2016 H2 Chemistry JC1 & 2 students post your questions here

last SG · 2016-02-03T02:17:14+00:00

Hello, would like to ask a questionAs to why c=c is stronger than c-c bond, is it because of the additional pi bond, or because the sigma bond in c=c is shor...

Shinyphua SG

Question 3 part a

UltimaOnline SG

Question 3 part a

Eh ur FB private izzit? Can't see the pic lah. Try using Twitter or smth else.

Ng.keebin SG

When you treat nitrobenzene with Sn in conc HCl to form phenylamine, why do you still need to add NaOH after that?

UltimaOnline SG

When you treat nitrobenzene with Sn in conc HCl to form phenylamine, why do you still need to add NaOH after that?

Because in the presence of acidic conc HCl, the product is the protonated conjugate acid phenylammonium ion (which is water soluble due to ion - permanent dipole interactions and hence difficult to separate from the other species present), which must thus be deprotonated by excess NaOH(aq) to obtain the desired product phenylamine, which must then be extracted via steam distillation.

The experimental details of steam distillation are beyond A levels, but Cambridge can (and have in past A level papers) set challenging questions on the chemistry concepts underlying practical procedures (including steam distillation). Students seriously intending to score a distinction A grade for your A levels, are strongly advised to study the following webpages.

---------------------------------------------------------------------------------

Source : http://www.chemguide.co.uk/organicprops/aniline/preparation.html

Sodium hydroxide solution is added to the product of the first stage of the reaction. The phenylamine is formed together with a mixture of tetrahydroxozincate(II) and hexahydroxozincate(IV) complex tin ions formed from the tetrachlorozincate(II) and hexachlorozincate(IV) complex tin ions present during the acidic HCl stage. The phenylamine is finally separated from this mixture. The separation is long, tedious and potentially dangerous (and thus not experimentally carried out at A levels, but questions can still be asked on the procedure), involving steam distillation, solvent extraction and a final distillation.

---------------------------------------------------------------------------------

Source : http://chemistry.stackexchange.com/questions/10531/preparation-of-phenylamine-aniline

This is the procedure I (a Chemistry undergraduate student) carried out in the lab :
1.Reduction of nitrobenzene by using tin and HCl acid
2.After the completion of the reaction, NaOH added to neutralize excess acid and dissolve any Sn(OH)2(s)
3.Steam distillation is carried out to obtain suspension of phenylamine
4.Salting carried out where NaCl salt is added (called 'salting'). Aqueous layer discarded
5.Solvent extraction : dry ether added and mixture shaken vigorously. Aqueous layer discarded
6.Distillation under reduced pressure (ie. partial vacuum) carried out and solid phenylamine left in flask

My questions:
1. Instead of carrying out distillation under reduced pressure at the end of the procedure to remove the ether, could I have just carried out simple distillation?
2. Why is steam distillation suitable in part 3? Could I have used vacuum distillation here?

(See webpage for the answers to this Chem undergrad's qns).

---------------------------------------------------------------------------------

Source : http://rod.beavon.org.uk/phenylamine_prep.htm

•What is the reaction between the tin and hydrochloric acid?
•Sn + 2HCl → Sn2+ + 2Cl- + H2

•What is the reducing agent, and to what is it oxidised?
•The reducing agent is Sn2+, which is oxidised to Sn4+.

•What is the initial precipitate, and what does it become on addition of excess sodium hydroxide?
•Tin(IV) hydroxide Sn(OH)4 (s). This is amphoteric, and with more alkali forms the soluble Sn(OH)6 2- hexahydroxozincate(IV) complex ion.

•Why is more water added to the solution?
•This enables a technique called steam distillation. In the large-scale preparation steam is blown through the mixture; in this one the steam is generated in situ. Phenylamine distils over with the steam; any unchanged nitrobenzene and the inorganic materials do not.

•Why is the initial distillate cloudy?
•It is a mixture of water and phenylamine which formas a cloudy emulsion

•What is the clear condensate?
•Water; by now all the phenylamine has distilled over.

•What is the purpose of adding sodium chloride?
•Phenylamine is significantly soluble in water, but very much less so in saturated sodium chloride solution. This process is called 'salting out'.

•What is the purpose of adding ethoxyethane?
•Solvent extraction; phenylamine is much more soluble in ethoxyethane than it is in water.

•Why are two portions of 4 cm3 used, rather than a single 8cm3 portion of of ethoxyethane?
•The theory of solvent extraction is a branch of chemical equilibria. It can be mathemically shown (and can be asked in data-based A level exam questions) that any solvent extraction is more effective if a given volume of extracting solvent is used in several portions rather than in a single one.

•Suggest a reason for using KOH as a drying agent, rather than the more conventional calcium chloride or sodium sulphate.
•The use of potassium hydroxide would also eliminate any traces of hydrochloric acid in the phenylamine.

•Why must all flames in the laboratory be extinguished?
•The vapour of ethoxyethane is very dense and will creep along bench-tops over a considerable distance. It is possible for an explosion to ensue even if the source of ignition is several metres away from where the ethoxyethane is being used.

