Two vessels containing 4dm^3 of argon and 10dm^3 of water vapour respectively at 150 degree celsius and 101 kPa were connected together.The connected vessels were allowed to cool to room temperature.What was the final pressure in the connected vessel?
Ans:20.3kPa
i used p1v1 + p2v2 = p3v3,couldn't get the answer..
Solution (method) :
Use PV=nRT to find the number of moles of argon only (before connecting the vessels).
After connecting the vessels (ie. new combined volume), apply PV=nRT with n = number of moles of argon only.
Solve for P.
H2 Organic Chem
Qn : What is the reaction that occurs when benzene is mixed with benzoyl chloride? Draw the mechanism.
Ans : Electrophilic (aromatic) substitution and nucleophilic (acyl) substitution.
(Note : the words in brackets above are not required for the H2 syllabus, but it's better to include them.)
Mechanism :
1. Benzene nucleophile attacks acyl carbon; acyl/carbonyl pi bond shifts up to become a lone pair on O.
2. Loss of proton H+ from benzene intermediate restores aromaticity; lone pair on the +ve formal charged O atom shifts back down to restore the acyl/carbonyl pi bond; Cl leaves as Cl- ion.
Product is diphenylmethanone aka diphenyl ketone aka benzoylbenzene aka benzophenone (main product) and hydrogen chloride (byproduct).
H2 Chemistry Organic Chem Qns.
greenapplejellybean asked :
hello everyone, i'm taking a levels this year (more specifically next week), i have a question for chem...
LiAlH4 in dry ether, and H2 with Ni catalyst, are both reducing agents. But from what I know, LiAlH4 can't reduce C=C double bonds and H2 with Ni catalyst can't reduce carboxylic acids and esters. Can someone clarify why this is so? And can H2 with Ni catalyst reduce things like aldehydes, ketones etc? Thanks in advance for any help rendered.
UltimaOnline replied :
The reason why LiAlH4 cannot reduce alkenes is because the way LiAlH4 works as a reducing agent is that it functions as a source of H- hydride ion nucleophiles. Alkenes are also nucleophilic due to their localized pi bonds. Hence both being nucleophiles will naturally repel away each other (think of it as : guys only like girls, guys do not like guys; nucleophiles only like electrophiles, nucleophiles do not like nucleophiles).
Yes, H2 with nickel catalyst (at temperatures above 140 deg C) can reduce alkenes, aldehydes, ketones and amides. Under usual conditions (ie. H2 syllabus) it cannot readily reduce carboxylic acids though.
FacadesofLight asked :
hi guys, hope u all can help me clarify these concepts.
1.) In Benzene ring, if the -OH group is attached to 1 and 4 position, upon addition with aq Bromine where would the bromine be directed.
We all know OH is electron donating and activates the ring and put it at 2 4 6 position. But isnt it relative concept?
If you add br with respect to position 1 OH, the bR will be 2 substituted. but if you "invert" the paper, it will be 3 substitution. can any chemistry expert help me clarify this? thanks.
2.) Which one would produce the monosubstitution or tri 2 4 6 substitution upon reaction with Phenol. Is it aq bromine or liquid bromine. Because i remember my chem teacher in school said that addition of aq bromine is mono sub and liquid br is 2 4 6 sub, but the chem notes im reading say the other way round. can anyone help me to clarify these concepts? thanks
UltimaOnline replied :
The phenolic OH group is electron donating by resonance, and is hence both an activator and ortho-para directing. Therefore, with aqueous bromine, all 4 available positions (ie. ortho & para with respect to both OH groups) will under electrophilic aromatic substitution of Br onto the benzene ring.
The strongly polar aqueous (ie. water) solvent will induce a stronger temporary dipole in the non-polar Br2 molecule, allowing for electrophiilc aromatic substitution to occur more readily (ie. the benzene ring nucleophile finds it easier to attack the Br2 electrophile because it has a larger magnitude of separation of partial charges); and consequently tri-substitution (onto the ortho, ortho and para positions) of the halogen occurs onto the phenolic benzene ring.
With a non-polar solvent such as CCl4, there is only a weak temporary instantaneous dipole on the Br2 molecule (ie. the benzene ring nucleophile finds it harder to attack the Br2 electrophile because it has a smaller magnitude of separation of partial charges), and hence only mono-substitution will occur (onto either the ortho or para position).
2010 Cambridge H2 Chem P3 Qns :
[QUOTE=midoforce]what's the bond angle between NO2 and O3, must be exactly accurate, not a range[/QUOTE]
NO2 : 134.3 deg
O3 : 116.8 deg
If (assuming) the Cambridge Mark Scheme is reasonable and fair, they will accept any reasonable angle approximately 120 deg (eg. for NO2, based on the trigonal planar electron geometry), provided you gave valid rationale (eg. "the single unpaired electron on the N atom takes up less space than a lone pair, hence the bond angle will be expected to be greater than 120 deg, eg. 130 deg").
Originally posted by TenSaru:Glutamic acid: HOOCCH2CH2CH(NH2)COOH (they gave skeletal but i cant draw it here)
There are 3 pKa values associated with glutamic acid: 2.1, 4.1, 9.5
Make use of these pKa values to suggest major species present in solutions of glutamic acid with the following pH values
pH1
pH3
pH7
pH11
pH 1
HOOCCH2CH2CH(NH3+)COOH
pH 3
HOOCCH2CH2CH(NH3+)COO-
pH 7
-OOCCH2CH2CH(NH3+)COO-
pH 11
-OOCCH2CH2CH(NH2)COO-
The idea is to, as you go from acidic environment to alkaline environment, deprotonate step-by-step one at a time (initially in very acidic conditions all groups are protonated; next deprotonate 1st the alpha carboxylic acid, then 2ndly the R group carboxylic acid, then 3rdly the ammonium group).
The alpha carboxylic acid is stronger than the R group carboxylic acid, due to the electron-withdrawing by induction effect of the (+vely charged) amine group.
-------------------------------------
Originally posted by Garrick_3658:
- Draw dot-and-cross diagram for NO2 molecule. It contains a dative covalent bond.
