How hard can Vika own?

...


Vika did the P4 Chemistry test while juggling 5 chickens with his bare hands, calculating the value of π to 67 decimal places and writing a literature script on Alexander The Great. When he had finished, he got a couple of hours left and decided to redo the test in 9 different languages, saving the polar bear populations in the North Pole, and playing 8 Guitar Hero songs with the monitor facing away from him and scoring perfect on each songs. In the end, when everyone has finished, Vika ended up finishing 3 bestselling books on Chemistry that the Cambridge professors actually bought for reference to mark the test and improve the world's study programs. Not to mention that he was blindfolded and tied onto his chair with an Osmium chain for the whole time.

Damn you Vika.


Note:
-I'm obviously not Vika.
-This is the first post in the month of June 2009,..posted this right after I arrived at home on the day we did Chem. P4.
-Osmium is the heaviest metallic element.

Chemistry, Again, Homework Help.

Well, we were on hiatus.

I will give you some idea on how to design an experiment to determine the hydration of sodium carbonate (find the X in Na2CO3.XH2O).

Basically, I propose two methods of doing it:

1. Make a solution of known concentration and do a titration.
2. Heat the solid chemicals.

Method No. 1 is well known and comprehensible but requires more detailed explanation and precautions.
Method No. 2 is easier to write and describe, in terms of experiment and precautions. However, this method may require us to know a certain terms and apparatus.

I'll explain the first method first.

The first method is fairly simple. You just need to dissolve a known amount (in gram) sodium carbonate in distilled water and commence a titration with hydrochloric acid.

Since in titration, hydrochloric acid will only react with sodium carbonate, hence by finding the concentration of sodium carbonate reacted (in g/dm3), we can determine the mass of the water in the sodium carbonate, hence the hydration number (or whatever it is called).

The second method works exactly like our Paper 3 Question Number 2. Here, you need to heat with Bunsen Burner a known amount of hydrated sodium carbonate. This way, the water inside the sodium carbonate will be evaporated. The decrease in mass will signify the mass of water evaporated and hence we can find how much water originally in the crystal by comparing the mass of sodium carbonate with the water inside it.

Some apparatus required for these experiments:

• Test tubes, as the container for sodium carbonate
• Pipette, to transfer exactly 25 mL of solution of sodium carbonate for method 1
• Burette, to do the titration.
• Beaker, to contain the result 25 mL of solution of sodium carbonate.
• White tile
• Bunsen Burner, for method 2
  
• Boiling tube, to contain the sodium carbonate in method 2
• Anything to hold boiling tube while it is heated
• Weighing machine, to measure the mass of materials in method 2
• Stirrer and funnel to transfer the chemicals carefully.

I would say that is all for apparatus. Any extra advice would be appreciated.

Now for the chemicals:

• Hydrochloric acid, 0.125 mol/dm3, 250-400 mL
• Aqua destilata, a cool name for distilled water
• Sodium carbonate, of course. It must be pure.
• Methane or ethane, to fuel our bunsen burner.
• Indicator, any will do I think.

That is all for chemicals. Any extra addition will be deeply appreciated.

For the method No. 1, we just need to write what we usually do plus add some precautions and details, such as add X drops of indicator and indicate when we need to stop. For the table, just copy from any old past paper.

For method number 2, what I can say is that we need to continuously reheat and cool it down. We need to do this until the mass is finally constant, that is when all water is evaporated.

Suggestion for table.
Method 1:

Method 2:


So that is all I can help you. Good luck and if you have any problem you can approach Wind, Ms. Ine Medyawati or Kawapada.

Thank you.






(As a side note, if you can check the freeexampapers.com, you can see the paper 3 O/N 2001 and 2003 to get more reference)

Tips For Chemistry P1

Here we are, approaching the last test for Chemistry AS and also happen to be the most difficult of all the three.

