Chemistry - Got Some Questions (2 Viewers)

[jazz519]

View attachment 31581
Need some help with Q12.

0.018 absorbance is 1.2 ppm

This is the diluted solution, undiluted is 10x as concentrated as it says diluted by a factor of 10

so [Ca2+] undiluted = 12 ppm

Whenever the question is to do with precipitation and it doesn't look there is a direct answer, it is probably to do with Ksp calculations. As here there is a saturated solution.

CaSO4(s) < -- > Ca2+(aq) + SO4 2-(aq)
Ksp = [Ca2+][SO4 2-]

From data sheet Ksp for calcium sulfate is 4.93 x 10^(-5)

But before you can just substitute the Ca2+ concentration, you need to change it to a mol/L value as that is the type of conc used in that equation

[Ca2+] = 12 ppm = 12 mg/L (as ppm is equivalent to mg/L)
[Ca2+] = 0.012 g/L

To convert g/L to mol/L we divide by the molecular mass, as we want to change g to mol (i.e. n=m/MM)

[Ca2+] = 0.012 / 40.08 = 2.994 x 10^-4 mol/L

Now just sub it in and the answer comes out:
4.93 x 10^(-5) = (2.994 x 10^(-4)) [SO4 2-]
[SO4 2-] = 0.165 mol/L

 

[Life'sHard]

Need some explanations.
1)

Is it A because the halogenation reaction only substitutes chlorine to the double bonded areas and the products formed are exactly the same?

2)

3)

I keep getting 13.2 but answer says it's incorrect.

 

[jazz519]

Need some explanations.
1)
View attachment 31646
Is it A because the halogenation reaction only substitutes chlorine to the double bonded areas and the products formed are exactly the same?

2)
View attachment 31645
3)
View attachment 31647
I keep getting 13.2 but answer says it's incorrect.

1) Using word substitutes isn't correct here as it's an addition reaction. The HCl is added across the double bond. There's only 1 product because if it you add Cl to carbon 2 or carbon 3, it doesn't matter as the final product will still be 2-chlorobutane if you name it using IUPAC rules.

2) Write eqn: NaOH(aq) + HNO3(aq) --> NaNO3(aq) + H2O(l)
Find both moles: n(NaOH) = 0.0025 moles, n(HNO3) = 0.0025 moles
no limiting reagent, so n(H2O) = 0.0025 moles
q = mcdeltaT
m = 25 + 25 = 50 g = 0.05 kg, c = 4.18, deltaT = 0.66 C

q = (0.05)(4.18)(0.66) = 0.13794 kJ

deltaH neut = -q/n(h2o)
deltaH neut = -0.13794 / 0.0025 = -55.176 kJ/mol

3) H2SO4(aq) + 2KOH(aq) --> K2SO4(aq) + 2H2O(l)
find moles: n(h2so4) = 0.0016 moles, n(KOH) = 0.00875 moles

find OH- and H+ moles: n(H+) = 0.0032 moles (as h2so4 as 2H+)
n(OH-) = 0.00875 moles (as koh only as 1OH-)

limiting reagent is H+

n(OH-) excess = 0.00875 - 0.0032 = 0.00555 moles

c(OH-) = n/v = 0.00555 / (45/1000)
c(OH-) = 0.1233... mol/L
pOH = -log[OH-] = 0.9089...

pH = 14-0.9089... = 13.09

Your mistake here is in not accounting for the fact that H2SO4 as 2H+ in it per molecule. You can't just say excess of something = one substance - another substance, without accounting for the molar ratios

 

[Life'sHard]

1) Using word substitutes isn't correct here as it's an addition reaction. The HCl is added across the double bond. There's only 1 product because if it you add Cl to carbon 2 or carbon 3, it doesn't matter as the final product will still be 2-chlorobutane if you name it using IUPAC rules.

