H2 Chem - Organic Chemistry

gohby SG

Hi UltimaOnline,

I have a question on organic synthesis.

http://img.photobucket.com/albums/v700/gohby/Chemistry/Organic%20Chemistry_zpsqovillgi.jpg

I understand that P will react with aqueous sodium hydroxide (nucleophilic substitution), resulting to the substitution of the 3 Cl groups to 3 OH groups. However I do not understand how/why the ketone in Q is formed.

UltimaOnline SG

Hi Gohby,

Due to close proximity of the 2 OH groups in geminal diols (in the intermediate between P and Q), intramolecular proton transfer (can be catalyzed by acid or base) followed by elimination of a H2O molecule with thermodynamically favourable positive entropy change can readily occur.

As predicted by Gibbs free energy formula, (positive entropy change) is encouraged by heating (as specified in the question), which also provides the activation energy for the proton transfer and elimination to occur (subsequent to the SN2 and SN1 hydrolysis nucleophilic substitutions).

Furthermore, heating also removes H2O (in gaseous state) from the solvated reaction mixture, and concordantly as predicted by Le Chatelier's principle, shifts the position of equilibrium over to the aldehyde/ketone side.

If large amounts of water are added at lower temperatures, the position of equilibrium can be made to shift back to the germinal diol side.

Additional info : 2 other factors that can affect the position of equilibrium between geminal diols and ketones/aldehydes, are : presence of strongly electron-withdrawing by induction groups (eg. ninhydrin and chloral hydrate) increasing the electrophilicity of the carbonyl C atom for nucleophilic hydrolysis thus favouring the hydrolysis product the geminal diol ; and ring strain due to angle strain (eg. cyclopropanone and its geminal diol hydrate) favouring the less severely strained tetrahedral sp3 gemial diol species, over the more severely strained trigonal planar sp2 ketone/aldehyde species.

While this reaction isn't specified in the basic H2 syllabus, but with sufficient clues given in the question (ie. the molecular formula, with which you can work out the degree of unsaturation to clue you into the structure of the final product, yet another useful tool which H3 / Olympiad / BedokFunland JC students can use as an advantage over Singapore JC H2 Chem students), such questions can, and has indeed (eg. in recent Singapore-Cambridge A level H2 Chem papers) been asked of H2 Chem students.

Bonus qn for your students : what would you get if you started with a geminal tri-halide (instead of a geminal di-halide)?

gohby SG

Hi UltimaOnline,

Thank you for your wonderful and detailed explanation. :)

I think if it were to be a geminal triol there would be a carboxylic acid formed since there would be a presence of a C=O bond and a hydroxyl group left, right?

However, what is the basis for intramolecular proton transfer? Is it because the C connected to the hydroxyl groups is very δ+ and thus the lone pair of electrons from one of the hydroxyl groups forms a second bond with the δ+ carbon, kicking out the OH- group, which will abstract the H atom that was attached to the first hydroxyl group?

UltimaOnline SG

You're totally welcome, Gohby :)

Correct, hydrolysis of a geminal tri-halide will generate a carboxylic acid.

Intramolecular proton transfers are usually catalysed by external acids and bases. But regardless of catalyst or not, such reactions can proceed simply because OH groups are amphiprotic, ie. they can function either as a Bronsted-Lowy acid or as a Bronsted-Lowry base.

See the (uncatalyzed) mechanism I drew for the intramolecular proton-transfer reaction :

Hi UltimaOnline,

Thank you for your wonderful and detailed explanation. :)

I think if it were to be a geminal triol there would be a carboxylic acid formed since there would be a presence of a C=O bond and a hydroxyl group left, right?

gohby SG

Hi UltimaOnline,

I have some questions here:

http://img.photobucket.com/albums/v700/gohby/Chemistry/hci_zpsu2otlaus.jpg

Remarks: I think the side products referred to is the combination of the free radicals during FRS. How do I proceed from here?

http://smg.photobucket.com/user/gohby/media/Chemistry/chiralcentres_zpsv6dx78qi.jpg.htm

Remarks: Referring to the compound (ignore the orange dots), the answer suggests that it reacts with liquid bromine in tetrachloromethane to form a compound with 12 chiral centres. I have indicated the possible chiral centres after the reaction with bromine (as indicated by the orange dots), but I have only counted 9.

3. Thiols are organic compounds containing the -SH functional group. They are sulphur analogue of alcohols. Some common reactions undergone by thiols are shown as follows.

