BedokFunland JC's A Level H2 Chemistry Qns (Part 2)

last SG · 2010-07-26T23:40:35+00:00

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However i checked several sources online, and it is mentioned that groups like

-CH2OH, CH2Cl, CH2OR are ortho and para directing which is really a contradiction to what you've said.

Could it be the "hyperconjugation" effect as mentioned in some of the sources i checked?

please help me out, im confused

If the OH, Cl and OR groups are directly bonded to the benzene ring, then because these substituents are electron donating by resonance (since they have a lone pair on the atom directly bonded to the benzene ring), even though they are electron withdrawing by induction (since O and Cl are more electronegative than C), be aware that resonance effects outweigh inductive effects unless the substituent is a halogen*, accordingly these 3 groups are ortho-para directing.

Summary : as long as you donate electrons by resonance, you're ortho-para directing. Halogens are ortho-para directing even though they're deactivating (ie. deactivates the benzene ring making it a weaker nucleophile). Halogens are deactivating because they withdraw by induction more strongly than they donate by resonance*. But because halogens donate by resonance, they're still ortho-para directing.

* To understand why halogens withdraw by induction more strongly than they donate by resonance, you have to consider two cases :

Case #1 : Fluorine.
F is the undisputed king of electronegativity. Hence, it is unsurprising that even though it donates well by resonance, it withdraws even more strongly by induction.

Case #2 : Chlorine, Bromine and Iodine.
Cl, Br and I are not so strongly electronegative. Cl and Br are both less electronegative than Nitrogen; and Iodine is (something that most JC students don't realize) even less electronegative than Carbon. If their (electron-withdrawing) inductive effects are so weak, then why don't their (electron-donating) resonance effects outweigh their weak inductive effects? Answer : because their (electron-donating) resonance effects are even weaker! Why? Consider the sideways or side-on overlap of unhybridized p orbitals of the halogen and the C atom (of benzene) to form the pi bond (by resonance). Notice that in the case of Cl, it would involve the sideways or side-on overlap of the 3p orbital of Cl with the 2p orbital of C. Because 3p orbitals are significantly more diffused than 2p orbitals, the sideways or side-on overlap is ineffective and the pi bond formed is weak. Therefore, Cl is poorly electron-donating by resonance. And accordingly, Br and I, are naturally even worse electron-donaters by resonance.

Hyperconjugation refers to a 3rd modality in which groups may donate electrons, in addition to by induction and by resonance. Specifically, this term is used to describe how alkyl groups donate electron density to stabilize carbocations. This is the more correct reason why a tertiary carbocation is more stable than a secondary carbocation is more stable than a primary carbocation is more stable than a (pop quiz : what's one level below primary? kindergarten? nope, its...) methyl carbocation. But for H2 'A' level purposes, you don't have to use this term "hyperconjugation", you can use the simpler concept of "induction" to explain why alkyl groups stabilize carbocations and why alkyl groups on a benzene ring are both (weakly) activating and also ortho-para directing.

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ForgiveMe asked :

1. Ticl3 is colored , TiF4(ionic bond) is not. How come TiF4 have no 3d electron?

2. Also, regarding the shape of complexes, when there are 4 ligands, how do u know it's sq planar and tetrahedral?

3. How do I know how many valence electron does. Nickel have , when it has variable oxidation state ? 0.o?

4. First order reaction with respect to halogenoalkane and zero order with respect to NaOH, suggests that the reaction is SN1. How come ah? Is there a link?

Q1. Ti is [Ar] 4s2 3d2, hence Ti4+ is [Ar] 4s0 3d0, hence no d-d* transitions.

Q2. This topic on complex ions is very complex, most of it involves concepts and calculations beyond A levels (eg. crystal-field theory, ligand-field theory, Jahn-Teller effect, etc). For A levels, you have no choice but to memorize a list of complex ions and their shapes. You can check out Jim Clarke's A level Chem website, and Rod Beavon's A level Chem website, for a list of complex ions commonly tested at A levels.

Q3. You should be able to work out the electron configuration for ground state atoms and ions, based on the element's position in the periodic table.
Ni is [Ar] 4s2 3d8
Ni+ is [Ar] 4s1 3d8
Ni2+ is [Ar] 4s0 3d8
Ni3+ is [Ar] 4s0 3d7
Ni4+ is [Ar] 4s0 3d6
etc

Q3. Yes of course there's a link. If you don't see the link, that means either you don't truly understand kinetics, or you don't truly understand organic chem mechanisms, or (most likely) both. (Unfortunate fact is, the vast majority of JC students don't truly understand chemistry even after taking it at A levels, they just memorize notes. So it's not your fault, it's just the Sg education system).