2) Write eqn: NaOH(aq) + HNO3(aq) --> NaNO3(aq) + H2O(l)
Find both moles: n(NaOH) = 0.0025 moles, n(HNO3) = 0.0025 moles
no limiting reagent, so n(H2O) = 0.0025 moles
q = mcdeltaT
m = 25 + 25 = 50 g = 0.05 kg, c = 4.18, deltaT = 0.66 C

q = (0.05)(4.18)(0.66) = 0.13794 kJ

deltaH neut = -q/n(h2o)
deltaH neut = -0.13794 / 0.0025 = -55.176 kJ/mol

3) H2SO4(aq) + 2KOH(aq) --> K2SO4(aq) + 2H2O(l)
find moles: n(h2so4) = 0.0016 moles, n(KOH) = 0.00875 moles

find OH- and H+ moles: n(H+) = 0.0032 moles (as h2so4 as 2H+)
n(OH-) = 0.00875 moles (as koh only as 1OH-)

limiting reagent is H+

n(OH-) excess = 0.00875 - 0.0032 = 0.00555 moles

c(OH-) = n/v = 0.00555 / (45/1000)
c(OH-) = 0.1233... mol/L
pOH = -log[OH-] = 0.9089...

pH = 14-0.9089... = 13.09

Your mistake here is in not accounting for the fact that H2SO4 as 2H+ in it per molecule. You can't just say excess of something = one substance - another substance, without accounting for the molar ratios

Thanks!

 

[Life'sHard]

I understand how this is an ester due to infrared and 13C NMR becuase of the sharp peak, 170 and 68 ppm chem shift. I am having difficulty in distinguishing between the 4 different carbon environments and using the hydrogen NMR to determine it. Can someone help me with this.
This is the solution.

 

[CM_Tutor]

The doublet / septet combination is classic for an isolated isopropyl group, -CH(CH₃)₂. The position of the CH signal at 5 ppm points to a neighbouring O atom. The singlet methyl must also be isolated, hence CH₃-C(=O)-O-CH(CH₃)₂. The ¹³C signal at 22 ppm points to two carbons in that environment, all others being 1, so that is consistent with an isopropyl. The 6H doublet must be 2 x CH₃ or 3 x CH₂ and with one neighbour. The 1H septet must have 6 neighbours, again pointing to 2 x CH₃ or 3 x CH₂.

 

[Life'sHard]

The doublet / septet combination is classic for an isolated isopropyl group, -CH(CH₃)₂. The position of the CH signal at 5 ppm points to a neighbouring O atom. The singlet methyl must also be isolated, hence CH₃-C(=O)-O-CH(CH₃)₂. The ¹³C signal at 22 ppm points to two carbons in that environment, all others being 1, so that is consistent with an isopropyl. The 6H doublet must be 2 x CH₃ or 3 x CH₂ and with one neighbour. The 1H septet must have 6 neighbours, again pointing to 2 x CH₃ or 3 x CH₂.

Last couple of questions. What is relative peak area? Is this just extra info that's not really useful in current syllabus and this q? Also, just for clarification what exactly are the 4 unique environments? Thanks

 

[CM_Tutor]

Last couple of questions. What is relative peak area? Is this just extra info that's not really useful in current syllabus and this q? Also, just for clarification what exactly are the 4 unique environments? Thanks

Relative peak area gives you the number of H atoms in each environment. This substance, 2-propyl acetate, has 3 hydrogen environments:

(CH₃)₂CH-O-C(=O)-CH₃

with an intensity ratio of 6:1:3 as the two methyls on the isopropyl have 6 x H atoms, the methine (CH) on the isopropyl has 1 x H atom, and the methyl on the acetate group has 3 x H atoms.

Each environment gives rise to its own signal in the spectrum:

• CH₃'s on the isopropyl group appear at 1.2 ppm because they are bonded to only hydrocarbon chain. It appears as a doublet as it has 1 x H neighbour on the adjacent carbon. It is large as it has to split an intensity of 6 over only two lines
• CH on the isopropyl group appear at 5 ppm because it is bonded to the highly electronegative O atom of the ester linkage. It appears as a septet as it has 6 x H neighbours on the adjacent carbon. It is so tiny because it has to split a total intensity of 1 over 7 lines.
• CH₃ on the acetate group appear at 2.0 ppm because the adjacent carbonyl group has more electron density than a hydrocarbon chain, but not as much as an O atom, so the shift isn't as far downfield. It appears as a singlet as it has no H neighbours adjacent. It is the same size / height as the isopropyl methyls as it has an intensity of 3 but only one lines (hence the same size as an intensity of 6 over two lines).

There are four carbon environments:

(CH₃)₂CH-O-C(=O)-CH₃

again with isopropyl methyls being equivalent, making for one signal being twice as tall as the others.

 

[Life'sHard]

Relative peak area gives you the number of H atoms in each environment. This substance, 2-propyl acetate, has 3 hydrogen environments:

(CH₃)₂CH-O-C(=O)-CH₃

with an intensity ratio of 6:1:3 as the two methyls on the isopropyl have 6 x H atoms, the methine (CH) on the isopropyl has 1 x H atom, and the methyl on the acetate group has 3 x H atoms.