I CH3CH2SH + CH3CH2Br → (CH3CH2)2S + HBr

II CH3CH2SH + NaOH → CH3CH2S−Na+ + H2O

III 2CH3CH2SH + Br2 → CH3CH2−S−S−CH2CH3 + 2HBr

Which of the following statement(s) comparing thiols with alcohols is/are true?

1 Thiols are stronger nucleophiles than alcohols.

2 Thiols are stronger reducing agents than alcohols.

Remarks: How do I tell that statements 1 and 2 are correct from the reactions? I can deduce that -SH is a stronger nucleophile than -OH because of the S is less electronegative than O but is there any way I can deduce that from the equations given?

http://img.photobucket.com/albums/v700/gohby/Chemistry/mechanism_zpsr6ml2hcu.jpg

Remarks: Is the rate determining step step 3 because the charge is on oxygen (as opposed to the carbon on step 2) and it is the bulkiest amongst all the nucleophiles.

Thank you for your help. :)

UltimaOnline SG

Hi Gohby,

Q1. You proceed by drawing (or if you're good enough, visualizing) the mechanism, ie. with curved arrows, and not just writing the mechanism, as poorly taught in Singapore JCs to H2 students (the majority of whom just blindly memorize the written mechanisms without deeper understanding). If you're good enough, always draw out the full mechanism in the A level exams. Then the answer to this MCQ will become understandable and self-evident.

Q2. There will indeed be 12 chiral C atoms. You missed out the new sp3 C atoms that used to be the sp2 C atoms of the alkene.

Q3. Thiols nucleophilically substitutes away the halide leaving group more readily (ie. lower Ea and greater yield at equilibrium) than alcohols do, as shown in reaction 1. Thiols are able to reduce Br2 to Br-, a reaction not seen with alcohols, as shown in reaction 2.

Q4. Step 3 may not actually be the true rate determining step. Don't trust JC prelim papers' qns and ans. Regardless, in step 3 the reactant behaves as a Bronsted-Lowry base rather than a nucleophile, and hence steric hindrance (ie. being bulky) is irrelevant. Also, for the reactant in step 2, the -ve charge isn't only on the C atom, but is delocalized by resonance over to the carbonyl O atom (that's why it was possible to deprotonate the aldehyde in the first place).

gohby SG

Hi Gohby,

Q1. You proceed by drawing (or if you're good enough, visualizing) the mechanism, ie. with curved arrows, and not just writing the mechanism, as poorly taught in Singapore JCs to H2 students (the majority of whom just blindly memorize the written mechanisms without deeper understanding). If you're good enough, always draw out the full mechanism in the A level exams. Then the answer to this MCQ will become understandable and self-evident.

Q2. Your image link has an error.

Q3. Thiols nucleophilically substitutes away the halide leaving group more readily (ie. lower Ea and greater yield at equilibrium) than alcohols do, as shown in reaction 1. Thiols are able to reduce Br2 to Br-, a reaction not seen with alcohols, as shown in reaction 2.

Q4. Step 3 may not actually be the true rate determining step. Don't trust JC prelim papers' qns and ans. Regardless, in step 3 the reactant behaves as a Bronsted-Lowry base rather than a nucleophile, and hence steric hindrance (ie. being bulky) is irrelevant. Also, for the reactant in step 2, the -ve charge isn't only on the C atom, but is delocalized by resonance over to the carbonyl O atom (that's why it was possible to deprotonate the aldehyde in the first place).

Hi UltimaOnline,

Thank you for your response.

1. My school taught me the FRS mechanism in full with the homolytic fission steps etc. However, as you've correctly pointed out, some schools (even some top schools!) do not teach the mechanism in full. Will students be penalised if they do not display the mechanism along with the arrows? I've always thought that in a mechanism the movement of electrons have to be accurately displayed but this requirement seems to be negated in FRS.

2. Oops, here is the link:

http://img.photobucket.com/albums/v700/gohby/Chemistry/chiralcentres_zpsv6dx78qi.jpg

4. This is an MCQ question whereby one of the choices is: " Step 1 is the rate determining step for this reaction", and the answer suggests that this choice is wrong. Hence I tried to figure out why that was so and hypothesised that perhaps step 3 is the rate determining step. (Thanks for clarifying btw). So how do I know if the choice "Step 1 is the rate determining step for this reaction" is wrong?