In the SN1 mechanism, the rate determining (ie. slow) elementary step involves only the electrophile (the alkyl halide), not the nucleophile (the OH- ion). Therefore, the rate equation is :
rate = k [electrophile]^1 [nucleophile]^0
which is same as
rate = k [electrophile]^1
which is same as
rate = k [alkyl halide]^1
which is same as
rate = k [alkyl halide]

Note : "alkyl halide" means "halogenoalkane".

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blacktoast94 posted :

i cant' do this super basic qn. :(
why CH3CH(CL)COOH and CH3CH(OH)COOH are capable of dissolving in an organic solvent?
i thought the interactions between them are id-id interactions hence don't release sufficient energy to overcome the H-bonding between the solute molecules. :/

First of all, organic doesn't mean non-polar. It means "carbon containing". Many organic solvents are at least slightly polar and/or capable of limited hydrogen bonding (eg. ethers cannot donate hydrogen bonds but can accept hydrogen bonds).

Secondly, even if enthalpy change may be slightly unfavourable (for some solutions), but often enough, the favourable entropy change may outweigh the unfavourable enthalpy change, to give a favourable Gibbs Free Energy change for the dissolving process.

Thirdly, Both carboxylic acids mentioned above have an ethyl group that is capable of instantaneous dipole - induced dipole van der Waals interactions (do not use acronyms like id-id, it's a bad habit) with the alkyl groups of the organic solvent (which you didn't specify).

Remember that as the alkyl group gets longer, the overall polarity of the molecule decreases significantly, even with a polar group such as OH or COOH present.

Fourthly, the Cl atom of the chlorocarboxylic acid is large and has significantly polarizable electron charge clouds, allowing for significantly favourable instantaneous dipole - induced dipole van der Waals interactions with the organic solvent.

Fifthly, some of the hydroxycarboxylic acid's capacity for hydrogen bonding has been used up for intramolecular hydrogen bonding, consequently resulting in less extensive intermolecular hydrogen bonding. In effect, there is much less hydrogen bonding between molecules of the hydroxycarboxylic acid required to be overcome by the organic solvent.

Sixthly, it really does depend [B][U]a lot[/U][/B] on the exact organic solvent used. Did you know, that phenol, phenylamine and benzoic acid are a lot more soluble in ether solvent, than in either aqueous (water) solvent or CCl4 (non-polar) solvent? This is because both (ethers and the 3 compounds mentioned) are both less polar than water, but more polar than CCl4 solvent.

As a general rule, partially polar compounds (ie. capapble of some hydrogen bonding, but has a large, non-polar benzene ring or hydrocarbon chain, such as phenol, phenylamine and benzoic acid) are more soluble in ether than water, while ionic species (eg. phenoxide ion, phenylammonium ion, and benzoate ion) are more soluble in water than ether.

Why? Because ionic species can form ion - permanent dipole interactions with water, which are stronger than the hydrogen bonding between water molecules. (You should be aware however that hydrogen bonding is generally stronger than ion - induced dipole interactions).

Cambridge could very likely set such an exam quesiton on "Separation of Organic Compounds by controlling the pH of Ether and Aqueous Solutions" in the upcoming A level exams.

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Blacktoast asked :

when BECL2 and CH3NH2 form a dative bond, how many hydrogen bond it is capable of forming?

"How many H bonds" vs "How many types of H bonds" are 2 different qns.

Furthermore, whether these are H bonds between molecules of the same species, of H bonds between these species and another species, eg. a solvent (eg. water), are again 2 different qns.

The correct answer, to the qn "how many H bonds may be formed between the molcule that is generated when 2 methylamine molecules complex with 1 beryllium chloride molecule" is ZERO.

This is because the N atom would have gained a +ve formal charge and have no more available lone pair for H bonding, and therefore cannot have electrostatic attraction with the partial +ve H atoms of a neighbouring complexed molecule of the same species. The Cl atoms are indeed partial -ve, but Cl is not electronegative enough to qualify for H bonding (at least for the 'A' level H2 syllabus).

So the only interactions between the complexed molecules are permanent dipole - permanent dipole van der Waals, not hydrogen bonding.

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Dixoria asked :

Im pretty confused between the terms hydrolysis, hydration as well as solvation.

Does hydration = solvation?

So far some hydrolysis reactions i've encountered includes
1)reactions with water (Inorganic chem- complete hydrolysis of SiCl4 to form acidic soln)
2)acidic/basic hydrolysis of esters/amides
3)Partial hydrolysis(polarisation of O-H bond in [Mg(H2O)]2+ to form [Mg(H2O)5(OH)] + H+)

Is hydrolysis a chemical reaction whereas solvation is not?
And is (3) a considered a chemical reaction?
Why are there so many different type of "hydrolysis" reactions?