• Refer back to the very first post in this blog. Those tricks are specially useful for the last 10 questions.
• Master your stoichiometry well. I repeat. Master your stoichiometry.
• Standard enthalpy change of formation: it only forms one mole. 2H2+O2-> 2H2O is not one mole. H2+0.5O2->H2O is one mole. (This sentence may contains a lot of grammatical error.)
• Iodine is solid at r.t.p and bromine is liquid at r.t.p

These two structures are two different chemicals. Sorry for the horrible drawing. The point is the position of the -OH group. (Important in chirality)

• C6H5 is a phenyl group and they are nonreactive.
• Either you find the odd one out or find the most correct choice of answer.
• When you see a double bond in an organic compound, do not easily mark it as an alkene. It can be either carboxylic acid/carbonyl (aldehyde and ketone)/ester.
• If the question starts with a general fact, please skip them and go directly to the question. If you do need to read them, please do not treat it like the "Did You Know" section in magazines. This is an exam.
• Haber process is exothermic, in case you forgot.
• Precipitate is still an ion. Yes it is a solid but it doesn't it need to be a pure element?
• Whatever it is, Group I base and alkali is stronger than the Group II base and alkali, provided that they are in the same period.
• Remember all of the bond angle. One method to memorize them is to remember two numbers: 109.5 and 2.5 . Every addition of lone pair, just subtract 109.5 by 2.5 and repeat it for further addition of lone pair.
• All alkane bond angle is roughly 109.5 (degree). Molecules that have a double bond will have a bond angle of 120 (degree). Molecules that have one/two lone pairs follows the trend of either nitrogen/oxygen.
• If the questions have a mathematic calculation (except stoichiometry), unless you are confident with your math. skill, skip them first.
• And again, Group I or II oxides will not dissociate into its element and oxygen gas with only using Bunsen burner.
• Enthalpy change=Bond making-bond breaking
• Pray a lot for this test. I may not be a religious person but at least faith will give us hope. And hope will give us something to cling on to success.
  

Chemistry Paper 2: Blind Spot Part 3

I'll skip nucleophile and electrophile with inorganic chemistry first because of the higher demand of the latter chapter.

For convenience, I will use point form in this section.


Since the Periodic Table tool is malfunctioning, we will use the one on top of this section.

The Basic

• Basically (this word is getting redundant), all of this chapter is about the electrons and protons. That's it!
• This is what I always assume. Imagine an element which possess 7 protons, 7 neutrons and 7 electrons. Then let's say I magically add 1 proton, 1 neutron and 1 electron (nothing less). This changes will affect their chemical and physical characteristics, hence its uses and chemical reactions . But we know that this kind of thing is virtually impossible. Just to give an example on how electrons and protons (and neutrons) is the crucial point in inorganic chemistry.
• In this part, we need to understand first the effect of the electron, especially their position at the outermost shell. As we all know, all chemical reactions depend on the transfer of electron.
  
• Like we all know, as the proton number increases, the number of electron increases.
• As the number of electron increases, hence the number of shell and orbital will also increase, as the electron need a space to move right?
• So as the number of shell increases, wouldn't the size of the atom will increase?
• So as the size of the atom increases, wouldn't the atomic radius increases?

Okay, grasp this concept first before we continue.

The Important Point

• Get back to the important point. So when atomic radius increases, the distance of its valence electrons from its nucleus will increase.
• Hence its attraction will decrease as the distance gets larger, right? (Important point)
  
• When the distance gets larger, the ionization energy will decrease as there is less attraction so less energy is required to remove them, right?
• Like we discussed before, as the proton number increases, so does electron and their shell. Down the group, they all have more proton numbers so their electron will increase too and so their shells and their atomic radius.
• Factors that affects ionization energy is not only their atomic radius. Size of their nucleus (their positive charge) and the shielding effect of electron will also play.
  
• So for group I-III, they will become more reactive down the group because it is easier for them to remove their outermost electrons down the group. However, the non metal which belongs to group V-VII accepts electron rather than donating them. To accept electron, they rely on their attractive force of their nucleus. This is another important point. Down the group, their reactivity will decrease as the distance between the nucleus and the outermost electron will become larger and hence their attraction will decrease.
• Another note for ionic radii. A lot of people often have misconception about this. You may notice in your textbook that, unlike atomic radius, ionic radius of elements have uneven trend. This is the reason. For positive ions, they will remove their outermost electron to form a positive ion. When we remove the electron, we will also remove its shell/orbital hence we will also reduce its size and atomic radius. The opposite things happen for negative ions. As they receive electrons, they will need to create shell/orbital to place their newfound electron. When this happen, the size of the atom will get larger and ionic radius will become larger. (Assuming that all elements will form cation, their ionic radii will decrease throughout the period).
• Electronegativity of elements decrease down the group. The reason is because of their attraction gets weaker as atom gets larger.
• Ionization energy is uneven across a period. This is due the fact that electron occupies different orbital and there is repulsion effect of electron when they occupy the same orbital.
  
