BedokFunland JC's A Level H2 Chemistry Qns (Part 1)

last SG · 2008-09-19T13:00:14+00:00

'O' & 'A' Level Qns. Qn 1. When 1.01g sample of potassium nitrate(V) is heated above its melting point, oxygen gas is evolved. The mass of the sample dec...

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'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.

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'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. <<<

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^_^

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'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.

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(drinking chocolate silently...)

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'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.

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'O' Level Chem.

anpanman posted :

>>> 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!"

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'O' & 'A' level Qn.

(inspired by a cold rainy morning on 24 Dec 08)

Qn : Why are rainy days so cold?

Ans : 3 reasons.

Firstly, when it rains, there is cloud cover which blocks out the sun (and infra red waves from solar radiation results in heat, or an increase in temperature).

Secondly, when it rains, the rain is often accompanied by strong winds, or in increase in wind velocities. This enhances the evaporating rate, which brings us the the third reason.

Thirdly, during and after a rain, the abundance of liquid water in the environment evaporates (a continuous process) to gaseous water into the atmosphere. This is an endothermic process, as energy (in the form of heat) from the environment is absorbed by the liquid water, in order to overcome the strong hydrogen bonding (for 'O' level students, say "intermolecular forces of attraction") between water molecules in close proximity in the liquid state, so that as the water molecules gain sufficient energy and vibration to overcome the hydrogen bonding attractive forces to move further away (from the rest of the water molecules in the liquid state), it is able to locate itself as an individual particle, ie. it is now in the gaseous state, as water vapour.

This continuous endothermic process of liquid water evaporating draws heat away from the environment, and hence exerts a cooling effect on the environment. (Think of your hand as the thermometer; in an exothermic process such as combustion a.k.a. burning, your hand feels hot, conversely an endothermic process such as water or alcohol on your skin evaporating makes your skin feel cold.)

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'O' & 'A' Level Qn.

Compound A liberates carbon dioxide gas with calcium carbonate, but produces only a small percentage of carbon dioxde gas (when reacted with calcium carbonate) as compared with hydrochloric acid (of the same molarity as an aqueous solution of A). Compound B decolourizes acidified potassium manganate(VII). Both compounds A and B have a molar mass of 74g each, and both compounds contain only 3 elements. When an equimolar mixture of A and B is boiled (together with a few drops of sulfuric acid as catalyst), an equilibrium mixture is obtained, which contains a new compound C, which is sweet smelling. Draw compounds A, B and C.

Answer :
A - CH3CH2COOH

B - CH3CH2CH2CH2OH

C - CH3CH2COOCH2CH2CH2CH3

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'A' Level Qn.

>>> i know that enthalpy change of soln= enthalpy change of hydration - L.E. but how do i represent it in this diagram..? <<<

That's a lousy formula, which most JC students dogmatically memorize without understanding.

A better formula, which makes obvious sense, and one which would immediately self-explain how to draw the Hess Law energetics cycle diagram, would be :

Solution Enthalpy = Lattice Dissociation Enthalpy* + Solvation Enthalpy (when the solvent is water, solvation enthalpy is called "Hydration Enthalpy", for obvious reasons).

Afterall (ie. the reason why this formula is superior and intuitive), the 1st step is the endothermic process of investing energy to overcome the electrostatic attraction between the oppositely charged ions, in order to produce the gaseous cations and anions. Thereafter, the 2nd step is the exothermic process in which ion-dipole interactions or 'bonds' are formed, between the ions and solvent molecules (which are usually polar water molecules). Hence the formula makes perfect sense, and the Hess Law energetics cycle diagram becomes self-explanatory.

(*Lattice Dissociation Enthalpy is endothermic; Lattice (association) Enthalpy is exothermic. Same magnitude, since we're talking about the same process of overcoming-vs-forming a lattice formation, but in reverse to each other.)

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'O' & 'A' Level Qn.

>>> Explain why application of pressure to ice at 0 degree Celsius makes it melt. Why do most other solids not show the same behaviour? <<<

Why ice melts under pressure :

Water in solid state exists as a fixed crystalline lattice structure, thanks to hydrogen bonds between water molecules that result in fixed (and precise) geometries with fixed (and precise) hydrogen bond lengths and intermolecular distances. Putting pressure would disrupt such a fixed crystalline lattice structure (that we perceive as a solid state), and hence the chaotic disorderly structure of water that results, is what we perceive as the liquid state.

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'O' Levels & 'A' Levels

Qualitative Analysis Question

Firstly, I want to discuss with everyone some procedures carried out in this experiment involving Qualitative Analysis. A few questions is in store too.

Objective: Identify element(s) present in solid R and solution S

- the Anion present in S

Solid R contains 2 elements. The metal found in R is also present in solution S

1a Put about half of Sample R into test tube and heat strongly until no further change is observed. Next, wait for residue to cool off.

