h2 chem qns

absol SG

1. Solid NaCl and conc. H2SO4 can be used to generate HCl in situ for the reaction with alkenes to produce choloroalkanes.

(iii) explain why 2-chloropropane is formed as the major prod. instead of 1-cholorpropane. (2 marks)

^i say its because of markonikov's rule but sounds dumb to me...or is it because the c=c with more H attached to it is more electron rich due to positive inductive effect of H, hence it is better at attracting the positive part of the addend ie H of HCl?

(iv) can a high yield of 2-iodopropane be obtained by a similar method by replacing NaCl with NaI? explain your answer with appropriate equations. (3 marks)

^dont know lol.

2. Hydrazine is sometimes used as a rocket fuel based on its reaction with oxygen. in this question you may assume that hydrazine is used in aqueous form in a fuel cell using dilute sulphuric acid as the electrolyte.

hydrazine = N2H4 (aq) in this case.

then my question is...since this is a fuel cell, it would be like 2 nonpermeable/seperate compartments right, 1 containing O2 gas and the other with N2H4, but why would they still need h2so4 as electrolyte?

reduction eqn: N2(g) + 4H+ + 4e double arrow N2H4(aq)

3. Au+ + e -> Au Ered=+1.69V

Au3+ + 3e -> Au Ered = +1.50V

using the above data explain why gold does not tarnish easily in air.

^i tried to explain something like because it has extremely negative Eox values, hence there are little oxidising agents strong enough with Ered > 1.5 to 1.69 to be able to start the redox reaction with Ecell>0, is that correct?

4. C2H6 + Br2 -> C2H5Br + HBr

(ii) explain why is this described as a free radical substituition reaction

(iii) explain why this reaction is an example of homogeneous catalysis.

5. this 1 involves 3 compounds M, N and Q.

M has a benzene ring with the branch CH2OH attached to it.

N has a benzene ring with the branch COOH attached to it.

Q has a benzene ring with a branch CH3, and taking that carbon substituted to be carbon number 1, on carbon number 4, there is an OH substituent on the benzene ring.

(i) describe and explain how the acidities of M, N and Q compare with each other

^i think this qns max 3 marks.

6. an industrial use of HCl is to make chlorine dioxide, which is manufactured on a massive scale for the bleaching of wood pulp and purification of water supplies. the reaction eqn is as follows:

NaClO3 + 2HCl -> ClO2 + 1/2 Cl2 + NaCl + H2O

Discuss the changes in oxidation no. of cholrine that occurs during this reaction.

^this qns i catch no ball. as in dunno what is happening in the eqn...i think its disproportionation of Cl in NaClO3, but...can Cl in HCl disproportionate also?

ty.

yiha093 SG

based on sec 4 limited knowledge as compared to h2

fuel cell no electrolyte how to allow flow of ions and hence oxid and reduct which gives out electrons to discharge electricity?

taking a jab at 1iv . i guess its because of reactivity or eqn that proves the yield.. but then again, not plausible ~.~

qn 6 i use english only i interpret it as * state reactant what oxi state - to what product state*

? _ ?

Wheref SG

For Question 1 :

I believe that you know that the reaction of conc. H2SO4 with NaCl gives u NaHSO4 and HCl right? This is under Group 7 chapter. This reaction only gives you these 2 products.

However, for H2SO4 and NaI, you will eventually get

H2SO4 + NaI ---> NaHSO4 + HI

8HI + H2SO4 --> 4I2 + H2S + 4H2O

The HI produced are oxidised again (due to them being strong reducing agents, thus tends to undergo oxidation) resulting in 2 gaseous products, the I2 and H2S products.

In conclusion, you will not get a high yield of 2-iodopropane.

UltimaOnline SG

Wheref is correct.

As for the other part of Qn :
As predicted by Markonikov’s Rule, the reaction pathway involving the more stable secondary carbocation intermediate resulting in the generation of 2-chloropropane, will be the major product. The reaction pathway involving the less stable primary carbocation intermediate resulting in the generation of 1-chloropropane, will be the minor product.

As for the rest of the questions, I’ll give my further comments later tonight (only if there is still any need for that, seeing how there are lots of helpful folks around).

UltimaOnline SG

Q2. Even if the two electrodes are compartmentalized, they need to be linked with a salt bridge, and an electrolyte capable of conducting electricity (in the form of mobile ions) still need to be present. If the two compartments were truly separate, charge would buildup and current would not flow.

Q3. Correct. Use oxidation potentials together with reduction potentials (for any given redox reaction) to consider feasibility/spontaneity.

Q4. The mechanism involves free radicals. It’s “homogenous” because the catalyst is in the same state as the reactants. A possible catalyst for this reaction would be a halogen radical.

Q5. Benzoic acid is most acidic, followed by para-methylphenol, followed by phenylmethanol. Explain using stabilities of conjugate bases after proton dissociation, and/or polarity/strength of conjugate base – proton bond before proton dissociation. Use electron donating/withdrawing by resonance/induction arguments.

Q6. Effectively, disregarding spectator ions, the O.S. of chlorine changes from +5 and -1 on LHS, to 0 and +4 on RHS. So it is still considered disproportionation as chlorine is simultaneously reduced (from +5) and oxidized (from -1).

absol SG

@ 1(iv)

wow, i seriously didnt even think so far. shit..havent finish memorising group 7 notes, and i didnt know HI will be oxidised when reacted with H2SO4 again =(.

@Q2,

N2H4 will undergo oxidation to produce H+ ions which are mobile charge carriers as given in e equation, so in abscence of the electrolyte H2SO4 will the electrochemical cell actually still work?(since i think its sort of like a simultaneous thing)

@Q5,

i still dont understand TT. both phenylmethanol and para-methylphenol are alcohols, so i understand that they are near neutral. but how do u explainusing stability of conjugate base in this case? as in, both will experience the positive inductive effects of CH3 also....so how do u actually determine which conjugate base more stable here?

@Q6

are there any err..common disproportionation eqns we need to memorise by heart in H2? like the last yr chem paper 2, last qns on transition metal, i totally didnt even know about the disproportionation of Cu+ ions...if there are a few that needs to be memorised, please tell me...haha. dont want to get stuck during an a lvl exam ):

all the other questions unnamed im fine with them, thanks for e help so far.

UltimaOnline SG

Q2
In the Hydrazine fuel cell, hydrazine is oxidized to molecular nitrogen at the anode (release electrons to the external wire), and molecular oxygen is reduced to water at the cathode. You’ll need enough H+ present at the beginning, it’s unrealistic and inefficient to expect the cathode to rely solely (eg. if you used deionized water) on the new H+ generated at the anode from the oxidation of hydrazine.

Q5
Phenols are more acidic than alcohols, since the conjugate base (phenate or phenoxide ion) is stabilized by having the negative formal charge of the C6H5O- delocalized by resonance into the benzene ring. The conjugate bases of alcohols are alkoxide ions, which are particularly unstable (thus are strong bases and nucleophiles) because the alkyl group is electron donating by induction and hence intensifies and destabilizes the negative formal charge of the RO-.

Q6
Yes, you need to memorize the disproportionation redox reactions of molecular chlorine with aqueous sodium hydroxide at 15 deg C (to generate sodium chloride, sodium chlorate(I), and water) and at 70 deg C (to generate sodium chloride, sodium chlorate(V), and water).