- FO2 does not exist, but ClO2 does. By considering possible types of bonding in both compounds, suggest reasons for the difference.
- When CuO is stirred with NH3(l), it dissolves to give a coloured solution. Oxide ion is acting as a Bronsted base. Suggest the equation for the rxn and colour of soln.
-------------------------------------
Structure of NO2 :
NO2 has an unpaired electron (free radical) on N, which is doubly bonded to an O atom, and singly datively bonded to another O atom. The N atom has a +ve charge, the singly bonded O atom a -ve charge, and the doubly bonded O atom has no charge. N lacks a stable octet in NO2, and the molecule is a free radical.
-------------------------------------
Why FO2 doesn't exist :
Reason #1
The problem with FO2, is that F (being in period 2), cannot expand its octet (since it doesn't have vacant, energetically accessible 3d orbitals to use). Hence, it cannot form a similar structure as ClO2, which has expanded its octet (4 bond pairs + 2.5 lone pairs on Cl = 11 electrons in terms of an expanded octet).
Reason #2
For FO2 to have single bonds between the F and two O atoms, F will gain a 2+ charge and each O atom with gain a -ve charge, and this separation of charge is destabilizing, since F is more electronegative than O.
Reason #3
In addition, in such a model, F atom would have an unpaired electron, making the molecule an unstable free radical.
Reason #4
And furthermore, the F atom wouldn't have a stable octet in such a model, which defeats the purpose of forming the structure in the 1st place.
-------------------------------------
CuO reacting with NH3 :
The oxide anion (ie. the base) abstracts a proton from NH3 molecule (ie. the acid), generating first the OH- ion and NH2- ion. These two anions, together with four other NH3 molecules, act as ligands to donate dative bonds to the Cu2+ ion, generating an octahedral complex ion.
Another possibility is that the OH- abstracts a further proton from another NH3 molecule (since liquid ammonia used means NH3 is present in large excess), generating H2O and another NH2- ligand. In such an octahedra complex ion, two ligands are NH2-, one is the H2O generated, and 3 are NH3 molecules.
The resulting complex ion should still be deep blue in colour, or a slight variant* of the deep blue colour of the usual tetraaminecopper(II) complex ion, since the Cu2+ ion remains as Cu2+, and the ligands are similar*.
(* Wikipedia out "spectrochemical series" for "strong field vs weak field" ligands)
CuO + 6NH3 --> Cu(NH2)(OH)(NH3)4
or
CuO + 6NH3 --> Cu(NH2)2(H2O)(NH3)3
'A' Level H2 Qn on Benzene :
wintersolstice posted :
1. The following shows the hydrogenation of cyclohexene.
C6H10 + H2 ---> C6H12
a)Calculate the enthalpy change for the hydrogenation of cyclohexene using data booklet.[2][I]my answer is -124 kJ mol^(-1)....
[/I]
b) In 1865, Friedrich Kekule described benzene as a ring of alternating single and double bonds.
From experimental determination, the enthalpy change of hydrogenation of benzene was found to be -205 kJ mol^(-1).
Explain, in terms of structure of benzene, why this value is not three times your answer in (a). Use an energy level diagram to illustrate your answer. [3]
[I]
i was thinking.....is it becoz the bond energy values provided in the data booklet is for gaseous compound ......but cyclohexene is a liquid.......so enthalpy change for vapourisation was not included.......so the value differ..????[/I]any help is much appreciated.
thanks.
Ultima Online replied :
Your reason may not be applicable, because the question did not specify the states for both species (benzene and cyclohexene), for which the values are as given. Even if it were applicable, it would be a minor reason, and you would have missed the point of the question, which is about RESONANCE or AROMATICITY.
(Nonetheless, your point about liquid vs gaseous state is indeed the correct answer for many similar questions on "why the discrepancy?", but just not (the correct answer) for this particular question.)
Now, the concepts of RESONANCE or AROMATICITY are very important fundamental concepts in Organic Chemistry, but are poorly dealt with in the H2 syllabus. Usually only H3 Chem students and/or Chem Olympiad students have a better understanding of it. Students taking Chem at solely the H2 or H1 level generally have a much more simplified understanding of RESONANCE (in some JCs the teachers don't even introduce the term "RESONANCE" to H2 students), if at all.
If you (ie. all JC students reading this) are serious about taking Science in the University (specifically Chemistry-based courses, eg. Medicine, Life Sciences, etc), I recommend you read up on RESONANCE, eg. on Wikipedia, from recommended textbooks, or ask your school/private tutor to teach you about it.
Back to your question, the correct answer involves a brief description and explanation on how the pi electrons of benzene are DELOCALIZED by RESONANCE throughout the ring (ie. the benzene ring is AROMATIC) ; include a diagram showing the overlapping unhybridized 2pz orbitals to form a continuous ring above and below the plane of the C-C sigma bonds ; state that all C-C bonds in benzene are of equal length and strength (with enthalpy/energy values between that of single and double bonds) ; draw the 2 resonance contributors (ie. both with alternating single and double bonds) and the resonance hybrid of benzene (ie. benzene with the circle in the middle) ; state that the capacity for delocalization of the pi electrons around the benzene ring by RESONANCE (aka its AROMATICITY) confers a greater stability upon the benzene ring (since it's pi electrons are now delocalized instead of being concentrated like alkenes; consequently alkenes are stronger nucleophiles and undergo electrophilic addition while benzene is a weaker nucleophile and undergoes electrophilic aromatic substitution instead); consequently with greater stability benzene has less energy and therefore has a less exothermic enthalpy of hydrogenation (compared to the mythical cyclohex-1,3,5-triene; ie. the non-resonance model of benzene, which some JCs' teachers erroneously label as "the kekule structure/model of benzene" ) .
An energy level diagram (such as this one, on Jim Clarke's webpage) will illustrate the last point nicely. Note that your question specifies that such an energy level diagram is required, to gain full credit/marks.
(Ok, you're not required to write all the points above for just 3 marks, but other H2 Chem exam questions on benzene may require some of the other points I've made above; so if you're serious about scoring a distinction in Chemistry, and/or studying Science in the University, then you shoud read up on RESONANCE and related concepts and do your best to understand all the points I've made).
