PAG 5 Synthesis of an Organic Liquid: 2-chloro-2-methylpropane

PAG 5 converts a tertiary alcohol into a haloalkane. 2-Methylpropan-2-ol, (CH₃)₃COH, is shaken with concentrated hydrochloric acid at room temperature and the hydroxyl group is replaced by a chlorine atom to give 2-chloro-2-methylpropane, (CH₃)₃CCl.

The reaction is a nucleophilic substitution that goes through a tertiary carbocation, which is why no heating and no catalyst are needed: a primary alcohol would not react under these conditions.

(CH₃)₃COH(l) + HCl(aq) → (CH₃)₃CCl(l) + H₂O(l)

The reaction itself takes twenty minutes. The marks are in what follows: separating the crude product from the aqueous layer, washing out the acid, drying, distilling the fraction that boils at 50–52 °C, testing the distillate for chloride and calculating the percentage yield.

Marks come from the written papers (practical skills are examined in every paper) and the practical endorsement, so every step of the method must come with its reason.

Three skills run through the whole practical: using a separating funnel (which layer to keep, when to remove the stopper, how to vent) and simple distillation (where the thermometer bulb sits, which way the water flows, what to collect).

The third is a qualitative test (hydrolysis, acidify with nitric acid, silver nitrate).

The same three skills appear in PAG 5 (oxidation of an alcohol) and PAG 4 and PAG 7 (analysis of inorganic and organic unknowns), so learn them here properly.

The quantities below are the standard ones: 10.0 cm³ of 2-methylpropan-2-ol and 35 cm³ of concentrated hydrochloric acid.

The acid is in large excess (about 0.39 mol against 0.105 mol of alcohol), so the alcohol is the limiting reagent and the yield is calculated from it.

Every row carries the reason, because the reason is what the mark scheme rewards.

When the shaking stops, the mixture settles into two layers. The upper layer is the crude 2-chloro-2-methylpropane and the lower layer is aqueous.

Students are asked to justify this, and the answer has two parts. The chloroalkane is immiscible with water because it cannot form hydrogen bonds with water molecules.

It is less dense than the aqueous layer (0.84 g cm⁻³ against more than 1.0 g cm⁻³ for the acid solution), so it floats.

Running off the lower layer has three marked manipulations. Remove the stopper first: with it in place no air can enter and the liquid stops flowing after a moment.

Open the tap slowly and watch the boundary. Close the tap as the boundary reaches it, so that the organic layer stays in the funnel.

Venting is the other technique. After sodium hydrogencarbonate is added, carbon dioxide is produced and the pressure would eject the stopper.

Stopper the funnel, invert it so the tap is uppermost, open the tap for a second to release the gas, close it, shake gently, and vent again.

Only with the funnel upright and the stopper out is liquid run through the tap.

Left: stopper out, tap open, the lower aqueous layer running into the conical flask until the boundary reaches the tap. Right: the funnel inverted and vented through the tap after the hydrogencarbonate wash.

The dried liquid is distilled and the fraction boiling at 50–52 °C is collected: the data-book boiling point of 2-chloro-2-methylpropane is 51 °C and the range is ±1 °C around it.

The apparatus is a pear-shaped flask with anti-bumping granules, a still head carrying the thermometer, a Liebig condenser sloping down to a receiver adaptor, and a receiving flask standing in iced water.

The thermometer bulb is level with the side arm, the opening into the condenser, so that it measures the temperature of the vapour that is actually distilling over.

A bulb in the liquid reads the liquid temperature, which rises as the alcohol concentrates and tells you nothing about what is in the receiver.

Cooling water enters the condenser at the lower connection, nearest the receiver, and leaves at the upper one: the jacket stays full and the coldest water meets the vapour that has already partly cooled, giving efficient counter-current cooling.

Heat with a water bath or an electric heating mantle, never a Bunsen burner: the product is highly flammable and a water bath is enough because it boils below 100 °C.

