Pre-Year 2 Preparation Guide
Don’t spend your whole summer vacation forgetting all of the chemistry you’ve learned
…a little bit of preparation will help your performance in Year 2
…a little bit of summer reading will expand your independent learning skills (which employers are very keen on)!
1. Summer Reading List 2. Revision of Year 1 Topics 3. A Quick Guide to (Essential) Differentiation & Integration (6F5Z2007) 4. Symmetry & Basic Crystallography (6F5Z2007) 5. NMR Spectroscopy – Preparation (6F5Z2006) 6. Essential 1st Year Organic Chemistry (6F5Z2006) 7. Biological Processes & Pharm Chem (6F5Z2009)
SUMMER READING LIST
PREPARATION FOR YEAR 2
Dr. Lindsey J. Munro [email protected] 1
Dr. Lindsey J. Munro [email protected] 2
Summer Reading List Preparing for your 2nd Year
To stop you from getting bored in the long, long holidays, here’s a copy of your summer reading list…in case you are near to a library – to help you find out which bits of chemistry that you like! All good preparation for career (& exam) success in the future.
Recommended Books for Yrs 2 & 3:
We arrange a book pack of the 3 recommended texts which will cover Years 2 & 3 from OUP at a reduced price of ~£122 (20% reduction & the cost works out at ~12-17p / day) – based on 2016. Usually you can purchase these from Portland Bookshop (MMU Student Union) or Blackwell Bookshop
“Physical Chemistry” by Atins & de Paula (Newest: 11th edition - 2017) + “Organic Chemistry” by Claydon, Greeves & Warren (Newest: 2nd edition - 2012) + “Inorganic Chemistry” by Weller, Overton, Rourke & Armstrong (Newest: 7th edition - 2018)
Chemistry3 will still be a useful introductory textbook – but you will also need the depth of information presented in the other recommended texts.
“Maths for Chemistry” will help explain & provide lots of practice questions for the maths that you will encounter in Years 2 & 3.
General Comments
In general, the following series of books are all good for getting a more in-depth idea about one topic – but I would not recommend that you buy them. They are all well written & light! See if you can find copies in a library.
I’ve noted how many copies of each book that there are available in the MMU library – please don’t take out all of these books & keep them sitting in a dark corner for the whole vacation. Just take one book – and try reading it. (NB: Not in one sitting!) Return any books as soon as possible. You might pick a topic that you’ve found especially difficult this year or the subject you’ve liked the best. Revising your notes from this year of anything you’ve found particularly topic is always a good start! Let me know which books you like the best (or least) & if you have any other recommendations.
Royal Society of Chemistry (RSC) Tutorial Texts: http://pubs.rsc.org/bookshop/search?searchtext=Tutorial+texts
Oxford Chemistry Primers: http://ukcatalogue.oup.com/category/academic/series/chemistry/ocp.do
Open University/RSC "The Molecular World": http://www.rsc.org/molecularworld (The OU is particularly good at writing texts that enable you to teach yourself = Key Grad Skill.)
Supplementary Book “A Guide to Modern Inorganic Chemistry” by Steven M. Owen & Alan T. Brooker – There are 2 copies in the library – but if you ever find cheap used copies on Amazon or e-bay then I would recommend buying it.
Dr. Lindsey J. Munro [email protected] 3
Summer Reading Do not buy these books – just get them from the library
1. Foundations of Organic Chemistry by Hornby & Peach (Oxford Chemistry Primers) - 4 copies
- Practice of functional group, nucleophile & electrophile identification - Revise 1st year notes - Revise Chemistry3 (especially if you don’t get one of the copies of the primer) - Highly recommended - essential prior to Year 2 Organic Chemistry in 6F5Z2006 = Core 1
2. “Physical Chemistry: Understanding our Chemical World” by Paul Monk - 11 copies in the library - Written by a former MMU lecturer.
- Easy to read guide to physical chemistry – introducing you to key concepts by trying to answer questions about things that we observe in the world around us…like “Why do we sneeze?”, “How is smoke in horror films made?” - Especially Chapters 3, 4 – 10 - Useful for Thermodynamics & Phase Equilibria lectures – Year 2 - 6F5Z2007 = Core 2 - Highly recommended.
3. “Molecular Symmetry & Group Theory” by Alan Vincent - 6 copies (1 – 1977, 5 – 2001) in the library
- A unique book that actually gets you to teach yourself a topic in a step-by-step guide. - It gives you a good head-start getting your head around symmetry elements – don’t be put off by the maths…and try building molecules using Scigress if you need help visualizing the structures. - Useful for Symmetry lectures – Year 2 & Spectroscopy lectures - 6F5Z2007 = Core 2 - Highly recommended.
