VCE Chemistry Question Thread

[lzxnl]

These are the possible substitution reactions:

Alkane + Cl₂ --> Chloroalkane + HCl (i.e. A hydrogen atom has been replaced with a chlorine atom on one of the side chains). The lost H atom bonds to the other Cl atom to form HCl.

Alkane + H₂O --> Alkanol + H₂ (i.e. A hydrogen atom has been replaced by an -OH group on one of the side chains). The lost H atom bonds to the other H atom to form H₂.

Chloroalkane + H₂O --> Alkanol + HCl (i.e. A chlorine atom has been replaced with an -OH group on one of the side chains). The lost Cl atom bonds to the other H atom to form HCl.

Hence it is possible for an -OH group to be substituted in place of a H atom.

I hope this helps!

Eeeeeeerrrrrrrmmmmmmmmmm..........

I have NEVER heard of hydrogen production by replacing a C-H hydrogen with an alcohol. For starters, C-H bonds are incredibly strong and breaking those often requires rather extreme conditions.
The reason why alkanes can react with chlorine gas is because of the UV light involved. The UV light breaks the molecular chlorine gas covalent bond perfectly in half to release highly reactive chlorine atoms (radicals) containing seven valence electrons. It is only through something as reactive as those chlorine atoms that we can break the C-H bond and replace it with a C-Cl bond. Furthermore, this homolytic cleavage of the Cl2 molecule is only possible due to the weakness of the Cl-Cl bond, arising from the size of the atoms and the repulsions between the nearly full valence shells.

Water, on the other hand, has none of this. You can't split a water molecule into a hydrogen atom and a hydroxyl radical because the OH bond is so much stronger than the Cl-Cl bond (the bond is polarised, giving it ionic character, while the distance between the atoms is also smaller). In addition, breaking of the OH bond won't be symmetric either. Oxygen, being more electronegative, will take the electrons from the covalent bond, leaving a positively charged hydrogen ion and a hydroxide ion. Neither of these is remotely reactive enough to react with the C-H alkane bond.

As for the third reaction, this reaction proceeds via a different mechanism. The C-Cl bond in a chloroalkane is polarised due to the higher electronegativity of the chlorine. The carbon is thus slightly electron-deficient and will thus attract the negative oxygen in water/hydroxide ion. As the oxygen atom approaches the carbon atom to make a bond, the incoming oxygen electrons push out the electrons in the C-Cl bond, leaving you with a Cl- ion and the C-OH bond. As the oxygen was originally bonded to two hydrogens and wants to bond again, one of the hydrogens drops off as a proton and, depending on the conditions, may combine with the Cl- ion to form HCl.

For substitution reactions, why exactly does like OH- swap with CL in an Alkane for example. Like why can't is swap with a hydrogen in the Alkane instead.

The main types of substitution reactions are as follows:
Nucleophilic substitution (the last type; a nucleophile is an electron-rich atom like the oxygen atom in water)
Electrophilic substitution (an electrophile is an electron-poor atom that wants more electrons like a C-F carbon)
Radical substitution (the first type, in which a free radical, or a species with an odd number of electrons like the chlorine atom, reacts and substitutes for something)

[ally12579]

Increase:
1. Presence of contaminants in the precipitate
2. Precipitate is incompletely dried.
3. Co-precipitations - other ions in the solution forming preciptates.

Decrease:
1. The slight solubility of precipitate.
2. Precipitate is lost during filtration.

Thanks 

[keltingmeith]

The main types of substitution reactions are as follows:
Nucleophilic substitution (the last type; a nucleophile is an electron-rich atom like the oxygen atom in water)
Electrophilic substitution (an electrophile is an electron-poor atom that wants more electrons like a C-F carbon)
Radical substitution (the first type, in which a free radical, or a species with an odd number of electrons like the chlorine atom, reacts and substitutes for something)

I have a feeling this is slightly outside the 3/4 curriculum.

[Maths Forever]

Eeeeeeerrrrrrrmmmmmmmmmm..........

I have NEVER heard of hydrogen production by replacing a C-H hydrogen with an alcohol. For starters, C-H bonds are incredibly strong and breaking those often requires rather extreme conditions.
The reason why alkanes can react with chlorine gas is because of the UV light involved. The UV light breaks the molecular chlorine gas covalent bond perfectly in half to release highly reactive chlorine atoms (radicals) containing seven valence electrons. It is only through something as reactive as those chlorine atoms that we can break the C-H bond and replace it with a C-Cl bond. Furthermore, this homolytic cleavage of the Cl2 molecule is only possible due to the weakness of the Cl-Cl bond, arising from the size of the atoms and the repulsions between the nearly full valence shells.

Water, on the other hand, has none of this. You can't split a water molecule into a hydrogen atom and a hydroxyl radical because the OH bond is so much stronger than the Cl-Cl bond (the bond is polarised, giving it ionic character, while the distance between the atoms is also smaller). In addition, breaking of the OH bond won't be symmetric either. Oxygen, being more electronegative, will take the electrons from the covalent bond, leaving a positively charged hydrogen ion and a hydroxide ion. Neither of these is remotely reactive enough to react with the C-H alkane bond.

As for the third reaction, this reaction proceeds via a different mechanism. The C-Cl bond in a chloroalkane is polarised due to the higher electronegativity of the chlorine. The carbon is thus slightly electron-deficient and will thus attract the negative oxygen in water/hydroxide ion. As the oxygen atom approaches the carbon atom to make a bond, the incoming oxygen electrons push out the electrons in the C-Cl bond, leaving you with a Cl- ion and the C-OH bond. As the oxygen was originally bonded to two hydrogens and wants to bond again, one of the hydrogens drops off as a proton and, depending on the conditions, may combine with the Cl- ion to form HCl.

