BedokFunland JC's A Level H2 Chemistry Qns (Part 2)

last SG · 2010-07-26T23:40:35+00:00

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A BedokFunland JC Original Qn

Question : Which molecule is more acidic? 2-hydroxypropanoic acid, or 3-aminopropanoic acid? Give three different reasons to justify your answer.

Solution :

2-hydroxypropanoic acid is a stronger acid compared to 3-aminopropanoic acid.

Reason #1 - the conjugate base of 2-hydroxypropanoic acid is stablized by the formation of an intramolecular hydrogen bond between the COO- carboxylate group and the OH hydroxy group. This hydrogen bond in the conjugate base is particularly strong (and hence particularly stabilizing), due to a formal (ie. not just partial) negative charge on the COO- carboxylate group.

Or, from the perspective of the pre-proton dissociation, the COOH proton dissociation enthalpy is made less endothermic, due to the weakening of the carboxylic O-H bond, in turn because the carboxylic O-H sigma bond electrons are inductively withdrawn away, due to the intramolecular hydrogen bond between a lone pair on the (partial negative) carboxylic O atom (hydrogen bond acceptor), and the (partial positive) H atom of the hydroxy group (hydrogen bond donor).

No such intramolecular hydrogen bond exists to stabilize the conjugate base of 3-aminopropanoic acid, because the inter-atomic distance is too large for the formation of a similar intramolecular hydrogen bond.

(Note : Cambridge will require the A level student to explicitly draw out and label the intramolecular hydrogen bond that's responsible for the greater acidity.)

Reason #2 - Although the OH hydroxy group and the NH2 amino group, are both electron withdrawing by induction, and therefore both exerting stabilizing effect on the conjugate base; however because oxygen is significantly more electronegative compared to nitrogen, accordingly the OH hydroxy group is more strongly electron withdrawing by induction, compared to the NH2 amino group. Consequently the 2-hydroxypropanoate anion conjugate base is more effectively stabilized by induction, compared to the 3-aminopropanoate anion conjugate base.

Reason #3 - Although the OH hydroxy group and the NH2 amino group, are both electron withdrawing by induction, and therefore both exerting stabilizing effect on the conjugate base; however inductive effect decreases rapidly with distance, or number of intervening sigma bonds. Accordingly, because the OH hydroxy group in 2-hydroxypropanoate ion is closer to the COO- carboxylate group, compared to the NH2 amino group in 3-aminopropanoate ion; consequently the 2-hydroxypropanoate anion conjugate base is more effectively stabilized by induction, compared to the 3-aminopropanoate anion conjugate base.

Because of these three different or separate reasons, 2-hydroxypropanoic acid is a stronger acid compared to 3-aminopropanoic acid.

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Oxygen in water or in any other molecule can make two H-bonds per molecule with itself the reason Two lone pairs on Oxygen. But flourine on the other hand only make only one H-bond per molecule despite the fact that f contains 3 lone pairs......Is it due to high electronegativitiy of F than O?? Or any other plausible explanation.

In H-F molecules, because the F atom has 3 lone pairs, so in theory, it is possible for the F atom to be engaged in 3 hydrogen bonds simultaneously.

However, this does not happen for 2 reasons :

The 1st reason : each time a lone pair is used up to form a hydrogen bond, the other lone pairs (on the F, O, or N atom) are withdrawn away by induction (towards the hydrogen bond) and become less available to participate in it's own hydrogen bond.

Then why can the O atom of water use both of its lone pairs to participate in 2 hydrogen bonds simultaneously, as in solid ice? This is more possible for water than for hydrogen fluoride, because water has 2 H atoms that (being less electronegative) can donate its (bonding) electrons inductively toward the (more electronegative) O atom, resulting in the lone pairs of the O atom being more available to participate in their own hydrogen bonding. In contrast, hydrogen fluoride only has 1 H atom that donates inductively. With less 'income', the F atom is forced to be more 'miserly' with it's limited 'shopping money'.

The 2nd reason : each hydrogen fluoride molecule only has 1 partial positive H atom to participate in intermolecular hydrogen bonding, while in contrast each water molecule has 2. As an analogy : the classroom of water has 2 guys (lone pairs) and 2 girls (partial positive H atoms); the classroom of ammonia has 1 guy (lone pair) and 3 girls (partial positive H atoms); the classroom of hydrogen fluoride has 3 guys (lone pairs) and 1 girl (partial positive H atom).

How many heterosexual (non-gay, non-lesbo) relationships can each class have? Obviously, water will have the most number of intermolecular hydrogen bond relationships, and therefore water has a higher boiling point than either ammonia or hydrogen fluoride.

While water as solid ice can gain maximum thermodynamic stability by forming 4 hydrogen bonds per water molecule in a tetrahedral geometry (eg. in ice 1-cubic); in contrast for hydrogen fluoride, the maximum thermodynamic stability (ie. most efficient way of arranging themselves to achieve maximum number of hydrogen bonds for the entire population) is achieved by forming zig-zag chains of HF in the liquid and solid states, with each HF molecule having 2 primary (semi)-permanent hydrogen bonds within each chain (as shown below); and between the chains, secondary (and weaker, more temporary) hydrogen bonds and/or permanent dipole - permanent dipole Keesom van der Waals interactions may occur.

https://en.wikipedia.org/wiki/Hydrogen_fluoride

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The radius of ion and atom decrease in order : P 3-, S2-.Ar, Cl-.