Jim Clarke's 'A' Level Chemistry website on Benzene :
http://www.chemguide.co.uk/basicorg/bonding/benzene1.html
http://www.chemguide.co.uk/basicorg/bonding/benzene2.html
Rod Beavon's (yeah, the UK lecturer now famous in Singapore because his question was used in the Singapore-Cambridge 2010 'A' Level H2 Chemistry Exam question on Planning) 'A' Level Chemistry website on Benzene :
http://www.rod.beavon.clara.net/benzene.htm
Wikipedia on Benzene :
http://en.wikipedia.org/wiki/Benzene
I'll now leave you with the above webpages for your perusal and enjoyment. Have fun!
'A' Levels H2 Chemistry
midoforce posted :did my chem As recently just want to ask the following questions:
1) is skeletal drawings accepted as "structural formula"
eg. qn 1d) in chem p3, i put all my answers in skeletal form.. will minus mark?2) is state symbols required when they ask to suggest a balanced equation? i know for certain processes such as to show combustion or solution etc you need to state, what about inorganic questions? must u state without them asking for it? (state symbols)
1 question says "suggest a balanced equation between barium ozonide and water" and in the description says barium ozonide is a solid, reacting with water to form oxygen to form alkaline solution, must i state the state symbols in my equation?
3) is writing [cu(nh3)4]2+ acceptable? or must it be [cu(nh3)4(h20)2]2+ no matter what
4) for dot and cross dative bond, i drew dative bonds as :-> (electron pair with an arrow) is it wrong?
1) Unless the question specified "full" or "displayed" structural formula, skeletal drawings are accepted. Don't worry.
2) State symbols are not required unless specified by the question. Don't worry.
3) The abbreviated form [Cu(NH3)4]2+ is accepted. Don't worry.
4) Technically it is wrong (this is one of many areas which different teachers, even those within a same JC, teach differently and some teach wrongly). It is wrong because an electron pair is either a lone pair or a bond pair, not both at once. The (misguided) teachers who teach the wrong way you described (ie. dative bonds drawn as :--->) imagine the straight arrow functions as a curved arrow, which is used in Organic Chemistry to show the electron-flow process or mechanism of a chemical reaction. In Physical Chemistry (by contrast), the lone pair has already become a dative bond, and should no longer be shown as a lone pair (or else the donor atom, if it is an element of period 2, may be misinterpreted to imply its octet has been violated).
But don't worry, Cambridge will accept it and won't penalize you for it. Usually Cambridge will ask you to show dative bonds in the form of a dot-&-cross diagram, in which case the dative bond must be shown as a dot-dot (or cross-cross).
H2 Chemistry - Topic : Electrochemistry
Originally posted by Audi:In this topic, the formula for Ecell is Ecell= E(reduction half cell) - E (oxidation half cell). So usually we look at the standard electrode potential, the one with the more +ve value undergoes reduction, fit into formula to get ans. Besides questions asking us to determine whether a rxn is feasible (that one dont have to see which one is more +ve or -ve, we look at the reaction equation to see which rxtant undergoes oxid. or reduction), I chanced upon this question which wasnt properly explained to us in school and which I dont understand why we DONT have to see which eqn is MORE +ve or -ve to determine whether they undergo oxid or reduc. Here goes:
When aq solution of Copper II sulfate and potassium iodide are mixed, a brown ppt is formed. On addition of aq sodium thiosulfate to this mixture, the brown colour disappears and a white ppt REMAINS (means it is formed previously and stays there or the white ppt has just formed and stayed there)?
What is the std cell potential of the rxn of ppotassium iodide with copper II sulfate?
During lecture we werent given much explanation regarding this. The teacher just said well, I2 was formed and Cu+ was formed...
then write this equation
2I- ---> I2 + 2e- +0.54V
Cu2+ +e --->Cu+ +0.15V
Ecell = (+0.15) - (+0.54) = -0.39V
From the 2 equations of cuz i can see that Cu2+ undergoes reduc and I- undergoes oxidation. But why in this case we look at the equation to determine which one undergoes oxid or reduc? Why not the numbers? (quite silly question but still want to ask)
If the brown colour I2 disappears eventually, why do we still use the equation 2I- ---> I2 + 2e- ?? shouldnt it be the opposite way round?
If possible, can someone write for me the entire equations of this whole reaction? (for better understanding) thanks
UltimaOnline replied :
In the 1st reaction :
Cu2+ reduced to Cu+, and I- oxidized to I2
Ppt formed is copper(I) iodide, which is white; but because iodine is brown, the result appears to be a brown ppt (which is really a mixture of two separate species, one white one brown).
In the 2nd reaction :
S2O3 2- oxidized to S4O6 2-, and I2 reduced to I-.
Hence the brown colour dissapears (I2 reduced to colourless I-), while the white ppt (copper(I) iodide) remains.
The following calculations assume standard molarities for all species (per molar quantities specified by equation), which of course isn't actually the case (based on the stoichiometry of the balanced redox eqns; furthermore the following reaction would not be feasible or spontaneous in the forward direction with a negative cell potential). So the following is "just for argument's sake".
For 1st reaction :
2I- ---> I2 + 2e- -0.54V oxidation potential at anode
Cu2+ + e- ---> Cu+ +0.15V reduction potential at cathode
E cell = Reduction potential at Cathode + Oxidation potential at Anode
= (+0.15) + (-0.54) = -0.39V
For 2nd reaction :
2S2O3 2- ---> S4O6 2- + 2e- -0.09V oxidation potential at anode
I2 + 2e- ---> 2I- +0.54V reduction potential at cathode
E cell = Reduction potential at Cathode + Oxidation potential at Anode
= (+0.54) + (-0.09) = +0.45V
H2 Chem Qns.
Originally posted by anpanman:Some questions to ask.
1. Use of the data booklet is relevant to this question
The standard enthalpy change of formation of methane is given in the equation below.
C(graphite) + 2H2 (g) -> CH4(g) standard enthalpy change of formation = -75kJ/mol
what is the enthalpy change of atomisation of graphite?
