'A' & 'O' Level Chemistry Qns (A Collection) replied by UltimaOnline @ Tue, 25 Nov 2008 00:36:46 +0800

UltimaOnline — Tue, 25 Nov 2008 00:36:46 +0800

'O' Level Chem.

>>> We encounter questions regarding the reduction of metal oxides by a reducing agent, such as carbon. The "standard" equation goes something like this

Metal oxide + carbon ---(heat)---> metal + carbon dioxide

However, i discovered that there were some reactions such as reduction of zinc oxide and lead(II) oxide which produce CO instead of CO2. I would appreciate if someone could explain. Thanks. <<<

I replied :

Zinc is a relatively (everything's relative... just ask Einstein) reactive metal, which means that it won't be that easy to grab away zinc's girlfriend ("oxygen"). Carbon can grab away oxygen from zinc, but it's harder for carbon monoxide to do so.

Carbon : "Eh gimmie ur oxygen gf, or else I call my gang whack u!"

Zinc Oxide : "Ok ok, I give u face this time... but I'll remember ur face... u be careful..."

Carbon (now in monoxide form, ie. Carbon monoxide) to Zinc (who has lost his oxygen) : "Muhahaha! Ok, now ur brudder's turn... tell your younger didi zinc oxide to give up his oxygen too!"

Zinc & his brudder Zinc Oxide : "Oi! U already got one oxygen liao, don't be too greedy hor! Don't see we all quiet quiet good to bully, think we sick kitten izzit! Knn u nvr die b4 izzit, I tell you..."

Carbon monoxide : "Ok lah! ok lah! Kao peh lah!"

'A' & 'O' Level Chemistry Qns (A Collection) replied by UltimaOnline @ Sat, 15 Nov 2008 14:16:15 +0800

UltimaOnline — Sat, 15 Nov 2008 14:16:15 +0800

'A' Level Qn.

Someone posted :
>>> What is the major organic product of the reaction of 2,3-Dimethyl-1,3-butadiene with Br2? <<<

Solution / Comments :

For the mechanism, the first step is nucleophilic attack resulting in electrophilic addition (where the pi bond acts as the nucleophile); the second step is nucleophilic attack resulting in nucleophilic addition (where the Br- (or OH- if aqueous conditions are used) acts as the nucleophilie).

Four points to consider :

1) The solvent used matters. With a non-polar, inert solvent, you'll get the addition of bromine only. Using aqueous bromine, you have a competing hydroxide ion nucleophile that will attack the cyclic bromonium carbocation intermediate, in effect adding not just Br, but also OH, across the double bonds.

2) If we regard the cyclic bromonium ion as a non-cyclic carbocation, (eg. imagine you're adding H-Br instead of Br-Br), then the more stable carbocation would be the one with more electron donating (non-electronegative-substituted) alkyl groups. If both double bonds are added across simultaneously, and based on this argument alone, you would end up with 1,4-dibromo-2,3-dihydroxy-2,3-dimethyl-butane.

3) However, because oxygen is significantly more electronegative than bromine, the resulting electronic instability of 1,4-dibromo-2,3-dihydroxy-2,3-dimethyl-butane (with the two partial +ve Cs being adjacent), may see the emergence of 2 other slightly more stable isomers (at least in regard to the two partial +ve adjacent Cs), 1,3-dibromo-2,4-dihydroxy-2,3-dimethyl-butane and 2,3-dibromo-1,4-dihydroxy-2,3-dimethyl-butane.

4) Let's consider sterics. Br is huge, but OH is small. But Br has to be added on first, since the pi-bond acts as a nucleophile (and attacks the +ve charged Br). So the next step of OH- nucleophile attacking the carbocation (or bromonium carbocation), might preferentially add on to the terminal C's, to minimize steric repulsion (or van der Waals repulsion), so this argument tends towards the isomer 2,3-dibromo-1,4-dihydroxy-2,3-dimethyl-butane.

The answer (single major product) isn't directly provided by this post, but hopefully these 4 points above might have possibly helped you understand the processes or considerations involved a little better.

'A' & 'O' Level Chemistry Qns (A Collection) replied by UltimaOnline @ Sat, 15 Nov 2008 02:19:47 +0800

UltimaOnline — Sat, 15 Nov 2008 02:19:47 +0800

(drinking chocolate silently...)