The C−H bond lengths of four hydrocarbons are given in the table below.

compound structural formula C−H bond length /nm

methane CH4 0.110

ethane CH3CH3 0.110

ethene CH2=CH2 0.108

ethyne CH≡CH 0.106

Which of the following helps to explain the shortest C−H bond length observed in ethyne?

A Ethyne is a linear molecule.

B The carbon orbital used in formation of the C−H bond in ethyne has the greatest s orbital character.

C An sp‐sp overlap is observed between the two carbon atoms in ethyne.

D The carbon‐carbon triple bond in ethyne is the strongest.

Remarks: The answer is B. Is C wrong because whilst true, it doesn’t explain the length of C-H bonds relative to the others? What is wrong with D?

UltimaOnline SG

Q1. Writing the mechanism is the minimum required to get the marks. Better to draw out the curved arrows. For FRS, Cambridge doesn't require it and Singapore JCs don't teach the full curved arrow mechanisms, simply because they don't think H2 Chem students are smart enough to draw the full mechanisms, hence it's reserved for H3 Chem, Olympiad Chem, and *ahem* BedokFunland JC students to draw out the full curved arrow mechanisms for FRS and other reactions (eg. nucleophilic acyl substitutions) not required by the basic H2 Chem syllabus.

Q2. There will indeed be 12 chiral C atoms. You missed out the new sp3 C atoms that used to be the sp2 C atoms of the alkene.

Q4. By looking at the rate equation (which will usually be given for the H2 syllabus). If the kinetic orders for each reactant aren't given or can't be deduced (eg. table of kinetic data), then the question must be such that you can make reasonable assumptions and deductions (comparing the stabilities of the reactants vs products, and concordantly the Ea required), but these are beyond the expectations of the H2 syllabus, and particularly for this reaction, it's gonna be a pretty close toss up between the 3 elementary steps, so if Cambridge asks this, they'll definitely give additional clues (most probably kinetics order data), and/or make the different Ea required for the 3 elementary steps more obviously varying and distinctive.

Q5. Option D (and for that matter, C as well) is wrong because it focuses on the C-C bond instead of the C-H bond.

gohby SG

Hi UltimaOnline,

Thank you for your help. :) I have some questions here:

Q4: This is a YJC Prelim 2012 question:

http://img.photobucket.com/albums/v700/gohby/Chemistry/yjc_zpssoqaqd1f.jpg

Am I right to say that given the information at hand it is not possible to discard or accept choice 3?

___________________________________________

http://img.photobucket.com/albums/v700/gohby/Chemistry/DHS_zpsovsugf6p.jpg

Answer: B

Remarks: I would have thought that compound Z undergoes side chain oxidation to yield the first compound in A and 1,2,4,5-benzenetetracarboxylic acid.

II: http://img.photobucket.com/albums/v700/gohby/sajc_zpsxe9phzxj.jpg

Answer: 2 and 3.

Remarks: Why is 1 wrong? Doesn’t aqueous bromine decolourises when it is added to both compounds?

III: Chlorofluoroalkanes are commonly used as aerosol propellants. However, they cause depletion to the ozone layer when they rise into the stratosphere. It has thus been suggested that fluoroalkanes should be used instead. Which of the following could be a possible reason for the suggestion?

A Fluoroalkanes are less volatile than chlorofluoroalkanes and making it more difficult to reach the stratosphere.

B Fluorine radicals may be produced, but unlike chlorine radicals, do not react with ozone.

C Fluorine radicals are not produced as the C‐F bonds are stronger than the C-Cl bonds.

D Fluorine radicals may be produced, but unlike chlorine radicals, are not regenerated after reaction with ozone.

Answer: C

Remarks: Is A wrong because the boiling point of the halogenoalkane has got no relation to its ability to reach the stratosphere? B is wrong because fluorine radicals are extremely reactive. However, how do I know that D is wrong? I thought that due to the reactivity of fluorine radicals it is unlikely to be reformed at the end of the reaction. How could I be so sure that fluorine radicals are NOT produced just because C-F bonds are stronger than C-Cl bonds?

IV: The thermal decomposition of calcium ethanoate produces its metal carbonate and a carbonyl compound.

(CH3CO2)2Ca → CaCO3 + (CH3)2CO

When a mixture of calcium ethanoate and calcium methanoate was heated, a mixture of three carbonyl compounds X, Y and Z were obtained. Both X and Y give a silver mirror with Tollen’s reagent but not Z. Both X and Z give yellow precipitate in aqueous alkaline iodine but not Y.

