Topic 10

Topic 10 on a page

Everything from this topic in one place, for the last read before an exam. If anything here is unfamiliar, go back to the full page.

Definitions to learn word for word

Nucleophile: an electron-pair donor, usually with a lone pair and a negative or partial negative charge (Nucleophiles, electrophiles and polarity).
Heterolytic fission: a covalent bond breaks and both electrons go to one atom, giving a positive and a negative ion (Mechanisms and curly arrows).
Primary, secondary, tertiary: the carbon bonded to the halogen (or OH) has one, two or three carbon neighbours (Primary, secondary and tertiary halogenoalkanes).
Molecular ion: the ion formed when a molecule loses one electron. Its m/z is the relative molecular mass (Mass spectra).

Halogenoalkane reactions (bromoethane as the example)

Reagent and conditions	Product	What the reagent does
aqueous KOH, warm, reflux	ethanol	OH⁻ is a nucleophile (substitution)
ethanolic KOH, reflux	ethene	OH⁻ is a base (elimination)
excess NH₃ in ethanol, sealed tube, heat	ethylamine	NH₃ is a nucleophile (substitution)
KCN in ethanol, reflux	propanenitrile	CN⁻ is a nucleophile; the chain is one carbon longer
AgNO₃ in ethanol	ethanol and a precipitate of AgBr	water is the nucleophile

Silver halide colours: AgCl white, AgBr cream, AgI yellow.

Mechanisms

Hydroxide: one step. A curly arrow from a lone pair on O to the δ+ carbon, and one from the C–Hal bond to the halogen. The product is an alcohol and a halide ion.
Ammonia: two steps. Step 1 gives $\ce{RNH3+}$ and a halide ion. Step 2 uses a second NH₃ to remove H⁺, giving the amine and $\ce{NH4+}$ .
Arrows start at a lone pair or a bond and end at an atom. They show electron movement.

Rates of hydrolysis

Class: tertiary > secondary > primary.
Halogen: iodo > bromo > chloro, explained by bond enthalpy (C–Cl 346, C–Br 290, C–I 228 kJ mol⁻¹), not by polarity.
Core Practical 5: AgNO₃ in ethanol at the same temperature, time to hide a cross, rate ∝ 1 ÷ time. The end point is a judgement by eye.

Alcohol reactions

Reaction	Reagent and conditions	Product and observation
Combustion	oxygen	CO₂ and H₂O (plenty of air); CO or C if air is short
Chloroalkane	PCl₅, dry, room temperature	chloroalkane, POCl₃ and steamy white fumes of HCl (test for –OH)
Bromoalkane	KBr and 50% concentrated H₂SO₄, reflux	bromoalkane
Iodoalkane	red phosphorus and iodine, reflux	iodoalkane
Elimination	concentrated H₃PO₄, heat	alkene and water

Oxidation with acidified potassium dichromate(VI)

Alcohol	Method	Product	Test
Primary	distil as it forms	aldehyde	Benedict's or Fehling's, warm: brick-red precipitate
Primary	reflux with excess	carboxylic acid	sodium carbonate: fizzing
Secondary	heat	ketone	no reaction with Benedict's
Tertiary	heat	no oxidation	stays orange

Dichromate goes orange to green whenever it oxidises something. In equations the oxidising agent is written [O]. Boiling temperatures rise in the order aldehyde < alcohol < acid, because of hydrogen bonding.

Techniques

Reflux: vertical condenser, open at the top, water in at the bottom and out at the top, anti-bumping granules. It lets you heat a volatile mixture for a long time without loss.
Distillation: thermometer bulb level with the side arm, receiver open to the air. It separates by boiling temperature.
Separating funnel: stopper off before the tap is opened, the less dense layer is on top. Run off the lower layer.
Drying: add an anhydrous salt and swirl until the liquid is clear.
Purity: the boiling temperature should be sharp and at the expected value.

Spectra

Technique	Gives
Mass spectrum	highest m/z = $M_\text{r}$ ; fragments such as 15 CH₃⁺, 29 C₂H₅⁺, 31 CH₂OH⁺, 43 CH₃CO⁺, 45 $\text{COOH}^{+}$
Infrared	O–H 3230–3550 (alcohol, broad); O–H 2500–3300 (acid, very broad); C=O 1680–1750; C–H 2850–2950; C–Cl 600–800; C–Br 500–600

Method for an unknown: molecular ion, then infrared, then fragments. Say which absorptions are absent as well.

The ten traps

Drawing a curly arrow that starts at an atom or ends in the middle of a bond.
Using the wrong role for OH⁻: aqueous gives substitution (nucleophile), ethanolic gives elimination (base).
Explaining the iodo > bromo > chloro order by polarity. It is bond enthalpy.
Forgetting the second ammonia molecule in the mechanism, or the excess in the first.
Counting the wrong carbons for primary, secondary or tertiary: count carbon neighbours of the carbon that carries the halogen or OH.
Giving the nitrile one carbon too few.
Distilling when the acid is wanted, or refluxing when the aldehyde is wanted.
Saying a tertiary alcohol is oxidised, or that dichromate "turns blue".
Sealing the top of the condenser or the receiver, and putting the condenser water in at the top.
Writing a mass-spectrum fragment without its + charge, or taking the tallest peak as the molecular ion.

