(a)  Questions (i) – (vi) refer to Compounds A-C, shown below.

(i)        (½ mark) Name the functional group present in Compound C __k__et___one____

(ii)       (½ mark) Classify the alcohol functional group in Compound B as primary,

secondary, or tertiary:                                             1° / 2° / 3°

(circle one)

(iii)      (1 mark) One of Compounds A-C can exist as enantiomers. Identify this

compound:                                    Compound A / Compound B / Compound C

(circle one)

(iv)      (1 mark) Briefly justify your answer to part (iii)

Compound B is the only compounds that contains a stereogenic centre / a carbon atom bonded to 4 different groups

(v)       (1 mark) Given that Compounds A and B have similar shapes and molecular

weights, which would you expect to have the highest boiling point?

Compound A / Compound B  

(circle one)

(vi)      (1 mark) Briefly justify your answer to part (v)

Compound B contains a polar functional group, meaning that it forms hydrogen bonds and/or dipole-dipole interactions (happy with either for answer).

(vii)     (1 mark) Give the systematic name for Compound C

butan-2-one

______________________________________

(b)   Questions (i) – (v) refer to the reaction scheme below.

(i)   (3 marks) Complete the reaction scheme by providing the missing reagents and structures in the boxes provided.

(ii)   (1 mark) Reaction 1 is a/an          addition / elimination / substitution / oxidation   

(circle one)

(iii)  (1 mark) Reaction 2 is a/an          addition / elimination / substitution / oxidation

(circle one)

(iv)  (1 mark) Reactions 3 and 4 are both examples of the same reaction class.

Give the name of this reaction type:

Acyl substitution

________________________

(v)   (1 mark) The functional group in Compound E is called a:

Carboxylic Acid

___________________

QUESTION 2:  (17 marks TOTAL)

One of the reactions in organic chemistry is oxidising methanol to methanoic acid using potassium dichromate as the oxidising agent. The full equation for this reaction is given below.

3CH3OH(aq) + 2K2Cr2O7(aq) + 8H2 SO4(aq) → 3HCOOH(aq) + 11H2O(aq) + 2Cr2(SO4)3(aq) + 2K2 SO4(aq)

(a)       (2 marks) Show that the molar mass of K2Cr2O7 is 294.18 g mol- 1 .

= 2 × 39.09 + 2 × 52.00 + 7 × 16 = 294.18 g mol-1

(b)       (2 marks) Calculate the amount in moles in 19.0 g of K2Cr2O7 .

n =  19.0 g ÷  294.18 g mol-1 = 6.459 × 10-2 mol

(c)       (3 marks) Calculate the mass of CH3OH that can be oxidised by 19.0 g of K2Cr2O7 .

6.459 × 10−2 mOl

2

m(CH3OH) = 0.0969 mol × 32.042= 3.10 g

(d)        Sulfuric acid is necessary for K2Cr2O7 to act as a strong oxidising agent.

(i)     (2 marks) Calculate the amount in moles of H2 SO4 that need to be used in this experiment when 19.0 g of K2Cr2O7 is used.

n(H2SO4) =                      × 8 = 0. 258 mOl

(ii)    (1 mark) If the amount in moles of H2 SO4 you calculated above is dissolved in

150 mL of water in this experiment. Calculate the concentraion of this H2 SO4 solution.

C = 0.258 mol / 0.150 L = 1.72 mol L-1

(e)      (3 marks) Suppose 325 mL of 0.150 mol L- 1 NaOH is needed for your experiment.

How would you prepare this if all that is available is a 1.01 mol L- 1 NaOH solution?

n(NaOH needed = 0.150 mol L-1 × 0.325 L = 0.04875 mol

volume of NaOH needed from 1.01 mol L-1 solution =  0.04875 mol / 1.01 mol L-1

= 0.04827 L = 48.27 mL and add water until the volume is 325 mL

QUESTION 2 [Continued]

(f)       Sodium carbonate, Na2CO3, commonly known as washing soda is a key component of

laundry soaps.A student dissolved some washing soda in 250 mL volumetric flask and made to the mark with deionised water. 25.0 mL of this solution was titrated with         0.542 mol L- 1 HCl solution. It required 12.55 mL of HCl to reach the end point of the  titration.