4. d- and f-block chemistry by C. J. Jones (RSC Tutorial Text) - 1 copy in library
- Transition metal chemistry lectures – Year 2 - 6F5Z2006 = Core 1
5. “Basic Atomic & Molecular Spectroscopy” by J. Michael Hollas (RSC Tutorial Text) - 1 copy in library
- Spectroscopy lectures (Phys. Chem.) – Year 2 - 6F5Z2008 = Analytical Techniques (Core 3)
Other libraries – Check!! In the UK the MMU library can issue you with a SCONUL card which allows you to visit (but not borrow) books from all UK university libraries.
Dr. Lindsey J. Munro [email protected] 4
REVISION GUIDE
PREPARATION FOR YEAR 2
Dr. Lindsey J. Munro [email protected] 5
Dr. Lindsey J. Munro [email protected] 6
Revision Guide Preparation for Year 2
During the long, long vacation you will probably be ready to start doing some thinking and getting ready for your second year. There are lots of good reasons to do this: Your second year marks count towards your final degree classification. When you apply for jobs during your final year, employers will look at these marks to assess whether they should hire you – so make sure they are good!! Many people find the 2nd year of the course a big jump up from the 1st year - therefore, it is important to revise your notes from last year in core topics, since we will assume that you already know & remember the content of these courses, without having to look them up. If you “just” managed to pass the course (anything less than 100%!), then look in particular at topics that you found difficult and think about how you could have improved your performance in both coursework & examinations. Independent study and practising questions on your own are essential elements to developing the problem-solving skills and knowledge required by employers to succeed. Before asking a question, make sure you have searched fully for the answer.
Hard work now = Interesting & challenging job later
Core 2 (Yr 2) = 6F5Z2007 Essential Preparation Physical + Inorganic
Guide to Differentiation & Integration (Separate Section) Essential 2
G Cp - If you don’t know how to deal with mathematical functions such as or dT (or if you T T do vaguely but it makes you want to cry just thinking about it) then working through this simple guide will be your key to success in the Thermochemistry module in CH2112. - Even if you got an A* at A-level you need to be fluent in this mathematics by the start of Year 2 … so yes, you should go through the guide too…
- Warning: If you don’t try this guide, you will find the Thermochemistry course more tricky than it really is – it really is just a question of practising so that you are familiar with the content. (Practising = Read / do questions in guide 10 – 30 times … not just once.)
- Highly recommended by this year’s 2nd years.
Monk & Munro’s “Maths for Chemistry” book (Chapters 14-20) - All exercises in all of these chapters … will help you gain the expertise & proficiency that you will need in the relevant mathematics required for this course & brush up on your independent learning skills as well … which will make your future employers v. happy!
Symmetry, Structures of Simple Solids & Basic Crystallography ESSENTIAL REVISION & PREPARATION Guide (Separate Section) - Chapters 3, 6 & 8 – “Inorganic Chemistry” by Weller et al. - Outline of key skills required to practise over vacation - Online Resources - Scigress Modelling Exercises
Notes from 6F4Z2004 – Fundamental Chemistry Concepts 2 Thermodynamics & Kinetics (Dr. Edge) - Especially the key principles, equations (memorised) & how to apply these.
Dr. Lindsey J. Munro [email protected] 7
Core 1 (Yr 2) = 6F5Z2006 Essential Preparation Organic + Inorganic Review of Organic Chemistry required for 2nd year (Separate Section) - When learning organic chemistry, each new element builds on the foundation of all of the previous knowledge. It is essential that you know how to do the following WITHOUT looking at a text book at the start of the year. You MUST be fluent in the language of organic chemistry before term starts in September:
o Name compounds o Identify functional groups, sp2, sp3 carbons, electrophiles & nucleophiles o Curly arrows - Should be able to draw mechanisms (without looking at notes) - Predict products of reactions based on Year 1 knowledge of chemistry o A wide range of reactions (nucleophilic addition, SN1, SN2, E1 and E2 etc.) o Carbonyl group chemistry o Grignard reagents - Overview of each - Additions to the C=O group of Grignard reagents o Reactions of Alkenes & Alkynes – Markovnikov’s rule o Relative stability of carbocations & how that relates to electrophilic additions to alkenes (Markovnikoff & anti-Markovnikoff additions) o Identification of primary, secondary, tertiary & quaternary carbons o R & S for naming enantiomers o Double bond equivalents
- Read this review, memorise the key elements now & throughout the term. - This information has to be second nature to you - Otherwise you will struggle with the key elements during lectures and fall behind.