+693 +843 +1129 +2587
tried to solve using all the methods i know but cant get the ans. only same result i got was 2587, which i think is wrong anyw.
2. The decomposition of H2O2 is known to be first order reaction
2H2O2 -> 2H2O + O2
the rate constant is found to be 4.95X10^-2 min-1. if initial concentration of H2O2 is 4.0mol dm-3, what will be the concentration of H2O2 after 42 min?
2.0 mol/dm3
1.0 mol/dm3
0.5 mol/dm3
0.25 mol/dm3
3. is SiCl4 a simple molecular compound? thanks everyone
Q1.
Atomization enthalpy of graphite + atomization enthalpy of H2 + 4 x (bond formation enthalpy of C-H) = -75kJ/mol
Atomization enthalpy of H2 given in data booklet.
Bond formation enthalpy of C-H given in data booklet.
Make Atomization enthalpy of graphite the subject.
Q2.
From rate constant, calculate half-life.
Then apply the formula :
Molarity (desired) / Molarity (initial) = (1/2)^n
where n = no. of half-lives.
H2 Chem Qns.
Originally posted by anpanman:Some questions to ask.
1. Use of the data booklet is relevant to this question
The standard enthalpy change of formation of methane is given in the equation below.
C(graphite) + 2H2 (g) -> CH4(g) standard enthalpy change of formation = -75kJ/mol
what is the enthalpy change of atomisation of graphite?
+693 +843 +1129 +2587
tried to solve using all the methods i know but cant get the ans. only same result i got was 2587, which i think is wrong anyw.
2. The decomposition of H2O2 is known to be first order reaction
2H2O2 -> 2H2O + O2
the rate constant is found to be 4.95X10^-2 min-1. if initial concentration of H2O2 is 4.0mol dm-3, what will be the concentration of H2O2 after 42 min?
2.0 mol/dm3
1.0 mol/dm3
0.5 mol/dm3
0.25 mol/dm3
3. is SiCl4 a simple molecular compound? thanks everyone
Q1.
Atomization enthalpy of graphite + atomization enthalpy of H2 + 4 x (bond formation enthalpy of C-H) = -75kJ/mol
Atomization enthalpy of H2 given in data booklet.
Bond formation enthalpy of C-H given in data booklet.
Make Atomization enthalpy of graphite the subject.
Q2.
From rate constant, calculate half-life.
Then apply the formula :
Molarity (desired) / Molarity (initial) = (1/2)^n
where n = no. of half-lives.
H2 Chem qns
Daniel asked :
When 0.192g one of the following oxobromides, SiO2Br8, Si2OBr6, Si3O2Br8, Si4O3Br10, was completely reacted with water, all its bromine was converted to bromide ions and produced 0.392g of a cream precipitate when an excess of silver nitrate was added. Which oxobromide was reacted?
2) What is the difference in structure between AlF3 and AlCl3?
3)Hydrogen Sulphide reacts with water accordin to the following eQuilibrium eQuation
H2S(aQ) + 2H20(l) ---> 2 H3O+(aQ) + S2-(aQ)
Which of the following would not cause the position of eQuilibrium to shift when added to the eQuilibrium mixture?
A)Water
B)Ammonium Sulphate
C)Sodium Carbonate
D)Lead(II) Nitrate4)A cell is set up by connecting a Cu2+/Cu half-cell and an acidified Cr2O72-/Cr3+ half-cell
Which of the following correctly describe the effect of the change to emf of the cell?
A Replacing copper with an alloy of copper and silver increase emf
B Adding H2SO4(aQ) to reduction half-cell decreases emf
C Adding NaOH(aQ) to oxidation half-cell decrease emf
D Adding KNO3(aQ) to reduction half-cell does not affect emf
Ans to Q1. Find the no. of moles of each silicon oxobromide (assuming the stated sample mass). Find the no. of moles of the AgBr ppt. Apply stoichiometry (of the formulae of all 4 oxobromides) to see which fits.
Ans to Q2. The former has a giant ionic lattice structure (large magnitude of electronegativity difference), the latter exists as simple covalent molecules (small magnitude of electronegativity difference).
Ans to Q3.
A)Water, doesn't shift position of equilibrium since it's already present in large excess (since the state symbols of other reactants and products are aqueous).
B)Ammonium Sulphate, acidic hence donates protons to water to generate H3O+, consequently shifting position of equilibrium to the LHS.
C)Sodium Carbonate, basic hence accepts protons from H3O+ to generate H2CO3 which exists in equilibrium with and hence decomposes into CO2 and H2O; consequently lowering the molarity of H3O+ and thereby shifting position of equilibrium to the RHS.
D)Lead(II) Nitrate, increases the ionic product of PbS until it exceeds solubility product of PbS, thereby causing S2- to precipitate out as PbS and thereby lowering the molarity of S2- ions; consequently shifting position of equilibrium to the RHS.
Ans to Q4.
A) This change makes the average oxidation potential at the anodic compartment less positive (because silver, being even less reactive than copper, has a oxidation potential even less positive than copper), consequently emf decreases. Hence wrong option.
B) This change shifts the position of equilibrium to the RHS (in the cathodic compartment), making the reduction potential at the cathode more positive; consequently emf increases. Hence wrong option.
C) This change causes Cu2+ to be precipitated out as Cu(OH)2, thereby lowering the molarity of Cu2+(aq), shifting the position of equilibrium to the RHS (in the anodic compartment) making the oxidation potential at the anode more positive; consequently emf increases. Hence wrong option.
D) While K+ and NO3- are inert, the additional water added lowers the molarity of both Cr2O7 2- and Cr3+, but to an equal extent (because both are aqueous species); hence reduction potential at cathode does not change; consequently emf does not change. Hence correct option.
'A' levels / 'O' levels / Scholarstic Assessment Test
Equal volumes of 0.1mol/dm3 of sodium phosphate and 0.1mol/dm3 of magnesium nitrate were mixed together, to precipitate out magnesium phosphate solid.
a) Determine the limiting reactant.
b) Calculate the molarity of aqueous sodium cations remaining at the end of the reaction.
