2016 H2 Chemistry JC1 & 2 students post your questions here
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Originally posted by UltimaOnline:
Yes, though focus on the H bonding, not the van der Waals.
Option C is wrong, because MgO while basic, isn't sufficiently soluble (due to highly endothermic lattice dissociation enthalpy) to generate a strong alkali (ie. high pH). Even option A is incomplete or partially incorrect, as aqueous alkalis at room temperature cannot dissolve SiO2 (due to an even higher endothermic lattice dissociation enthalpy, in turn due to SiO2 being a giant covalent lattice structure). Molten (not aqueous) alkalis, or (at the very least) concentrated alkalis at high temperatures, are required.Thanks UltimaOnline! :)
How do I compare the basicity between hydrazine and hydrogen peroxide then. Although the lps on H2O2 belong to oxygen and are thus less available for donation, there are 4 electron pairs on hydrogen peroxide compared to 2 on hydrazine - doesn't that matter?
How exactly does increasing the concentration of [OH]- increase the feasibility of the reaction with silicon dioxide?
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Originally posted by gohby:
Thanks UltimaOnline! :)
How do I compare the basicity between hydrazine and hydrogen peroxide then. Although the lps on H2O2 belong to oxygen and are thus less available for donation, there are 4 electron pairs on hydrogen peroxide compared to 2 on hydrazine - doesn't that matter?
How exactly does increasing the concentration of [OH]- increase the feasibility of the reaction with silicon dioxide?
No prob Gohby :)
No of lone pairs doesn't matter as much as electronegativity (afterall, no matter how many lone pairs you have, you'll still incur a unipositive formal charge upon donating a dative bond to the proton). Thus amines and hydrazines are more nucleophilic and more Bronsted-Lowry basic, compared to alcohols and peroxides.
Molarity affects both kinetics and equilibria (which are themselves inter-connected, ie. Kc = kf / kb), and thus thermodynamic feasibility and favourability. -
JC2 students, how long it takes you to solve the following deductive elucidation question on esters, is indicative of your grade in the A level exams. Time yourself to find out how prepared you are for the upcoming A levels.
If you can correctly solve all of it within 10 minutes, you're an A grader.
If you can correctly solve all of it within 20 minutes, you're a B or C grader.
If you can correctly solve most of it within 30 minutes, you're a D or E grader.
If you can correctly solve at least half of it but need more than 30 minutes, you're a S grader.
If you still can't solve at least half of it even after spending more than 30 minutes, you're a U grader.
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Hello UltimaOnline,
HCI/13/P1/Q19
How does L exhibit geometric isomerism?
Thank you
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Originally posted by gohby:
Hello UltimaOnline,
HCI/13/P1/Q19
How does L exhibit geometric isomerism?
Thank you
Hi Gohby,
Cycloalkenes with at least 8 membered rings or larger, are able to exhibit geometric isomerism without excessively destabilizing ring strain caused by angle strain. If the ring was too small (eg. cyclohexane), only the cis isomer would exist but not the trans (otherwise ring strain caused by angle strain would be excessively destabilizing), and hence geometric isomerism doesn't exist for small cycloalkenes.
BedokFunland JC Challenge Qn (for H2 Chem students aiming for the A grade) :
Is the cis or trans geometric isomer more stable for cycloalkenes with a ring between 8 to 11 atoms? How about 12 atoms or larger? Explain your answer for each case.
I won't post the answer here. Interested students can go ask your school teacher or private tutor. -
Hi! May i ask how to solve this qns?
Modified 2010 Alevel P3 Q3 (d)
Alcohol J, CxHyOH, is a volatile fungal metabolite whose presence when detected in air can indicate hidden fungal attack on the timbers of a house.
