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As we are focusing on Physical Natural Sciences in this series, this blog post includes a maths-related, physics-related and chemistry-related Natural Sciences interview script.

### Question 1 (Maths)

Question: The gamma function is defined as . You are given that as for all natural numbers n. By using integration by parts, deduce a property of the Gamma function. Can you think of a similar function? Calculate [Top tip: keep in mind all the facts they have given you. They don’t give you them for no reason, so they must be things you’ll be using later]

Student: Since I am integrating by parts, I will be integrating one part and differentiating the other. I know how to differentiate and integrate both and , but I think I will choose to differentiate because that should make it simpler?

Student: Integration by parts states that . In this case I’ll have that and . So then [pausing] I think I see now why I was told that as for all n, because I’ll use that in evaluating the first part. [thinking] Actually, the whole first term is going to disappear since its zero at infinity and at zero…

Interviewer: Continue.

Student: With the second term on the RHS, I just seem to have the same problem that I started with and since I don’t know what z is, it will just keep going if I keep using integration by parts?

Interviewer: What do you mean that you have the same problem as you started with? Can you expand on that?

Student: Well, I started off with , and even though the integration by parts has reduced the power by one, I’ve still got in there.

Interviewer: So….

Student: So are you saying I should plug in should plug in into the RHS to get rid of the integral?

Interviewer: Don’t ask me; have a go and see.

Student: Ah. I think I get that . Is that the property you were looking for?

Interviewer: Yes, that’s it. Can you think of another function that has a very similar property?

[silence as the student thinks]

Student: I suppose in the simplest example a function that is always zero will have this property?

Interviewer: Well, yes, that’s true, but I was more thinking of a non-trivial function that also shows this property.

Interviewer: It’s going to be something that has a nice recursive definition like above….

Student: Well, it looks quite similar to the factorial function since

Interviewer: Now can you have a go at working out

Student: So , but that integral looks a bit inaccessible. [thinking] Maybe I should use , and then I just need to work out a simpler case like , ?

Interviewer: Go on….

Student: I guess ideally I want because that’s just 1, so I thinks its best to work out but then I’m not sure what is?

Interviewer: Are you sure about that? What do you think it’ll be?

Student: , so if x is really big, that must mean gets really close to zero? Then that would mean that

Interviewer: Good, continue.

Student: So, using our formula from before:

Interviewer: That’s right. So how do you think Gamma and factorial are related.

Student: Well I know that 0! = 1! = 1 and 3! = 6, so it seems like the pattern is going to be

Interviewer: Yes, that’s exactly right! In fact, the Gamma function can be seen as a way of extending the factorial function, which is only defined for natural numbers, to real and even complex numbers.

#### Further Hints:

• If you haven’t studied reduction formulae for integration, the interview could hint at this by asking: “Do your two integrals look similar?” or, more explicitly, “Can you write your second integral using the gamma function?”.
• When it comes to calculating , if you haven’t realise you should be using the recursion formula, it might be prompted: “Is there another way that you could write using something else you’ve derived?”
• If you haven’t spotted the similarity between the factorial and gamma functions, the interviewer could prompt you by asking “What is in terms of ”, or something similar in the hopes that you recognise the are also characteristic of the factorial function.

#### Extending the Question:

An interesting extension to this question could be, “Using what you know about how different functions grow, reason why as ?”. This should test your understanding of the relative growth of the exponential compared to powers of x.

This limit is actually something that you meet and prove in your first-term Analysis course at Oxford, and is very useful.

#### What the Question is Testing:

Similar to previous questions, this tests how well you react to new material. The wording in the question “Deduce a property..” was purposefully left somewhat ambiguous as it does not give you so much of an end goal, but instead tells you what method to use to find a useful property. Much of your independent work studying maths at university will be curated like this; using techniques you know to find out things you don’t.

#### Related Topics from University:

Factorials are important in the topic of  probability, although the gamma function itself is only briefly mentioned in your first-year probability course. You will study the Gamma distribution, though, which is a general case of the chi-squared distribution – something you may have met at A-level.

### Question 2 (Physics, also related to Chemistry)

Question: Suppose you have two neutral molecules. How do they interact as the distance between them varies?

Student: Well, I know there are four fundamental forces in physics: the weak force, the strong force, the electromagnetic force and gravitational force. The weak force and strong force are only relevant on extremely small distance scales; would it be OK if I discount them?

Interviewer: Yes, please ignore those forces for the purposes of this question.

Student: Ok, so now I just need to consider the effects of the gravitational and electromagnetic force, which in classical physics have an infinite range…

Interviewer: Let me interrupt you there. Why do you think we should include the gravitational force in our model?

Student: Well, initially I thought it would be relevant because gravity acts over all distance scales. (Pause.) But having thought it over, I realise now that the Coulomb force is several orders of magnitude stronger than the gravitational force, so I guess it would be sensible to ignore gravity for this question.

(Interviewer nods in agreement)

Student: Ok, in that case let’s just consider the electrostatic force acting between these molecules. I know that molecules consist of protons surrounded by an orbiting electron cloud in which the electrons are free to move. So even though each individual molecule is neutral overall, attractive Van der Waals forces between the two molecules will be created due to the induced electron-dipole attraction.

Interviewer: What about the forces between the protons?

Student: Well, the protons are both positively charged, so there will be a repulsive Coulomb force between them… But, at the same time, for significantly large distances the molecules would just look neutral.