Ans :
a) Magnesium nitrate.
b) 0.15 mol/dm3
'A' Level Qn.
for the electrophilic addition of aqueous bromine to propene, why does the Br add to the first carbon atom rather then the second carbon atom?
Here's my guess. Someone correct me if I'm wrong. Bromine enters the first carbon atom rather than the second one due to steric hindrance. If you take observe the structure of propene, the first carbon atom is surrounded by only hydrogen atoms while the second carbon atom is surrounded by 2 alkyl groups. So if bromine enters the second carbon atom, the resulting molecule will be unstable.
Good attempt, but errorneous. Since alkenes are trigonal planar about their sp2 hybridized carbons, there is little steric hinderance for the incoming electrophile.The correct reason is that of electronics : of the two possible carbocation intermediates, the secondary carbocation is more stable than the primary carbocation (alkyl groups are electron donating by induction and hyperconjugation). Consequently, the pathway with the more stable carbocation requires a lower activation energy, the rate of reaction is faster, more of this pathway's product is formed, and hence we label this product as the major product.
This is the basis for Markovnikov's rule (which may fail in more complex cases, eg. addition of hydrogen halide to 3,3,3-trichloropropene. Draw the full mechanism to figure out why).
daniel798 persisted :
since the secondary carbocation is more stable, wouldnt it mean the bromine will add to the second carbon to form a secondary carbocation. then H2O will add to the first carbon atom?
I patiently continued :
No, to generate the secondary carbocation, the Br+ electrophile will be added onto the terminal C atom, and not to the secondary C atom.
When aqueous solvent is used, the propene nucleophile will attack the electrophilic bromine atom (due to inter electron repulsion between the electron-rich alkene and the Br2 molecule, a temporary dipole is induced in the Br2 molecule rendering it electrophilic), consequently forming either a secondary carbocation (major product pathway) or a primary carbocation (minor product pathway).The water nucleophile (note : although JCs commonly teach the nucleophile being the hydroxide ion nucleophile, but this is technically inaccurate as the molarity of water nucleophile far exceeds the molarity of hydroxide ions, even if the hydroxide ion is a much stronger nucleophile than water) then attacks the carbocation intermediate, consequently losing a proton (to rid itself of the destabilizing positive formal charge on the O atom).
Accordingly, the secondary alcohol would be the major product, and the primary alcohol would be the minor product.
.
'A' Level Qn.
Audi asked :
For each of the following reactions A and B,
(i) identify the 2 acids and 2 bases present
(ii) suggest, with reasons, which ion or molecule is the stronger acid, and which is the stronger base.
A : NH3 + H2O <----> NH4+ + OH- Kc =1.8 X 10^-5 mol/dm3
B C6H5O- + CH3CO2H <---> C6H5OH + CH3CO2 Kc=1.3 X 10^5 mol/dm3
Solution :
For A, NH3 is the base, H2O is the acid, NH4+ is NH3's conjugate acid, OH- is H2O's conjugate base. Since position of equilibrium lies to the left, OH- ion is a stronger base than NH3, and NH4+ is a stronger acid than H2O.
For B, phenoxide ion is the base, ethanoic acid is the acid, phenol is phenoxide ion's conjugate acid, and ethanoate ion is ethanoic acid's conjugate base. Since position of equilibrium lies to the right, phenoxide ion is a stronger base than ethanoate ion, and ethanoic acid is a stronger acid than phenol.
As an exercise in Organic Chem, you should also be able to come to the same conclusion for which are the stronger acids & bases, by looking at electronegativities, induction and resonance considerations of the charges (ie. analysis of electronics) of LHS vs RHS species, as an alternative to using Kc values. If you do this correctly, the conclusions should concur with the Kc values.
'A' Level Qn.
Audi asked :
Calculate the pH of a solution which is 0.40 moldm-3 w.r.t ethanoic acid and 0.20mol dm-3 w/r/tsodium ethanoate
[Ka for ethanoic acid = 1.80 X 10^-5 mol/dm3)
Calculate the change of pH of 1.0 dm3 of the above solution in (a) when 0.050 mol of solid NaOH is added (assume no change in vol)
What is the change of pH when 0.050mol of NaOH is aded to 1.0 dm^3 of pure water?
Solution :
I recommend you do not use the Henderson-Hasselbach equation. Instead just use the formulaic definition of Ka, which (like any Kc) is basically the molarities of RHS over the molarities of the LHS, for the proton dissociation equation of the weak acid.
Do the ICE table for CH3COOH <---> H+ + CH3COO-. You'll notice that for a buffer system, the equilibrium molarities of the weak acid and its conjugate base are approximated back to initial molarities, which means you can skip the ICE table if you wish. Your ICE table should have units as molarities.
Plug the Equilibrium molarities into the Ka formula, make molarity of protons [H+] the subject, and hence find pH. (Ans : 4.44)
Do the ICF table for adding the sodium hydroxide. Usually we use moles as units for the ICF table, but since the volume of solution does not change in this instance, we can work directly in molarities.
Your ICF table for CH3COOH + OH- --> CH3COO- should have Initial molarities 0.4, 0.05 and 0.2. Your Final molarities should be 0.35, 0, 0.25.
Plugging these values into your Ka formula (for acidic buffers, you can use Ka of the acid or Kb of the conjugate base), and making molarity of protons the subject, you obtain pH = 4.5986 or 4.6 to 3 sf. Hence the change in pH is approx 0.1586 or 0.159 to 3 sf.
As to the adding of NaOH(s) to pure water (pH 7 at rtp), notice that NaOH is a strong alkali. Hence, no Kb value is required. The pOH is 1.3 hence pH is 12.7, hence change in pH is 5.7.
'A' level H2 Chem Qn (Acid-Base Equilibria)
Given 150 cm3 of 2.0 mol/dm3 of ammonia solution (pKa of NH4+ = 9.25) at r.t.p., what volume of 0.5 mol/dm3 of hydrochloric acid is needed to prepare a buffer solution with pH = 9.5?