The receiver stands in a beaker of iced water because the product is volatile, and the system is open to the air at the receiver; a sealed apparatus would burst.

Discard the first few drops (traces of water and acid), collect while the thermometer stays in the 50–52 °C range, and stop when the reading rises (alcohol, boiling point 82 °C, is coming over) or falls (no more vapour reaching the bulb).

Simple distillation: thermometer bulb level with the side arm, water in at the lower connection and out at the upper, anti-bumping granules, electric heating and a receiver open to the air. Add a beaker of iced water under the receiver and collect the 50–52 °C fraction.

Both techniques use a Liebig condenser and both are assessed in PAG 5.

In reflux the condenser stands vertically above the flask, so vapour condenses and drips back; the mixture can be heated for a long time without losing volatile reactants, solvent or product.

In distillation the condenser slopes downwards, so the condensed vapour runs away into a receiver and is separated from the mixture.

Reflux: the condenser is vertical and condensed vapour drips back into the heated flask. The coil-type condenser shown works like the straight Liebig condenser usually drawn in exams.

The distillate is a covalent liquid, so silver nitrate alone shows nothing: there are no free chloride ions. The chloroalkane must first be hydrolysed.

Put a few drops of distillate in a test tube, add about 2 cm³ of ethanol and a few drops of sodium hydroxide solution, and warm in a water bath.

Ethanol is used because the chloroalkane does not dissolve in water; ethanol dissolves both the chloroalkane and the aqueous reagents, so the molecules can meet.

(CH₃)₃CCl(l) + OH⁻(aq) → (CH₃)₃COH(aq) + Cl⁻(aq)

Because the haloalkane is tertiary, the hydrolysis is fast even with water alone: (CH₃)₃CCl(l) + H₂O(l) → (CH₃)₃COH(aq) + HCl(aq).

So a simpler version of the test warms the distillate with ethanol and aqueous silver nitrate directly.

Either way, the next step is to add dilute nitric acid until the solution is acidic, then silver nitrate solution. A white precipitate of silver chloride confirms chloride ions:

Ag⁺(aq) + Cl⁻(aq) → AgCl(s)

Why acidify at all? If the solution is still alkaline, silver ions give a brown-black precipitate of silver oxide, Ag₂O, which masks the test.

Why nitric acid? Hydrochloric acid would add chloride ions and give a white precipitate whatever the sample contained (a false positive).

Sulfuric acid gives a white precipitate of silver sulfate with concentrated silver nitrate solutions. Nitric acid adds nothing that reacts with silver ions.

To confirm the precipitate is silver chloride rather than silver bromide or iodide, add dilute ammonia solution: silver chloride dissolves in dilute ammonia, silver bromide only in concentrated ammonia and silver iodide in neither. Silver chloride also darkens in sunlight.

Hydrolyse with ethanol and warm sodium hydroxide, acidify with dilute nitric acid, add silver nitrate: a white precipitate of silver chloride that dissolves in dilute ammonia solution.

A positive test proves only that the distillate contains chlorine bonded to carbon. Identity and purity come from the whole set of evidence: the distillate boiled at 50–52 °C, it is immiscible with water, and it gives chloride ions on hydrolysis.

The alcohol and its product have very different physical properties, and the exam asks you to explain both differences in terms of intermolecular forces.

Be careful with the boiling point explanation. The chloroalkane has the larger Mr, so its London forces are actually stronger than the alcohol’s; what it lacks is hydrogen bonding.

Write: “2-chloro-2-methylpropane cannot form hydrogen bonds between its molecules; the dipole–dipole and London forces present are weaker than the hydrogen bonds between alcohol molecules, so less energy is needed to separate them and its boiling point is lower.”

Solubility follows the same logic. 2-Methylpropan-2-ol forms hydrogen bonds with water, so it mixes with water and with the acid at the start of the reaction.

