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+?
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