Halogenoalkanes

Question 1

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

Answer 1

Absorbs (harmful) ultraviolet / uv (light / radiation)

Question 2

) A molecule of CFC-11 breaks down in the upper atmosphere to form a
chlorine free radical.
Give the equation for this reaction.

Answer 2

CCl3F → *CCl2F + *Cl

Question 3

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

Answer 3

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

Question 4

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

Answer 4

absorbs infrared radiation
M1 idea of IR being taken in
1
M2 molecule has polar bonds
M2 accept polar molecule

Question 5

Reaction 1 occurs via a nucleophilic substitution mechanism.
Explain why the halogenoalkane is attacked by the nucleophile in this reaction

Answer 5

) Bromine is more electronegative than carbon
Allow difference in electronegativity if polarity of bond shown
M1
C is partially positive / electron deficient
M2 and M3 can be awarded from diagram that shows nucleophilic
attack
M2
Lone/electron pair (on the nucleophile) donated to the partially positive car

Question 6

For cynaide nucleophile what atom are the lone pair of electrons on

Answer 6

carbon not nitrogen

Question 7

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.
Percentage yield

Answer 7

Amount 1-bromo-2-methylpropane
(= (2 × 1.26) / 136.9 = 2.52/136.9) = 0.0184 mol
Correct answer scores 3 marks; answer to at least
2sf and any individual marks for M1/2 should be at
least 2sf; answers that are a factor of 10x out score
2;
1
M2 mass of 2-methylpropan-1-ol expected
(= 0.0184 × 74.0) = 1.36 g
Allow ECF through the question
1
M3 % yield = 100 × (0.895/1.36) = 65.7% (65-67%)
1
Alternative method:
M2 amount of 2-methylpropan-1-ol produced
= 0.895/74.0 = 0.0121 mol
M3 % yield = 100 × (0.0121/0.0184) = 65.7%
(65‑67%)