The product cannot hydrogen bond with water, so as it forms it separates as a second layer: that is why the mixture turns cloudy and why the separating funnel works.

The boiling point is also the purity check. A pure liquid distils over a narrow range, 1 to 2 °C, at its data-book value.

Impurities widen the range and usually raise the temperature, because the more volatile component distils first and the mixture becomes richer in the less volatile one.

A sample that comes over at 50–52 °C, is immiscible with water and gives a positive chloride test is accepted as 2-chloro-2-methylpropane.

The question gives the volume and density of the alcohol used and the volume and density of the product collected.

The route is always the same: volume × density gives mass; mass ÷ Mr gives moles; the 1 : 1 equation gives the theoretical moles of product; multiply by the product’s Mr for the theoretical mass; compare with the actual mass.

Quote every value with its unit and keep three significant figures until the last step.

Data: 10.0 cm³ of 2-methylpropan-2-ol, density 0.78 g cm⁻³, Mr 74.1; 35 cm³ of concentrated hydrochloric acid (about 11 mol dm⁻³, so about 0.39 mol, in excess); 6.0 cm³ of 2-chloro-2-methylpropane collected, density 0.84 g cm⁻³, Mr 92.6.

A yield of 50 to 60 % is typical and the examiner wants specific reasons why it is below 100 %.

Some alcohol is left unreacted or is held in the calcium chloride complex. Product dissolves slightly in the aqueous washings.

Product is lost as vapour because it is volatile. Liquid stays behind on the drying agent, in the funnel and on the glassware.

The first drops of distillate are discarded and some product is left in the flask so that it is not distilled to dryness.

• Keeping the wrong layer. Here the product is the upper layer (0.84 g cm⁻³), but decide from the density every time; a bromoalkane sinks.
• Not venting. Hydrogen chloride vapour, the volatile product and then carbon dioxide all pressurise the funnel. Invert and open the tap; do not pull the stopper out of a pressurised funnel.
• Leaving the stopper in while running off the lower layer, then blaming the tap.
• Washing with sodium hydroxide. Hydroxide ions hydrolyse the tertiary chloroalkane back to the alcohol; use sodium hydrogencarbonate.
• Distilling too fast or too hot, so the alcohol (boiling point 82 °C) comes over and the “product” boils over a wide range.
• Thermometer bulb in the liquid rather than level with the side arm.
• Receiver not cooled. The product boils at 51 °C; an uncooled open receiver in a warm fume cupboard loses it as vapour.
• Adding silver nitrate before hydrolysis, or before acidifying, or acidifying with hydrochloric acid. No free chloride without hydrolysis; brown silver oxide in alkali; a false positive with hydrochloric acid.
• Yield working without units or with early rounding, and “side reactions” as the reason for a low yield with no reaction named.
• Reflux wording for a distillation. The condensate is collected in the receiver; it does not return to the flask.

Say

“The tertiary alcohol forms a stable tertiary carbocation, so it reacts with concentrated hydrochloric acid at room temperature.”

“The chloroalkane is the upper layer because it is immiscible with water and less dense (0.84 g cm⁻³).”

“Sodium hydrogencarbonate neutralises the acid; hydroxide would hydrolyse the product.”

“The bulb is level with the side arm to measure the temperature of the vapour distilling over.”

“Dilute nitric acid is added because it neutralises hydroxide ions without adding halide ions.”

Do not say

“The OH is replaced by chlorine” (a chlorine atom, not Cl₂). “Add acid before silver nitrate” without naming nitric acid. “The organic layer is always on top.” “Weaker intermolecular forces” without saying that hydrogen bonding is missing. “Some was lost” or “human error” as the reason for a yield below 100 %.

Watch for

State symbols on every equation, including (l) for the alcohol and the product. Units on every line of the yield calculation and three significant figures until the end.

The 50–52 °C range is a ±1 °C window around the data-book value of 51 °C. Density decides the layer; hydrogen bonding decides the boiling point and the solubility.