• Take period 3 as an example. Sodium electron configuration is [Ne] 3s1, while magnesium's is [Ne] 3s2. Aluminium electronic configuration is [Ne] 3s2 3p1. The p orbital is located farther than the s orbital and it is at higher energy level (less energy required to remove electron at higher energy level). Use this logic to understand why first ionisation energy of aluminium is lower than magnesium, the second ionisation energy of aluminium is higher than magnesium and the third ionisation energy of magnesium is higher than aluminium.
  
• Again we will use period 3 as an example. Phosphorus electronic configuration is [Ne] 3s2 3p3, while sulphur's is [Ne] 3s2 3p4. Refer back to "Chemistry Paper 2: Blind Spot Part 2". The extra electron in sulphur is located at the px orbital. Inside the px orbital in sulphur, there are two electrons whilst in phosphorus there are only one electron. The two electrons will create a repulsion effect and it will decreases the ionisation energy. Hence phosphorus ionisation energy is higher than sulphur.

Group II

• Group II sulphates had its solubility decreases down the group. Oxide, on the other hand, becomes more soluble down the group.
  
• All of Group II nitrates are soluble and all of its carbonate are insoluble.
• Group II thermal stabilities increases down the group which means down the group, the metal carbonates and nitrates will be more stable towards heating down the group. Memorize all the three of them fully.
• Yes I know that Group II oxides are able to dissociate. But only at high temperature. What do I mean by high is 3000 C to 4500C.
  

Group VII

• There is a lot to remember. Firstly, down the group, Group VII reactivity decreases and its colour becomes darker. Their extent and ease of reaction will also decrease means that when it reacts with something, the yield will not be maximum (if it is supposed to form hydrogen halide, the result will be a mixture of hydrogen halide and the halogen)
• Hydrogen halides bond energy will decrease down the group. Hence it will become thermally less stable down the group and their reactivity will increases. Be careful on this part. Hydrides and elements are two different things. Many people (including myself) made mistakes when the questions ask about their states of matter (Iodine is solid, bromine is liquid at r.t.p) result after reaction (when you put a hot iron metal inside compound, example) and their reactivity with acids or salt (NaX).
• Each halides ion can be oxidised by the halogen above it. Example, I-(aq) can be oxidised to Br2 or Cl2.
• About oxidation, down the group it is easier to oxidise halide ions to halide element.
  

Thank you for reading this post. I hope that it will be useful for you guys.

Chemistry Paper 2: Blind Spot Part 2

Next is the ever elusive atomic structure and chemical bonding. The points that we often missed about these topics are the shape of orbital, shape of sigma and pi bond, what is instantaneous dipole or permanent dipole.

Orbital is the region where we can find the electron most of the time. Remember that electron moves randomly at a specific area? Orbital is simply the part of that specific area where we it has the highest probability of having the electron.

For now we only need to know the shape of s orbital and the p orbital.



Picture of s orbital



Picture of p orbital which consists of px, py and pz orbital. Note their orientation.

The difference of 1s orbital with 2s orbital is their size. 1s orbital is smaller than 2s orbital and the same goes for p orbital. It is just about their size.

When we draw the electron in the box diagram, take note of the placement of the arrows and their directions.


Move on to the chemical bonding. Important points here are the shape and how to draw sigma and pi bond. Another important point is the understanding between the difference of temporary dipole-dipole and permanent dipole-dipole.


This is the picture of s orbital overlapping another s orbital forming sigma bond (example H-H)



This is the picture of s orbital overlapping p orbital forming sigma bond (example H-Cl)



This is the picture of p orbital overlapping with another p orbital forming sigma bond (example Cl-Cl)

We can also draw a "sausage" (ellipse) for the px px sigma bond.

Pi bond is formed when two p orbitals overlap sideways to produce regions of electron density above and below the axis joining the two nuclear centers. Pi bond is formed when a species forms double bond or triple bond.