Each environment gives rise to its own signal in the spectrum:

• CH₃'s on the isopropyl group appear at 1.2 ppm because they are bonded to only hydrocarbon chain. It appears as a doublet as it has 1 x H neighbour on the adjacent carbon. It is large as it has to split an intensity of 6 over only two lines
• CH on the isopropyl group appear at 5 ppm because it is bonded to the highly electronegative O atom of the ester linkage. It appears as a septet as it has 6 x H neighbours on the adjacent carbon. It is so tiny because it has to split a total intensity of 1 over 7 lines.
• CH₃ on the acetate group appear at 2.0 ppm because the adjacent carbonyl group has more electron density than a hydrocarbon chain, but not as much as an O atom, so the shift isn't as far downfield. It appears as a singlet as it has no H neighbours adjacent. It is the same size / height as the isopropyl methyls as it has an intensity of 3 but only one lines (hence the same size as an intensity of 6 over two lines).

There are four carbon environments:

(CH₃)₂CH-O-C(=O)-CH₃

again with isopropyl methyls being equivalent, making for one signal being twice as tall as the others.

I still don't see how the isopropyl methyls are equivalent. From the solution above, one looks closer to the oxygens than the other does... Is there a proper way to determine whether they are equivalent or not? Or is it mostly a given for esters that the methyl groups at the ends are equivalent?

 

[CM_Tutor]

I still don't see how the isopropyl methyls are equivalent. From the solution above, one looks closer to the oxygens than the other does... Is there a proper way to determine whether they are equivalent or not?
View attachment 32675

Remember that the molecule is three dimensional and can rotate around the C-O bond

 

[Life'sHard]

Remember that the molecule is three dimensional and can rotate around the C-O bond

Ok ok I see it now. Completely forgot these were 3d after doing so many 2d diagrams lmao. Thanks!

 

[CM_Tutor]

Get a straw. Stick three toothpicks at one end, 120 degrees apart when looking along the straw but pointed away from the end (like three slant edges of a pyramid). Paint two red and one green.

Hold the straw in your palm with your thumb sticking up like the O atom. One red stick is closer than the other. Now spin the straw... on average, all three toothpicks are as close to your thumb as each other.

 

[Life'sHard]

1)

Why can't it be C?

2)

Is equillibrium expression another way of saying ?

3)

Why isn't it A?

 

[CM_Tutor]

Question 1: Buffers will resist changes in pH but they have a limited capacity, determined by how much of each species in the buffer is present. Large amounts of acid or base will overwhelm a buffer... and anyway, buffers don't prevent changes in pH, they resist changes in pH. If I add so much as a drop of acid with a pH lower than that of the buffer to the buffered solution, it must result in a decrease in pH, even if a minute one.

Question 2: Yes, they mean the expression for K. The question doesn't make sense as asking for the equation for the reaction.

Question 3: The signal from the w hydrogens will be a singlet as their neighbour, the carbonyl, has no hydrogen atoms and 0 + 1 = 1 lines.
The signal from the x hydrogens will be a doublet as hydrogen z is its neighbour, and 1 + 1 = 2 lines.
The signal from the y hydrogens will be a singlet as their neighbour is an O atom, hence no hydrogens, and 0 + 1 = 1 lines.
The signal from the z hydrogens will be a triplet as hydrogen z has the two x hydrogens as its neighbour, and so 2 + 1 = 3 lines.

w will give an intensity = 3 singlet around 2.1 or so ppm
x will give an intensity = 2 doublet around 2.5 ppm, perhaps - higher than w, anyway, given the two methoxy groups nearby
y will give an intensity = 6 singlet around 4 to 5 ppm, given the adjacent oxygen.
z will give an intensity = 1 triplet around 5 ppm or higher, given two adjacent oxygens, it will be higher than y, anyway.

If I take the three lines of z as having a height ratio of 1:2:1, then we know that a total height of 4 matches 1 x H, so

w will have a height of 12

x will have two lines of height 4, and

y will have a height of 24.

 

[CM_Tutor]

I got this, would this also be correct? Since it matches with the graphs

View attachment 32712

No, it would not, for two reasons, one minor, one major.

The minor one is the C atom adjacent to the C of the carboxyl group is missing an H atom.