Question: X is CH3CHO and Z is CH3COCH3. However, what is Y, and how do I work out what is the equation involved from the equation given?

UltimaOnline SG

Q4. Step 1 is probably the rate determining step (although the -ve formal charge is delocalized over both the O and C atom, but in the first place the H bonded to the C atom isn't acidic enough, so requires higher Ea), but you (ie. even as professional chemists, let alone as A level students) cannot say for sure without carrying out kinetics experiments. For A levels, Cambridge will only ask such qns (that A level students cannot say for sure), only in conjunction with appropriate option combinations, eg. Cambridge reasons that even if the A level student cannot know (based on the syllabus) if option 2 is correct or wrong, but because the student should know option 1 and 3 are correct, but because there's no such answer as "1 and 3 only", hence Cambridge expects the student to reason (using logic rather than chemistry) that option 2 must also be correct.

QI. Yes if your suggested answer is one of the options, go for it. But if not (since the exact type of oxidation isn't specified and is obviously not in H2 syllabus), just choose the best, most feasible, answer.

QII. Yes, in both cases. However, with the benzene ring, an additional observation (steamy fumes) is also observed with Br2(aq).

QIII. Always choose the best option. The strength of C-F bond as a reason is specified in the basic H2 syllabus. Fluoroalkanes are NOT less volatile than chlorofluoroalkanes. You cannot know that D is wrong (since that's beyond the H2 syllabus), but you should know that C is right.

QIV. (I'll reply on this later, gotta go work now...)

UltimaOnline SG

QIV. Y is methanal. Mathematical permutations and combinations (note a reactant can react with another molecule of itself). Reactants are 1H+1C=O and 1C+1C=O. Hence products are 1C (methanal), 2C (ethanal) or 3C (propanone). Don’t worry too much about the equation or mechanism. This qn is meant to be solved via pattern recognition (and sufficient test data provided), as the reaction mechanism is beyond the H2 syllabus.

Ztutor SG

Calling all tutors, in need of students? Use Utor now. Your one stop app to find students.

http://bit.ly/1K2OrQA

gohby SG

Hi UltimaOnline,

I have some “application” questions this time.

Q1: http://img.photobucket.com/albums/v700/gohby/Chemistry/Hofmann_zpsbhaiqwk4.jpg

Remarks: The answer is A. Firstly, am I right to say that a quaternary, rather than a tertiary amine intermediate would be formed instead? Secondly, without understanding the Hofmann elimination mechanism, how would I be able to arrive at option A? If I were to try to extrapolate from the example given, A wouldn’t have been the answer since the example shows that nitrogen has been eliminated.

Q2: http://img.photobucket.com/albums/v700/gohby/Chemistry/ether_zpsld41to9s.jpg

Remarks: The answer is C. Is this a reasonable question to ask in the syllabus (since ethers are not in the syllabus)? I do not understand why there would be nucleophilic substitution taking place for the CH2Br but elimination occurring instead for CH2CH2Cl.

Q3: http://img.photobucket.com/albums/v700/gohby/Chemistry/Claisen%20condensation_zpsqydyie04.jpg

Remarks: The answer is A. Why? Is there a way to arrive at the answer without understanding (the sophisticated) mechanism for Claisen condensation?

Q4:

http://img.photobucket.com/albums/v700/gohby/Chemistry/Least%20likely_zpsdxb98tfr.jpg

Remarks: The answer is C. A is likely to occur since it’s part of the nucleophilic addition reactions for carbonyl compounds. B is likely to occur since it’s FRS. I presume C is unlikely to occur because the positive charge is stabilised by resonance. However, I would have chosen D - D is a primary halogenoalkane, so SN2 nucleophilic substitution reactions are more likely, but D shows the first step of the SN1 reaction. Though I understand that the benzene ring is a bulky substituent that would favour the SN2 reaction happening instead, how would I be able to compare the likelihood of D or C occurring?

Lastly, wrt your response on QI in the previous post,

QI. Yes if your suggested answer is one of the options, go for it. But if not (since the exact type of oxidation isn't specified and is obviously not in H2 syllabus), just choose the best, most feasible, answer.

I would have thought that A was the best and most feasible answer since 1,2,4,5-benzenetetracarboxylic acid is one of the choices. I would have at least entertained B as a feasible option if not for the presence of the ketone. I did not find it feasible that the alkane can inexplicably be oxidised to a ketone..