Na2CO3 (aq)  + 2HCl(aq)   →  2NaCl (aq) + CO2(g) + H2O(l)

(i)        (1 mark) Calculate the amount in moles of HCl used in this titration.

= 0.542 mol L_1   ×  12.55  × 10-3 L = 6.80 × 10-3 mol

(ii)       (2 marks) Calculate the amount in moles of Na2CO3 in 25.0 mL sample.

n(Na2CO3 ) =   =  3. 40  ×  10 −3 mOl

(iii)      (1 mark) Calculate the concentration of Na2CO3 in the solution.

C(Na2CO3 ) =                        =  0. 136 mOl L−1

QUESTION 3:  (10 marks TOTAL)

(i)        (3 marks) Use the information given above to calculate rHo for the reaction.

 rH。= [20 × (-393.5) + 16 × (-285.8) – (-636)] kJ mol-1

= [-7870 – 4572.8 + 636] kJ mol-1

= - 11806.8 kJ mol-1

(ii)       (1 mark) If the arachidonic acid is combusted in a calorimeter would the

temperature of the calorimeter stays the same, increase or decrease?

Justify your answer

(iii)      (2 marks)  If the reaction above produced H2O(g) rather than H2O(l), rHo would

be:

less  negative            unchanged            more negative        (Circle one)

Justify your answer.

(iv)      (2 marks)  Using your answer for (a)(i) above, calculate the mass of arachidonic

acid, C20H32O2, needed to warm a 400 kg bear from 5oC to 25oC. Assume that the average specific heat of bear flesh is 4.18 J g- 1 。C- 1. M(C20H32O2) = 304 g mol- 1 .

q = - 400 × 1000 g × 4.18 J g-1 。C-1 × (25 -5)。C = 3.344 × 107 J = 3.344 × 104kJ n = - 3.344 × 104 kJ / - 11806.8 kJ mol-1 = 2.832 mol

mass =  2.832 mol × 304 g mol-1 =860.93 g = 861 g

QUESTION 3 [Continued]

(2 marks)  Calculate rHo for Reaction 1.

S(s)    +    3/2O2(g)           SO3(g)  SO3(g          SO2(g) )    +   ½ O2(g) S(s)    +    O2(g)         SO2(g)

Ho   =  -790/2 = -395 kJ mol-1             Ho   =  196/2 = 98 kJ mol-1                  Ho   =    -395 + 98 = - 297 kJ mol-1

QUESTION 4:  (10 marks TOTAL)

(a)       H2 S is a flammable gas with a characteristic odour of rotten eggs. It burns in oxygen to

form SO2 and water.

2H2 S (g)  + 3O2(g)  → 2SO2(g) + 2H2O(g)

In one experiment SO2 gas is formed at a rate of 0.36 mol L- 1s- 1 .

(i)        (1 mark) Calculate the rate of consumption of H2 S in this Experiment.

0.36 mol L-1s-1

(ii)       (2 marks) Calculate the rate of consumption of O2 .

Rate of conSumption of 02  =    × 3 = 0.54 mol L−1S −1

(iii)      (2 mark) What is the overall rate of this reaction.

0.36mol L-1 s-1/ 2 or 0.53 mol L-1 s-1 / 3 = 0.18 mol L-1 s-1

(iv)      The rate law for the above reaction is determined to be

Rate = k[O2]2 [H2 S]2

a.        (2 mark) What happens to the rate of reaction if the concentration of O2 is doubled and concentration of H2 S is halved? Justify your answer.

b.      (3 marks) Calculate the rate of this reaction when the concentrations of H2 S and O2 are 0.063 mol L- 1 and 0.053 mol L- 1, respectively, and the  rate constant is 2.02 × 102 mol-3 L3 min- 1 . Show your working.

rate = k[H2S]2[O2]2

= 2.02 × 102 mol-3 L3 min-1 × [0.063 mol L-1]2[0.053 mol L-1]2 = 2.25 × 10-3 mol L-1 min-1