- It is highly likely that you will be tested on this in the 1st week of term. - Highly recommended by this year’s 2nd years.
Recommendations from the current 2nd Years: Revise the following mechanisms (so that you know how to identify the key features – such as functional groups, electrophiles, sp, sp2 and sp3 carbons, draw the arrows and complete the mechanism WITHOUT reference to a textbook – this must be second nature to you): Electrophilic addition (in particular alkenes & alkynes) Nucleophilic addition (especially carbonyl chemistry) Electrophilic substitution Nucleophilic elimination Chemistry3 by Burrows, Holman, Parsons, Pilling & Price (Chapters 19-24 = excellent start)
Notes from 6F4Z2003 – Fundamental Chemistry Concepts 1 Organic Chemistry - Especially the key principles, terminology, functional groups & mechanisms (see above).
Inorganic Chemistry - Transition Metals & Coordination compounds (Chapter 24)
Notes from 6F4Z2004 – Fundamental Chemistry Concepts 2 Atomic & Molecular Structure - Molecular Orbital Theory - Make sure that you can quickly draw, explain & interpret the MO diagram for CO, NO etc.
Thermodynamics – Review all key concepts Kinetics - Review all key concepts
Dr. Lindsey J. Munro [email protected] 8
Required Preparatory Reading Stereochemistry: Chirality, optical activity, enantiomers, racemic mixtures, absolute configuration R/S and E/Z nomenclature. Chemistry3, Chapter 10 “Isomerism and Stereochemistry”
Spectroscopy: Basic NMR theory and intepretation of 13C NMR spectra, IR spectroscopy & interpretation of IR spectra, UV spectroscopy, recognition of functional groups in organic molecules. Chemistry3, Chapter 11 “Molecular Spectroscopy” Chemistry3, Chapter 13 “Molecular Characterisation”
Core 3 (Yr 2) = 6F5Z2008 Essential Preparation Physical + Analytical Reactions in Solutions – Electrochemicals Cells (6F4Z2004) (Dr. Rego) - Especially the key principles, equations (memorised) & how to apply these.
Essential Preparation for 6F5Z2009 Biological Processes & Pharmaceutical Chemistry Recommended reading list – see Section 5 Industrial Chemistry (6F4Z2006) - Specifically the Pharmaceutical Chemistry topics (Dr. Caprio & Dr. Birkett)
Molecular Modelling (6F4Z2004) (Dr. Munro) - Know the basics of how to use the Scigress modelling code
Make sure you bring your 1st year notes & revision guides back with you - we will assume that you will be using them daily during term time.
Dr. Lindsey J. Munro [email protected] 9
Dr. Lindsey J. Munro [email protected] 10
ESSENTIAL REVISION & PREPARATION
GUIDE TO DIFFERENTIATION & INTEGRATION
Dr. Lindsey J. Munro [email protected] 11
A Quick Guide to Differentiation & Integration (Only 1 hour required)
Ignore the rumours – calculus (differentiation & integration to you & me) really isn’t the hardest topic in the universe – nor is it an unsurpassable mountain that you will never ever understand. If you follow this guide you will no longer fear the very mention of these words. You will be able to understand the basics so that you don’t feel like you have to lie down & take a nap every time you G T2 1 see any of the following: or dT . T T p T1 Still awake? Excellent – such fortitude of spirit is exactly what is needed. Now, those of you who are thinking you don’t need to revise this topic or you can’t be bothered or that you are never going to understand it – think again! This v. short guide has been produced in response to previous survivors of the second year – including the ones who get 1sts – so no trying to escape – I’m expecting you to have tried to follow, understand & remember this guide. The emphasis is on the word “tried” (at least a couple of times). By the end of your 1st term, you will have seen these symbols so much that it will be second nature to you – it’s just a question of practice. There is, however, one basic test that you have to pass in order to follow this guide – so brace yourselves: can you add & subtract 1 from another number? If yes, then let’s crack on with making differentiation easier…if not, then practise adding and subtracting…& then start with this guide tomorrow.
Differentiation
Differentiation is just a mathematical function like addition, subtraction or taking a log. It is used to calculate: rates of reaction (kinetics) slopes of graphs entropy and other thermodynamic quantities.
If y = a xn and we want to find the rate at which y changes as x changes (i.e. the slope of the graph of y versus x) then we need to: dy “differentiate” y (i.e. the equation) with respect to x to get . dx
Note: This does not mean “dy” divided by “dx”.