Solution (pun intended) :
Let x be the volume of HCl added to achieve a pH of 9.5
NH3 + H+ ---> NH4+
I (mols) 0.3 | 0.5x | 0
C (mols) -0.5x | -0.5x | +0.5x
F (mols) 0.3 - 0.5x | 0 | 0.5x
[NH3] = (0.3 - 0.5x) / (0.15 + x)
[NH4+] = (0.5x) / (0.15 + x)
At pH 9.5, pOH = 4.5 and [OH-] = 10^-4.5
Hydrolysis of NH3 : NH3 + H2O <---> NH4+ + OH-
Kb = ( [NH4+] [OH-] ) / [NH3]
1.7783 x 10^-5 = [(0.5x) / (0.15 + x)] [10^-4.5] / [(0.3 - 0.5x) / (0.15 + x)]
x = 0.21596 dm3 = 216 cm3 (to 3 sf)
'A' level H2 Chem Qn (Acid-Base Equilibria)
Q1. Which of the following is true for pure (ie. non-acidic, non-alkaline) water at (i) 274 K? (ii) 299 K?
a) pH = 7 b) pH < 7 c) pH > 7 d) pOH = 7 e) pOH < 7 f) pOH > 7
Q2. Is a solution with pH = 7 considered acidic, neutral or alkaline, at (i) 274 K? (ii) 299 K?
Answers :
Q1.(i) - pH > 7 and pOH > 7
Q1.(ii) - pH < 7 and pOH < 7
Q2.(i) - acidic
Q2.(ii) - alkaline
Topic : Acid-Base Equilibria
Regarding the Henderson-Hasselbalch equation, it's not that you shouldn't use it, it's that you need never use it. You can of course, use it if you want. But it's *never* compulsory to use it.
Units for Molarity (ie. molar concentration) = M = mol/dm3.
0.2 moles of sodium benzoate is dissolved in 10 mL of 0.1 M benzoic acid solution. Calculate the pH. (Kb of benzoate ion = 1.5x10^-10).
Solution (pun intended) :
Assuming you used solid sodium benzoate and the volume of solution didn't change, then we have a buffer solution consisting of weak acid benzoic and its conjugate base benzoate. Since it is a buffer system, to calculate pH we can consider either the hydrolysis of the base (ie. Kb of benzoate), or the proton dissociation of the acid (ie. Ka of benzoic acid).
Let's consider the hydrolysis of the benzoate ion.
Benzoate ion + water <--> Benzoic acid + hydroxide ion
I (molarity) 20 | - | 0.1 | 0
C (molarity) -x | - | +x | +x
E (molarity) 20-x | - | 0.1+x | x
Since x is much smaller than initial molarities of both weak acid and its conjugate base (since this is a buffer solution), hence equilibrium molarities are approximated back to initial molarities
Kb = [benzoic acid] [hydroxide ion] / [benzoate ion]
1.5 x 10^-10 = [0.1] [OH-] / [20]
[OH-] = 3 x 10^-8
pOH = 7.52
pH = 6.48
Also, note that because the only reaction between benzoic acid (an acid) and benzoate ion (a base) is a Bronsted-Lowry acid-base proton transfer reaction (don't say "neutralization reaction" that's 'O' level terminology that shouldn't be used at JC/Poly level). In such a reaction, a proton is transferred from the Bronsted acid to the Bronsted base.
benzoic acid + benzoate ---> benzoate ion + benzoic acid.
As you can see, the final moles (and hence the molarities) of benzoic acid and benzoate ion do not change. If it helps you to understand, you can take it that BOTH the weak acid AND its conjugate base both undergo hydrolysis to an equal extent, and hence cancel out each other's effect.
And since the moles and molarities (of both weak acid and conjugate base) do not change, we have the accurate values for molarities of both the weak acid and its conjugate base, and therefore by using either the Kb (of the conjugate base) or Ka (of the weak acid), we are considered to already have taken everything (ie. proton dissociation from the weak acid AND hydrolysis of the conjugate base) in our calculations. This is because Ka and Kb are related by Kw.
'A' Level H2 Chem Qns. Topic : Organic Chemistry mechanisms.
These questions test whether the candidate is able to apply fundamental concepts of OC (Organic Chem) to deduce mechanisms involving (either as reactants or products) compounds not usually given much attention to, in the H2 Chem syllabus. Examples of such compounds are ethers and epoxides.
Draw the mechanisms for :
1) Propene + aqueous bromine; followed by alcoholic KOH (heat under reflux). Product : 1,2-epoxypropane.
2) Ethoxyethane + hydroiodic acid (heat under reflux). Immediate products : ethanol and iodoethane. Final products (with excess HI) : iodoethane and water.
3) Epoxyethane + aqueous acid (heat under reflux). Products : ethan-1,2-diol.
4) Epoxyethane + methylammonium chloride (heat under reflux). Products : 2-hydroxy-N-methyl-ethylamine.
5) 1,2-epoxypropane + ethyl magnesium bromide (Grignard reagent), followed by hydrolysis. Hint : alkyl addition occurs at the less sterically hindered carbon atom. Product : pentan-2-ol
Solution :
1) Propene nucleophile attacks Br+ electrophile; H2O nucleophile attacks secondary carbocation; proton lost to rid O+ of +ve formal charge; basic OH- ion abstracts proton from hydroxy group of 1-bromo-propan-2-ol; -ve formal charged O atom nucleophilically attacks (ie. donates dative bond to) electrophilic C atom (bonded to the electronegative halogen) without violating C's octect; thanks to the halogen which leaves as a halide ion (halogens are good leaving groups because halide ions are stable, due to their large ionic radius and therefore low charge density).
2) Immediate products : O atom of epoxide is protonated, giving rise to a +ve formal charge on O atom and a strong +ve partial charge on the adjacent C atom, rendering it electrophilic. Halide ion nucleophile then attacks the electrophilic C atom, causing heterolytic cleavage of the O-C bond, ridding the O atom of its +ve formal charge.
Final products : the ethanol further reacts with hydroiodic acid as follows : the hydroxy group is protonated, upgrading a poor leaving group (because hydroxide ion is unstable) to an excellent leaving group (because water is stable). Iodide ion nucleophile attacks electrophilic C atom and water is eliminated.