Thank you very much, really appreciate your help! :)

UltimaOnline SG

Hi UltimaOnline,

I have some “application” questions this time.

Q1: http://img.photobucket.com/albums/v700/gohby/Chemistry/Hofmann_zpsbhaiqwk4.jpg

Q2: http://img.photobucket.com/albums/v700/gohby/Chemistry/ether_zpsld41to9s.jpg

Q3: http://img.photobucket.com/albums/v700/gohby/Chemistry/Claisen%20condensation_zpsqydyie04.jpg

Remarks: The answer is A. Why? Is there a way to arrive at the answer without understanding (the sophisticated) mechanism for Claisen condensation?

Q4:

http://img.photobucket.com/albums/v700/gohby/Chemistry/Least%20likely_zpsdxb98tfr.jpg

Lastly, wrt your response on QI in the previous post,

Thank you very much, really appreciate your help! :)

Such higher order-type qns are designed to be do-able only by distinction students, and as such for these qns, the boundaries between in-syllabus and beyond-syllabus is blurred.

Many of these qns will require the ability to draw mechanisms based on understanding (not blind memorization), which rules out the majority of Singapore JC H2 Chem students.

Typically, only students who actually understand and can draw mechanisms for reactions not previous encountered (ie. H3 Chem, Olympiad Chem, and BedokFunland JC students), can correctly answer these questions with a correct understanding of the underlying mechanisms.

True, some (not all) of such qns can still be guesstimated using pattern-recognition, but that would be more of an IQ test (or for most JC students, a random game of roulette) rather than actually understanding the underlying chemistry involved.

Q1. With the mechanism drawn, the answer becomes obvious. SN2, deprotonation, SN2, E2. Go ahead and draw out the mechanism to explain to your students.

Q2. Dehydrohalogenation to generate an alkene group cannot occur for the methyl side-chain as the adjacent (benzene ring) C atom does *not* have a H atom present to be eliminated.

Q3. This qn requires understanding and drawing out of the mechanism (you should go ahead and draw out the full mechanisms to explain to your students, but advise them to skip such qns in the actual A levels Paper 1 first, and go back to them only after they've completed the rest of the paper, in the interest of exam-smart time-management). Since the qn specified a mixed / crossed (instead of self) Claisen condensation, it could either be ethyl ethanoate as nucleophile attacking ethyl pentanoate as electrophile, or ethyl pentanoate as nucleophile attacking ethyl ethanoate as electrophile. Either way, all 4 options given in the qn are wrong. Draw out the mechanism and you'll see the structures of the 2 possible products.

Q4. Option C shows the tetrahedral sp3 hybridized intermediate of a benzene ring during electrophilic aromatic substitution. The Br- must compulsorily act as a Bronsted-Lowry base to carry out deprotonation, to restore the benzene ring's aromaticity (temporarily lost during electrophilic aromatic substitution), to re-acquire the resonance stabilization energy as dictated by thermodynamics and Gibbs free energy. This is in the H2 syllabus under "Candidates must be able to explain why the benzene ring undergoes electrophilic aromatic substitution instead of electrophilic addition".

QI. 1st of all, your suggested 1,2,4,5-benzenetetracarboxylic acid is NOT a product under option A. 2ndly, 1,2,4,5-benzenetetracarboxylic acid wouldn't be a product in any case, as oxidation of akyl side chain of benzene ring requires a benzylic H atom, which isn't present here. As for "alkane inexplicably oxidised to a ketone", it's not so inexplicable when you consider the side-chain oxidation of benzene ring by KMnO4 (which *is* within the H2 syllabus), extrapolated out in this qn's molecule. So it's not "alkane inexplicably oxidised to a ketone" per se, but rather "alkyl side chain of benzene explicably oxidised to a ketone".

No problem, Gohby, you're totally welcome. Keep your questions coming (thousands of silent forum lurkers, quite certainly both students as well as teachers/tutors, also follow and benefit from our discussions here).

gohby SG

Thank you for your help UltimaOnline :)

I have some more questions here:

Q5:

http://img.photobucket.com/albums/v700/gohby/Chemistry/crackin_zps1k1zqfbz.jpg

Remarks: How do I establish which equation is wrong?

Q6:

http://img.photobucket.com/albums/v700/gohby/Chemistry/imine_zpsgxwtogli.jpg

Remarks: The answer is 1 only. However for option 3, when I add NaOH to the mixture, according to the equation NH3+H2O ⇄ NH4+ + OH-, wouldn’t the concentration of ammonia increase, thereby increasing the yield of imine?