General rules for differentiation y with respect to x: Learn these off by heart.
n dy n1 1. If y = x , then n x dx
n dy n1 2. If y = ax where a is a constant, a n x dx dy 3. If y = a where a is a constant, 0 dx dy (Proof - write y = a as y = ax0 - since x0 = 1, n = 0, thus a 0 x01 0 ) dx dy a 1 4. If y = ln (ax) - do not panic - dx ax x dy dv du 5. If y = uv then: uv = “1st x differential of 2nd + 2nd x differential of 1st” dx dx dx (Product Rule applies to the product of two functions u & v which are both functions of x.)
Dr. Lindsey J. Munro [email protected] 12
So (apart from the example with the natural log (ln) and products), all you have to do is: look at the expression for y write down the constant, a (if present) multiply by the power multiply by x raised to the old power minus 1 Easy peasy!
Here are some quick examples - so you know what lies ahead.
dy dy 1. y = x5 5x51 5x 4 2. y = x 4 4x 41 4x3 dx dx
dy 1 dy 2 3. y = x 3 3x31 3x 2 4. y = 2x 21 2x 3 dx x 2 dx x3
Hint: (y = x –2)
Can you see the pattern?
Remember: 1 1 x n and x x 2 x n
Your Go - 1 Differentiate the following expressions: (Answers at the back) 1 1. y = x 7 2. y = x 8 3. y = x 4
3 4. y = 5x 7 5. y = 5x 8 6. y = x 4
Let’s make it a bit more complicated In real life, y is rarely so simple. It is often equal to a series of terms added together - but all you do is differentiate each term in turn.
dy 1. y = x 7 + x 3 + 3 7x6 3x 2 0 7x6 3x 2 dx 6 dy 18 2. y = 2x 4 + + 20 2 4x3 6 3x 4 0 8x3 x 3 dx x 4
Your Go - 2 Differentiate the following expressions: 1 1. y = x 6 - x 2 + 210 2. y = x 5 + + 0.6 x 6 4 3. y = x 3 4. y = x ln(6x) x 2
Dr. Lindsey J. Munro [email protected] 13
Where am I actually going to see anything like this?
You will see lots of equations involving differentials in your course – although probably not all with y and x as the variables.
You may need to investigate the rate at which y changes with respect to changes in x. y For macroscopic changes (ie. big changes that can be observed), this is written: x dy For microscopic changes (ie. tiny changes that cannot be observed), this is written: dx
Sometimes the change in a variable (such as free energy G) is dependent on changes in both pressure (written as dp) and temperature (written as dT). This would be written:
G G dG dp dT p T T p
Arrrgghhhh – this looks horrible. But let’s think about what it means:
G δG is used rather than dG since G is dependent on more than one variable (p and T). p T Rate of change of G with respect to p = The slope of the graph of G against p. The T subscript just means “whilst holding the temperature variable constant”.
So each term contributes to the total change in G:
G Term 1: dp Rate of change of G with respect to p x the actual change in p p T
G Term 2: dT Rate of change of G with respect to T x the actual change in T T p
If you still find that difficult to understand: - all we are doing is taking the slope of a graph & multiplying by the change in the x coordinate to find the change in the y coordinate – which is pretty simple maths.
G G Maxwell’s relations tell us that: V and S p T T p so if we look at graphs of G against p or T for a substance (in gas, liquid and solid states), we can understand more about its thermodynamic properties.
Differentiation will help you check any integrations you do.
Dr. Lindsey J. Munro [email protected] 14
Product Rule You will only see the Product Rule in one derivation during this course and in the Physical Chemistry course in Year 3. However, it is important that you know this rule, since at first sight it looks completely counter-intuitive and could lead to one (or maybe two) moments of confusion during lectures!
First let’s look at a few really simple examples of the Product Rule, in order to prove to you that it really is true:
1. y = x7 x x3 u = x7 (“1st function”) and v = x3 (“2nd function”) du dv we can easily calculate: 7x6 and 3x2 dx dx Substituting into the formula: dy dv du uv = “1st x differential of 2nd + 2nd x differential of 1st” dx dx dx gives us: dy x7 37 x 2 x 3 x 6 dx which simplifies to become: dy 10x9 dx
This is fairly trivial example since we can see that y = x7 x x3 = x10 and therefore using the 1st rule of differentiation: .