When 0.10cm^3 of liquid J was dissolved in an inert solvent and an excess of sodium metal added, 10.9cm^3 of hydrogen gas (measured at 298 k) was produced according to the following equation:
CxHyOH+Na ---> CxHyONa+1/2 H2
When a 0.10cm^3 of liquid J was combusted in an excess of oxygen in an enclosed vessel, the volume of gas (measured at 298K) was reduced by 54.4cm^3. The addition of an excess of NaOH caused a reduction in gas volume of 109cm^3 (measured at 298K)
Use the data to calculate the values of x and y in the molecular formula for J, CxHyOH.
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Originally posted by CKTR:
Hi! May i ask how to solve this qns?
Modified 2010 Alevel P3 Q3 (d)
Alcohol J, CxHyOH, is a volatile fungal metabolite whose presence when detected in air can indicate hidden fungal attack on the timbers of a house.
When 0.10cm^3 of liquid J was dissolved in an inert solvent and an excess of sodium metal added, 10.9cm^3 of hydrogen gas (measured at 298 k) was produced according to the following equation:
CxHyOH+Na ---> CxHyONa+1/2 H2
When a 0.10cm^3 of liquid J was combusted in an excess of oxygen in an enclosed vessel, the volume of gas (measured at 298K) was reduced by 54.4cm^3. The addition of an excess of NaOH caused a reduction in gas volume of 109cm^3 (measured at 298K)
Use the data to calculate the values of x and y in the molecular formula for J, CxHyOH.
Balance the equation in terms of x and y. Coefficients should be 2 and (2x + (y-1)/2) --> 2x and (y+1).
If the alcohol was gaseous, based on the redox acid-base reaction between the monoprotic alcohol and electropositive sodium metal, determine the volume of gaseous alcohol (ie. if the liquid alcohol was present in the gaseous state) to be 21.8cm3. This is done out of convenience, to work in gaseous volumes instead of moles. If you prefer, you can convert everything to moles instead of working in gaseous volumes.
Hence, based on CO2 generated, and based on the stoichiometry of the balanced equation, we have 21.8x = 109, hence x = 5.
Reduction of 54.4 cm3 implies : [Used Up gaseous volume] - [Generated gaseous volume] = 54.4 cm3.
Alternatively, reduction of 54.4 cm3 implies : -[Used Up gaseous volume] + [Generated gaseous volume]= -54.4 cm3.
Used up gaseous volume refers to oxygen (since the alcohol was actually liquid, ie. zero gaseous volume), which is [ (21.8 / 2)(10 + (y-1)/2) ] cm3, based on the stoichiometry of the balanced equation.
Generated gaseous volume refers to carbon dioxide, ie. 109 cm3.
Hence y = 11. -
I have got another question.
A 0.0250 mol sample of an insoluble metal oxide is known either to exist as MO or M2O3, metal M is the metal. The sample was dissolved in 150cm^3 of 1.3mol dm^-3 hydrochloric acid. The resulting solution was made up to 500 cm^3 with distilled water. 25.0cm^3 of the solution then required 36.40 cm^3 of the 0.200 mol dm^-3 aqueous sodium hydroxide for neutralisation.
(A) Calculate the no. of moles of HCl acid that reacted with the metal oxide. (I got 0.0494 mol)
(B) Hence deduce the possible identity of the metal oxide (having trouble with this)
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Originally posted by CKTR:
I have got another question.
A 0.0250 mol sample of an insoluble metal oxide is known either to exist as MO or M2O3, metal M is the metal. The sample was dissolved in 150cm^3 of 1.3mol dm^-3 hydrochloric acid. The resulting solution was made up to 500 cm^3 with distilled water. 25.0cm^3 of the solution then required 36.40 cm^3 of the 0.200 mol dm^-3 aqueous sodium hydroxide for neutralisation.