'A' & 'O' Level Chemistry Qns (A Collection) replied by UltimaOnline @ Fri, 14 Nov 2008 11:43:44 +0800

UltimaOnline — Fri, 14 Nov 2008 11:43:44 +0800

'A' Level Qn.

Many students, as taught by the JCs, do not understand many aspects of Chemistry, they just blindly memorize. As a teacher and tuition teacher, I always I always encourage students to understand the *meaningfulness* behind anything they're required to memorize. Here, students should appreciate the *purpose* of hybridization, and why the mathematical formula to determine hybridization works.

Consider Ethyne and Ammonia.

1) Ethyne.
To minimize electron repulsion, the VSEPR molecular geometry is linear, and the VSEPR electron geometry is also linear (since there are no lone pairs). To achieve linear VSEPR electron geometry, hybridization of C orbitals must be sp (without hybridization, the bond angles would result in significant electron repulsion and an unstable geometry). This leaves 2 p orbitals (s s p p --> s p p p --> sp sp p p) for overlap (with the other C's 2p orbitals) to form the 2 pi bonds (over the sigma bond; hence triple bond between the 2 Cs).

2) Ammonia.
The molecular VSEPR geometry is trigonal pyramidal, but the electron geometry is tetrahedral. Hence hybridization is sp3 (to achieve tetrahedral electron VSEPR geometry, to minimize electron repulsion).

A related question to this :

Identify and explain whether PH3 is a stronger or weaker nucleophile and base, as compared to NH3.

The solution, is to recognize that P is one electron shell larger than N; consequently the electron repulsion (between electron pairs, regardless lone or bond) is less in PH3 as compared to NH3; consequently there is less need and therefore less extent of hybridization of electron orbitals; consequently the bond angles in PH3 are almost orthogonal as compared to (slightly less than, due to greater repulsion between lone pair and bond pairs) tetrahedral bond angles in NH3; which implies that the lone pair in PH3 is largely the s orbital as compared to sp3 orbital in NH3; consequently the s orbital lone pair of PH3 is less available for nucleophilic attack and/or accepting a proton, as compared to NH3.

'A' & 'O' Level Chemistry Qns (A Collection) replied by UltimaOnline @ Thu, 13 Nov 2008 22:57:09 +0800

UltimaOnline — Thu, 13 Nov 2008 22:57:09 +0800

'A' Level Qn.

This question was first asked by a NYJC dragon boater student of mine, Calvin Seah YY. Due credit goes to him, because it is indeed an excellent question (even if I was the one who figured out the answer... no 'A' level student has so far been able to solve Calvin's Riddle on his/her own, without my help). In fact, it's such a fascinating question that makes for such an excellent challenge, that I won't reveal the answer here.

I'll reveal the answer to Calvin's Riddle only to students who come for my tuition. Heh. Just kidding lah. If you think you know the answer, do NOT post it here, but Private Message me. If you're right, I'll confirm it and say, "Congrats". If you're wrong, I'll tell you, "Keep on persevering, the Esplanade wasn't built in a day."

Calvin's Riddle

(aka Calvin's Cunningly Cryptic Chemistry Conundrum Challenge)

-------------------------------------------------------------

According to the oxidation states of Mn in KMnO4 and Mn2+, the d orbitals of the former should be empty and the latter should be partially filled. So why is the former "purple" and the latter "colourless"?

-------------------------------------------------------------

Solution :

To find out, either join my tuition ('A' & 'O' Level Chemistry & Biology), or figure it out yourself and Private Message me to check your answer. Have fun!

'A' & 'O' Level Chemistry Qns (A Collection) replied by UltimaOnline @ Tue, 11 Nov 2008 19:18:30 +0800

UltimaOnline — Tue, 11 Nov 2008 19:18:30 +0800

'A' level Qn :

Explain how chlorine has a bleaching effect.

Solution :

Did you know...

The bleaching effect of chlorine, is actually due to the hypochlorous acid (formed when chlorine reacts with water), rather than the chlorine itself?

The bleaching effect is really a result of oxidation; and hypochlorous acid is a stronger oxidizing agent than chlorine itself.