The use of the term "hydration" by organic chemists in "hydration of alkenes" is anomalous and technically improper, and physical and inorganic chemists close one eye to the improperiety of this usage by organic chemists.

Apart from "hydration" of alkenes, in all other branches of chemistry, the difference between "hydrolysis" and "hydration" is :

Hydration is a subset of solvation, and refers to physical interaction between the solute and solvent.

Hydrolysis refers to chemical reaction with water. However (as with hydration), the term hydrolysis is also often loosely used.

For instance, in the hydrolysis of alkyl halides, the nucleophile used is often the OH- ion rather than water (OH- is a stronger nucleophile since the O atom has a -ve formal charge, whilst in water the O atom only has a -ve partial charge).

While the early use (in some branches of chemistry) of the term "hydrolysis" stemmed from the "lysis" of either water or the compound (reacting with water), but the term "hydrolysis" has now evolved to a more functional meaning of "chemical reaction with water".

For the difference between solution and solvation (or hydration, if the solvent is water), read Jim Clarke's webpage on "Solubility of Group II compounds" which will be helpful for 'A' level H2 Chem students :
http://www.chemguide.co.uk/inorganic/group2/problems.html

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I'm really confused with whether Potassium or Lithium is a stronger reducing agent.

I've read about Standard Reduction Potential(althought I don't understand much) and it says that Lithium with the lowest reduction potential is the stronger reducing agent.However, according to the Reactivity Series, Potassium is more reactive than Lithium and thus is a stronger reducing agent.

Can anyone clarify this with me?Thanks in advance.

It so happens that this particular comparison you've brought up (or to be precise : Lithium), is problematic. Yes, K is indeed more reactive than Li (due to less endothermic 1st ionization energy of K due to greater distance between the +vely charged nucleus and valence shell, and greater shielding effect from greater no. of intermediate electron shells).

However, it is also true that the standard oxidation potential of Li is more positive than that of K. This may indeed be correctly interpeted as an apparent contradiction, at A levels. As it turns out, the reason for this contradiction has to do with the high charge density of the Li+ ion, resulting in extraordinarily strong ion-permanent dipole interactions with water, in the aqueous state. This phenomenon additionally favours the oxidation of Li to Li+ in the aqueous state (over K to K+), even though the 1st ionization energy of Li to Li+ is significantly more endothermic (compared with K to K+) in the gaseous state.

Therefore, for 'O' level purposes, it would suffice to simply state that because K is indeed more reactive than Li, hence K is a stronger reducing agent than Li. For 'A' level purposes, the high achieving student should be able to describe this apparent contradiction and explain it both ways.

Other than this particular problematic case of Li, you (both 'O' and 'A' level students) can use the Standard Redox Potentials accordingly, as per the example below :

Because a reducing agent is oxidized during the reaction, thus it is more appropriate to compare oxidation potentials (rather than reduction potentials). Since you're taking the reverse direction to these equations given, accordingly you should take the -ve sign for these potentials.

Eg. Based on the image below, since the reduction potential of Ag+ to Ag is +0.8V, hence the oxidation potential of Ag to Ag+ is -0.8V.

Let's put aside Lithium for the moment, and compare Na versus K.

Since the oxidation potential of Na to Na+ is +2.71V, while the oxidation potential of K to K+ is +2.92V, we see that K thus has a more positive potential (ie. greater potential = greater tendency = greater inclination = greater propensity) to be oxidized, compared to Na.

Therefore K is a stronger reducing agent than Na.

Now try it out for yourself, say, is bromine or iodine a stronger oxidizing agent? This time round, since oxidizing agents are being reduced, you should compare reduction potentials, not oxidation potentials.

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A pure sample of N2O4(l) is introduced into an evacuated vessel. The vessel, of constant vol, is heated to a constant temp such that the equilibrium below is established.
2NO2 (g) <--> 2NO (g) + O2(g)
The value of the pressure p is then found to be 20% greater than if only NO2 (g) were present.

What is the mole fraction,x, of oxygen in this equilibrium mixture?
A) 0.17
B) 0.20
C) 0.33
d) 0.40

2008 Nov P1 Q7

CS Toh's website : http://www.post-1.com/step-by-step/?a=solutions

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Forgive me posted :

How come the species with high boiling point will be more easily liquified?
Liquified means solid to liquid right? Or from gas to liquid? O.o Thanks in advance ! :)

"Liquified" means "to become liquid" (from either gas or solid).

Ask yourself : What's a liquid? It's the state of matter when the molecules are so close together that we call it a liquid instead of a gas (but still moving fluidly rather than vibrating about fixed positions, as per the solid state). Then ask yourself : why or how do molecules of a liquid manage to stay so close together? It's because of the intermolecular van der Waals /or hydrogen bonding that holds them close toegther?