This is an example of py orbital forming a pi bond. Notice the shape of the orbital after the pi bond is formed.

Move on to instantaneous dipole or permanent dipole. This is one of three known intermolecular bonds that we have learned (I have absolutely no idea how many are there).

So basically both of these things are due to the effect of electrostatic attraction of the dipole that is created due to 2 different things:
(-) Random movement of electrons for the temporary/instantaneous dipole.

(-) The effect of polarization caused when an element with higher electronegativity bonded with another element of lower electronegativity for the permanent dipole.

The fact that movement of electron is extremely random will sometimes cause the electron to be positioned in such way:


Here we can notice at the second picture had all of its electron positioned at one side of the atom. Then, a millisecond later it would disperse right away. Then a few milliseconds later it would be positioned as the second picture and so on an on (remember, it is moving at speed of light).

When the electrons are positioned as in the second picture, the ion will be polarized for a while and it will produce a dipole with negative side at the left side of the atom (where the electrons gather). So the left side will be positively charged for a while isn't it?


This is what we called as the temporary dipole-dipole/ instantaneous dipole-dipole/ van der Waals force.

This bond is important only when all other bonds are not present, permanent dipole dipole and hydrogen bond. This is important in alkane and monoatomic element such as noble gases.

Take a look at HCl :



Here, we can see that Cl atom, being much electronegative than H, will polarize H and will cause it to become slightly more positive, whereas Cl will become slightly more negative. Thus when there are two HCl molecules come into contact:


The interactions (the vertical line) is the result of the attraction which is the permanent dipole. Basically, if it is a polar molecule, it will have the permanent dipole bond as the stronger intermolecular force.

Hydrogen bond is a special case. As defined in Wikipedia, hydrogen bond results from a permanent dipole force between hydrogen atom(s) bonded to nitrogen , oxygen or fluorine (thus the name "hydrogen bond", which must not be confused with a covalent bond to hydrogen). It is much stronger than the the permanent dipole force since the polarization here is much stronger (electronegativity of F is greater than Cl).

Consider this. Ethanoic acid (vinegar) is able to form a hydrogen bond where as ether such as methoxymethane (CH3OCH3) can't form any hydrogen bond.

Any question, feel free to ask.
All comments will be deeply regarded.

Thank you for reading this.

Chemistry Paper 2: Blind Spot Part 1

The title is quite aggravating.

So a few more days and we are going to have our P2 chemistry. Though not as difficult as its P1, still it poses a lot of challenge, especially if we want score higher.

Some points to remember is that the questions here are divided into several parts. What I mean is that each questions will usually correspond with one of the major topics in chemistry. So for a better mark, always skip the part that you feel incapable (example: organic chemistry) to do and search through the part that you are confident with (example: thermodynamic), don't waste time doing it linearly. Strange thing is that organic chemistry always seems to appear while the other parts comes out randomly (serious). So a good strategy is that to master your organic chemistry.

I bold "usually" with a reason. The setters may elaborate two topics in one question (periodic table and chemical bonding) or perhaps split one topic in two questions (isomerism and organic chemistry reactions). Study all parts, no matter what.

Some topics that resurges back are the electrolysis, atomic structure and chemical bonding.

Inorganic chemistry had become a frequent topic lately (periodic table, group II, group VII).

The topic that keeps freaking people is the nucleophile and electrophile; especially their reactions and how to differentiate them. Their mechanism, fortunately, is easily memorized once we know which is which.

When asked about electrolysis, the most common format of the question is:
(-) Draw a diagram and describe the electrolysis of bla, bla, blah
(-) Reaction at cathode and anode
(-) The products and their uses

To answer the first part, you can only memorize the diagram that we have learned, namely for the extraction of aluminum from aluminum oxide, electrolysis of brine and purification of copper. Note that we do not need to draw sophisticated diagram, a simple one will do.


This is a simple diagram for the extraction of aluminium.


And this is the picture for the electrolysis of brine. Do not forget to draw the diaphragm!

To determine the reaction at cathode or anode, all we have to do is to determine the cation (positive ion) and anion (negative ion). This is the important point. Cathode is the negative pole whilst anode is the positive pole. Therefore reaction at cathode involves cation (positive ion) while reaction at anode involves anion (negative ion).