2. We will practise this one more time: 1 1 yx9 u = x2 and v = x9 x2 x2 du 2 dv we can easily calculate: 2x3 and 9x8 dx x3 dx Substituting into the formula gives us: dy 12 9xx89 dx x23 x which simplifies to become: dy92 x89 x 9x6 2 x 6 7 x 6 dx x23 x
1 This is fairly trivial example since we can see that y x97 x and therefore using the 1st rule x2 dy of differentiation: 7x6 . dx
Your Go - 3 Differentiate the following expressions using the Product Rule: 1 1. y = x6 x x2 2. yx8 x6
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Application:
In the derivation of the Gibbs-Helmholtz equation, we must differentiate the following expression by T (rather than x – but the principle is exactly the same): G 1 1 y = G u = T 1 and v = G TT T du 1 dv dG we can easily calculate: 1T 2 and dT T 2 dT dT Substituting into the formula gives us: d G/ T 11dG G dT T dT T 2 dG We know that G = H – TS, therefore S dT d G/ T 1 G S dT T T 2 GH We can substitute in for S in terms of G: S (by rearrangement of G = H – TS) T d G/ T 1 GHGGHGH dT T T T2 T 2 T 2 T 2 T 2
d G/ T H Leaving us with a very useful expression, dT T 2 from which we can derive the Gibbs-Helmholtz equation by integration:
GG 11 21 H TTTT2 1 2 1
Clearly your next step has to be to learn about integration!
Dr. Lindsey J. Munro [email protected] 16
Integration
Right so you’ve seen how easy differentiation is - integration is just the opposite of differentiation (just as subtraction is the opposite of addition). Integration is used to calculate: Area under a graph (either general expression in terms of “x” or between specific limits)
y
x
A simpler relationship for a system represented by a differential equation.
Indefinite Integrals dy If a x n and we want to find an equation relating y and x then we need to: dx “integrate” the expression above with respect to x which is written as y a x n dx . Since there are no limits, we will just get a general expression for y in terms of x. This is called an indefinite integral.
Note: should never ever be referred to as “big S” or “that squiggly thing” – it is the integral sign.
General rules for integration of simple functions Learn these off by heart. (NB: c is just a constant.)
x n1 1. Basic: x n dx c n 1
a x n1 2. Multiply by constant, a: a x n dx c n 1
3. Constant, a: a dx a x c
a 4. Complicated case: dx aln(x) c - You will see this often in thermodynamics. x
So (apart from the example with the natural log (ln)), all you have to do is: add 1 to the power divide by the new power add a constant No problem!
Here are some quick examples – so you know what lies ahead – (c is a constant).
41 5 31 4 4 x x 3 x x 1. x dx c c 2. x dx c c 4 1 5 3 1 4
21 3 31 2 2 x 5x 1 x x 1 3. 5x dx 5 c c 4. dx x 3dx c c c 2 1 3 x 3 3 1 2 2x 2
Dr. Lindsey J. Munro [email protected] 17
Whenever, you integrate an expression – CHECK that when you differentiate the result it is the same as the original.
Check: 5 4 x dy x 4 1. y c 5 x Yep – it works! 5 dx 5 Now you should check Examples 2 – 4.
Your Go – 4 Integrate the following expressions: 1 1. y x7 dx 2. y 3x8dx 3. y x5 dx x 4
2 3 1 4. y dx 5. y dx 6. y 5 dx 5 x x x
Integrating with limits (definite integrals)
Sometimes we want to find the integral of something between 2 different limits. For instance we might want to know the area under the graph y = 2x 2 + 4x + 6 between x = 3 and x = 10. In this case, we want to calculate the value of the expression with x = 10 (so substitute in 10 as the value of x) and then subtract from this the value of the expression when x = 3 (by substituting in 3 as the value of x) – this will leave us with the value for the expression integrating between the limits 3 and 10. Confused? Just look at the mathematics below and all will become clear(er):
10 3 2 10 2 x x 2x 4x 6dx 2 4 6x c 3 3 2 3
3 10 2x 2 2x 6x c 3 3 3 3 210 2 2 3 2 210 610 c 2 3 6 3 c 3 3 2 926 3 c 54 c 2 872 3 Note that the constant c in each bracket cancels out. This always occurs when the integral has limits. Therefore, in these cases, we don’t bother to write in the constant c (as shown in the answers to the questions below).