So you realize that the stronger the intermoleculer van der Waals or hydrogen bonding, the easier it is to keep the molecules together in the liquid state (as opposed to weaker van der Waals which allow the molecules to fly apart into the gaseous state), which is to "liquify" (or "liquefy") it.

And if the intermoleculer van der Waals or hydrogen bonding is strong, what can you say about the boiling point? Higher, of course! Because it takes more energy to overcome such intermolecular attractions in order to boil the species (eg. H2O, NH3, etc).

Now do you see the link between boiling point and how readily a particular species is liquified? Chemistry is all about seeing links. If you can see the links between every topic of Chemistry (physical chemistry, inorganic chemistry, organic chemistry), then you would not only truly understand and appreciate Chemistry, you will enjoy it so much that getting an A grade will be (relatively) effortless.

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Forgive me asked :

Which following sew give best yield of 2 bromo-4- nitro methyl bezene?
A) alkylation -nitration - bromination
B) alkylation bominaton nitration

Ans is a becos it's major product. Bt how do I know it's major product . And after I add nitro into my methyl benzene , should I do 2-4 or 3-5 ??which one should i follow?

Alkyl groups are electron donating by induction (actually hyperconjugation, but this term is not required at H2 level), hence they are mildy activating and are ortho-para directing.

Halogen groups withdraw electrons by induction more strongly than they donate by resonance, hence they are mildly deactivating but because they donate by resonance, they are still ortho-para directing.

Nitro groups withdraw electrons by both induction (due to a +ve formal charge on N) as well as by resonance (due to the N=O double bond allowing the nitro group to withdraw by resonance without violating N's octet). Accordingly, nitro groups are strong deactivators, and are meta directing.

Activators win deactivators, and ortho-para directors win meta directors.

But both option Also form the same product . How u know Which is one gt major whic one dun have?

If you carry out option B : alkylation bominaton nitration, notice that because Br is large and poses significant steric hinderance, the major product will see the Br+ electrophile substituting away the para (rather than ortho) proton. In contrast, in option A : alkylation -nitration - bromination, because the nitro group is also fairly large and will more likely take up the para position, the subsequent Br+ electrophile will be forced to take up the ortho (relative to the methyl) position.

Q2) which pair of reagent from
Carbocations
A) CH3COCH3 + HCN
B) CH2=CHCH3 + HBr
C) CH3CH2Cl + NaOH

Isnt it all suppose to form carbocation? Why only b?

Q2.
A) CH3COCH3 + HCN, upon nucleophilic addition of the CN- nucleophile, the intermediate generated is an alkoxide ion (ie, the -ve formal charge is on the O atom, the C atom does not have a +ve formal charge).
C) CH3CH2Cl + NaOH, because this is a primary alkyl halide, it undergoes SN2 rather than SN1. (Due to both sterics and electronics reasons).

Q4) if the decay rate of a radioactive isotope z decease from 1800 coun per
Min to 180 count per min in one week , wad is the half life?

Q4.
Molarity (desired) / Molarity (initial) = (1/2)^n
n = 3.322 half-lives = 1 week
Hence half-life = 0.301 weeks

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is Na2O an alkali or not? cos an an alkali is a soluble base so is NaOH an alkali along with Na2O im veey confused with it. pls help me clarify what is what

An alkali ( noun ) is any solution ( adjective : "alkaline solution" ) which contains a higher concentration of OH- ions over H+ ions.

Bases which are able to undergo hydrolysis ( hydrolysis = "chemical reaction with water" ) to generate soluble products are called "soluble bases" because the solid base appears to dissolve in water (it is really undergoing a chemical reaction with water, ie. hydrolysis, to generate soluble products).

For instance, Na2O or sodium oxide, contains the strongly basic O2- dinegative oxide ion, which in a bid to stabilize itself, abstracts protons H+ ions from water, generating 2OH- ions.

O2- + H2O ---> 2OH-

Since NaOH is soluble (unlike say, MgO), the resulting solution is alkaline, ie. has a higher concentration of OH- ions over H+ ions, and has pH greater than 7 (note that pH = 7 holds true for neutral solutions (only) at 25 deg C; therefore at 25 deg C, any pH greater than 7 is considered alkaline).

There are 3 different definitions of acids (and therefore, of bases). These are : Bronsted-Lowry, Lewis and Arrhenius. For 'O' levels, the Arrhenius definition "an acid dissolves in water to generate H+ ions" is often used, even though the other two definitions (Bronsted-Lowry and Lewis), used at 'A' levels, are more correct and useful.

To summarize :

A (Bronsted-Lowry) base is any species that accepts protons H+ ions. A 'soluble' base undergoes hydrolysis ('chemical reaction with water') to generate a solution containing a higher concentration of hydroxide OH- ions over protons H+ ions, which is called an "alkali" or "alkali solution" or "alkaline solution".

