because its role here is to strip out a $\text{H}_2\text{O}$ molecule from the ethanol chain, not to donate a proton as it would in a normal acid–base reaction. That is why the syllabus insists you name both the reagent ("concentrated $\text{H}_2\text{SO}_4$, acting as a dehydrating agent") and the approximate temperature (443 K) — omit either and the mechanism described is incomplete.

Why ethanoic acid reacts with carbonates like any acid. Even though ethanoic acid is a weak acid — it does not ionise completely in water — it still contains an ionisable $\text{–COOH}$ hydrogen, which is enough to react with the base-like carbonate ion and liberate $\text{CO}_2$. Weakness affects how much it ionises and how vigorously it reacts, not whether the acid–carbonate pattern applies at all.

Why esterification needs a catalyst but carbonate reaction does not. Esterification is a slow equilibrium reaction between the –COOH of the acid and the –OH of the alcohol; concentrated $\text{H}_2\text{SO}_4$ here plays a third role (distinct from its dehydrating role above) — it catalyses the joining of the two groups and also helps pull away the water formed, pushing the reaction toward the ester. Without it, esterification barely proceeds at a usable rate.

Ethanoic acid is treated first with sodium hydrogen carbonate, and separately with ethanol in the presence of concentrated sulphuric acid. Write both balanced equations, name the test for each product, and explain what each reaction demonstrates about ethanoic acid.

Step 1. Reaction with sodium hydrogen carbonate follows the general acid–bicarbonate pattern: salt + water + carbon dioxide.

Balance check: left side has 2 C, 4 H (from the acid) + 1 H (from bicarbonate) = 5 H total counted with the salt/water split correctly, 1 Na, 3 O (acid) + 3 O (bicarbonate); right side reproduces the same atom counts across $\text{CH}_3\text{COONa}$, $\text{H}_2\text{O}$ and $\text{CO}_2$ — the equation is balanced as written.

Step 2. Test for the gas: pass it through limewater; the limewater turns milky, confirming $\text{CO}_2$. This demonstrates ethanoic acid's ordinary acidic behaviour, shared with every acid you have studied.

Step 3. Esterification with ethanol, catalysed by a few drops of concentrated $\text{H}_2\text{SO}_4$:

The product, ethyl ethanoate, has a distinct sweet, fruity smell — this is the test for a successful esterification, since the reactants themselves (a pungent acid and a solvent-smelling alcohol) do not smell fruity.

Step 4. Compare: Step 1's reaction is a general acid property — any acid reacting with a carbonate would give the same $\text{CO}_2$ test. Step 3's reaction is a specific organic property of the carboxylic acid group, unique to acids like ethanoic acid reacting with alcohols; mineral acids such as HCl do not undergo esterification.

Ethanoic acid therefore shows both general acid reactivity (with carbonates, giving $\text{CO}_2$) and the specific esterification reactivity characteristic of carboxylic acids (with alcohols, giving a fragrant ester).

• Calling ethanoic acid a "strong acid" — it is a weak acid; you must know it reacts as an acid (giving $\text{CO}_2$ with carbonates) but you are not expected to explain its strength using pKa or ionisation theory.
• Naming the dehydration reagent as just "sulphuric acid" — it must be concentrated $\text{H}_2\text{SO}_4$, in excess, at about 443 K; a dilute or unstated-condition answer is marked incomplete.
• Confusing esterification (acid + alcohol → ester + water, conc. $\text{H}_2\text{SO}_4$ as catalyst) with saponification (ester/fat + NaOH → soap + alcohol) — they are reverse-direction-related but examined as two distinct named reactions.
• Treating vinegar as pure ethanoic acid — vinegar is a dilute solution, about 5–8% ethanoic acid, not the concentrated acid itself.

Ask the tutor: Want a quick drill on matching each reagent — sodium, hot concentrated sulphuric acid, or a carbonate — to the right reaction and test? Ask the AI tutor to quiz you.