It is Ok for the decreasing of P, S and Cl but Ar. Why it is biger than Cl-

For A level purposes, just write the following and you'll get full marks :

All of the species mentioned above are isoelectronic; as proton number increases, the nuclear charge and effective nuclear charge (since for isoelectronic species the shielding effect is exactly the same), also increases; hence the ionic/atomic radius decreases as proton number increases, from P3- to S2- to Cl- to Ar.

The tricky bit here, is that noble gases (particularly the lower periods) do not chemically combine with other atoms readily. Hence, while P, S and Cl have 4 different values for atomic radii : namely atomic, covalent, ionic and van der Waals ; however for Ar, being a noble gas element, experiments can ascertain only its van der Waals radius. This results in different values quoted from different sources, which confuse students.

For A level purposes, you need not concern yourself with these matters. Cambridge has no wish to confuse students with such a despicable low blow, (at least ostensibly) preferring to assess students on their fundamental chemistry concepts such as the simple relationship between proton number, electron configuration, effective nuclear charge and (simple) atomic radius.

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Another BedokFunland JC original question

If methylamine were allowed to react with an equimolar mixture of methyl iodide and ethyl iodide, assuming the reaction came to completion, generating only tertiary amines as the products, and ignoring steric and electronic factors (ie. assume both alkyl halides are equally electrophilic), if you were playing a game of Roulette at the Chemistry Casino, what would your odds of winning be, if you were to place your bet on the identity of a product (picked randomly from a randomized sample of the products) as :

a) trimethylamine

b) dimethyethylamine

c) diethylmethylamine

d) triethylamine

Answers (in percentages) :

a) vīgintī quīnque (latin), venticinque (italian), ni-juu-go
(japanese)

b) quīnquāgintā (latin), cinquanta (italian), go-juu (japanese)

c) vīgintī quīnque (latin), venticinque (italian), ni-juu-go
(japanese)

d) [100 - (a + b + c) ] %

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Hm. I want the Lewis structure of IF3 because I wonder that whether F can borrow only 1 e from I ? Can you draw it in dot-dot Lewiss

In terms of formal charge, F didn't 'borrow' any electrons. Its 3 lone pairs are its own, and half of its bond pair belongs to F, the other half belongs to I. For formal charge, 3 lone pairs and 1 bond pair = 7 valence electrons, thus F has no formal charge.

In terms of stable octet, F and I share both electrons in the bond pair, so the F atoms have a stable octet, while the I atom has an expanded octet. For stable octet, 3 lone pairs and 1 bond pair = 8 valence electrons for F, and 2 lone pairs and 3 bond pairs = 10 valence electrons for I, which has vacant, energetically accesible orbitals to accomodate an expanded octet.

In terms of oxidation state, which is formal charge + electronegativity consideration, for every I-F bond present, the I atom loses 1 electron, and each F atom gains 1 electron. Afterall, that's the definition of electronegativity, the capacity to attract electrons due to effective nuclear charge. I has 0 formal charge, but 3 bond pairs with more electronegative F atoms, and hence the OS of the I atom is 0 + +3 = +3. The OS of each F atom is 0 (formal charge) + -1 (electronegativity consideration) = -1.

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Phenolphthalein is an example of an organic molecule that does not contain a transition metal ion, and yet is coloured (indeed, its capacity for colour is the basis for its function as an acid-base indicator). Suggest how colour arises in (coloured) organic molecules that do not contain a transition metal ion.

Solution :

Protonating or deprotonating phenolphalein alters or modifies, and thus changes the colour of, the molecule's conjugated system (ie. a system of connected p-orbitals with delocalized electrons in compounds with alternating single and multiple bonds).

Conjugated systems have the capacity for colour, because similar to the d-d* electron transitions of a transition metal ion, electrons of organic molecules' conjugated systems, are also able to transit between the different energy levels of a conjugated system's pi molecular orbitals, absorbing light in the various wavelengths in the process (which provide the energy required for the pi to pi* electron transitions), and reflecting wavelengths of a complementary colour.

http://en.wikipedia.org/wiki/Phenolphthalein

http://en.wikipedia.org/wiki/Conjugated_system

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NeverLess asked :

what's the rationale behind adding the volume of Na2S2O5 and HA together in order to get the concentration? I thought it's supposed to be just the no. of mole of NaA divide by the volume of NaA only, even after adding the two solutions together?

Your misconception is a common one amongst students (not just O level students, but many JC students still don't properly understand about this).

Once you've added the two solutions (eg. soln A and soln B) together, the new volume of the new solution of A = new volume of the new solution of B = old volume of A + old volume of B.

There is no such thing (ie. it's nonsense) to say "the new solution = x volume of soln A + y volume of soln B".