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Q1. You never realized, all these years, that the so-called "displacement" reaction is actually redox?!?!? Cl2 is reduced to Cl-, while oxidizing Br- to Br2!!!

Q2. Acid anhydrides are electrophilic. That's something which you should be able to figure out by analyzing its structure. Guys like girls, nucleophiles like electrophiles. Only nucleophiles have the 'balls' (the 'balls' usually refer to a lone pair, but can refer to a pi bond in the case of alkene nucleophile and benzene ring nucleophile) to attack electrophiles. A nucleophile such as phenol, can attack acid anhyrdide electrophiles in a similar fashion to attacking acyl chloride electrophiles, to generate a phenate ester.

Q3. Enthalpy of neutralization is defined as the enthalpy change per mole of water generated. So multiply by the coefficient of water in the equation.

Q4. Not required for H2 syllabus. If Cambridge wants you to apply it directly, the question will provide the formula.

Q5. If you keep adding excess strong base, then the pH will continue to rise, approaching (but never reaching) the pH of the strong base used. The OH- ion is a stronger base compared to the basic salt generated, which means if you're asked to calculate the pH at a point beyond the equivalence point (equivalence point = stoichiometric neutralization between acid and base, or oxidizing agent and reducing agent; end-point = point when a suitable indicator changes colour, it only approximates equivalence point; but for 'A' levels both terms may be used interchangeably), just consider the molarity of the stronger OH- ion from the excess NaOH added in, and ignore the contribution of OH- from the hydrolysis of the salt, because as predicted by Le Chatelier's principle, the hydrolysis of the salt is suppresssed by addition of extra OH- ions.

Q6. Yellow. Fe3+ is a clear-cut case. The case of Cr3+ is more probematic. For 'A' levels, you can just state the colour of the hydrated (ie. hexa-aquated) Cr3+ as green, even though strictly speaking and in theory, if all 6 ligands are water, the colour of (hexa-aquated) Cr3+ is pale violet. Problem is, in a percentage of any given sample, the counter anion (eg. Cl-, NO3-, SO4 2-, etc) will always substitute one or more of the water ligands, generating the stronger colour green, which will mask away the weaker pale violet colour. For more info on complex ions and their colours, check out Jim Clarke's website, and Rod Beavon's website.

Q7. Do not use the term "acylation" on its own, like some JCs badly teach. Because the term when used alone is ambiguous : it could refer to either Friedal-Crafts acylation, which involves first the reaction between a suitable Lewis acid catalyst such as AlCl3 and the acyl halide, followed by electrophilic aromatic substitution (recommended you always include the word "aromatic" to remind yourself that the benzene ring is the nucleophile in these reactions, but for 'A' levels Cambridge won't penalize you if you leave out the word "aromatic" in the term) where the benzene ring is the nucleophile. Or the term "acylation" could refer to nucleophilic acyl substitution, the mechanism for which is addition-elimination, and is the reaction responsible for generating esters, generating amides, hydrolyzing esters, hydrolyzing amides, and the reaction of nucleophiles with acid anhydride electrophiles, amongst other reactions.

If Cambridge asks you "what type of reaction is this? acid + alcohol ---> ester + water", the exam-smart candidate will give all 4 of these correct answers, to ensure you secure the full marks : esterification (because an ester is generated), condensation (because a water molecule or small alcohol is eliminated to join two molecules together intermolecularly, or two parts of the same molecule together if intramolecularly), nucleophilic acyl substitution (because it is nucleophilic substitution occuring on acyl group rather than an aliphatic group), and addition-elimination (which is the mechanism for nucleophilic acyl substitution, just as SN1 and SN2 are the mechanisms for nucleophilic aliphatic substitution).

To answer your question : with a Lewis acid catalyst, the reaction proceeds at room temperature, no heating is necessary.

Q8. Do not blindly follow Markovnikov's rule. It may fail sometimes (eg. in addition of HX onto 3,3,3-trifluroprop-1-ene). Rather, understand why it usually works (ie. comparative stability of the two alternative possible carbocation intermediates during the mechanism, which you may be asked to draw out in the 'A' level exam) and you can work out when it may fail.

Q9. Hydrolysis can be acidic (ie. proton-catalyzed) or alkaline (ie. base-promoted). Since esterification (using alcohol and carboxylic acids, instead of acyl halides) employs acidic conditions, therefore acidic hydrolysis of esters are also reversible (ie. it's actually the same reaction under the same conditions! it's only a matter of initial molarities of LHS vs RHS; ie. Qc vs Kc : at initial conditions, does position of equilibrium lie to the left or right?). Accordingly, alkaline hydrolysis of esters is not reversible.