Household bleach is the sodium salt of hypochlorous acid, sodium hypochlorite, of which the ClO- ion (hypochlorite ion aka chlorate(I) ion) undergoes disproportionation at higher temperatures to generate Cl- ion (chloride ion) and ClO3- ion (chlorate ion aka chlorate(V) ion) which is an even stronger oxidizing agent (compare the oxidation states of chlorine in Cl2 vs ClO- vs ClO3-).

>>> How Bleach Works

http://chemistry.about.com/od/chemistryfaqs/f/bleach.htm

An oxidizing bleach works by breaking the chemical bonds of a chromophore (part of a molecule that has color). This changes the molecule so that it either has no color or else reflects color outside the visible spectrum.

A reducing bleach works by changing the double bonds of a chromophore into single bonds. This alters the optical properties of the molecule, making it colorless.

In addition to chemicals, energy can disrupt chemical bonds to bleach out color. For example, the high energy photons in sunlight (e.g., ultraviolet rays) can disrupt the bonds in chromophores to decolorize them. <<<

'A' & 'O' Level Chemistry Qns (A Collection) replied by UltimaOnline @ Mon, 10 Nov 2008 09:46:18 +0800

UltimaOnline — Mon, 10 Nov 2008 09:46:18 +0800

'A' level 2008 Qn

The molecules of compond P, C7H15Br, are chiral. On treatment with NaOH(aq), P produces alcohol Q, C7H15OH, which does not react with hot, acidified Na2Cr2O7(aq). The elimination of HBr from compound P produces a mixture of 4 different isomeric alkenes with the formula C7H14, only two of which are geometrical isomers of each other. Suggest the structural formulae of compound P and the 4 alkanes.

Solution :

P is 3-bromo-2,3-dimethyl-pentane.

'A' & 'O' Level Chemistry Qns (A Collection) replied by UltimaOnline @ Sun, 09 Nov 2008 19:05:20 +0800

UltimaOnline — Sun, 09 Nov 2008 19:05:20 +0800

H2 'A' level 2008 paper :

List and explain the following alkyl halides (halogenoalkanes) according to their boiling points :

Alkyl bromide, alkyl iodide, alkyl chloride.

Solution :

There are two opposing factors, which you have to describe and explain in the exam to score full marks.

Electronegativity and hence strength of permanent dipole-dipole interactions would point to (highest boiling point) alkyl chloride, alkyl bromide, alkyl iodide (lowest boiling point).

However, no. of electrons and hence strength of induced dipole-dipole van der Waals interactions would point to (highest boiling point) alkyl iodide, alkyl bromide, alkyl chloride (lowest boiling point).

As it turns out, based on experimental evidence, van der Waals interactions trumps/owns/pwnz/outweighs permanent dipole-dipole interactions (specifically in THIS context of alkyl iodide versus alkyl bromide versus alkyl chloride; because we're talking about a lot of electrons here).

Hence, the correct answer is (highest boiling point) alkyl iodide, alkyl bromide, alkyl chloride (lowest boiling point).

'A' & 'O' Level Chemistry Qns (A Collection) replied by UltimaOnline @ Sat, 08 Nov 2008 23:31:57 +0800

UltimaOnline — Sat, 08 Nov 2008 23:31:57 +0800

'A' Level 2008 Qn.

Qn. In another separate experiment, a new alkane F, C6H14, was produced. When reacted with bromine under ultraviolet light, F produced only two isomeric monobromo compounds G and H, with the formula C6H13Br. Compound H was chiral. Suggest the structures of F, G and H, explaining your reasoning.

Solution :

The only C6H14 alkane that when monobrominated would yield only 2 isomers one of which is a chiral compound, is 2,3-dimethylbutane.

Notice that 2,3-dimethylbutane has a mirror plane of symmetry. Substituting away H on the 4 terminal C atoms (1st, 4th, 2 methyl and 3 methyl carbons) are equivalent. That leaves the 2nd and 3rd C atoms, which are also equivalent thanks to this mirror plane.

To obtain the chiral compound, substitute away H on (any of the 4) terminal carbons. The 2nd (or 3rd) C atom would then be the chiral carbon.

'A' & 'O' Level Chemistry Qns (A Collection) replied by UltimaOnline @ Sun, 02 Nov 2008 12:30:29 +0800

UltimaOnline — Sun, 02 Nov 2008 12:30:29 +0800

'A' levels.

