Organic Chem
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General advice for students in organic chemistry:
Focus on the knowledge component first, because there cannot be proper understanding without adequate knowledge. Thus, it is important that the student finds a way to remember the 50-odd reactions and the 6 mechanisms in this section. Cue cards provide an efficient way to achieve memory through repetition. We provide intensive revision classes during holiday to speed up this process.
Once the student has developed the adequate knowledge, he could able to apply the knowledge and appreciate the how the questions being ask.
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6. Carbonyl -FAQ
Q: Why don’t carbonyl compounds undergo nucleophilic substitution?
A: The carbonyl functional group is either bonded to two other carbon atoms (as in a ketone) or a
carbon atom and a hydrogen atom (as in an aldehyde). If the carbonyl compound is to undergo
nucleophilic substitution, then either the C—C bond or the C—H bond has to be broken. The
breaking of these two types of bonds is energetically demanding. In addition, if the C—C bond or
the C—H bond is broken, then the leaving group is a very strong nucleophile (H– for aldehyde or a
carbanion ion for ketone).
Q: Are aldehydes and ketones equally reactive towards nucleophilic addition?
A: No. Generally, an aldehyde is more reactive than an isomeric ketone. As aldehyde has a H atom and
an alkyl group bonded to the carbonyl functional group, a nucleophile would face less steric
hindrance when approaching the carbonyl carbon of an aldehyde as compared to a ketone. In
addition, there are two bulky electron-releasing alkyl groups bonded to the carbonyl group of ketone.
As a result of this electronic effect, the carbonyl carbon atom of ketone is less electron-deficient than
an aldehyde, thus less susceptible to nucleophilic attack.
Q: How is H–C≡N considered a nucleophile?
A: The nucleophile is actually the C≡N ion and not HCN. Although we normally represent the cyanide
ion as CN, it does not mean that the negative charge (which represents an extra electron in the
species) resides on the nitrogen atom. Thus, take note that the attacker is actually the more electronrich
carbon atom (as it is negatively charged) and not the nitrogen atom. Although the nitrogen atom
also possesses a lone pair of electrons (which means that it is also nucleophilic), but as the C atom is
less electronegative than N, thus the C atom is more likely to donate its lone pair of electrons. In
essence, HCN is just the source of the nucleophile. As HCN is a weak acid that partially dissociates
in water, the [CN-] is considerably low:
HCN + —–> H+ + CN- Ka = 5 × 10^-10 mol dm^-3
Q: How can HCN a weak acid? Isn’t the electronegativity difference between H and C small?
A: In HCN, the most electronegative N withdraws electron density away from the C atom, which in turn
withdraws from the H atom. This made the H atom relatively electron-deficient and hence susceptible
to extraction by a base.
Q: But shouldn’t the CN− formed be highly unstable as the negative charge resides on the C atom which
has a lower electronegativity value?
A: Yes, you are right that the electronegativity of C is not “good enough” to hold the negative charge.
But do not forget that there is a highly electronegative N atom bonded to the C atom, through
inductive effect, the N atom would withdraw electron density away from the C atom and help to
stabilise the CN− ion. In addition, the orbital that holds the extra electron on the C atom is sp
hybridised, containing 50% s-character. Since an s-orbital is relatively closer to the nucleus than a porbital,
an extra electron in a sp hybridised orbital is more strongly attracted by the nucleus than in a
sp2 hybridised orbital, which in turn is more strongly attracted than in a sp3 hybridised orbital. This
accounts for the stability of the CN− ion.
Q: Why can’t we just simply allow the ethanal to react with NaCN(aq) or KCN(aq)? What is the
purpose of ensuring that there is still the undissociated HCN present in the system?
A: HCN serves two roles in this reaction. Firstly, it provides the nucleophile CN albeit in low
concentrations. Secondly, it acts as a proton donor in the second step of the mechanism to generate
the final product (see Step 2 of the mechanism).
Q: If the solvent can act as a proton donor and NaCN as a source of CN-, can’t we forego the use of
HCN?
A: Now, the extraction of a H+ ion from a water molecule is much more difficult than from a H–CN
molecule. This is because the H–O (BE = +460 kJ mol1) bond is much stronger than the H–C bond
(BE = +410 kJ mol^-1). Such explanation also provides us with the understanding of why HCN,
though is a weak acid, is still a stronger acid as compared to H2O. So, in reality, extracting a H+ from
a water molecule to form the cyanohydrins is a minor reaction as compared to the extraction from a
HCN molecule.
Q: If Fehling’s solution can only oxidise aliphatic aldehyde while Tollens’ reagent can oxidise both
aliphatic and aromatic aldehydes, does it mean that Tollens’ reagent is a stronger oxidising agent?
A: Yes, Tollen’s reagent is a stronger oxidising agent than Fehling’s solution. This would also mean that
aromatic aldehyde is more resistant to oxidation than the aliphatic aldehyde (refer to Understanding
of Advanced Organic and Analytical Chemistry for further details).
Q: Is there a way to obtain the balanced equation from the basic?
A: Yes, if you can’t memorise it, just know how to derive it as follow:
RCOCH3 —> RCOO− + CHI3
RCOCH3 + H2O —> RCOO− + CHI3
RCOCH3 + H2O + 3I2 —> RCOO− + CHI3
RCOCH3 + H2O + 3I2 —> RCOO− + CHI3 + 3I−
RCOCH3 + H2O + 3I2 —>RCOO− + CHI3 + 3I− + 4H+
RCOCH3 + H2O + 3I2 + 4OH− —> RCOO− + CHI3 + 3I− + 4H+ + 4OH− (Alkaline medium used)
RCOCH3 + 3I2 + 4OH− —>RCOO− + CHI3 + 3I− + 3H2O
Another way to help remembering is that the I2 : OH− : I− : H2O ratio is 3 : 4 : 3 : 3.