Hi, as i’m doing my revision, I realised that I am unsure of the qns below. Thanks.

1) Redox
I understand that Cl2 and Br ion can undergo displacement reactions, but can they undergo redox as well?

2) Organic Compounds
Are we required to learn all the reactions of acid anhydrides in H2 chemistry? I only know that it reacts with alcohols & amines.

3) Chemical Energetics
Why do we need to multiply the stoichiometric coefficient of the limiting reagent for enthalpy change of reaction and not the enthalpy change of neutralisation?

4) Reaction Kinetics
I understand that the Arrhenius equation means that a lower Ka would result in a larger rate constant, k, hence faster reaction. However, how do we apply this equation to a kinetics question?

5) Ionic Equilibra
In a titration curve between a weak acid (CH3COOH) and Strong base (NaOH), I understand that salt hydrolysis occurs at the equivalence/end point. However, what happens after end point?

6) Transition Elements
I’m unsure of the real colour of [Fe(H2O)]3+, is it yellow or very pale violet?

7) Does the process of acylation (Electrophilic substitution) reaction requires heating? other than anhydrous AlCL3?

Cool Does Electrophilic addition of Br2 (aq) (E.A.) follow the markovnikov’s rule?

9) Why is the hydrolysis if ester an irreversible rxn and esterification a reversible rxn?

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To secure full marks for this qn (a common A level exam qn), you have to explain :
why formation of CaF2 is more favourable than formation of CaF.
and
why formation of CaF2 is more favourable than formation of CaF3.

The Universe is a shrewd businessman. It will automatically favour the reactions that generate the most profit, thermodynamically speaking, for itself.

While ionizing Ca to Ca2+ requires more investment compared to ionizing only to Ca+, but the returns when forming much stronger ionic bonds between Ca2+ and F- are much higher (compared to weaker ionic bonds between Ca+ and F-), and therefore significantly increasing the proft margin.

The investments refer to the endothermic ionization energies, the returns refer to the exothermic lattice enthalphies, the profit margins refer to the formation enthalpies.

Hence, the universe finds that it's a far more profitable business venture to manufacture CaF2 instead of CaF.

In the case of CaF3, even though it's true that the ionic bonds between Ca3+ and F- would be even stronger than Ca2+ and F-, and thus the returns would be higher, but because ionizing Ca2+ to Ca3+ requires removing and electron from an inner quantum shell, the endothermic investment is crazy expensive! So much so that the overall formation enthalpy would actually be endothermic! It's an unprofitable, ill-advised business venture, and (unlike say, Temasek Holdings) Mr Universe is way too smart to be duped into such wasteful, loss-making dealings.

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snowflyssb asked :

why does BrCl4- exist but not FCl4-? is it due to difference in electronegativity?

why phosphorus does not occur naturally as P(triple bond)P, as opposed to nitrogen?

UltimaOnline answered :

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forgiveme asked :

1.RBr->R+ + Br-(fast)
R+ +oh- -> ROH ( slow)
Does the rate eqn need to include br-?

2) why ch3ch2ch=chcl
Is inert towards hydrolysis?

Q1. No, the rate equation is 1st order wrt OH- and 1st order wrt R+, but since R+ is an intermediate rather than a reactant, you have to substitute R+ with RBr in the rate equation.

Q2. The resistance to hydrolhysis is due primiarly to the partial double bond character between C and X (where X = any halogen), making cleavage of the bond difficult under normal conditions. There are two alternative explanations for this (Cambridge will accept either explanation).

Resonance explanation : The lone pair on the halogen is delocalized by resonance to form a pi bond with the sp2 hybridized C (which it can do so without violating the C atom's octet, thanks to the existing pi bond which can delocalize away to form a lone pair on the neighbouring sp2 hybridized C atom), thereby giving the C-X bond, partial double character.

Orbital overlap explanation : Due to the overlap of the pi orbital of the double bond and the p orbital of the halogen, the C-X bond has double bond character.

An additional contributing factor or explanation, which has nothing to do with the strength of the C-X bond, has to do with the fact that in Chemistry, guys only like girls, similarly nucleophiles only like electrophiles. Nucleophiles hate other nucleophiles (since they're not homosexual), and therefore the electron-rich double bond of the alkene nucleophile repels away the electron-rich OH- nucleophile.

Both factors above (ie. partial double bond character of the C-X bond, due to resonanc and/or orbital overlap; as well as the repulsion between the two electron-rich nucleophilic species) result in a significantly increased activation energy required for nucleophilic substitution (including "hydrolysis", if the nucleophile is water or the OH- ion) to occur.

A temperature of approximately 350 deg C and a pressure of approximately 200 atm, is usually employed to hydrolyze halogenobenzenes (aka aryl halides) into phenol.