(A) Calculate the no. of moles of HCl acid that reacted with the metal oxide. (I got 0.0494 mol)
(B) Hence deduce the possible identity of the metal oxide (having trouble with this)
Due to experimental limitations, the values have to be rounded off, ie. 0.0494 rounded off to 0.0500 and hence the metal oxide must have been MO (eg. MgO or CaO or BaO, etc), since 0.0250 mol of MO consists of 0.0250 mol of diprotic dinegative oxide anions O2-, which accepts 0.0500 mol of H+ in a Bronsted-Lowry acid-base proton transfer reaction to generate 0.0250 mol of water, ie. 2H+ + O2- --> H2O -
An organic compound A, C9H11Br, on treatment with hot aqueous potassium hydroxide gave a compound B, C9H12O. B responded to oxidation in three different ways. With acidified potassium dichromate it yielded C, C9H10O. With sodium hydroxide and iodine it yielded D, C8H7O2Na and a yellow ppt. With hot, acidic potassium manga are (vii) it yielded E, C7H6O2.
Identify compound A-E and explain the above reactions.
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Helloooo chemistry
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Compound P has the molecular formula C8H9ClO. When P is reacted with PCl5, white fumes are evolved. Treatment of P with neutral iron (lll) chloride did not produce a rep purple coloration . p is optically active and when it is boiled with aq sodium hydroxide l, followed by acidification with dilute nitric acid, and addition of silver nitrate, a white precipitate is produced. When P is reflexes with aq sodium hydroxide, Q, C8H8O, which gives a reddish brown precipitate with Fehling's solution is obtained.
Suggest the structural formula of P and Q, explaining clearly your reasoning.
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Originally posted by LavXuan:
An organic compound A, C9H11Br, on treatment with hot aqueous potassium hydroxide gave a compound B, C9H12O. B responded to oxidation in three different ways. With acidified potassium dichromate it yielded C, C9H10O. With sodium hydroxide and iodine it yielded D, C8H7O2Na and a yellow ppt. With hot, acidic potassium manga are (vii) it yielded E, C7H6O2.
Identify compound A-E and explain the above reactions.
A is C6H5CH2CHBrCH3, and the rest follows accordingly -
Originally posted by LavXuan:
Compound P has the molecular formula C8H9ClO. When P is reacted with PCl5, white fumes are evolved. Treatment of P with neutral iron (lll) chloride did not produce a rep purple coloration . p is optically active and when it is boiled with aq sodium hydroxide l, followed by acidification with dilute nitric acid, and addition of silver nitrate, a white precipitate is produced. When P is reflexes with aq sodium hydroxide, Q, C8H8O, which gives a reddish brown precipitate with Fehling's solution is obtained.
Suggest the structural formula of P and Q, explaining clearly your reasoning.
P is C6H5CH2CHClOH
Q is C6H5CH2CHO -
1. Are peroxides or oxides more stable? If peroxides are unstable why do they form? Which Grp II metals form peroxides, and do they require varying degree of heat to form oxides? What is the general trend in forming oxides from peroxides?
2. Pg 76 of George Cheong Inorg Chem: 'The single covalent bond between oxygen atoms in peroxide is relatively weaker due to repulsion.' What and where is the repulsion?
3. Will all peroxides ultimately form oxides and if so, does that mean peroxide can be considered an 'intermediate'? (Pg 77 of George Cheong Inorg Chem)
4. Pg 93 of George Cheong Inorg Chem: If it is considered a disproportionate reaction with reference to BaO2, then what about water? OS of O in H20 is -2 which is the same as Ba(OH)2.
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Originally posted by Ephemeral:
1. Are peroxides or oxides more stable? If peroxides are unstable why do they form? Which Grp II metals form peroxides, and do they require varying degree of heat to form oxides? What is the general trend in forming oxides from peroxides?
2. Pg 76 of George Cheong Inorg Chem: 'The single covalent bond between oxygen atoms in peroxide is relatively weaker due to repulsion.' What and where is the repulsion?
3. Will all peroxides ultimately form oxides and if so, does that mean peroxide can be considered an 'intermediate'? (Pg 77 of George Cheong Inorg Chem)
4. Pg 93 of George Cheong Inorg Chem: If it is considered a disproportionate reaction with reference to BaO2, then what about water? OS of O in H20 is -2 which is the same as Ba(OH)2.