(my observation:Solid changes from brown to yellowish orange. Gas produced relights glowing splint. oxygen is released.)

2 To new sample R, add small volume of concentrated HCl and warm mixture gently and leave to stand for a few minutes.

(my observation: bubbled produced. Gas produced turns moist blue litmus paper red and is then bleached. Cl2 gas is evolved.)

3a To a portion of solution S , add NaOH until change is seen

(observation: white precipitate formed. )

3b Add NaOH in excess to mixture in (a)

observation: white ppt dissolves in excess NaOH to give colourless solution.

3c To a portion of mixture from (b), add a small piece of aluminium foil and warm gently

observation: bubbles prodiced. Red litmus paper turns blue. Ammonia gas produced and NO3- ion is present

4a To a portion of solution S, add aqueous potassium chromate(VI)

obsevation: solution S turns yellow. Yellow precipitate is produced. Hence S is reducing agent.

4b Pour small portion of mixture from (a) into boiling tube and add dilute HCl until boiling tube is almost 1/3 filled. Leave mixture to stand for a few minutes

observation : yellow precipitate turns white. Redox action has occurred.

What I find strange is that a delivery tube was given to us and not all made use of it. So I assume CO2 gas must have evolved at some point during experiment. Having said that, Qualitative Analysis experiments do not tell us which gas to test for. In this case, the gas to test for for question one is rather unclear. So how then do we know which method for testing should be applied?

And my friend said the elements present in solid R are lead and oxygen, whereas anion in solution S in NO3-. I think the answer regarding solid R's components are wrong. Can someone try to guess what it might be fromthe observations described above? They may not be entirely correct but I hope just reading through the procedure would give you a brief idea.

Please help me solve these queries of mine as I am rather eager to learn about the kind of methods for gas test I should use. I can't perform all the tests because this would take up time and maybe, the gas would have been diffused after some time.

Not forgetting, a big thanks!

Warning : This experiment may be a health hazard, and should not have been given to students.

S appears to be lead(II) nitrate, which when mixed with potassium chromate(VI), produces a yellow precipitate of lead(II) chromate (which is fatally toxic and carcinogenic).

R appears to be lead(IV) oxide (stock name) or lead dioxide (prefix name), PbO2(s), a brown solid, which decomposes to oxygen gas and a yellowish-orange solid Pb3O4(s) which is actually (PbO)2.PbO2(s), a complex of lead(II)oxide and lead(IV)oxide.

Because lead(IV) oxide is a strong oxidizing agent, chloride ions are oxidized to chlorine gas.

Again, this experiment involves fatally toxic and carcinogenic materials, and should not be conducted by students.

In regard to your question on which gas to test for, you have to test for any gas you suspect is present. You'll have to pick up clues from the other tests and their results, and make educated guesses and prioritize your tests accordingly, for maximum efficiency.

Thermal decomposition of the following solids may produce the corresponding gases (note : there are exceptions) :

oxides - oxygen (rarely)

hydroxides - water vapour

carbonates - carbon dioxide

sulfates - sulfur dioxide and oxygen

nitrates - nitrogen dioxide and oxygen

Efferverscence which occurs when you add a solution to a solid, without heating, may indicate hydrogen gas (eg. metal and acid) or carbon dioxide gas (eg. carbonate and acid).

When metal halids are heated with concentrated sulfuric acid (usually too hazardous for students to handle), steamy fumes of hydrogen halides and halogen gases may be produced.

Your explanation helps a lot. May I know in which part of the experiment indicate that Pb2+ ion was present in solution S and solid R?

The cation in question gave a white ppt with NaOH(aq), which dissolved in excess NaOH(aq). Hence, the cation could be either Al3+, Pb2+, or Zn2+.

Of these 3 cations, only Pb2+ qualifies as a cation of a transition metal* (that has partially filled d-orbitals), and hence only Pb2+ forms coloured compounds (eg. the yellow lead(II) chromate ppt).

(*Note : although zinc is a d-block metal, it is not a transition metal. It does not form coloured compounds.)

We Love Qualitative Analysis!

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'A' Level Qn.

is BeCl2 covalent or ionic ? if it is ionic compound with covalent characteristics do i treat it as covalent or ionic for doing mcq?

>>> is BeCl2 covalent or ionic ? <<<

It is covalent, due to the high electronegativity and high charge density (both these properties, relative to the rest of Group II metals). Its high electronegativity prevents complete loss of electrons to form cations which are stable on their own (required for formation of ionic compounds). Its high charge density attracts and causes lone pairs from chlorine to form bond pairs with itself.

As a gas, it is simple covalent. As a solid, it is a polymer macromolecule (with dative covalent bonds between the chlorine and beryllium atoms of adjacent molecules).

One interesting point to note, about beryllium complex ions, is that being in period 2, it does not have empty 3d orbitals with which to expand its octet. Consequently, beryllium complex ions are 4-coordinated, meaning they can at most have 4 bond pairs, eg. [Be(H₂O)₄]²⁺.