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'O' and 'A' Levels : Chemistry Calculations

4) Calculate the concentration of a solution of sodium hydroxide given that 25.0 cm³ of it required 18.8 cm³ of 0.0500 mol dm^-3 H₂SO₄.

H₂SO₄ + 2 NaOH ® Na₂SO₄ + 2 H₂O

5) Calculate what volume of 0.05 mol dm^-3 KOH is required to neutralise 25.0 cm3 of 0.0150 mol dm^-3 HNO3.

HNO₃ + KOH ® KNO₃ + H₂O

6) A 250 cm³ solution of NaOH was prepared. 25.0 cm³ of this solution required 28.2 cm³ of 0.100 mol dm^-3 HCl for neutralisation. Calculate what mass of NaOH was dissolved to make up the original 250 cm³ solution.

HCl + NaOH ® NaCl + H₂O

7) What volume of 5.00 mol dm^-3 HCl is required to neutralise 20.0 kg of CaCO₃?

2 HCl + CaCO₃ ® CaCl₂ + H₂O + CO₂

8) 3.88 g of a monoprotic acid was dissolved in water and the solution made up to 250 cm³. 25.0 cm³ of this solution was titrated with 0.095 mol dm^-3 NaOH solution, requiring 46.5 cm³. Calculate the relative molecular mass of the acid.

9) A 1.575 g sample of ethanedioic acid crystals, H₂C₂O₄.nH₂O, was dissolved in water and made up to 250 cm³. One mole of the acid reacts with two moles of NaOH. In a titration, 25.0 cm³ of this solution of acid reacted with exactly 15.6 cm³ of 0.160 mol dm^-3 NaOH. Calculate the value of n.

10) A solution of a metal carbonate, M₂CO₃, was prepared by dissolving 7.46 g of the anhydrous solid in water to give 1000 cm³ of solution. 25.0 cm³ of this solution reacted with 27.0 cm³ of 0.100 mol dm^-3 hydrochloric acid. Calculate the relative formula mass of M₂CO₃ and hence the relative atomic mass of the metal M.

11) A 1.00 g sample of limestone is allowed to react with 100 cm³ of 0.200 mol dm^-3 HCl. The excess acid required 24.8 cm³ of 0.100 mol dm^-3 NaOH solution. Calculate the percentage of calcium carbonate in the limestone.

12) An impure sample of barium hydroxide of mass 1.6524 g was allowed to react with 100 cm³ of 0.200 mol dm^-3 hydrochloric acid. When the excess acid was titrated against sodium hydroxide, 10.9 cm³ of sodium hydroxide solution was required. 25.0 cm³ of the sodium hydroxide required 28.5 cm³ of the hydrochloric acid in a separate titration. Calculate the percentage purity of the sample of barium hydroxide.

Partial Solutions :

Q4. Calculate moles of H2SO4. Since it's a diprotic acid, hence calculate moles of protons H+ involved, which is also the same as the moles of hydroxide ions OH- involved, which in turn is the same as the moles of NaOH involved. Now that you've the moles of NaOH, and the Qn gives you volume of NaOH, hence determine the molarity of NaOH, commonly written as [NaOH], using the formula : Molarity = Moles / Volume (in dm3).

Q5. Calculate moles of HNO3 present. Since both acid and base are monoprotic, hence determine moles of KOH required. Since molarity of KOH is given, hence determine volume required, using the formula : Molarity = Moles / Volume (in dm3).

Q6. Calculate moles of HCl neutralized, which would be the same as moles of OH- present in the 25cm3 NaOH(aq) aliquot, hence calculate the moles of OH- present in original 250cm3 of NaOH(aq) solution (from which the aliquot was obtained), thereafter multiply by the molar mass of NaOH to determine the sample mass of NaOH required by the qn.

Q7. Let x be the volume (in dm3) of HCl required. Given the molarity of HCl, calculate moles of HCl involved, in algebraic terms. Given sample mass of CaCO3, calculate the moles of CaCO3 (simply divide the sample mass by molar mass). Bearing in mind that the CO3 2- ion is diprotic, hence calculate the moles of HCl required for acid-base neutralization. Equate this value to the moles of HCl in algebraic terms found earlier, and solve for x.

Q8. Let x be the molar mass of the monoprotic acid. Given sample mass, find moles of the acid (simply divide sample mass by molar mass), in terms of algebraic variable x. Given 250cm3 volume, determine molarity of acid in the 250cm3 solution. Since a 25cm3 aliquot was used for titration, calculate moles of H+ (still in terms of x) involved in the titration. Given molarity and volume of NaOH(aq) used in the titration, calculate moles of NaOH involved, and equate to moles of H+ found earlier (in terms of x), and solve for x.