Q1 & 2. Oxides are more stable in the solid state, stabilized by stronger ionic bonding (but oxides, unlike peroxides, are too unstable to exist in the aqueous state, due to too high an anionic charge density). Peroxides are generally considered more unstable than oxides, because of 2 reasons : the -1 OS of oxygen, which being electronegative, prefers an OS of -2 ; the inter-electron repulsion between the (lone pairs of the) partially (if organic peroxides) or formally (if inorganic peroxides) negatively charged atoms of the O-O peroxo group weakens the (unpolarized and hence also weaker) covalent O-O bond. Stability is relative, and even though peroxides are relatively less stable, they're still relatively stable enough to exist, and thus as a thermodynamic manifestation of entropy, anything that can exist (ie. be generated) by a valid reaction pathway will exist ; it's only a matter of relative kinetics, stabilities, equilibria and formation abundance, amounts or concentrations. Theoretically, any metal ion can ionically bond (with varying degrees of covalent character) with the peroxide ion if present (you can look up the redox mechanisms for formation of peroxides if interested), and hence all metal peroxides are possible. But if the metal cation has high cationic charge density, the peroxide anionic charge cloud becomes polarized and distorted to a greater extent, hence decomposing more readily (ie. at a lower temperature).
Q3. If heated, yes. But you can dissolve the metal peroxide in water to generate hydrogen peroxide and the metal hydroxide (though the industrial manufacture of H2O2 uses more economical and controllable synthesis methods), or you can react the metal peroxide with carbon dioxide to generate oxygen gas and the metal carbonate, so you cannot say "all peroxides ultimately form oxides".
Q4. Water (ie. neither the H or O atoms) is not itself oxidized or reduced. Rather, it is the peroxo O-O group (atoms) which undergoes disproportionation, from OS of -1, to OS of 0 (in O2 gas) and -2 (in oxide anion). -
What chemical tests can be carried out to differentiate between bromobenzene, (bromomethyl)benzene and ethanoyl bromide? Reagents provided are aq NaOH, NH3, HNO3, HCL, AgNO3 and distilled water.
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Originally posted by Ephemeral:
What chemical tests can be carried out to differentiate between bromobenzene, (bromomethyl)benzene and ethanoyl bromide? Reagents provided are aq NaOH, NH3, HNO3, HCL, AgNO3 and distilled water.
Differentiate via ease of hydrolysis, the rate of which can be measured by the rate of AgBr(s) precipitation. Adding AgNO3(aq) to all 3 test substances will suffice.Read CS Toh's A Level Study Guide : "Relative Strength of the C-Hal Bond" page under the chapter of "Halogen Derivatives", and also "Acyl Chlorides" page under the chapter on "Carboxylic Acids & Derivatives" .
Alternatively, read the following pages on Jim Clark's A Level Chemistry website :
http://www.chemguide.co.uk/organicprops/haloalkanes/agno3.html
http://www.chemguide.co.uk/organicprops/arylhalides/reactions.html
http://www.chemguide.co.uk/organicprops/acylchlorides/background.html
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Hello UltimaOnline,
I would like to enquire about the following questions:
HCI 08/P1/Q31
Is 2 be expected solely from diagram which shows that C forms 3 bonds in C60, thus delocalised electrons in the unhybridised 2p orbitals exists?
How can I ascertain that 3 is not an answer, given that the write-up mentions that it was discovered in the produced formed when graphite was vapourised?
HCI 08/P1/Q32
Why is 2 not an answer? If we have the enthalpy change of atomisation of AlF3, then the need for the enthalpy change of formation of AlF3, enthalpy change of atomisation of aluminium and the enthalpy change of atomisation of fluorine would be obviated wouldn’t it?
Thank you! :)
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Originally posted by gohby:
Hello UltimaOnline,
I would like to enquire about the following questions:
HCI 08/P1/Q31
Is 2 be expected solely from diagram which shows that C forms 3 bonds in C60, thus delocalised electrons in the unhybridised 2p orbitals exists?