Q7:

http://img.photobucket.com/albums/v700/gohby/Chemistry/deuterium_zpsytr2ghdh.jpg

Remarks: The answer is 8. I could only find the 8 isomers if the question states that it was the molecular formula, instead of the structural formula that was shown (because if the propyl substituent group could be branched there would be more isomers). However since the question states that it is a part-structural formula, I would have to assume that the propyl substituent group is unbranched, am I right?

Q8:

http://img.photobucket.com/albums/v700/gohby/Chemistry/cream_zpsvuoebonr.jpg

Remarks: How do I ascertain what is Y and Z from the table?

UltimaOnline SG

Thank you for your help UltimaOnline :)

I have some more questions here:

Q5:

http://img.photobucket.com/albums/v700/gohby/Chemistry/crackin_zps1k1zqfbz.jpg

Remarks: How do I establish which equation is wrong?

Q6:

http://img.photobucket.com/albums/v700/gohby/Chemistry/imine_zpsgxwtogli.jpg

Q7:

http://img.photobucket.com/albums/v700/gohby/Chemistry/deuterium_zpsytr2ghdh.jpg

Q8:

http://img.photobucket.com/albums/v700/gohby/Chemistry/cream_zpsvuoebonr.jpg

Remarks: How do I ascertain what is Y and Z from the table?

No prob, Gohby :)

Q5. The qn is in error. What the AJC teacher meant was "in the absence of a catalyst" (ie. thermal cracking, not catalytic cracking). In which case, option C is the answer, as based on the mechanism (ionic vs free radical), branching in products are a feature of catalytic cracking (due to carbocation rearrangement via hydride and methanide shifts). To achieve the branched product in option C via thermal cracking involves the participation too many highly unstable methyl radicals (unlikely to be generated based on the long chain reactant), and/or highly unfeasible free radical rearrangements, both requiring overly high activation energies, and are thus both thermodynamically and kinetically unfeasible. Finally (tell your students this only after going through everything above and enjoying their worried looks), cracking isn't in the H2 syllabus, so Cambridge won't ask such qns in the 1st place.

Q6. Another problematic qn. There are 2 reasons that outweigh your (reasonable enough) assumption. 1stly, with the addition of NaOH, the overall yield of the imine decreases due as some of the aldehyde will be used up to undergo Aldol side reactions, hence shifting the position of equilibrium of the original reaction to the LHS. 2ndly, the mechanism to generate the imine requires a fairly neutral pH (adding NaOH makes it too alkaline for the proton transfers that need to occur during the mechanism for generation of the imine). Yes, you would say this is unfair to H2 students, which is why Cambridge won't set such a qn in the first place.

Looking at problematic Q5 and Q6, remember what I've always said about not trusting Singapore JC Prelim papers too much?

Q7. C3H7 side chains could refer to either propyl or iso-propyl.

Q8. The key phrase is "after some time". If you wait long enough, both the alkyl dibromide and the bromo acyl bromide will yield the same ppt. But if you only wait a short while (ie. after some time, not too long), the bromo acyl bromide will yield a greater mass of ppt, as acyl halides undergo hydrolysis instantaneously at rtp, while akyl halides require a much higher Ea, a much higher temperature, a much longer time.

gohby SG

Hi UltimaOnline,

Thank you for your help. :) I have another set of questions to ask -

Q9.

It has been noted that prolonged exposure of wine to the atmosphere will result in the wine becoming “vinegary”. Therefore, wineries attempt to prevent such exposure by storing the wine in a mixture of carbon dioxide and nitrogen. How does this mixture prevent the “vinegary” taste from occurring?

1 Oxidation of ethanol into ethanoic acid will be prevented.

2 The acidity of the wine will be decreased.

3 The amount of carbon dioxide dissolved in the wine will be increased.

Answer: 1 only

Remarks: I can infer that the vinegary taste arose as a result from oxidation, but how do I conclusively strike out the other options?

Q10.

H2NCH2CH2CH2CO2H

Which statements about the above compound are correct?

1 It is a 2-aminocarboxylic acid.

2 It is soluble in water due to zwitterion formation.

3 It migrates to the anode of an electrolytic cell at pH 12.

Answer: 1 and 2.