This is in contrast with magnesium (also in Group II), which are 6-coordinated, eg. [Mg(H₂O)₆]²⁺.

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'A' Level Qn.

wanna use this thread to ask another question..

what is the shape of CH3CH=CH2?

and double check is the shape of HNO3 trigonal planar? as the H ion is dative bonded to 1 of the O atom hence i should consider only the 3bond pairs of electrons around N atom ?

>>> what is the shape of CH3CH=CH2? <<<

You have to specify, "about which atom?". Look at how many lone pairs and bond pairs each atom has, to determine the geometries within the molecule.

Also, you should be able to distinguish between electron geometry (which tells you the hybridization of the atom) versus molecular geometry (which tells you the shape of the molecule).

For instance, what is the hybridization of N in the ammonia molecule? Since it has 4 electron pairs, the electron geometry is tetrahedral and the N is sp3 hybridized. But since one of the 4 electron pairs is a lone pair, the molecular geometry is trigonal pyramidal.

For CH3CH=CH2

Eg. the 1st C (sp3 hybridized) is tetrahedral, the 2nd C (sp2 hybridized) is trigonal planar, the 3rd C (sp3 hybridized) is trigonal planar.

>>> and double check is the shape of HNO3 trigonal planar? <<<

HNO3 or nitric(V) acid :

+ N = O

H--O

(+ve formal charge on N atom, since it is in Group V but only has 4 bond pairs (maximum octet as it has no empty d-orbitals to expand its octet); -ve formal charge on 1 O atom, since it is in Group VI but it has 3 lone pairs and 1 bond pair)

Conjugate base - NO3- or nitrate(V) ion :

+ N = O

(+ve formal charge on N atom, since it is in Group V but only has 4 bond pairs (maximum octet as it has no empty d-orbitals to expand its octet); -ve formal charge on 2 O atoms, since they are in Group VI but they have 3 lone pairs and 1 bond pair)

The fact is, to draw Kekule structures of a molecule or polyatomic ion correctly, you need to be familiar with the concept of formal charges, which is often neglected by many JC teachers.

Similarly, to have a proper understanding of Organic Chemistry, you need to be familiar with the concept of resonance, which again is often neglected by many JC teachers.

But if you come for my BedokFunland JC tuition, I will teach it to you (all my students), as these concepts (among many others) are fundamental and integral to a proper understanding and appreciation of Chemistry.

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'A' Level Qn.

Which is the better nucleophile? OH- or SH-?

Which is the better base? OH- or SH-?

Explain.

The OH- ion can act as either a base or a nucleophile, depending on the solvent. Explain.

Solution :

Often, a strong nucleophile is also a strong base, and vice-versa. However, occasionally you may encounter a species that is a strong nucleophile but is a weak/moderate base, or a strong base, but is a weak/moderate nucleophile.

Basicity measures stability of a species (a thermodynamic issue), while nucleophilicity measures the rate at which a nucleophile attacks an electrophile (a kinetic issue).

A base is a species that abstracts a proton (ie. H+ ion).

A nucleophile is an electron-rich species that has a lone pair available to donate to form a bond pair with an electron-poor species or electrophile.

Although oxygen is more electronegative compared to sulfur, however because the atomic radius of sulfur is greater (one additional electron shell) than that of oxygen, the negative charge is better stabilized in the SH- ion. Therefore, OH- is a stronger base compared to SH-.

And also because the atomic radius of sulfur is greater (one additional electron shell) than that of oxygen, the electron density in sulfur is more polarizable (ie. easier to shift electron density around) as compared to oxygen; consequently the lone pairs in SH- is more available for donation as compared to that in OH-. Therefore SH- is a stronger nucleophile compared to OH-.

As far as the H2 Chemistry syllabus is concerned, the OH- ion acts as a nucleophile under aqueous conditions, but acts as a base under alcoholic conditions. The reasons for this are as follows :

Water is a protic (ie. protons available for donation), polar solvent. Alcohol is an aprotic (ie. no protons available for donation), (relatively) non-polar solvent.

When a water molecule donates a proton to the OH- ion, the water molecule becomes a OH- ion and the OH- ion becomes a water molecule. So status quo is maintained. Hence it is primarily the fact that water is a strongly polar solvent, one which forms stabilizing ion-dipole interactions with the OH- anion, that contributes to the tendency of the OH- to function more as a nucleophile than as a base.

Under alcoholic solvent conditions, the OH- ion is not stabilized by the solvent; because the alcohol solvent can neither donate protons, nor form stabilizing ion-dipole interactions with the OH- ion (alcohol can only form weaker permanent dipole-induced dipole interactons with the OH- ion), hence the highly unstable OH- anion ion here, has a high tendency to function as a base, to abstract any available proton as soon as possible to stabilize it's negative charge.

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