How can I ascertain that 3 is not an answer, given that the write-up mentions that it was discovered in the produced formed when graphite was vapourised?
HCI 08/P1/Q32
Why is 2 not an answer? If we have the enthalpy change of atomisation of AlF3, then the need for the enthalpy change of formation of AlF3, enthalpy change of atomisation of aluminium and the enthalpy change of formation of fluorine would be obviated wouldn’t it?
Thank you! :)
Yo Gohby!Yes, the diagram is sufficient for students to interpret, deduce and infer this. But 'solely'? The more capable (ie. A grade) H2 Chem students will be expected to already be familiar with the various allotropes of C, including fullerenes (of which Buckminsterfullerene is the most well known) and their properties, even if not included in the basic H2 syllabus.
Because Buckminsterfullerene exist as simple, discrete molecules, hence electrons can only delocalize within the individual molecule, and not across the individual molecules, and thus Buckminsterfullerene cannot conduct electricity (which refers to a function at the macro-scale, not micro-scale; ie. a solid piece of graphite, being a giant covalent lattice, held in your hands can conduct electricity, but a solid piece of Buckminsterfullerene, being a large no. of individual C60 molecules held together by van der Waals forces, held in your hands cannot conduct electricity).
Then it wouldn't be called a Born-Haber cycle anymore, would it? (It would then be a generic enthalpy or energy cycle, but not Born-Haber cycle, which is a specific subset or type of enthalpy or energy cycle). The question specifies Born-Haber cycle. And also, specifically for A level H2 Chem purposes, atomization enthalpy is always applied to individual elements only. Thus Born-Haber cycles (especially for A levels) follows a fixed set of steps strictly. Clockwise : from the constituent elements, atomization of both constituent elements, IE for the metal, EA for the non-metal, lattice formation enthalpy, to get the compound in standard state. Anti-clockwise : from the from the constituent elements in standard state, formation enthalpy, to get the compound in standard state. So while you have a valid point about this being a lousy or debatable question, and unlikely to be asked by Cambridge (though Cambridge certainly has its fair share of lousy or debatable questions every year), nonetheless the best answer for this MCQ is still the one given by Hwa Chong.
No prob, Gohby ;)
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do you need to heat the reaction moisture before testing for ammonia test or can you just test for it straight with a litmus paper after the reaction?
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Originally posted by Ng.keebin:
do you need to heat the reaction moisture before testing for ammonia test or can you just test for it straight with a litmus paper after the reaction?
If you plunge the litmus paper into the solution, a red-to-blue color change may be due to OH- ions present from other basic sources, making the solution alkaline. Warming the solution and allow any ammonia or volatile amine present to vaporize, reach and hydrolyze on the moist litmus paper held above the solution to effect a red-to-blue color change, is thus a more reliable indicator for the presence of ammonia or a volatile amine. -
Why can't phenol react with PCl5?
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Originally posted by Ng.keebin:
Why can't phenol react with PCl5?
Eh you asked this qn before leh.Simple answer :
Because only aliphatic alcohols (and carboxylic acids) react with SOCl2 and PCl5, but not phenols (ie. OH group directly bonded to benzene ring).
Further answer :
Due to the sideways overlap of p-orbital of the O atom and the pi-orbital of the benzene ring, a lone pair on the O atom is delocalized by resonance to form a pi bond with the sp2 C atom of the benzene ring, giving the C-O bond in the resonance hybrid partial double bond character, which is thus strong and does not cleave readily. Hence all such nucleophilic substitutions on the benzene ring (which involves the potential leaving group having partial double bond character with the benzene ring in the resonance hybrid) are strongly resisted.