Your Go – 5 Integrate the following expressions: 1 2 4 1 1. y x3dx 2. y 3x 4 dx 3. y x6 dx 3 0 1 2 x
Dr. Lindsey J. Munro [email protected] 18
Application to Thermodynamics: In thermodynamics, you will often see that the entropy change of the system ΔS as we change the temperature from T1 to T2 is: T2 C S p dT T T1
Two scenarios arise:
-1 -1 a. If CP is constant (ie. Cp = 425 J K mol ) then if we can get a general form of the equation for this scenario by integrating:
T2 C T2 1 S p dT C dT T p T T1 T1
T2 C p ln(T ) T1 C p ln(T2 ) ln(T1 ) T 2 C p ln T1 (If you don’t understand how the last step is true then you need to revise the log laws.)
Example: -1 -1 A liquid is heated from a temperature of 300K to 500K. Given that Cp = 425 J K mol for this liquid, calculate the change in entropy.
(i) Assign labels to the values given in the question:
T1 = 300K T2 = 500K -1 -1 Cp = 425 J K mol
(ii) Starting from , do the derivation above to give:
T 2 (Memorise this equation.) S C p ln T1
(iii) Substitute values into the equation:
T 500 2 J K-1 mol-1 S C p ln 425 ln 217.1 T1 300
-1 -1 b. If Cp is dependent on T (ie. Cp = (4 T + 10) J K mol ) then if we need to deal with each situation on an individual basis by substituting in for Cp and then integrating.
Example: -1 -1 A liquid is heated from a temperature of 300K to 500K. Given that Cp = (4 T +10) J K mol for this liquid, calculate the change in entropy:
(i) Assign labels to the values given in the question:
-1 -1 T1 = 300K T2 = 500K Cp = (4 T +10) J K mol
Dr. Lindsey J. Munro [email protected] 19
T2 C p (ii) Starting from S dT , substitute in for T1, T2 and Cp T T1
T2 500 C p 4T 10 S dT dT T T T1 300 500 10 4 dT 300 T 500 4T 10 ln(T)300 4 500 10 ln(500) 4 300 10 ln(300) 805.1 J K 1mol 1
You will see integrals of this type v. often – so make sure you remember this result: p2 1 p T2 1 T dp ln 2 or dT ln 2 p p T T p1 1 T1 1
Additional Terminology
n 3 Gi add together G for each component 1 to n Gi G1 G2 G3 i1 i1
n 3 Gi multiply together G for each component 1 to n Gi G1 G2 G3 i1 i1
Additional Mathematics to Revise
Using a calculator incorrectly is one of the most common sources of errors – always insert lots of brackets: 20 + 5 should equal 5.84 = (20 (18 + (34) )) + 5 = 20 (18 + ) + 5 18 34
20 not 11.94 = 20 18 + (34) + 5 = 34 5 18
20 not 7.36 = 20 18 + (34 5) = 34 5 18
Clearly label variables – helps you to insert the numbers correctly into the equation. Memorise equations as you proceed through the course – they have to be 2nd nature. Write down equation to be used. Insert units – ensure all variables converted to correct units. (ie. J not kJ, K not oC)
Rearranging equations Logarithm rules
Dr. Lindsey J. Munro [email protected] 20
Answers
Your Go – 1 dy dy dy 4 1. 7x 6 2. 8x 7 3. 4x 5 dx dx dx x5 (y = x–4)
dy dy dy 12 4. 5 7x6 35x6 5. 58x7 40x7 6. 3 4x 5 dx dx dx x5
Your Go – 2 dy 1. 6x5 2x1 0 6x5 2x dx
dy 6 2. 5x 4 (6)x 7 0 5x 4 dx x7
dy 8 3. 4 2 x 3 3 x 2 3x 2 dx x3 1 (Remember: x 2 ) x 2
1 1 1 dy 1 2 6 1 2 1 1 1 1 1 4. x x 1 dx 2 6x 2 x 2x 2 x 2 x x 1 (Remember: x x 2 )
Your Go - 3 1. y = x 6 x x 2 u = x 6 and v = x 2 du dv therefore: 6x5 and 2x1 dx dx Substituting into the Product Rule formula gives us: dy x6 2 x x 2 6 x 5 2 x 7 6 x 7 8 x 7 dx
dy We can see that y = x 6 x x 2 = x 8 and therefore using the 1st rule of differentiation: 8x7 . dx
1 1 2. yx8 u = x6 and v = x 8 x6 x6 du 6 dv therefore: 6x7 and 8x7 dx x7 dx Substituting into the Product Rule formula gives us: dy1 6 8 x78 6 x 8x78 x 8 x 6 x 2 x dx x6 x 7 x 6 x 7
1 dy We can see that y x82 x and therefore using the 1st rule of differentiation: 2x . x6 dx Your Go – 4
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8 7 x 1. y x dx c 8
9 9 8 x x 2. y 3x dx 3 c c 9 3
1 x6 x 41 x6 x 3 x6 1 3. y x5 dx c c c x 4 6 4 1 6 3 6 3x3
2 1 x 51 x 4 2 1 4. y dx 2 dx 2 c 2 c c c x5 x5 5 1 4 4x 4 2x 4
3 1 5. y dx 3 dx 3ln(x) c x x
1 1 1 6. y 5 dx 5 dx 5x x 2 dx 1 x x 2 1 1 1 x 2 x 2 5x c 5x c 5x 2 x c 1 1 2 1 2
Your Go – 5 1 4 1 4 4 3 x 1 0 1 1 1. y x dx 0 0 4 0 4 4 4 4
2 2 x5 25 15 y 3x 4dx 3 3 3 96 3 18 3 2. 5 5 5 = 18.6 1 5 1 5 5
4 7 2 4 7 4 7 7 6 1 x x x 1 4 1 2 1 3. y x dx 3 2 2 2 2 x 7 2 2 7 2x 2 7 2 4 7 2 2
= 2322.4 (to 1dp)
CONGRATULATIONS!! You now know all the calculus necessary for your 2nd year …if you’ve found this a bit tricky then come back tomorrow & go through it all again. Feeling confident about calculus just requires practice.