Interviewer: So, going back to the original question: how would the two forces you’ve talked about combined vary with distance in this scenario?

Student: (long pause) I think it would be most helpful if I considered the overall electrostatic potential between the two molecules as a function of distance, where negative potential indicates an overall attractive force. Let’s say they start infinitely far apart; the potential between them would obviously be zero. As I bring them closer together, the attractive Van der Waals force would dominate for large-distance scales since the molecule appears neutral overall, and this Van der Waals force grow in magnitude until it reachesd itsa maximum.m; Aas a result, the potential would reach a corresponding minimum.

Interviewer: What about for even smaller values of ?

Student: Well, when the molecules are sufficiently close together that the electron clouds begin to overlap, the electron cloud repulsion and proton-proton repulsion would dominate. This repulsive force would tend towards infinity as the distance between the molecules tended towards zero; the electrostatic potential would therefore tend towards positive infinity.

Top tip: This is the crux of the question, and you probably would have to go through a lot more prompting to arrive at this conclusion!

Interviewer: Ok, let’s move on. The mathematical equation for the potential that you have just described is often modelled as the Lennard-Jones potential, and is given by the following (begins writing equation on paper)

Could you please sketch this on a set of axes for me?

Student: I’m guessing V stands for potential and that r is the distance. Can I just check that and are non-zero real constants?

Interviewer: Yes, that’s right.

Top tip: The fact that you’ve just gone through the theory should help you sketch the graph, although don’t worry if you get stuck at any stage. Remember to articulate your thoughts aloud, and go through the standard graph sketching procedure, i.e. behaviour as , , stationary points etc.

Student: … So I think it looks like this:

Interviewer: Could you please find the distance at which this potential is minimised?

Student: Ok, so I need to differentiate this function and set the derivative equal to 0:

Since a, r and are all non-zero, I can divide through by

:

Hence , meaning that the minimum potential occurs at .

Interviewer: Thank you, that’s all from me.

#### Extending the question

• What happens when the two molecules are released from a small distance away from ?
• What is the physical significance of the distance ?
• Oscillations about would release EM energy. In what region of the EM spectrum would you expect to observe the energy emitted?

### Question 3 (Chemistry)

Question: Can you draw the mechanism for Friedel-Crafts Acylation?

Student: Yes, this is an electrophilic aromatic substitution reaction, with aluminium trichloride (or another lewis acid) as the catalyst. This forms a carbocation that the double bond can then attack before losing a hydrogen to regain aromaticity.  [As always in chemistry: draw it out! It’s easier to understand.]

Interviewer: Would you expect the reaction to occur again? Creating a polysubstituted benzene?

Student: (pauses to think) No, I don’t think it would. Ketones are mesomerically electron-withdrawing, meaning that the benzene ring is deactivated and electrophilic aromatic substitution is less favourable.

Interviewer: What makes you say that it’s electron-withdrawing?

Student: Well, it has a pi system which is conjugated with that of the benzene. Therefore, it has a mesomeric effect. Oxygen is very electronegative and so it will pull electron density from the benzene ring pi system.

Interviewer: Can you draw a resonance diagram to show that? [This is deceptively difficult]

Student: Yes, it might be something like this?

Interviewer: Okay, and can you draw the mechanism for Friedel-Crafts alkylation? Would you expect this reaction to occur multiple times?

Student: (draws the reaction below) I think that this would occur multiple times as alkyl groups are electron-donating mesomerically and inductively. Therefore, I would expect the reaction to get progressively faster as it becomes more favourable.

Interviewer: Which positions do you think would be the most preferable for electrophilic substitution to occur? [Top tip: If you’re unsure, try drawing out resonance forms]

Student: Due to resonance, alkyl groups are o-/p- (ortho and para) directing because it can stabilise positive charges at these positions through delocalisation in the pi system. However, there are also steric effects, and so if the alkyl group is bulky then it’s more likely that you would only get para substitution.

Interviewer: What about under thermodynamic conditions? With high temperature? What trisubstituted product would form? [Top tip: This bit of the question is difficult, don’t fret if you don’t understand it straight away]

Student: Well, in this case everything is reversible, and so I would expect that you would form the meta- substituted product because this has the least steric clashing. (shown below)

Interviewer: Thanks, that’ll be all.

Further Hints

(If student doesn’t know Friedel-Crafts) The reaction is an addition of a carbocation to a benzene ring, could you suggest a mechanism for that?

(If student doesn’t understand preferential sites for substitution) What if we draw out the mechanism for a substitution at two different positions? Can you think of why one would be preferred over the other?

Extending the Question

If you wanted to add an OH group and an acyl group to a benzene ring, in which order would you add them?

Would low temperature stop polysubstitution of the benzene ring for alkylation? Why?

What the Question is Testing

This question tests your knowledge of aromatic chemistry. However, even the best students won’t know the answers from their course all the way through. The tutors are trying to test your ability to think through answers by using curly arrows to justify the conclusions that you might know from your A-level course. It’s important to take your time and draw things out if you’re unsure.

Related Topics at University

This is hugely important in pharmaceutical chemistry, and comes up in aromatic chemistry in first and second year, as well as biological chemistry. It’s a large topic that also ties into heterocyclic chemistry (furans, pyrroles etc.), which you can look up if you’re interested.

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