Qn.

Describe the bond enthalpies of the halogen molecules down the periods. Explain any anomaly present in the trend.

Ans.

(Quote Data Booklet values here.)

Down the group, due to increasing number of electron shells, the bond lengths increase and hence bond strengths and bond enthalpies decrease.

An anomaly presents itself in that the bond enthalpy of F-F is significantly lower than the bond enthalpies of Cl-Cl and even Br-Br.

The reason for this anomaly, is that due to the very short bond length of F-F, the two F atoms are brought exceedingly close together, that there is significant electrostatic repulsion between the lone pairs of both F atoms; consequently decreasing the effective bond enthalpy of F-F.

'A' & 'O' Level Chemistry Qns (A Collection) replied by UltimaOnline @ Tue, 28 Oct 2008 23:46:00 +0800

UltimaOnline — Tue, 28 Oct 2008 23:46:00 +0800

Solution to Qn 9.

Q9) When trichloroethanal is heated with aqueous sodium hydroxide, two products are formed in a single reaction pathway. One of the products is trichloromethane. Draw the mechanism for this pathway and hence identify the other product.

Solution :

1) Hydroxide ion nucleophile attacks delta +ve carbonyl carbon.

2) Electron density shifts up; pi-bond pair becomes a lone pair on oxygen.

3) When electron density shifts back down (lone pair becomes pi-bond pair, reforming the carbonyl group), the trichloromethyl group leaves as an anion.

4) The trichloromethyl anion is capable of being a leaving group (ie. it is stable enough to exist on its own if only briefly), only because of the 3 electron withdrawing Cl groups.

5) However, carbanions are still more unstable compared to alkoxide or alkonoate ions, hence the next step is

6) proton transfer (ie. acid-base reaction) from methanoic acid to the carbanion (ie. lone pair on carbanion becomes bond pair with proton (H+); bond pair between proton (H+) and oxygen becomes lone pair on oxygen), forming

7) trichloromethane and sodium methanoate (electrostatic attraction between Na+ and HCOO-; however note that these remain aqueous meaning dissolved/solvated in water; ion-dipole interactions with polar water molecules).

'A' & 'O' Level Chemistry Qns (A Collection) replied by UltimaOnline @ Thu, 23 Oct 2008 11:57:28 +0800

UltimaOnline — Thu, 23 Oct 2008 11:57:28 +0800

'A' level Qn.

hydrogen iodide, hydrogen bromide, hydrogen chloride, hydrogen fluoride.

Arrange (and explain your arrangement) the above 4 hydrogen halides according to

a) their boiling points, from lowest to highest

b) their acidity when dissolved in water, from strongest to weakest

Solution :

The lowest to the highest boiling points :

Hydrogen chloride (-85.1°C) , hydrogen bromide (–66.38°C), hydrogen iodide (–35.36 °C), hydrogen fluoride (19.54 °C).

Hydrogen fluoride has the highest boiling point due to strong hydrogen bonding between molecules (fluorine is the most electronegative element in the periodic table, hence is capable of the strongest hydrogen bonding).

From hydrogen chloride down the Group to hydrogen iodide, the boiling points increase due to the increasing numbers of electrons in the molecules that in turn leads to an increase in the frequency and strength of (induced dipole - induced dipole) van der Waals forces.

The strongest to weakest acids :

HI > HBr > HCl > HF

This is because the bond energies for the hydrogen halides further down the group are lower, so they dissociate more easily (see Data booklet values below). Furthermore, the conjugate base halide ions increase in stability down the Group, due to the negative charge being spread out over a larger ionic radius (directly related to number of electron shells).

Bond energies

H-F 562 kJ/mol

H-Cl 431 kJ/mol

H-Br 366 kJ/mol

H-I 299 kJ/mol

'A' & 'O' Level Chemistry Qns (A Collection) replied by UltimaOnline @ Thu, 23 Oct 2008 06:30:45 +0800

UltimaOnline — Thu, 23 Oct 2008 06:30:45 +0800

'A' level Qn :

How is it that transition metals can function as catalysts?