Remarks: I think the answer should be 2 and 3 instead. Isn’t the compound a 3-aminocarboxylic acid? Additionally, at pH 12, the carboxylic acid will react with the base to form a carboxylate ion, which should migrate to the positive terminal (anode) of the electrolytic cell.

Q11.

Which reactions yield a carbon compound incorporating deuterium, D?

1 CH3COCH3 --(NaOD, I2, D2O)---->

2 CH3CO2CH3 --(NaOD, D2O) ----->

3 CH3CH2CN --(NaOD, D2O) ------->

Answer: 1 and 2 only.

Remarks: Why would the product in reaction 1 contain deuterium? Are the products of reaction 1 CH3COO- and CHI3?

Q12.

SAJC/PRELIM2009/P1/Q23

Which of the following pairs of compounds can be attacked by cyanide ions via

nucleophilic reactions?

A CH3CH2Cl and CH3COCH2CH3

B CH3CH2Cl and CH3COOCH3

C C6H5Cl and CH3COCH2CH3

D C6H5Cl and CH3COOCH3

Answer: B

Remarks: I know that C and D are wrong because of chlorobenzene, but I would have assumed that the ketone in A can be attacked by cyanide ions nucleophilically as per nucleophilic additions of carbonyl compounds. However, am I also right to say that the C in the ester group is also susceptible to nucleophilic attacks as it is rather δ+?

gohby SG

Q7. C3H7 side chains could refer to either propyl or iso-propyl.

I thought about C3H7 possibly referring to iso-propyl, but given that the question mentioned the formula is a part-structural formula, if C3H7 were to be iso-propyl, wouldn't it have been referred to as CH(CH3)CH3 instead?

UltimaOnline SG

Hi UltimaOnline,

Thank you for your help. :) I have another set of questions to ask -

Q9.

1 Oxidation of ethanol into ethanoic acid will be prevented.

2 The acidity of the wine will be decreased.

3 The amount of carbon dioxide dissolved in the wine will be increased.

Answer: 1 only

Remarks: I can infer that the vinegary taste arose as a result from oxidation, but how do I conclusively strike out the other options?

Q10.

H2NCH2CH2CH2CO2H

Which statements about the above compound are correct?

1 It is a 2-aminocarboxylic acid.

2 It is soluble in water due to zwitterion formation.

3 It migrates to the anode of an electrolytic cell at pH 12.

Answer: 1 and 2.

Q11.

Which reactions yield a carbon compound incorporating deuterium, D?

1 CH3COCH3 --(NaOD, I2, D2O)---->

2 CH3CO2CH3 --(NaOD, D2O) ----->

3 CH3CH2CN --(NaOD, D2O) ------->

Answer: 1 and 2 only.

Remarks: Why would the product in reaction 1 contain deuterium? Are the products of reaction 1 CH3COO- and CHI3?

Q12.

SAJC/PRELIM2009/P1/Q23

Which of the following pairs of compounds can be attacked by cyanide ions via

nucleophilic reactions?

A CH3CH2Cl and CH3COCH2CH3

B CH3CH2Cl and CH3COOCH3

C C6H5Cl and CH3COCH2CH3

D C6H5Cl and CH3COOCH3

Answer: B

Q9. Option 2 is wrong because CO2 being an covalent, electrophilic, acidic oxide, will actually (directly) increase, not decrease the acidity (of course indirectly, when the acidity is increased as a result, the bacterial population is consequently reduced, thereby making the pH overall still less acidic compared to if the bacteria thrived and oxidized the ethanol to ethanoic acid. But that's beyond H2 Chem syllabus, and that'll just be a sadistic "heads-I-win-tails-you-lose" MCQ setup by the examiner). It is the bacterial oxidation of ethanol to ethanoic acid (in the presence of atmospheric oxygen), which is really what gives the sour vinegary taste, rather than dissolving CO2 to generate carbonic(IV) acid (which is what makes soft drinks fizzy, eg. Coca Cola). Option 3 is wrong because it's true but irrelevant. As mentioned earlier, hydrolysis of CO2 will generate carbonic(IV) acid, which increases the acidity but does not *directly* (sadistic examiners notwithstanding ; indirectly it's arguable that it does) help to prevent the oxidation of ethanol to ethanoic acid in presence of atmospheric oxygen (and it is the ethanoic acid that gives the vinegary taste). The primary purpose of forcing CO2 and N2 in is to eliminate O2. Of course, if it were as simple as that, then it'll be better just to use N2. The CO2 present is additionally meant to reduce the pH to levels more acidic than that required for bacteria (which oxidize ethanol to ethanoic acid) to thrive. And it sure doesn't hurt that the dissolved CO2 or carbonic(IV) acid adds a nice fizzy touch to the alcoholic beverage.