The products for each electrolysis are aluminum (extraction of aluminum, duh), sodium hydroxide, chlorine and hydrogen gas (electolysis of brine) and copper (purification of copper). I suppose I don't need to discuss their uses because most textbooks had already discussed them and if you can't find them in textbook, Google is everybody's best friend (or Yahoo maybe).

Some extra tips for P3

Here are some extra tips:

• In case (just in case), the first part of the question is not about titration, don't panic. Usually, if not titration, it will be about rate of reaction. Unlike biology, here we need to use the data that we get in practical and make a graph out of it. Marks will mostly be counted in the quality of the graph, just like Physics. The good (OR bad) things about this kind of practical, instead of calculation and stoichiometry, you will be asked to comment on the experiment or suggest/improve a hypothesis. This is like the fusion of both Biology and Physics. And yes I do realize some of the questions asked has an extremely nerdy instructions (believe me, I took 15 minutes to fully grasp the concept on some questions) but don't worry. Usually for this kind of experiment, you will be asked to vary certain values (concentration/mass) while keeping the other values constant. O/N 2007 P3.2 is a good example (freeexampapers.com) (thx to Kenny Buntara). Here you are asked to determine the rate of reaction H2O2 and we need to vary the concentrations of H2O2. So those long and confusing instructions is just basically tell us on how to vary the concentrations. See, if we limit the volume of H2O2 to be only 40 cm3 and you add 20 cm3 of water and 20 cm3 of H2O2, it's just dilution, nothing special. How to make it more concentrated? Simple. Add less water but more H2O2, but make sure the total volume is still 40 cm3 (10 cm3 of water+30 cm3 of peroxide).
  

• Now about qualitative analysis. On some occasion, we are also required to determine unknown organic compound (holy s***!). But don't worry, on scope of our knowledge, we only need to know how to identify :
  
• Alkane
• Alkene
• Alcohol
• Halogenoalkane
• Aldehyde
• Ketone
• Organic acid (carboxylic acid)
• And we don't need to know what exactly it is (you don't need to know that it is pentane or pentanol, you just need to know whether it is alkane or alcohol)
  
• Now here's the catch. At our current level, it is impossible for us to handle combustible material such as alkane and alkene. Anyway, to identify alkane and alkene is to use liquid bromine and source of UV light. So if the question doesn't show such reagents in the question, you can easily forget about alkane and alkene. Halogenoalkane, to identify this is actually the same as identifying Cl-, Br- or I-. So that will leave us out with Alcohol, Carboxylic acid, Aldehyde and Ketone.
• Start with an easier one first, identifying carbonyl compound. The only way to identify them is to use DNPH (2,4-Dinitrophenylhydrazine) and it will produce this precipitate:

left picture is negative test, right pic. is positve test

• To identify between aldehyde or ketone, we will use Tollen's reagent (silver dissolved in ammonia) or Fehling's reagent (basically it looks like Benedict's reagent). Picture on the book at page 354 shows a good picture of positive Tollen's test.
  

The picture labelled 2 shows the positive result of Fehling's test.

• Take note though. Sometimes if our preparation is a little bit messed up, carboxylic acid will sometimes give positive result for DNPH. I suggest that you clean everything after usage. If the result is still the same, I suggest do the Tollen/Fehling test first to identify the aldehyde.
  
• Now identifying alcohol and carboxylic acid. The simplest way to identify them is to put carbonate/bicarbonate (magnesium/calcium is fine) inside the two unidentified liquid. The one that gives off bubble is the carboxylic acid. It's like adding baking powder with antiseptic alcohol and vinegar (ethanoic acid). The one that will react is the acid. Note that both of them will react with reactive metal (not salt) such as magnesium.
  
• If you are still not assured, usually in the test, they will ask us to make an ester (add carboxylic acid with alcohol, boil then add water). Remember it will gives sweet smelling odour (if you ever play those balloon with straw, then you can imagine that smell).
• And remember to put all of your evidence that indicates that those tests shows this compound is "this" or "that". Don't take the risk!
• Last form of the test is usually measuring of enthalpy change. Usually they will ask us to react something and record the temperature. The only tips that I can get here is that you need to be extremely, extremely, d*** accurate! Although the marks are not really high, but their marks for accuracy is really and impossibly strict. Let's say that the examiner get a value of 2.7 and their limit of accuracy is 0.05 while some of our equipments can't measure until that degree of accuracy. Even slightest difference will give us wrong answer, such as finding the average of 2.7 and 2.8. This accuracy limit is also applied during our calculation. So our only hope is to get the measurement right. Fortunately, their marks are not really high and they are quite fair, such that they can miraculously design the test that it is possible for us to get the marks in one part but not the other.