•Why is the water run out of the condenser?
•Phenylamine has a sufficiently high boiling temperature for an air condenser to be efficient at condensing the vapour.

•Why is the phenylamine kept condensing on the thermometer bulb before allowing distillation to proceed?
•This allows the thermometer to come into thermal equilibrium with the vapour - a general necessity in distillation, but needing more care than usual if the boiling temperature of the distillate is high. Phenylamine boils at 184oC.

hykel SG

https://twitter.com/_hykelchem/status/707837536685588480?s=09

I don't really know how to answer both question 7 and 8. Please help!

UltimaOnline SG

https://twitter.com/_hykelchem/status/707837536685588480?s=09

I don't really know how to answer both question 7 and 8. Please help!

For the HCJC qn, only C is possible, since the halogen radical only abstracts H atoms in propagation steps, leaving behind alkyl radicals which can form new C-C bonds with other alkyl radicals in termination steps. Options A, B and D require cleavage of C-C bonds, and/or other alkane molecules such as methane and ethane to be available. Try drawing out the reaction mechanisms, you'll find only C is possible.

For the MJC qn, the 2 factors which determine product distribution during free radical substitution are : no. of H atoms substitutable, and stability of alkyl radical intermediates (aka reactivities of pri vs sec vs tert H atoms). Based on no. of H atoms substitutable, ratio of 1-bromo-2-methylpropane to 2-bromo-2-methylpropane should be 9 : 1. Based on stability of alkyl radical intermediates, ratio of 1-bromo-2-methylpropane to 2-bromo-2-methylpropane should be 1 : 6. Hence, actual ratio obtained (ie. combining the 2 factors mathematically by multiplication) = 9 x 1 : 1 x 6 = 9 : 6 = 3 : 2

qeducation SG

When you treat nitrobenzene with Sn in conc HCl to form phenylamine, why do you still need to add NaOH after that?

After the addition of Sn in conc HCl, it will form C6H5NH3+, as phenylamine being a base, will react with conc HCl to form a salt. Thus, the addition of NaOH is to turn the salt C6H5NH3+ back into phenylamine.

-Quintessential Education

Find out more: http://qeducation.sg/

hykel SG

https://twitter.com/_hykelchem/status/710759882488549376?s=09

I'm not sure how to get compound A and B. Please help :)

UltimaOnline SG

https://twitter.com/_hykelchem/status/710759882488549376?s=09

I'm not sure how to get compound A and B. Please help :)

Step 3 is nucleophilic acyl substitution (addition-elimination) to get amide group. Step 4 is nucleophilic aliphatic substitution (SN2) to get ether group.

In step 6, thionyl chloride converts carboxylic acid group to acyl chloride group.

For A grade H2 Chem, H3 Chem and Olympiad Chem students, here is the SOCl2 mechanism by James Ashenhurst : http://www.masterorganicchemistry.com/2011/12/03/reagent-friday-thionyl-chloride-socl2

For A grade H2 Chem, H3 Chem and Olympiad Chem students who are seriously intending to study Chemistry, Chemical Engineering, Biochemistry, Pharmacy, Dentistry or Medicine in the University, here is a more advanced discussion on the stereochemistry of the SOCl2 mechanism :

Image source : https://en.wikipedia.org/wiki/SNi

If you have difficulty understanding the diagram above, James Ashenhurst explains the reaction in detail here : http://www.masterorganicchemistry.com/2014/02/10/socl2-and-the-sni-mechanism/

Ephemeral SG

Hello! Keebin and I did the organic chem deductive elucidation killer Q13 and we got a few answers as attached -

When O C4H6O6 is heated with concentrated sulfuric(VI) acid, two possible products M and G with molecular formulae C4H2O4 and C8H8O10 are formed as a result of intramolecular and intermolecular esterifications, respectively. Elucidate O, M, G.

https://twitter.com/123chem2016/status/711185220322463745

M4 is obtained frm O1 as well. Are they all possible? As for the G compounds, degree of unsaturation for G3 is 7 instead of 5. Technically G1 can also under elimination (of water) to form another C=C...

Benzaldehyde is less reactive than methanal. Why is that so? Why does delocalisation of electrons through resonance reduce the electrophilicity of carbonyl carbon and make it less reactive? If the the electrons are delocalised over to the O atom, then doesn't it increase the partial positive charge on carbonyl C and hence make it more susceptible to nucleophilic addition?

When Grignard reagents react with carbonyl, a covalent bond is formed between O and MgX. Why can't it be an ionic bond?

UltimaOnline SG

Hello! Keebin and I did the organic chem deductive elucidation killer Q13 and we got a few answers as attached -

https://twitter.com/123chem2016/status/711185220322463745

When Grignard reagents react with carbonyl, a covalent bond is formed between O and MgX. Why can't it be an ionic bond?