Solution :

Transition metals and their compounds function as catalysts either because of their capacity for variable oxidation states (hence allowing them to readily transfer electrons; this provides an alternative pathway with lower activation energy (as compared to the uncatalyzed pathway), hence increasing the rate of reaction); or because of their capacity to adsorb other substances on to their surface and activate them in the process (transition metals can bond to a wide variety of ions and molecules, and can have different number of bonds, so the reactant molecules can be held in place while the reaction occurs).

A case of the former would be the redox reaction (catalyzed by Fe²⁺/Fe³⁺ ions) between peroxydisulphate(VI) ions S₂O₈^2-, and iodide ions I-; to generate sulfate(VI) ions SO₄^2- and iodine molecules I₂.

Overall Redox : 2I- + S2O8 2- --> I2 + 2SO4 2-

Step 1 : S2O8 2- + 2e- <---> 2SO4 2-

Reduction potential = +2.01V

Step 2 : Fe3+ + e- <---> Fe2+

Reduction potential = +0.77V

Step 3 : I2 + 2e- <---> 2I-

Reduction potential = +0.54V

The uncatalyzed reaction (steps 1 and 3) is energetically feasible (according to redox potential values); however, the reaction is slow because the negatively charged S2O8 2- and I- anions repel each other.

In the presence of Fe2+/Fe3+ ions, a alternative pathway (also energetically feasible, but) with lower activation energy (since the reactants are oppositely charged ions) occurs :

S2O8 2- + 2Fe2+ --> 2SO4 2- + 2Fe3+

2Fe3+ + 2I- --> I2 + 2Fe2+

A case of the latter would be the hydrogenation reaction between ethene and hydrogen in the presence of a nickel catalyst.

For heterogenous solid metal catalysts in the hydrogenation reaction, the accepted mechanism today is called the Horiuti-Polanyi mechanism.

Binding of the unsaturated bond, and hydrogen dissociation into atomic hydrogen onto the catalyst
Addition of one atom of hydrogen
Addition of the second atom

'A' & 'O' Level Chemistry Qns (A Collection) replied by UltimaOnline @ Tue, 21 Oct 2008 19:32:24 +0800

UltimaOnline — Tue, 21 Oct 2008 19:32:24 +0800

Soln to Q8.

Q8) Using the following information, draw the mechanism for the reduction of an amide to an amine, using LiAlH₄ in dry ether.

[Description : after the hydride ion attacks the δ⁺ carbon and the pi bond electron density gets kicked up to become a lone pair on the oxygen; the oxygen then attacks AlH₃ (again because Al in AlH₃ lacks a full octet), and subsequently becomes a leaving group as the lone pair on the nitrogen forms a pi bond with the δ⁺ carbon; finally, another hydride ion attacks the δ⁺ carbon, and the pi bond electron density shifts back to become a lone pair on the nitrogen.]

Solution :

'A' & 'O' Level Chemistry Qns (A Collection) replied by UltimaOnline @ Mon, 20 Oct 2008 13:08:47 +0800

UltimaOnline — Mon, 20 Oct 2008 13:08:47 +0800

'O' level & 'A' level Qn.

For instance, given that average molar mass of Cl is 35.5, and that there are 2 isotopes, Cl-35 and Cl-37, find the relative frequency of each isotope.

Solution :

(x/100)(35) + ((100-x)/100)(37) = 35.5

x = 75

Hence 75% of all chlorine atoms have a molar mass of 35g, and 25% of all chlorine atoms have a molar mass of 37g.

'A' & 'O' Level Chemistry Qns (A Collection) replied by UltimaOnline @ Mon, 20 Oct 2008 12:20:28 +0800

UltimaOnline — Mon, 20 Oct 2008 12:20:28 +0800

'O' and 'A' level Qn.

Methane was burnt in an incorrectly adjusted burner. The methane was converted into a mixture of carbon dioxide and carbon monoxide in the ratio of 99:1, together with water vapour. What will be the volume of oxygen consumed when 'y' dm3 of methane is burnt?

