AQA A level Chem 3.3 Halogenoalkanes

State the benefit to life on Earth of ozone in the upper atmosphere. (1)

Suggest one reason why the use of CFCs was not restricted until several years after Rowland and Molina published their research. (1)

One of these reasons:

• lack of evidence that ozone was being depleted

• lack of alternatives to CFCs

• commercial interest to continue to use CFCs

• hard to obtain international agreement (1)

CFC-11 is a greenhouse gas that can contribute to global warming.

State and explain how CFC-11 is able to contribute to global warming (2)

absorbs infrared radiation (1)

molecule has polar bonds (1)

Which compound is not formed by reacting 3-bromo-3-methylhexane with warm, ethanolic potassium hydroxide?

A 2-ethylpent-1-ene

B 3-methylhex-1-ene

C 3-methylhex-2-ene

D 3-methylhex-3-ene (1)

B (1)

The question below refers to the reaction of 1-bromopropane with a solution of potassium cyanide in aqueous ethanol.

What is the organic product of this reaction?

A propylamine

B butylamine

C propanenitrile

D butanenitrile

D (1)

The question below refers to the reaction of 1-bromopropane with a solution of

potassium cyanide in aqueous ethanol.

The reactions of 1-bromopropane and 1-chloropropane with potassium cyanide

in aqueous ethanol occur at different rates under the same conditions.

Which row correctly shows the compound that has a faster rate of reaction and the correct reason for this?(1)

A (1)

Which species could act as a nucleophile?

A BH₃

B NH₄⁺

C PH₃

D SiH₄ (1)

C (1)

When 2.0 cm³ of 1-bromo-2-methylpropane (Mr = 136.9) were reacted with

an excess of sodium hydroxide, 895 mg of 2-methylpropan-1-ol (Mr = 74.0)

were obtained.

The density of 1-bromo-2-methylpropane is 1.26 g cm⁻³

Calculate the percentage yield for this reaction. (3)

Amount 1-bromo-2-methylpropane (= (2 × 1.26) / 136.9 = 2.52/136.9) = 0.0184 mol (1)

mass of 2-methylpropan-1-ol expected (= 0.0184 × 74.0) = 1.36 g (1)

100 × (0.895/1.36) = 65.7%

A student compares the rates of hydrolysis of 1-chlorobutane, 1-bromobutane and 1-iodobutane.

The suggested method is:

• add equal volumes of the three halogenoalkanes to separate test

tubes

• add equal volumes of aqueous silver nitrate to each test tube

• record the time taken for a precipitate to appear in each test tube.

State and explain the order in which precipitates appear. (2)

Silver iodide then silver bromide then silver chloride (1)

bond strength C−I < C−Br < C−Cl (1)

When 2-bromobutane reacts with ethanolic potassium hydroxide, two structurally isomeric alkenes are produced, one of which shows stereoisomerism.

Explain why two structurally isomeric alkenes are formed (2)

H is lost from different carbon atoms (1)

H removes from C₁ and C₃ to give two isomers (1)

Reaction of 2-bromobutane with potassium hydroxide can produce two types of product depending on the solvent used. In aqueous solution, the formation of an alcohol, E, is more likely but in ethanolic solution the formation of alkenes is more likely.

Name alcohol E and draw its structural formula. By reference to the structure of the halogenoalkane, explain why the initial step in the mechanism of the reaction producing the alcohol occurs.(5)

Alcohol =butan-2-ol (1)

Appropriate structure for CH₃CH(OH)CH₂CH₃ (1)

C-Br bond is polar (1)

Lone pair on OH^-(1)

Attacks the C^S+(1)

The boiling point of iodomethane (CH₃I) is higher than that of fluoromethane (CH₃F) even though electronegativity of iodine is less than of fluorine.

Explain why iodomethane has the higher boiling point by considering the forces that act between CH₃I molecules and comparing these forces with the forces between CH₃F molecules (3)

Iodine is a bigger molecule than fluorine so the van der Waals forces between CH₃I molecules are stronger than CH₃F (1)

The dipole-dipole forces between CH₃F is greater than CH₃I (1)

As van der Waals forces are the strongest, CH₃I takes more energy to break and therefore has a higher boiling point (1)