3) O atom of epoxy group protonated; water nucleophile attacks electrophilic C atom; O atom loses proton to rid itself of the +ve formal charge.
4) O atom of epoxy group protonated; methylamine nucleophile attacks electrophilic C atom; N atom loses proton to rid itself of the +ve formal charge.
5) Grignard reagent (ie. carbon / alkyl nucleophile) attacks the less sterically hindered C atom (ie. the 1st C atom not the 2nd). The C-O bond heterolytically cleaves to generate the pent-2-oxide ion, which abstracts a proton from water (ie. hydrolysis) to generate the final pentan-2-ol product.
'A' levels / Challenging 'O' levels
Chemistry Calculations Qn.
Titanium(IV) oxide TiO2, is heated in hydrogen has to give water and a new titanium oxide. TixOy. if 1.598g of TiO2 produced 1.438g of TixOy, what is the formula of the new oxide?
Solution :
No of moles of Ti before redox = No of moles of Ti after redox
this implies :
1 (No of moles of TiO2) = x (No of moles of TixOy)
this implies :
1 (1.598 / 79.9) = x (1.438 / (47.9x + 16y))
this implies :
x / y = 2 / 3
Therefore, formula of titanium oxide = Ti2O3
'A' Levels / Challenging 'O' Levels
Chemistry Calculations Qn.
1) A mixture of MgSO4.7H2O and CuSO4.5H2O is heated until a mixture of anhydrous salts is formed. if 5.0g of the mixture give 3.0g of the anhydrous salts, calculate the percentage by mass of MgSO4.7H2O in the mixture.
2) A solution of acid contains 7.30g of HOOC(CH2)nCOOH per dm^3 of solution. 20.0cm^3 of this acid was titrated against NaOH, using phenolpthalein indicator. in the titration, 25.0cm^3 of the alkali was used. the sodium hydroxide contained 1.36g of hydroxide ion per dm^3. Calculate the value of n in the molecular formula and therefore name this acid.
Solution :
Q1. Let x and y be the moles of CuSO4 and MgSO4 present. You can write up to 3 mathematical equations.
?x + ?y = 2g
and
?x + ?y = 3g
and
?x + ?y = 5g
Choose any 2 of these equations, and solve simultaneous equations for x and y.
Q2. From mass concentration of OH- solution, find molarity of OH- solution. Hence find moles of OH- ions used in titration. Hence find moles of protons (ie H+ ions) neutralized. Hence find moles of the (diprotic) dicarboxylic acid present. Hence find molarity of the (diprotic) dicarboxylic acid. Find molar mass of the (diprotic) dicarboxylic acid, in terms of algebraic variable n. Hence find (continuing in terms of algebraic variable n) mass concentration of the (diprotic) dicarboxylic acid. Therefore equate this mass concentration value to 7.30g/dm3. The acid is hexane-1,6-dioic acid, or simply hexanedioic acid.
A levels H2 Chem
Topic : Organic Chem
For (C6H5)CH=CHCH3,
when this compound undergoes nitration, which position does the propene direct the NO2+?also when this molecule undergoes hydration in H3PO4, where does the H+ and the OH- add since both carbon atoms have 1 R group. When adding to an alkene, does length of R group affect the product or only the number of R groups affect the product?
when an alkyl halide undergoes elimination, under sodium hydroxide in alcoholic medium,
CH3CH2X -----> CH2=CH2 + HXthen,
HX + OH- -----> H2O + X-
why does HX react with OH- ion when NaOH is in alcohoic medium, so there aren't any OH- ions present?
also when reacting alkyl halide with ammonia in alcoholic medium,
CH3CH2X + NH3(alc) -----> CH3CH2NH2 + HXwhy doesn't HX react with NH3 here?
Regard the alkene group as an alkyl group (by induction, it is neither electron donating nor withdrawing; by resonance it can be both electron donating and withdrawing). In other words, a weakly activating, ortho-para directing substituent.
Accordingly, the product of the electrophilic aromatic subtitution is generated as the NO2+ electrophile substitutes away a proton on either the para position (since a propenyl group presents significant steric hinderance) or the ortho position (since there are two ortho positions versus one para position). For A levels, you need to give both ortho and para isomeric products.
During hydration of the alkene double bond, the more stable carbocation intermediate will be the carbocation atom (ie. C with +ve formal charge) bonded directly to the benzene ring, because the benzene ring can stabilize the carbocation by both induction/hyperconjugation and resonance, while the terminal methyl group can stabilize the carbocation by induction/hyperconjugation only.
Therefore, the major alcohol product will have the OH group bonded to the C atom next to the benzene ring, since a more stable intermediate means lower activation energy means faster rate of reaction means more of *that* product formed, which we identify by use of the label "major product".
As for length of alkyl chain, yes the greater the length, the greater its electron-donating by induction capacity. Although this is not usually tested at H2 Chem level, so just FYI.
The equation you quoted for elimination of HX from alkyl halides using alcoholic OH- ions, is misleading. If you know the mechanism (I encourage all H2 students to approach OC (Organic Chem) from an understanding-of-mechanism instead of blind-route-memorization approach), then you would realize that the equations given are misleading.
The OH- ion in alcoholic solvent has a greater propensity to behave as a base rather than a nucleophile (since the alcoholic solvent affords less hydrogen bonding stabilization to the OH- ion compared to aqueous water solvent). Understand that bases accept protons in a bid to stabilize themselves. And so the OH- ion under alcholic solvent functions as a base (and carries out elimination) rather than as a nucleophile (to carry out substitution).
A Note on Nucleophilicity versus Basicity : Comparing F- with I-, you should be able to articulate that F- is a stronger base (since higher charge density = more unstable = more desperate to protonate itself) while I- is a stronger nucleophile (since greater ionic radius = electron charge clouds more polarizable = more willing to donate dative bonds to electrophiles).