Even deeper further answer :
In addition, depending on the solvent used (eg. nucleophilic or not) and any other reagents present (eg. pyridine in SOCl2 reaction), let us consider the following 6 nucleophilic substitution mechanisms for the nucleophilic substitution reaction between phenol and SOCl2 or PCl5 : SN1, SN2, SNi (nucleophilic substitution with internal return), and Nucleophilic Aromatic Substitutions via either SNAr (addition-elimination) or via the benzyne intermediate (elimination-addition) or via SRN1 (free radical nucleophilic aromatic substitution).
SN2 isn't possible for phenols because the electrophilic C atom is itself part of the benzene ring, which hence poses an insurmountable steric hindrance (ie. it is geometrically, sterically and electrostatically impossible for the electron-rich Cl- nucleophile to approach the electrophilic C atom from within the electron-rich benzene ring itself as required by the SN2 mechanism) and concordantly, generating a geometrically unstable and thermodynamically infeasible pentavalent transition state (bear in mind this is nucleophilic substitution, not nucleophilic addition).
SN1 and SNi for arenes are highly thermodynamically unfavorable because upon attempted elimination of the leaving group (ie. the addition product of the phenolic group with the SOCl2 or PCl5 reagent), the positive formal charge on the consequent (highly unstable and thus thermodynamically infeasible to form) aryl carbocation species is highly destabilizing for 2 reasons : the positive formal charge would be on a sp hybridized C atom (ie. high % of s orbital character results in increased electronegativity which exacerbates the destabilizing effect of a positive formal charge), and furthermore the positive formal charge can *not* be delocalized by resonance (which is only possible for benzylic carbocations and carbanions, ie. when the formal charge is on the C atom outside, not within, the benzene ring), as the ring strain due to angle strain (ie. destablization due to significant deviation from the ideal bond angles of the sp hybridized C atom as predicted by VSEPR theory) prevents formation of an allene resonance contributor.
Nucleophilic aromatic substitutions do not readily occur and have high activation energy barriers (do you know why? Cambridge may ask you to suggest reasons for this). SNAr (addition-elimination) mechanism requires (preferably 2 or 3) strongly electron-withdrawing by both induction and resonance groups (preferably NO2 nitro groups) at the ortho and para positions (in salicylic acid, the ortho COOH group is indeed electron-withdrawing by induction and resonance, but not as strongly as NO2 groups because a formal positive charge on the N atom is more strongly electron-withdrawing by induction compared to a mere partial positive charge on the C atom).
The benzyne intermediate (elimination-addition) mechanism requires a very strong base for deprotonation of benzene (usually employing the amide NH2- ion, in which case the final product would be phenylamine rather than chlorobenzene) and the benzyne intermediate is itself highly unstable due to reasons discussed earlier (ie. destabilizing ring strain caused by angle strain due to significant deviation from the ideal bond angles about a sp hybridized C atom as predicted by VSEPR theory resulting in an unusually weak triple bond and hence being thermodynamically unfavorable), with equally unstable cumulene and biradical resonance contributors.
SRN1 (free radical nucleophilic aromatic substitution) requires a free radical initiator, and for which halogens are the leaving group to be eliminated rather than the incoming nucleophile. -
My Qs come from George Cheong Organic Chem book:
1. Must a tertiary alcohol be attached to 3 R groups? What if there is a double bond and hence there is only 2 R-groups?
2. Is water slightly acidic at rtp? (Pg 187) Do we just take it as water is neutral at rtp when answering Qs (for acid-base equilibria)?
3. (Pg 190) Why is organic solvent required to dissolve all 3 compounds? Why can't the experiment be carried out just as it is?
4. I don't understand Example 7.7 on pg 191. What do the pKa values mean and how do we use them to solve the Q?
5. Is formation of ester with excess alcohol and conc h2so4 at 140 degree celsius in syllabus?
6. Why is H3PO4 a weaker oxidising agent? What are the relative strength of some of the oxidising agents? (Pg 194)
7. (pg 196 fig.7-5) Why doesn't Et-O-H form protonated ethanol when it abstracts a H atom?