Chemistry Solutions 9701/12/O/N/15

Ref: 9701/12/O/N/15 #3

Three substances have the physical properties as shown in the table.

What could be the identities of U, V and W?

Solution

D

It is not possible to identify the ionic compound V from the data given.
For U, the choice is Zinc since it is insoluble in water. Na reacts with water to produce NaOH.
A boiling point of 58 degrees Celsius indicates that W is a liquid at room temperature. Hence SiCl4 is chosen as HCl is a gas at room temperature.
Since U=Zinc and W=SiCl4, the answer is D.




Ref: 9701/12/O/N/15 #17
A student observed the reactions when sodium chloride and sodium iodide were each reacted separately with concentrated sulfuric acid and with concentrated phosphoric acid. Some observations are recorded in the table.

Which deduction can be made from these observations?
A. Concentrated phosphoric acid is a stronger oxidising agent than concentrated sulfuric acid.
B. Concentrated phosphoric acid is a stronger oxidising agent than iodine.
C. Concentrated sulfuric acid is a stronger oxidising agent than chlorine.
D. Concentrated sulfuric acid is a stronger oxidising agent than iodine.

Solution
D

Concentrated H2SO4 and H3PO4 both react with sodium chloride and sodium iodide to form colourless acidic gases (HCl and HI respectively). Both H2SO4 and H3PO4 are weaker oxidising agents than chlorine, hence they are not able to oxidise the chloride ions in HCl.
Concentrated H2SO4 is a stronger oxidising agent than iodine, hence it is able to oxidise the iodide ions in HI to iodine gas (purple vapour).
Concentrated H3PO4 is a weaker oxidising agent than iodine, hence it is unable to oxidise the iodide ions in HI.


Ref: 9701/12/O/N/15 #24

The depletion of the ozone layer in the upper atmosphere reduces the Earth's natural protection from harmful ultraviolet radiation.
Which compound would cause the most depletion of the ozone layer?


A. CCl3F
B. CF4
C. CO2
D. SO2


Solution
A


C-Cl has the lowest bond energy, so it is the easiest bond to break. Chlorine free radicals are formed when exposed to ultraviolet radiation. (For questions like this, always look for the molecule with the most C-Cl bonds.) These chlorine free radicals will go on to destroy the ozone layer according to the following steps:

Ref: 9701/12/O/N/15 #25


Compound X has been investigated for use as an artificial sweetener.

The two C-Cl bonds can be hydrolysed by hot NaOH(aq). The C-O-C bonds cannot be hydrolysed by hot NaOH(aq).


What are the numbers of specified types of -OH groups before and after hydrolysing the two C-Cl bonds?

Solution

C

The circled blue -OH groups are secondary alcohols. The green -OH group is the primary alcohol. The red -OH group is the tertiary alcohol.


Ref: 9701/12/O/N/15 #37


Compound Q is obtained by adding H2O across the double bond in compound P.

Which statements about these two compounds are correct?
1. P shows cis-trans isomerism.
2. Q contains two chiral centres.
3. Q is a tertiary alcohol.


Solution
C. 2 and 3 only

We highly recommend that you take a look at the explanation on optical isomers for skeletal formulae over at Chemguide.
1 is wrong because the two carbon atoms in the double bond are part of the ring structure. Hence, geometrical isomerism is "restricted" --- the groups around each carbon atom cannot be rotated to give cis-trans isomers.




Ref: 9701/12/O/N/15 #40


Compound Z is heated with concentrated acidified potassium manganate(VII). This produces an equimolar mixture of CO2 and CH3COCH2CH2CH(COCH3)CH2CO2H.
What could be the structural formula of Z?

Solution
A. 1, 2 and 3

The C=C double bonds are broken when the alkene undergoes vigorous oxidation.
Primary carbon atom with a double bond will form CO2.
Secondary carbon atom with a double bond will form carboxylic acid.
Tertiary carbon atom with a double bond will form ketone.