Dr. Lindsey J. Munro [email protected] 22
ESSENTIAL REVISION & PREPARATION
Symmetry, Structures of Simple Solids & Basic Crystallography (6F5Z2007)
Dr. Stuart Langley [email protected] 23
Preparation for Symmetry, Structures of Simple Solids & Basic Crystallography (6F5Z2007)
One topic that students find tricky in Year 2 is visualising the symmetry of different molecules and crystal structures. Don’t worry you can fix this! The recommendations below are designed to help you practise these skills so that you can really do well in this core module.
“Chemistry3” by Burrows et al.: Chapter 6 (Solids) - This is a useful concise introduction to these topics.
In addition, the Yr 2 recommended textbook “Inorganic Chemistry” by Weller, Overton, Rourke & Armstrong has three excellent chapters, which will act as a nice introduction
1. “Inorganic Chemistry”: Chapter 3 (The Structures of Simple Solids) - Sections 3.1 – 3.10: Description of the structure of metallic & ionic solids
2. “Inorganic Chemistry”: Chapter 6 (Molecular symmetry) - It is fully recommended that when reading chapter 6, you use a molecular bonding kit, or some blue-tack and cocktail sticks, to build the molecules that are being focused upon so that you can actually have a go at performing the symmetry operations. One of the hardest elements associated with molecular symmetry operations is being able to visualise them, which many people struggle with. So don’t make it hard for yourself, make some models!
3. “Inorganic Chemistry”: Chapter 8 (Physical Techniques in Inorganic Chemistry) - Sections 8.1 and 8.2 (on X-ray and neutron diffraction).
“Molecular Symmetry & Group Theory” by Alan Vincent will also be of use here – the library has 6 copies.
It’s a pretty unique textbook in that you learn something & then immediately have to put it into practice – so by the end of the book you have taught yourself how to apply these skills – which is really useful. Please bear in mind that the library has only six copies available … so you will need to move fast!
Online Resources:
Inorganic Chemistry by Weller et al. – Free online resources global.oup.com/uk/orc/chemistry/ichem6e/ - Videos of chemical reactions - 3D rotatable molecular structures for each Chapter (including Chapters 3, 6 & 8) - Uses chemtube3d www.chemtube3d.com Instructions for how this supports the text is given on first page of each chapter). There is a whole section on symmetry (found within the structure and bonding tab), which provided you have Java enabled on your device will allow to “spin” molecules around in order to see, in 3D, what the symmetry transformation looks like.
Dr. Stuart Langley [email protected] 24
Molecular Modelling Activities using SCIGRESS
The Scigress annual licence tends to run out at the end of July. Therefore, in order to do these exercises using Scigress, you may need to download the latest version / new licence via the S: drive – available via myMMU: S:\Faculty of Science & Engineering\School Of Science & The Environment\Chemistry\Software\ Scigress_20xx
SCIGRESS has three relevant exercises that will support and further your understanding of these two areas:
Experiment 11 – Crystal Lattices and Close Packing Sites Experiment 12 – Identification and Execution of Symmetry Operations Experiment 13 – The Relationship Between Infrared Spectra and Molecular Geometry
Dr. Stuart Langley [email protected] 25
Overview
In this exercise you will enter lattice parameters obtained from crystallographic studies into a SCIGRESS document so as to represent the crystal structure of a compound in different ways. This will help to identify close packing sites which, in turn, aids in understanding why different compounds crystallize in different lattice types.