Mechanism of Elimination of HX :
The alcoholic OH- ion abstracts ("plucks") a beta proton; the sigma bond between the beta carbon and beta proton becomes a pi bond between the alpha and beta protons; without violating the alpha carbon's octet thanks to halogen which leaves as a halide ion (remember that halogens are good leaving groups because halide ions are stable because of large ionic radius and thus low charge density; of the halogens only F is a poor leaving group because F- is unstable due to high charge density).
Accordingly, if you wish to represent the elimination reaction with an equation (although the full mechanism is always preferred).
CH3CH2X(alc) + OH-(alc) -----> CH2=CH2(alc) + H2O(alc) + NaX(alc)
Write the above equation, and *not* the two step equation you quoted (ignore your JC's notes, don't use or buy or study *any* JC's lecture notes, every JC's notes are different and many are often filled with nonsense, just use CS Toh's A level Study Guide instead for the A levels).
Similarly, based on the mechanism, the nucleophilic attack of ammonia on alkyl halides should be written as :
CH3CH2X(alc) + NH3(alc) -----> CH3CH2NH3+X-(alc)
or
CH3CH2X(alc) + NH3(alc) -----> CH3CH2NH3+(alc) + X-(alc)
Note that the positive formal charge is on the N atom on the ammonium salt, because the N atom has 4 bond pairs and 0 lone pairs, but N is in Grp V. To obtain the amine (ie. the conjugate base form), simply deprotonate the ammonium salt using aqueous NaOH or KOH.
CH3CH2NH3+X-(aq) + Na+OH-(aq) -----> CH3CH2NH2(aq) + H2O(aq) + Na+X-(aq)
or
CH3CH2NH3X(aq) + NaOH(aq) -----> CH3CH2NH2(aq) + H2O(aq) + NaX(aq)
'A' Level H2 Chem.
Topic : Balancing Redox Eqns
Q1.
Fe2+ + MnO4- => Mn2+ + Fe3+
Write the reduction half-equation :
MnO4- ---> Mn2+
Balance the heteroatoms, ie. Manganese. Already balanced.
Balance the O atoms.
MnO4- ---> Mn2+ + 4H2O
Balance the H atoms by adding protons.
8H+ + MnO4- ---> Mn2+ + 4H2O
Balance the charges by adding electrons.
8H+ + MnO4- + 5e- ---> Mn2+ + 4H2O
Write the oxidation half-equation :
Fe2+ ---> Fe3+
Balance the heteroatoms, ie. Iron. Already balanced.
Balance the O atoms. No need.
Balance the H atoms. No need.
Balance the charges.
Fe2+ ---> Fe3+ + e-
To obtain lowest common multiple for electrons,
multiply oxidation half-equation by 5
5Fe2+ ---> 5Fe3+ + 5e-
Add LHS + LHS, RHS + RHS for both half-equations :
5Fe2+ ---> 5Fe3+ + 5e-
+
8H+ + MnO4- + 5e- ---> Mn2+ + 4H2O
=
5Fe2+ + 8H+ + MnO4- ---> 5Fe3+ + Mn2+ + 4H2O
Q2.
Cr2O7 2- + C2O4 2- => Cr3+ + CO2
Write down the reduction half-equation.
Cr2O7 2- ---> Cr3+
Balance the heteroatoms, ie. Chromium.
Cr2O7 2- ---> 2Cr3+
Balance the O atoms.
Cr2O7 2- ---> 2Cr3+ + 7H2O
Balance the H atoms.
14H+ + Cr2O7 2- ---> 2Cr3+ + 7H2O
Balance the charges.
6e- + 14H+ + Cr2O7 2- ---> 2Cr3+ + 7H2O
Write the oxidation half-equation.
C2O4 2- ---> CO2
Balance the heteroatoms, ie. Carbon.
C2O4 2- ---> 2CO2
Balance the O atoms. Already balanced.
Balance the H atoms. None present.
Balance the charges.
C2O4 2- ---> 2CO2 + 2e-
To obtain lowest common multiple for electrons,
multiply oxidation half-equation by 3
3C2O4 2- ---> 6CO2 + 6e-
Add LHS + LHS, RHS + RHS for both half-equations :
3C2O4 2- ---> 6CO2 + 6e-
+
6e- + 14H+ + Cr2O7 2- ---> 2Cr3+ + 7H2O
=
3C2O4 2- + 14H+ + Cr2O7 2- ---> 6CO2 + 2Cr3+ + 7H2O
'A' Level H2/H1 Chem
Challenging 'O' level Chem
Topic : Chem Calculations
A mixture of methane and ethane with combined sample mass of 13.43g is completely burned in oxygen. If the total sample mass of CO2 & H2O produced is 64.84g, calculate the percentage by volume and percentage by mass, of the two alkanes in the mixture.
Solution :
CH4 + 2 O2 ---> CO2 + 2 H2O
C2H6 + 7/2 O2 ---> 2 CO2 + 3 H2O
Let x and y be the moles of CH4 and C2H6 burnt.
x(12+4) + y(12+12+6) = 13.43g ---------- (equation 1)
and
x(12+16+16) + 2x(2+16) + 2y(12+16+16) + 3y(2+16) = 64.84g ---------- (equation 2)
From equation 1,
16x + 30y = 13.43
x = (13.43 - 30y)/16
Substitute into equation 2,
44(13.43 - 30y)/16 + 36(13.43 - 30y)/16 + 88y + 54y = 64.84g
44(13.43 - 30y) + 36(13.43 - 30y) + 16(88)y + 16(54)y = 16(64.84)
590.92 - 1320y + 483.48 - 1080y + 1408y + 864y = 1037.44
-128y = -36.96
y = 0.28875 moles of C2H6
x = 0.29797 moles of CH4
Percentage by volume of CH4 = percentage by moles of CH4 (assuming constant temperature and pressure) = moles of CH4 / total moles of both gases
= 0.29797 / (0.28875 + 0.29797) = 50.8%
Percentage by volume of C2H6 = 100% - percentage by volume of CH4 = 49.2%
Percentage by mass of CH4 = sample mass of CH4 / total sample mass of both gases
= 4.76752 / (4.76752 + 8.6625) = 35.5%
Percentage by mass of C2H6 = 100% - percentage by mass of CH4 = 64.5%