To illustrate with concrete examples :

If you add 100cm3 of NaCl, to 100cm3 of KBr, there's only ONE new solution.

This ONE solution can be called any of the following :

a solution of Na+ ions
a solution of Cl- ions
a solution of K+ ions
a solution of Br- ions
a solution of NaCl
a solution of NaBr
a solution of KCl
a solution of KBr
best description to use : a solution containing Na+, Cl-, K+ and Br-.

And the volume of this ONE solution (any of the 8 labels above, refer to the SAME ONE solution), is simply 100 + 100 = 200 cm3.

Hence to find the new molarity (ie. molar concentration) of any of the species, Na+ or Cl- or K+ or Br- or NaCl or KCl or NaBr or KBr, is simply take the moles of that species, divided by the new volume of this ONE solution, which is 100 + 100 = 200 cm3, to obtain the new molarity of the species you're interested in.

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NeverLess replied :

Wow I get it now! To think I bore this huge misconception with me all this while! Thank you sir! Both for your effort and kindness to give this thorough explanation! THANK YOU!

-------------------------
No prob, you're totally welcome! :)

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A BedokFunland JC Original Qn

50cm3 of sulfuric(VI) acid of unknown molarity was mixed with 75cm3 of 0.45mol/dm3 of aqueous potassium hydroxide. 55cm3 of the resulting alkaline solution A was mixed with 80cm3 of 0.35mol/dm3 of phosphoric(V) acid. 90cm3 of the new resulting acidic solution B required 39.97cm3 of aqueous sodium carbonate(IV) for complete neutralization. In a separate experiment 625cm3 of the same sulfuric(VI) acid required 192.3cm3 of the same sodium carbonate(IV) solution for complete neutralization. Calculate the molarities of the sulfuric(VI) acid and the sodium carbonate(IV) solutions.

Answers :

[H2SO4] = 0.2 mol/dm3
[Na2CO3] = 0.65 mol/dm3

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Methyl alcohols and methyl ketones undergo iodoform reaction. Can methyl carboxylic acids undergo a similar reaction to generate iodoform?

The key to understanding any organic chemistry reaction, is its mechanism :

http://www.organicchem.org/oc2web/lab/exp/oxid/iodoform.pdf

http://www.organic-chemistry.org/namedreactions/haloform-reaction.shtm

http://www.demochem.de/D-Iodof-e.htm

Mechanism to oxidize methyl alcohols to methyl ketones (occuring spontaneously under the "alkaline aqueous iodine" conditions of the iodoform test) :

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

No iodoform generated with methyl carboxylic acids (or esters or amides).

Under the alkaline conditions of the iodoform test, carboxylic acids are deprotonated, generating carboxylate ions, which are thus anionic, rendering the alpha C's protons non-acidic.

The first step of the iodoform mechanism (for methyl ketones; for methyl alcohols there is an additonal prior step which oxidizes the methyl alcohol to methyl ketone) involves alpha deprotonation of the methyl ketone's alpha protons (by the OH- base) to generate the enolate ion nucleophile, which subsequently attacks the diiodine electrophile.

This substitution of the alpha protons with iodine atoms repeats itself until the CH3 group becomes the CI3 group, which is a far better leaving group because CI3- anion has a much lower charge density compared to the CH3- anion.

The penultimate step involves the OH- nucleophile attacking the methyl ketone's acyl/carbonyl C atom, eliminating CI3 as CI3-. The final step involves a proton transfer from the (newly generated) carboxylic acid, to the CI3-, generating the yellow precipitate of triiodomethane (aka iodoform).

However (considering the first step of deprotonation), if you were to use a methyl carboxylic acid (instead of a methyl alcohol or a methyl ketone), a dinegative anionic charge on the conjugate base is simply too destabilizing. Hence methyl carboxylate ions cannot be further deprotonated, and therefore the iodoform reaction cannot proceed with methyl carboxylic acids.

Methyl amides and esters, also do not undergo the iodoform reaction, because the carbonyl O atom already bears additional electron density by resonance (as the O atom of the ester, or N atom of the amide, already donates its lone pair by resonance to the carbonyl group). Hence methyl esters and amides also (like carboxylate ions), are unable to be further deprotonated, and therefore unable to undergo the iodoform reaction.

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Nov 2006 P2 Q.3(d)

CH2=CHCH2Br is more electrophilic compared to CH2=CHBr.

Suggest an explanation for this.

There are two reasons for this observation (which applies to both aryl halides and vinyl halides).

The main reason is :
Due to the sideways overlap of the p orbital of the Br atom and the pi orbital of the alkene functional group in CH2=CHBr, a lone pair on the Br atom is delocalized by resonance to form a pi bond with the sp2 hybridized C atom of the alkene, resulting in the C-Br bond having partial double bond character (ie. more endothermic bond dissociation enthalpy) in the resonance hybrid, which thus significantly increases the Ea required for nucleophilic substitutions.

The other reason is :

In CH2=CHBr, the incoming nucleophile (eg. OH-) will be repelled by the nearby (ie. close proximity) electron-rich nucleophilic alkene double bond functional group, which thus significantly increases the Ea required for nucleophilic substitutions.

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