Mixed test

Name $\ce{CH3CHBrCH2CH3}$ and say whether it is primary, secondary or tertiary.
Describe, with curly arrows in words, the mechanism of 1-chloropropane with aqueous KOH.
1-chlorobutane is hydrolysed more slowly than 1-iodobutane, although the C–Cl bond is more polar. Explain.
Three bottles each contain a primary halogenoalkane: a chloro-, a bromo- and an iodoalkane. Describe how to tell them apart.
10.0 g of 2-methylpropan-2-ol ( $M_\text{r}$ = 74.0) gives 8.9 g of 2-chloro-2-methylpropane ( $M_\text{r}$ = 92.5). Calculate the percentage yield.
Give the products when butan-1-ol is (a) distilled with acidified dichromate, and (b) heated under reflux with excess. Name a test for each product.
State what happens when butan-2-ol and 2-methylbutan-2-ol are each heated with acidified dichromate.
Write balanced equations for (a) ethanol with PCl₅, (b) propan-2-ol dehydrated with concentrated H₃PO₄, (c) the complete combustion of propan-1-ol.
A compound has a molecular ion at m/z 74, a peak at m/z 45, and an infrared spectrum with a very broad absorption from 2500 to 3300 cm⁻¹ and a strong absorption at 1715 cm⁻¹. Identify it.
List the infrared absorptions above 1650 cm⁻¹ that you would predict for butanal, $\ce{CH3CH2CH2CHO}$ .
In a reflux, explain why (a) a condenser is used, (b) the top of the condenser is open, (c) anti-bumping granules are added.
Explain why propanal can be separated from propan-1-ol by distillation as the reaction proceeds.

Answers cover these until you have tried the questions

2-bromobutane. The carbon bonded to Br has two carbon neighbours, so it is secondary.
The hydroxide ion is a nucleophile. A curly arrow starts at a lone pair on the oxygen of OH⁻ and points to the δ+ carbon bonded to Cl. A second curly arrow starts in the C–Cl bond and points to the chlorine atom, which leaves as Cl⁻. It is one step, and the product is propan-1-ol.
The rate depends on how easily the carbon–halogen bond is broken. C–I has the lowest bond enthalpy (228 kJ mol⁻¹) and C–Cl the highest (346 kJ mol⁻¹), so the C–I bond breaks more easily and 1-iodobutane is hydrolysed faster. Polarity would predict the opposite order, so it is not the deciding factor.
Warm each with silver nitrate in ethanol at the same temperature. The iodoalkane gives a yellow precipitate fastest, the bromoalkane a cream precipitate more slowly, and the chloroalkane a white precipitate slowest.
$n(\text{alcohol})$ = 10.0 ÷ 74.0 = 0.1351 mol. The equation is 1 : 1, so the theoretical mass = 0.1351 × 92.5 = 12.5 g. Percentage yield = 8.9 ÷ 12.5 × 100 = 71%.
(a) Distilled as it forms: butanal, tested with Benedict's (or Fehling's) solution, warm: brick-red precipitate. (b) Reflux with excess: butanoic acid, tested with sodium carbonate (or sodium hydrogencarbonate) solution: fizzing.
Butan-2-ol is a secondary alcohol: it is oxidised to butanone and the dichromate turns orange to green. 2-methylbutan-2-ol is tertiary: no oxidation, and the mixture stays orange.
(a) $\ce{CH3CH2OH + PCl5 -> CH3CH2Cl + POCl3 + HCl}$ . (b) $\ce{CH3CH(OH)CH3 -> CH3CH=CH2 + H2O}$ . (c) $\ce{2C3H7OH + 9O2 -> 6CO2 + 8H2O}$ (or $\ce{C3H7OH + 4.5O2 -> 3CO2 + 4H2O}$ ).
$M_\text{r}$ = 74. The very broad O–H (2500–3300) and C=O (1715) show a carboxylic acid. The peak at 45 is $\text{COOH}^{+}$ and C₃H₆O₂ has $M_\text{r}$ = 3(12.0) + 6(1.0) + 2(16.0) = 74.0. The compound is propanoic acid, $\ce{CH3CH2COOH}$ .
C–H (alkane) 2850–2950; C–H (aldehyde) 2700–2900; C=O 1680–1750, strong and sharp. There is no O–H.
(a) The condenser cools the vapour back to liquid, so the volatile compounds return to the flask and none is lost. (b) The top is open so that the apparatus is not a closed system: a sealed flask that is heated could burst. (c) Granules give smooth boiling, so there are no large bubbles or sudden bumping.
Propanal has no O–H and cannot form hydrogen bonds with itself, so it boils at 49 °C, far below propan-1-ol (97 °C). It vapourises as it forms and is condensed and collected before it can be oxidised further.