All your structures are wrong. The question specified the products are generated as a result of intra and inter molecular esterifications, and dehydration of alcohol to alkene is not involved (as verified by the degree of unsaturation). O1 (hence M1 and G1 as well) is/are wrong because geminal diols spontaneously dehydrate to generate aldehydes / ketones, with a negligible molarity of the geminal diol present at equilibrium (it's true that KMnO4 and K2Cr2O7 oxidizes aldehydes via their hydrolysis into geminal diols, but KMnO4 and K2Cr2O7 are able to thusly shift the position of equilibrium to the RHS as the geminal diol is oxidized into carboxylic acid). Your O2 (and thus M2, G2, M3, G3 as well) is/are wrong because your M2, M3 and G3 suffers from excessive ring strain due to angle strain (an approximately 90 deg 4-membered ring is already strained from sp3 C atoms ideally 109.5 deg, let alone sp2 vinylic C atoms ideally 120 deg C). Now that I've revealed your structures of O are wrong, there's only 1 possible structure left, which is the correct structure of O, and hence the correct M and G follows accordingly.

Because of both sterics and electronics : the electrophilic C atom in benzaldehyde suffers from greater steric hindrance (for the incoming nucleophile) compared to methanol ; in terms of electronics methanal has a valid resonance contributor in which the carbonyl C atom has a unipositive formal charge, while in benzaldehyde the unipositive formal charge is delocalized from the benzylic C atom over to the ortho, para and ortho C atoms in the benzene ring, consequently in their resonance hybrids, the magnitude of partial positive charge and hence electrophilicity, is significantly greater in the methanal C atom than in the benzaldehyde C atom.

The O-Mg bond in the intermediate R-O-Mg-X upon nucleophilic addition of the Grignard reagent, has significant covalent and ionic character both. Unless otherwise specified by the question, Cambridge will accept it whether you draw & describe the bond as ionic (with significant covalent character), or as covalent (with significant ionic character). It's actually better to draw it as ionic (ie. O- +Mg with formal charges between them), to show Cambridge that you're aware of the greater magnitude of electronegativity difference between O & Mg compared to C & Mg (which is more covalent than ionic).

WILLYNILLY SG

Hello! I love Chemistry.

UltimaOnline SG

A couple of JC students asked in the "Lurkers" H2 Chemistry thread a CS Toh question based on the hydrolysis of a germanium chloride. For the fun of it, I then decided to write a 2016 BedokFunland JC H2 Chemistry question based on the hydrolyses of germanium chlorides, for your entertainment and pleasure. If you can correctly answer my BedokFunland JC questions, it's indicative of getting an A grade for your Singapore-Cambridge H2 Chemistry A level exams. After all, I always get the distinction A grade myself every year, so I know exactly what it takes.

UltimaOnline SG

H2 Chemistry students are often confused between Iodometry and Iodimetry, the 2 commonly encountered types of redox titrations involving iodine. If you're uncertain of the difference, go check it up yourself now.

Tis the Age of the Internet,
Google by thy Sword,
Wikipedia be thy Shield

- BedokFunland JC

Liying98 SG

https://twitter.com/itsliyingg/status/715502704089403392

Qns 10 parts d and e

UltimaOnline SG

https://twitter.com/itsliyingg/status/715502704089403392

Qns 10 parts d and e

Compared to Singapore JC questions, Cambridge will be more specific and less ambiguous on exactly what they want you to comment on. Don't worry too much about silly Singapore JC questions in which different school teachers (even within the same JC, let alone across different JCs) routinely disagree with each other.

Formation enthalpy of Al2O3 is significantly (almost 3 times) more exothermic than formation enthalpy of MgO, despite the more endothermic 1st + 2nd + 3rd ionization enthalpies for Al compared to Mg, illustrating both the stronger ionic (with significant covalent character) bonds for Al2O3 compared to MgO (due to Al3+'s higher cationic charge density compared to Mg2+), as well as the greater thermodynamic and thermal stabilities of Al2O3 over MgO.

Al2O3 is expected to be more stable (ie. both lattice formation enthalpy and formation enthalpy more exothermic) compared to Al2S3, due to the higher anionic charge density of O2- compared to S2-.

Flying grenade SG

what's the difference betwen lattice formation enthalpy and formation enthalpy?

UltimaOnline SG

what's the difference betwen lattice formation enthalpy and formation enthalpy?

Former is from gaseous ions to solid lattice, under standard conditions. Latter is from elements to compound, under standard conditions.

Ephemeral SG

What is a simple chemical test to differentiate between 2,4,6-tribromophenol and 2,4,6-trichlorophenylamine? (Cannot use FeCl3)

Can aq NaOH or limited RX in ethanol be used?

UltimaOnline SG

What is a simple chemical test to differentiate between 2,4,6-tribromophenol and 2,4,6-trichlorophenylamine? (Cannot use FeCl3)

Can aq NaOH or limited RX in ethanol be used?

What would be your expected observations in your proposed answer? Then you'll know if your method works.

And instead of me spoonfeeding you guys all the answers, how about you 100+ lurkers post your own attempted answer? Ideally, let this forum be a discussion amongst you students helping each other out as well, and I'll come in only to guide and advise.

« Previous Page Next Page »