So this is the end of extra tips. Comments are much appreciated.

Special Posting: Practical Chemistry a.k.a. P3

Oke tinggal sehari atau dua hari lagi buat practical. Menurut gw, nih paper yg paling sulit, soalnya kesalahan dikit aja bisa bikin resultnya ngaco dan amburadul. Such small stupid mistakes can be irrelevant and sometimes unnoticeable. Example, using the same dropper for different unknown substances. Klo misalnya precipitatenya warnany bisa kuning atau merah atau ungu atau putih, I say it is still safe to combine them. Tp klo ada 3 unknown substance, trus ada 2 yg precipitare warnanya putih, good luck.

Usually they will provide two droppers, I suggest you use one for to add the acids and bases (you know, you will need to use droppers to add ammonium hydroxide or HCl) and use the other one to add the unknown substance to empty test tube. Terus siapin satu beaker kosong isinya distilled water buat bersihin droppernya. Jadi abis pake buat satu reagent lu pake droppernya buat ambil air trus buang lah airnya kemana. Minimal bisa buat reduce contamination.

Trus buat calculation. Inget biarpun titrasi lu ngaco, calculation nggk ada accuracy marknya. Jadi, sebisa mungkin, try to get full mark in calculation part. 5 marks mungkin kecil, tapi bisa nyelamatin kita. Klo perlu hapalin semua calculationnya biar lebih confident. Confidence is the key to succeed in this exam. Also in qualitative analysis, sometimes you can actually guess which substance is which even if you haven't finished the whole set of questions. You can actually save time here by writing the possible outcome before we actually test it. Jadi biarpun salah waktu kita coba, tp nulis observasinya bener, kita masih dapet mark.

Klo lu orang lebih confident, boleh juga coba kerjain qualitative nya dahulu. Jawaban di qualitaive lebih exact. Pengecualian kalau titrasinya menggunakan oxidising agent (KMnO4 dan yg lainnya). Klo kelamaan ditinggalin, nanti reagentnya bisa keoxidize duluan, resulting in inaccurate measurement.

Trus klo ada yang ambidextrous atau at least bisa kerja pake 2 tangan, gunakanlah dua tangan anda. Tp klo misalnya lu orang begitu pake 2 tangan langsung tangan kanannya stop bekerja, jangan dicoba.

Klo ada dilution (yg musti dimasukin ke dalam tube aneh terus dikocok), lu orang buka distilled water tubenya, terus tumpahin, terus klo udah level tertentu, baru pake tube kecilnya. Dan klo udah approaching the line, ganti pake dropper biar meniscus readingnya pas di line. Beda sedikit aja bisa deviate the result by a lot. Trus pas di shake, ati2 jgn sampe ada yg tumpah airnya (sedikit gpp), reason same as above.

Jangan lupa tabulate all of the results, even though if you are not asked to. Gw pernah ilang 2 mark gara2 nggk nulis table di bagian awal2. Jangan panik waktu ngerjain, kejadian dulu ada anak ngambil A-level, saking groginya sampe mecahin apparatusnya (nggk tau apparatusnya apa). Trus semua data harus 2 decimal places, klo nggk bisa ilang tuh 1 mark.

Klo bisa, hapalin semua perubahan warna saat titrasi. Klo misalnya diminta titrasi sampe warna pink pake methyl orange, harus bener2 exact warna itu. Always use white tile to let us differentiate the colour better. Klo bener2 kepepet (definisi kepepet: udah coba 3 kali tapi bedanya sampe 5 cm3 atau warnanya lu bener2 nggk tau) (*warning* very risky), cb ngintip temen2 yg lain atau invigilator punya practical. Glancenya of course jgn ketauan, main mata dikit aja. Sebaiknya klo bingung tanya dulu, klo nggk dihiraukan baru.... (anak baek jangan menyontek. Inget ini bener2 risky. Klo ketahuan bisa automatic failure, so jangan main api klo nggk mau kebakar). Trus klo misalnya Ms. Ine liatin lu orang punya practical, try to just ignore her face and body language and expression.