Solution :

Write and balance the equations :

Carbon dioxide
CH4 + 2O2 --> CO2 + 2H2O

Carbon monoxide
CH4 + 3/2 O2 --> CO + 2H2O

Based on stoichiometry and info given in qn, we find

Carbon dioxide
CH4 + 2O2 --> CO2 + 2H2O
0.99y 2(0.99y) 0.99y (don't care)

Carbon monoxide
CH4 + 3/2 O2 --> CO + 2H2O
0.01y 3/2(0.01y) 0.01y (don't care)

Total vol. of oxygen required =
vol of O2 needed for 1st equation + vol of O2 needed for 2nd equation
= 2(0.99y) + 3/2(0.01y)

'A' & 'O' Level Chemistry Qns (A Collection) replied by UltimaOnline @ Mon, 20 Oct 2008 11:51:29 +0800

UltimaOnline — Mon, 20 Oct 2008 11:51:29 +0800

'O' level and 'A' level Qn.

Draw the Dot-&-Cross diagram of Urea (Chemistry 'O' Level Qn), given that urea is CO(NH2)2.

Solution :

(Diagram above hosted by http://upload.wikimedia.org/wikipedia/commons/c/c0/Urea.png )

So your diagram should have :

between O and C, double bond, so "xoxo"

between C and each N, single bond, so "xo"

between N and each H, single bond, so "xo"

make sure your O has two lone pairs, ie. "xx" and "xx" (not bonded with C)

make sure your N has one lone pair, ie. "xx" (not bonded with H or C)

check that total no. of valence electrons (both bond pairs and lone pairs) = 12 pairs = 24 valence electrons

'A' & 'O' Level Chemistry Qns (A Collection) replied by UltimaOnline @ Sun, 19 Oct 2008 23:36:55 +0800

UltimaOnline — Sun, 19 Oct 2008 23:36:55 +0800

'O' Level Qn.

Question :

A sample of nitrogen gas contains the same number of atoms as found in 4.0g of methane gas. What is the mass of the sample of nitrogen gas?

Answer :

(2)(x/28) = (4/16)(5)

x = 17.5g

'A' & 'O' Level Chemistry Qns (A Collection) replied by ohyeah193720 @ Sun, 19 Oct 2008 18:54:52 +0800

ohyeah193720 — Sun, 19 Oct 2008 18:54:52 +0800

yup i got the answer(i got around 73.7% in 3 sig. fig.) thx a lot

'A' & 'O' Level Chemistry Qns (A Collection) replied by Garrick_3658 @ Sun, 19 Oct 2008 18:15:48 +0800

Garrick_3658 — Sun, 19 Oct 2008 18:15:48 +0800

kns why all of you last minute one. like me :P

http://sgforums.com/forums/2297/topics/333447

'A' & 'O' Level Chemistry Qns (A Collection) replied by ohyeah193720 @ Sun, 19 Oct 2008 17:57:52 +0800

ohyeah193720 — Sun, 19 Oct 2008 17:57:52 +0800

hello.

about qn :

3) A mixture of MgSO₄.7H₂O and CuSO₄.5H₂O is heated until a mixture of the anhydrous salts, is obtained. If 5.0g of the hydrated mixture when heated gives 3.0g of the anhydrous salts, calculate the % by mass of CuSO₄.5H₂O in the initial hydrated mixture.

Ans : 73.9%

how did you get this answer?i cannot get this answer.(instead i got the % of MgSO₄.7H₂O greater than that of CuSO₄.5H₂O )

'A' & 'O' Level Chemistry Qns (A Collection) replied by UltimaOnline @ Sun, 19 Oct 2008 16:09:33 +0800

UltimaOnline — Sun, 19 Oct 2008 16:09:33 +0800

Solution to Q7.

Q7) Whenever the quesiton states "using hot, concentrated, acidified KMnO4", what functional groups must you watch out for?

1) Acid-base (proton transfer) reaction.

Eg. a N atom with 3 bond pairs has a lone pair which can accept a proton (to form NH4+).

2) Acid hydrolysis.

Eg. Groups such as nitriles, amides, esters, etc.

3) Oxidation.

While K2Cr2O7 can only oxidize primary and secondary alcohols, and aldehydes; note that KMnO4 can also oxidize side-chains of benzene rings to benzoic acid; oxidative cleavage of alkene double bonds, etc.

'A' & 'O' Level Chemistry Qns (A Collection) replied by UltimaOnline @ Sun, 19 Oct 2008 14:31:54 +0800

UltimaOnline — Sun, 19 Oct 2008 14:31:54 +0800

'O' level & 'A' level Qn.