Q9. Find molar mass of hydrated ethandioic acid, in terms of algebraic variable n. Given sample mass, find moles of acid present (in terms of n). Given volume of 250cm3, find molarity of acid in 250cm3 solution, hence determine moles of diprotic acid neutralized in titration (ie. in 25cm3 aliquot). Since ethandioic acid is diprotic, hence determine moles of H+ neutralized in titration (in terms of n). Equate this expression to the moles of OH- neutralized in titration (use the formula : moles = molarity x volume, for the NaOH), and solve for n.

Q10. Let the molar mass of metal M, be x (grams). Accordingly, determine moles of M2CO3 present, in terms of x (use the formula : moles = sample mass / molar mass; note that sample mass is given, and molar mass will be in terms of x). Given 1dm3 solution, calculate molarity (still in terms of x). Given 25cm3 aliquot involved in neutralization, determine moles of CO3 2- ions involved in neutralization (use the formula : moles = volume x molarity). Bearing in mind the carbonate(IV) ion CO3 2- is diprotic, hence calculate moles of H+ neutralized (still in terms of x). Equate this expression to the moles of HCl used (use the formula : moles = molarity x volume), and solve for x.

Q11. Let the % by mass of CaCO3(s) present in the 1g sample be x. Hence determine the sample mass of pure CaCO3(s) present, which is simply (x/100)(1g). Divide this by the molar mass of CaCO3(s) (refer to the periodic table) to obtain the moles of CaCO3(s) involved in neutralization (in terms of x). Find total moles of HCl present (moles = molarity x volume). Find moles of excess HCl by subtracting twice the moles of CaCO3, from the total moles of HCl present. (Why twice? because tghe carbonate(IV) ion CO3 2- is diprotic!). Equate this expression (of moles of excess HCl, in terms of x) with the moles of NaOH(aq) used in titration (use the formula : moles = molarity x volume), and solve for x.

Q12. Similar to Q11, but more involved and convoluted.
Let the % purity, which is the same as % by mass of Ba(OH)2 in the 1.6524 g sample, be x. Hence find sample mass of pure Ba(OH)2 present, in terms of x grams. Hence find moles of Ba(OH)2 present (formula : moles = sample mass / molar mass). Next find moles of OH- present (still in terms of x), bearing in mind that Ba(OH)2 is a diprotic base. Find total moles of HCl used (formula : moles = 0.2M molarity x 0.1dm3 volume). Subtract away moles of OH- present from total moles of HCl present, to obtain moles of HCl in excess (still in terms of x).

Given molarity of HCl, find moles of HCl present in 28.5cm3. This would give you the moles of H+ involved in the other titration, which is the same as the moles of OH- involved in this other titration. Given that in this other titration, volume of OH- used was 25cm3, hence determine molarity of NaOH(aq). Hence determine moles of OH- present in 10.9cm3. Equate this value, to the moles of HCl in excess (in terms of x) found earlier, and solve for x.

UltimaOnline SG

hmm because

there is one question like..

H3C-Li H3C-K H3C-F H3C-MgBr H3C-OH

need to arrange from least polar to most polar..

So I did like C-Li , electronegativey is 1.57
C-K , 1.73
C-F : 1.43
C-Mg : 1.24
C-O : 0.89

so H3C-OH (least polar) , H3C,MgBr, H3C-F.H3C-Li,H3C-K (Most polar)

then 2nd question,

find the type of intra-molecular Bonds,

BeF2
CH4
CBR4
LIF

I use the electronegativity table to find out..
like more than 0.5 less than 2.0 = polar
less than 0.5 is non-polar
more than 2.0 is ionic...

3) CH4 , non polar
NH3 , 3.04-2.2 = 0.84 (polar)
Just that for CH3CH2OH --->iM SO lost
CH3NH2 also..

Ultima, I learn all this as above in just one lesson & they expect us to know all. we are not given resources also.

C-K isn't a covalent bond, it's ionic. Therefore you cannot apply electronegativity ratings, which are to determine the polarity of covalent bonds. But if your teacher wants you to go ahead anyway, then the C-K electronegativity DIFFERENCE (a keyword you kept missing in your posts) is approx 2.5 - 0.9 = 1.6, while the C-F electronegativity difference is approx 4.1 - 2.5 = 1.6. So based on these calculations, the bonds are approximately equally polar, which is misleading because one is more ionic in character, while the other is more covalent in character.

Yes, you're correct in your method to determine nature of intramolecular bonds.

CHCL3 and C2H5OH :

Between tricholoromethane molecules : permanent dipole - permanent dipole van der Waals (aka Keesom forces). Between ethanol molecules : hydrogen bonding. Between trichloromethane and ethanol molecules : permanent dipole - permanent dipole van der Waals (aka Keesom forces).

UltimaOnline SG

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