Q10. You're correct. Don't trust Singapore JCs.

Q11. In the reaction mechanism for the iodoform reaction, the 1st 3 hydroxide ions function as Bronsted-Lowry bases, while the 4th hydroxide ion functions as a nucleophile, generating RCOOD and CI3-. The CI3- then functions as a Bronsted-Lowry base to abstract a D+ ion from either RCOOD or the solvent D2O. Either is possible, and since the qn ensured both the hydroxide ions and solvent water consist of D instead of H, it's not an issue here.

Q12. Potentially, both ketones and esters may react with cyanide ions (being a strong nucleophile). However, as far as the H2 syllabus is concerned, only ketones (and not esters) react with cyanide ions. The rationale for this is that ketones (and aldehydes) are significantly more electrophilic compared to esters (or for that matter, carboxylic acids and amides as well). The underlying rationale for this rationale, which is not taught at either H2 or even H3 (and not well understood even by most JC teachers), is that the partial positive charge at the electrophilic carbonyl C atom in ketones is due to both induction and resonance (ie. the resonance hybrid has an electron-deficient carbonyl C atom as a consequence of the carbonyl O atom withdrawing electrons by resonance), while the partial positive charge at the electrophilic acyl C atom in esters is due only to induction (ie. the resonance hybrid does not have an electron-deficient acyl C atom as a consequence of the acyl O atom withdrawing electrons by resonance, due to the adjacent O atom that simultaneously donates electrons by resonance to the acyl C atom).

If at this point you wanna retort, "but the Singapore JC mark scheme said it is the ester that kena attacked leh, not the ketone!", then see my Q10 above.

Qn about Iso-Propyl. If it were full-structural formula : when C3H7 is to be iso-propyl, then it would have been referred to as CH(CH3)CH3 ; when C3H7 is to be propyl, then it have been referred to as CH2CH3CH3. The fact that it's only part-structural formula, means the parts of the molecule that are ambiguous (ie. C3H7) could refer to *either* propyl or iso-propyl. So considering both stereoisomers and structural isomers, there are several different (permutations and combinations of) molecules possible for the part-structural formula given in the question.

1 month left to A levels liao... HUAT AH!!! ;D

gohby SG

Hi UltimaOnline,

Your cogent explanations are amazing - thank you very much. :D

Which of the following will not be produced when 1-chloropropane is heated in ethanolic sodium hydroxide?

CH3CH2CH2ONa

CH3CH2CH2OCH2CH3

Answer: A.

Remarks: How is B produced during the reaction?

http://img.photobucket.com/albums/v700/gohby/Chemistry/compoundy_zpsd5n21kuf.jpg

Remarks: Choice 3 is correct. Is it due to electrophilic substitution with phenylamine, even though the amine is not a primary amine. Students would know that bromine water undergoes electrophilic substitution with phenylamine - would they be required to infer that it would undergo the same reaction even where the amine is not a primary amine?

http://img.photobucket.com/albums/v700/gohby/Chemistry/Cyanohydrin_zpsp7p1b7v6.jpg

Remarks: How do I know if C is correct? How do I know that the cyanohydrin produced does not undergo an SN1 reaction and reacts with ammonia thereafter to form an amine?

4. Re hydrolysis of halogenoalkanes/acyl chlorides/esters/amides (sorry this is long!):

(i) What affects the rate of hydrolysis - the strength of the C-X bond or the extent of δ+ on the C atom?

In alkyl chlorides, the rate of hydrolysis is dependent on the strength of the C-X bond and not the polarity of the C-X bond. However, when we are juxtaposing the rate of hydrolysis between different groups, such as acyl chlorides and (halogenoalkanes, for instance), it is mentioned that acyl chlorides are readily hydrolyzed due to the electron withdrawing O and Cl atoms intensifying the partial positive charge on the C atom.

(ii) How do I compare the rate of hydrolysis across different groups (esters, amides, acyl chlorides, phenoxide, alkoxides, halogenoalkanes)? I know esters and amides hydrolyse slower as compared to acyl chlorides from the conditions required to hydrolyse them, but what is the basis of the different rates of hydrolysis?