The major one is that the signal for that methine (CH) carbon, which is split into a septet, occurs at a chemical shift of something like 5. This is way too far downfield to be the effect of just an adjacent carbonyl or carboxyl, it points to a direct bond to an electronegative atom like an O. Similarly, the singlet methyl group that occurs at around 2.1 ppm, which is consistent with an adjacent carbonyl or carboxyl, would have a chemical shift of perhaps 4.5 ppm or more if it were bonded to an O atom as you have drawn.

¹H NMR gives four varieties of information:

• number of signals = number of environments
• chemical shift of each signal tells about the neighbouring environment in terms of electron density and shielding / deshielding
• splitting tells about the presence of NMR active nuclei on the neighbouring atoms; for HSC, we look only at splitting by neighbouring H's
• intensity / integration tells about the ratio of ¹H atoms in the environments, usually giving the number of H atoms.

You need to consider each. The difference between methyl 2-methylpropanoate (your structure) and 2-propyl acetate (the correct answer) is in the second point, the chemical shifts. Both compounds have 3 ¹H environments, in a 1:3:6 ratio, and with the splitting being (in the order of the intensity ratio) septet - singlet - doublet, so they are indistinguishable from the other three varieties of information, but on chemical shifts, they are clearly different:

methyl 2-methylpropanoate has a singlet (3H, d = 4.5 ppm), a septet (1H, d = 2.1 ppm), and a doublet (6H, d = 0.9 ppm).
2-propyl acetate has a septet (1H, d = 5 ppm), a singlet (3H, d = 2 ppm), and a doublet (6H, d = 1.2 ppm).

Note: Chemical shifts are approximations.

 

[Life'sHard]

But we aren’t given chemical shift data for H nmr in the hsc?

No I’m pretty sure chemical shift is part of H NMR and by extension a part of the HSC. It’s just part of the data provided. It’s similar to the carbon NMR chem shift.

 

[CM_Tutor]

But we aren’t given chemical shift data for H nmr in the hsc?

You are correct, the data sheet only gives ¹³C NMR chemical shift data. I have seen trials give a table of ¹H NMR data to support such questions and I suppose that the HSC would need to do so for you to answer questions like this.

 

[CM_Tutor]

There is a good table of ¹H NMR shifts in either the Sydney Grammar 2019 or 2020 Trials (or maybe in both).

...and they aren't going to give you any of the really odd results, like cyclooctadecanonaene, C₁₈H₁₈, which is a ring of 18 carbons alternating double bond - single bond - double bond - single bond etc all the way around the ring. The double bonds go cis, trans, trans, cis trans, trans, cis, trans, trans and the molecule is planar.

Its ¹H NMR spectrum at 120 degrees Celsius is a single signal at 5.45 ppm. When cooled to -60 degrees Celsius in THF-d₈ (solvent), the spectrum shows two environments, both singlets, in 2:1 intensity ratio. The chemical shifts are at 9.3 ppm and -2.9 ppm (yes, that is a minus).

 

[CM_Tutor]

Partly.

ethyl methanoate, H-C(=O)-O-CH₂-CH₃, has three hydrogen environments in intensity ratio 1:2:3.

methyl ethanoate, CH₃-C(=O)-O-CH₃, has two hydrogen environments in intensity ratio 3:3 = 1:1.

propanoic acid, CH₃-CH₂-C(=O)-OH, has three hydrogen environments in intensity ratio 1:2:3.

Distinguishing methyl ethanoate from the other two can be done by number of environments. Chemical shifts are needed for the other pair, as:

ethyl methanoate has a singlet (1H, d = 9.5 or so ppm), a quartet (2H, d = 4 ppm) and a triplet (3H, d = 1.5 ppm).

methyl ethanoate has a singlet (3H, d = 4 or so ppm) and a second singlet (3H, d = 2 ppm).

propanoic acid has a singlet (1H, d = 11 or so ppm) a quartet (2H, d = 2 ppm) and a triplet (3H, d = 1 ppm).

Based on these, C is propanoic acid, A is methyl ethanoate, B is ethyl methanoate, and my estimates for chemical shift (which I did without looking at the spectra) were pretty good, except the methine (CH) in ethyl methanoate isn't as far downfield as I guessed.

 

[CM_Tutor]

I would start by identifying the one with only two signals and no splitting.

For the other two, the quartet of B compared to C shows B has the CH₂ bonded to the O atom in an ester whereas C has it bonded to carbons and thus in the carboxylic acid.