Klo kerjain, tolong-tolong dengan sangat jangan pernah buru-buru kaya dikejar setan atau mau show off. With proper time planning and by realizing your ability, you should be able to allocate your time wisely and judgmentally. No need to rush with reason such as "later I don't have time to do the written work". Plan, plan, plan!

Jangan cheat your result, bcos most probably the examiner will notice. (I don't now actually how to cheat here though)

Trus klo ada pertanyaan yg regarding improve the quality of experiment atau why such experiment is blablabla, jgn panik. The best answer is always the simplest answer but we should not simplify our point. Pikir pake kepala dingin, klo mereka mintany simple modification, jgn pikir yg aneh aneh. Conduct experiment in vacuum is one of the most desperate answer. Liat sekitar lu dan liat apparatus2nya, trus pikir lagi. Klo nggk bisa lompatin aja daripada ngabisin waktu.

Last but not least doa. Biarpun udah disiapin seperfect mungkin, tapi selalu ada uncertainty. By conducting our experiment accurately and precisely, we should reduce those uncertainty by a great amount. But still, it is always good to have faith.

Organic Chemistry (Chiral Center)

My friends asked me to discuss about the organic chemistry especially regarding the last part of the Paper 1 which mostly depicts extremely funny and sometimes complex organic substance (example menthol or even organic acid with fancy names).

Consider this structure as an example:


The IUPAC name for this structure is 3,5-dihydroxy-3-methylpentanoic acid and also called mevalonic acid (and yes it is one of those fancy names).

Upon seeing such structure, the first thing that we will notice directly are the two alcohol groups and one carboxylic acid group. You will also notice one Carbon chiral center.


Most of the questions asked will be about:
(-) Reactions of the functional group, reaction mechanism and required reagent
(-) Nature of functional group and how to identify them
(-) Number of chiral center
(-) Prediction of result of reaction

There can be more questions that can be asked, but mostly those are the stuff they will talk the most.

(+) Let's start about chiral center. First point is that if the carbon has a double bond, it will not be the chiral center. This include alkene, cycloalkene, ketone, aldehyde and benzene.

For cyclic compound, refer to these pictures:


Arrows indicates carbon atom. Picture indicates position of carbon atoms but none of them are chiral center.


Now here comes the interesting stuff. The encircled carbon atom is a chiral but the one pointed with an arrow is not a chiral center. Why? Remember the definition of chiral center. The carbon atom must be attached with 4 different functional group.

Let us magnify the picture.


Sorry for the bad quality. Left picture is the magnified carbon atom which was arrowed, while the right picture is the encircled carbon atom.

Notice at the left picture, the central carbon is attached with 2 CH2. Since it is bonded with two same structure, it will not produce enantiomer (I will explain later, much later), therefore it is not chiral.

Central carbon at the right picture is bonded with 4 different groups, making it the chiral center.

That is the end of part 1, I hope you guys understand and please comment!

Posting Pertama: Perkenalan dan Tips

Atas paksaan dan kemauan, akhirnya saya membuat blog ini. Didedikasikan untuk membantu murid dengan program A-level yg mengalami kesulitan atau ada pertanyaan (mogah-mogahan bisa membantu yg lain, bkn cuma A-level).

N.B. : Due to the limited language skills of the author, some of the posts here will either be in Bahasa, a bilingual post or noticeable spelling and grammatical error.



Some tips for the last 10 questions of paper 1, some of you may know about this but still credits to Albert Lowis and Wesley P.D. for telling me this.

The last 10 questions of paper 1 are still MCQs, but we are given information (true/false info) to choose. Usually we are given 3 information and 4 choices of answers.

A: Information 1, 2, 3 are correct
B: Information 1, 2 are correct
C: Information 2, 3 are correct
D: Only information 1 is correct

Always check information 3 first.

If it is right, check number 1

• If it is right, answer is A
• If it is wrong, answer is C

If it is wrong, check number 2

• If it is right, answer is B
• If it is wrong, answer is D