The main types of substitution reactions are as follows:
Nucleophilic substitution (the last type; a nucleophile is an electron-rich atom like the oxygen atom in water)
Electrophilic substitution (an electrophile is an electron-poor atom that wants more electrons like a C-F carbon)
Radical substitution (the first type, in which a free radical, or a species with an odd number of electrons like the chlorine atom, reacts and substitutes for something)

Yes, this is correct, but I was only referring to the general formula. Of course you need a catalyst to enable these reactions to occur. I have excluded reaction 2, I realise that does not make sense.

My side notes are only to describe the rearrangement of these atoms during the substitution reactions.

Please refer to pages 145 to 151 of Heinemann Chemistry 2.

[thushan]

I have a feeling this is slightly outside the 3/4 curriculum.

Yeah nucleophilic/electrophilic is not at 3/4 level.

[lzxnl]

Yeah nucleophilic/electrophilic is not at 3/4 level.

Yeah I know, but it's a better list of reactions than what the textbook has

[Maths Forever]

Yeah I know, but it's a better list of reactions than what the textbook has

So at what level is nucleophilic and electrophilic substitution taught???

[keltingmeith]

So at what level is nucleophilic and electrophilic substitution taught???

I don't know about Melbourne, but my first memory of it was in first year organic (CHM1052/CHM1022). Granted, we didn't go too far into it, and reaction mechanisms are mainly taught in second year synthetic (CHM2911)

[Maths Forever]

I don't know about Melbourne, but my first memory of it was in first year organic (CHM1052/CHM1022). Granted, we didn't go too far into it, and reaction mechanisms are mainly taught in second year synthetic (CHM2911)

Thanks EulerFan101! I look forward to learning more about this!

[lzxnl]

Both nucleophilic and electrophilic substitutions are covered in CHEM10004 (Chemistry 2)

[stockstamp]

Hi,

Was hoping for some clarification here:

What volume of 0.100 M sulfuric acid would be required to neutralise a solution containing 0.500 g of sodium hydroxide and 0.800 g of potassium hydroxide?

Below is how I worked out the answer - would this be correct?

• NaOH (aq) --> Na⁺ (aq) + OH^- (aq)
• KOH (aq) --> K⁺ (aq) + OH^- (aq)
• n(NaOH) = n(OH^-) = 0.5/40 = .0125 mol
• n(KOH) = n(OH^-) = 0.8/56.1 = 0.0142603 mol
• n(OH^-)total = 0.0125 + 0.0142603 = 0.027603 mol
• Therefore n(H₃O⁺) required to neutralise is also = 0.027603 mol
• H₂SO₄ (aq) --> 2H₃O⁺
• n(H₂SO₄) required to neutralise = 1/2 n(H₃O⁺) = 0.0138015
• V(H₂SO₄) = n/C = 0.0138015/0.1 = 0.138 Litres

Is this correct? Am I working it out the right way?

any help appreciated!

[Eiffel]

Essentially, there are two precipitation reactions occurring:

NaCl (aq) + AgNO3 (aq) --> AgCl (s) + NaNO3 (aq)

KCl (aq) + AgNO3 (aq) --> AgCl (s) + KNO3 (aq)

For both reactions, Ag⁺ (aq) + Cl^- (aq) --> AgCl (s)


m (KCl) + m (NaCl) = 0.8870 g

m (AgCl) = 1.9113 g

n (AgCl) = m / M = 1.9113 / [107.9 + 35.5] = 0.013328 mol (5 significant figures)

Therefore, since n (AgCl) = n (Cl^-), n (Cl^-) = 0.013328 mol

n (Cl^-) = n (Cl) = 0.013328 mol

Therefore, m (Cl) = n x M = 0.013328 x 35.5 = 0.47316 g (to 5 significant figures)

% composition of Cl = [0.47316 / 0.8870] x 100% = 53.34% (to 4 significant figures)

% composition of Na and K = 100% - % composition of Cl

= 100% - 53.34% = 46.66% (to 4 significant figures)

I think this is correct. I hope it helps!

how did you know there was two precipitation reactions?

also, how do you convert between grams per litre and molarity (mols / l) and vice versa

[IndefatigableLover]

how did you know there was two precipitation reactions?

You know that they're two precipitation reactions due to the solubility rules that you would have learnt from Year 11 in Chemistry 1&2. People tend to forget about solubility rules but you should dedicate them to memory!

[Maths Forever]

how did you know there was two precipitation reactions?

also, how do you convert between grams per litre and molarity (mols / l) and vice versa

Hi Eiffel,

Since both compounds contain chloride ions, they will both form a precipitate with the silver ions when AgNO3 is added, in the form AgCl (s). Hence the precipitate is formed by chloride ions from both NaCl and KCl, as far as I can see.

To convert grams/litre to mols/litre (molarity), you need to divide by the molar mass of the given compound.

Recall, n = m / M

Please let me know if you still need help. 

[IndefatigableLover]

Quick question in terms of 'Concordant titres'...

Say if I did six trials and the results of those six (in chronological order) were:

18.70mL, 18.73mL, 18.57mL, 18.65mL, 18.65mL, 18.65mL

Which results would I use when calculating the average titre?