SRJC H2 Chem 2013 Prelim P3 Soln

2013 H2 Chemistry Prelim Examinations Paper 3 Suggested Solutions 1 Phosphorus is the thirteenth element in the world to be discovered and due to this is sometimes referred to as the Devil’s element. The reference is also made due to its versatility in forming chemical species with explosive of toxic natures. (a) One example of a highly reactive phosphorus-containing ion is phosphonium, PH4 +. (i) Write a balanced chemical equation for the reaction between phosphonium iodide and potassium hydroxide to form phosphine, PH3. PH4I + KOH  PH3 + KI + H2O (ii) Hence, suggest a suitable method of measuring the rate of reaction. Volume of gas produced / Pressure increase (iii) Draw a ‘dot-and-cross’ diagram showing the electrons (outer shells only) in the phosphonium ion, and use the VSEPR (valence shell electron pair repulsion) theory to predict its shape. You may find both a written description and a 3-dimensional sketch useful in your answer. There are 4 bond pairs of electrons around the P atom. To minimise repulsion, the 4 electron pairs are directed to the corners of a tetrahedron. Shape of PH4 +: tetrahedral (iv) The bond angle in ammonia is 107.8o. Predict the H – P – H bond angle in phosphine and justify your answer. [7] H–P–H bond angle: any specific angle between 90o and 107.8o N atom is more electronegative than P atom. The bond pair is pulled closer towards the N atom. Bond pair-bond pair repulsion is greater in NH 3. Hence, the bond angle in PH3 is smaller than that in NH3. FYI: Because PH3 is trigonal pyramidal in shape, the bond angle should also be larger than that in a molecule with a trigonal bipyramidal shape. Hence, bond angle in PH3 is larger than 90o. 1292

(b) Although phosphorus is present in numerous harmful chemicals, it is a useful feature in organophosphorus compounds. Ylides are a class of organophosphorus compounds that are well-known for their role in the Wittig reaction. The synthesis of methylpropene via the Wittig reaction is shown below. P                     PC H 3                      PC H 2 (i) Name the type of reaction that occurs in step I and outline the mechanism for the reaction. SN2 P CH3 Br           C H H H BrP            PC H 3 (ii) Suggest why 2-bromo-2-methylpropane is not favourable to be used as a reactant in step I. Steric hindrance posed by the three phenyl groups on the P atom of P . (iii) State the role of butyl lithium, C4H9Li, in step II. C4H9Li acts as a base. [5] C4H9Li II CH3Br I O III + Br– + – 1293

(c) Two resonance structures exist to stabilise the ylide used in step III above. PC H 2                       PC H 2 In this step, tetrahydrofuran, THF, is used as an aprotic solvent – one that lacks an acidic hydrogen. (i) Explain why, unlike phosphorus, nitrogen cannot form organic compounds similar to ylides. Nitrogen is a period 2 element and does not have available, low-lying d orbitals in its valence shell to accommodate electrons. Hence, it cannot expand beyond the octet structure. (ii) Explain why THF is a suitable solvent to employ in step III whereas water is not. [4] Water is a protic solvent that can release H+ ions to protonate the negatively- charged C atom of the ylide, thus curbing the Wittig reaction. An aprotic solvent like THF prevents neutralisation of the ylide. (d) -carotene is a food colouring that can be extracted from the pigmentation found in red-orange plants and fruits such as carrots. It can be synthesised using excess of an aldehyde and a diylide, a compound with two equivalents of phosphorus in its structure. -carotene Suggest the structures of the aldehyde and a symmetrical diylide with 6 stereoisomers that can be used to produce -carotene. [2] 1294

O   P P The 6 possible stereoismers C-C-C T-T-T T-C-C (Same as C-C-T) T-T-C (Same as C-T-T) T-C-T C-T-C   (e) Compounds X and Y are isomers that can undergo the Wittig reaction with the following ylide. PC H ( C H3) The products formed have the same molecular formula, C7H14, but only the product formed from X displays stereoisomerism. X and Y both do not react with ammonical silver nitrate. Draw the structures of X and Y. [2] X: O     Y: O [Total : 20] 1295

2 (a) Deuterium (symbol D or Hଵ ଶ ) was discovered in 1931. Deuterium accounts for approximately 0.0156% of all the natur ally occurring hydrogen in the oceans, while the most common isotope (hydrogen-1) accounts for more than 99.98%. Chemically, deuterium behaves similarly to ordinary hydrogen. (i) On the same diagram, sketch how a beam of singly positively-charged deuterium ions and a beam of hydrogen ions will behave in an electric field. In your diagram, indicate clearly the angle of deflection for each beam. Diagram should show charged deuterium ion deflecting less than hydrogen ion. Deuterium ion and hydrogen ion must deflect in the same direction. Since angle of deflection  e/m ratio:  of Deuterium ion is half of  of hydrogen ion (ii) Suggest the difference in the melting point and thermal stability of DC l, DBr and DI. For melting point: Number of electrons: DCl < DBr < DI Extent of van der Waals’ forces of attraction: DCl < DBr < DI Energy required to over come the VDW: DCl < DBr < DI Melting point: DCl < DBr < DI For thermal stability: Bond length: DCl < DBr < DI Energy required to overcome the covalent bond: DCl > DBr > DI Thermal stability: DCl > DBr > DI (iii) Chloride and iodide ions are known to react differently with concentrated deuterium sulfate, D2SO4. For the reaction involving iodide, it was observed that the yield of D I was significantly lower as compared to the yield of DCl for the reaction using chloride. It was also noted that a foul rotten egg smell was detected for the reaction involving the iodide. With reference to the Data Booklet and with the aid of equations , explain the above mentioned observations. [10] 1296

From the Data Booklet Cl2 + 2e ⇌ 2Cl - +1.36V I2 + 2e ⇌ 2I- +0.54V Eɵ value of I- is less positive, it is a stronger reducing agent as compared to Cl- thus I- has the ability to reduce the sulfur to a lower oxidation state For chloride: Cl ‐ + D2SO4  DSO4 - + DCl For iodide: (1) I- + D2SO4  DSO4 - + DI (2) 8DI + D2SO4  4I2 + D2S + 4D2O The foul rotten egg smell was due to the release of D2S. Reason for low yield of DI The yield for DI is lower due to further reaction with concentrated D2SO4 Or There are by-products in the reaction between I- and D2SO4 hence affected the yield of DI. (b) Deuterium can replace the normal hydrogen in water molecules to form heavy water, D2O. This difference increases the strength of water’s hydrogen-oxygen bonds which make it more difficult to undergo electrolysis. Some data between light water and heavy water are given below. Property D2O (Heavy water) H2O (Light water) Freezing point (oC) 3.82 0.00 Boiling point (oC) 101.4 100.0 Density at standard condition (g/mL) 1.1056 0.9982 Heavy water and light water can be tested using their freezing points and boiling points. Suggest with reasoning, how a scientist can deploy another physical method to differentiate the two types of water without the use of a temperature measuring device. [2] Freeze the heavy water and light water separately. Drop the heavy water ice cube and light water ice cube into a glass of light water. Heavy water having a higher density than light water will sink in the glass of light water. 1297

(c) During Worl