Qn : What is the difference between "peptide" and "amide"? Is there a difference between "polypeptide" and "polyamide"?

Ans : "polypeptide" equals "polyamide", but "peptide" not equals "amide".
The peptide bond refers specifically to the C-N bond within the amide group (or amide linkage). So a macromolecule (eg. protein, nylon, etc) with lots of amide groups (ie. a "polyamide") hence also has lots of peptide bonds (ie. a "polypeptide").

'A' & 'O' Level Chemistry Qns (A Collection) replied by BadzMaro @ Sat, 18 Oct 2008 07:11:12 +0800

BadzMaro — Sat, 18 Oct 2008 07:11:12 +0800

No.. NOT CHEMISTRY !! NOOO!!!!

'A' & 'O' Level Chemistry Qns (A Collection) replied by UltimaOnline @ Sat, 18 Oct 2008 03:40:56 +0800

UltimaOnline — Sat, 18 Oct 2008 03:40:56 +0800

Originally posted by tinuviel07:

hi ultima, any recommendations where you get your resources and references from?

Hi TinTin,

Are you asking for general resource recommendations for your own tuition work (eg. for 'A' and 'O' level Chemistry)? Or are you asking specifically about the resources I'm using (eg. perhaps possibly because my materials/questions may seem somewhat unusual or "off the beaten track/path")?

If the former, then the market for 'A' level (H2/H1) and 'O' level sciences are already flooded and spoilt for choice; just head down to Bras Basar complex (Popular bookstore for sanctioned/authorized publications; other bookstores for (unsanctioned/unauthorized) JCs and Sec Schs' Prelim papers).

Choose publications that you can work well with (browse through their notes, questions and answers and see how compatible this book is with yourself, how much you agree with their answers, how useful you feel the books are, etc).

If the latter, then it's probably because I've always felt (for the purpose of the greater good for mankind, for the Universe) the importance of going beyond convention, dogma, orthodoxy, conforminity; if any significant breakthrough towards higher ground and higher gain is hoped to be made.

Specifically in the education context, being keenly aware (being an ex-MOE teacher) of the limitations of the education system, specifically that in their forced entry into the rat race of paper chase qualifications, students in Singapore don't get to enjoy what they study (particularly tragic for those students who actually would like to be able to deepen their understanding and appreciation of their chosen subjects). And part of this is because (and while I do not fault this or claim to offer any feasible alternatives, because such is necessary based on practical considerations; I certainly would want to do my part to work beyond such limitations, for the sake of students) the MOE/SEAB 'A' level and 'O' level curriculum does not allow (primarily due to time constraints, as well as a so-called 'results oriented' culture; the irony of it) students to truly understand, truly appreciate, truly enjoy their subjects.

Back to the question (if taken to be the latter interpretation), again while I don't mean to deliberately evade the answer, I think it not necessary to specifically name names, publications, online resources, etc. Suffice to mention that I obtain a variety of materials I deem to be useful and helpful for my (tuition) students, both from excellent sources on the internet, as well as publications both local and overseas. The local sources, as I've mentioned, may be obtained from Bras Basar (Popular and other bookstores). The overseas sources, include (multiple copies of) University level (Chemistry and Biology) textbooks (by multiple authors) that were obtained from Amazon.com, Kinokuniya, Clementi Book Store, NUS bookshops, NTU bookshops, etc. All in all, I use metal filing cabinets (the ones used by Mindef and MOE) to house my own private academic library of materials, textbooks and resources, which my (tuition) students may freely use and (not-so-freely) borrow. Approximate cost of resources (over the years, in total) : Sg$7,000 to Sg$8,000.

>>> hi ultima, any recommendations where you get your resources and references from? <<<

In closing, I'm sure my reply was not what you were looking for, but I can only say every teacher (including tuition teacher) is unique and will work best with a unique set of materials/resources. And selecting and assembling your own private academic library is really lots of fun.

And as a colleague (being fellow tution teachers), here's looking forward to a great 2009 year ahead (for both students and tutors)!

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'A' & 'O' Level Chemistry Qns (A Collection) replied by UltimaOnline @ Sun, 19 Oct 2008 16:09:33 +0800

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'A' & 'O' Level Chemistry Qns (A Collection) replied by UltimaOnline @ Sat, 18 Oct 2008 03:40:56 +0800