Recommende d Exercises No recommended exercises.
Background
Many substances crystallize in lattices based upon close packing schemes. In close packing, each atom or ion is considered to be a hard sphere. Each layer of spheres is arranged to pack the spheres as close together as possible. This will be the case if the spheres are in contact and if each row is offset from the neighboring row by half the diameter of a sphere. Close packing in three dimensions is achieved if the spheres of any one layer lie directly above the depressions in the layer immediately beneath it. Within a layer, there are depressions in the center of a triangular array of three spheres in contact with one another. Two close packing patterns are possible. If the third layer is aligned above the first, then the third and first layers are identified as A layers. The offset layer sandwiched between them is a B layer. The resulting pattern is ABABAB, referred to as hexagonal close packing (hcp).
Dr. Stuart Langley [email protected] 26
An alternative to hexagonal close packing occurs if the third layer is offset from both the A and B layers, resulting in an ABCABC pattern known as cubic close packing (ccp).
ccp
In each close packing scheme, there are two types of spaces or holes between adjacent layers of spheres. A tetrahedral hole exists between a sphere and a triangular array of spheres in the layer directly below it. An octahedral hole exists between a triangular array of spheres in one layer and the triangular array centered directly below it in the adjacent layer.
Tetrahedral Hole in Center Octahedral Hole in Center
Metals tend to crystallize in close packed structures (ccp and hcp). Ionic compounds tend to crystallize in lattices based on close packed structures with the large ion (usually anion) in a close packed array and the counterion occupying holes (e.g. NaCl is ccp of Cl- with Na+ occupying octahedral holes).
Modeling Section
Aluminum Lattice. 1. Open the Workspace. 2. Select Action | Crystallize. A Crystal Shape box opens. Select the Main tab. 3. Scroll through the Space Group list and select F23. This identifies the crystal system (face centered cubic) of metallic aluminum. 4. Enter a=4.04958 in the Angles section of the dialog window. The values for b, c, α, β, and γ remain unchanged. 5. Select Infinite Lattice in the Build box.
Dr. Stuart Langley [email protected] 27
6. Click the Fractional Coordinates tab, and enter the following values from the table below for atoms 1 to 4. Click OK and a crystal structure appears in the Workspace.
atom symbol x y z 1 Al 0 0 0 2 Al 0 0.5 0.5 3 Al 0.5 0 0.5 4 Al 0.5 0.5 0
7. Select View | Crystal Boundaries | Show Lattice Boundaries (this option is turned on by default).
Viewing the lattice
1. Rotate and view the lattice from different angles. Find the arrangement of layers (ABC or AB). View the lattice looking from one corner of the cube to its most distant diagonal. 2. Find a tetrahedral hole. Select the atoms which surround the tetrahedral hole. Alter the atom size by changing the van der Waals scale to better view the tetrahedral hole. ❏ Select View | Styles | Atoms... In the resulting Atom Attributes dialog box, click on the Shape tab. Type in a new scale value (0-1). 3. Narrow the window so that the boundary around the lattice is small. Copy the lattice and paste into a word processing document. 4. Find an octahedral hole and select the atoms which surround the hole. As in the preceding step, narrow the window, copy and paste into the same word processing document. 5. Orient the lattice to show that the octahedral hole exists between two staggered triangular arrays of atoms. Paste the lattice into a graphics application, and draw lines between the atoms to identify the two triangular arrays. Paste the result into the word processing document.
Dr. Stuart Langley [email protected] 28
Sodium Chloride Lattice Many ionic lattices result from a close packed array of the large ion with the small ion occupying either octahedral or tetrahedral holes. Proceed with the NaCl lattice as in the case of the aluminum lattice. 1. Select P4(2)32 from the space group list. 2. Edit the cell parameters by entering a = 5.64056. 3. Open Fractional Coordinates and enter the following values:
atom symbol x y z 14 Na 0.5 1 0.5 1 Na 0 0 0 15 Cl 0.5 0.5 0.5 2 Na 0.5 0.5 0 16 Cl 0.5 0 0 3 Na 0.5 0 0.5 17 Cl 0 0.5 0 4 Na 0 0.5 0.5 18 Cl 0 0 0.5 5 Na 1 1 1 19 Cl 0.5 1 0 6 Na 1 0 0 20 Cl 1 0.5 0 7 Na 0 1 0 21 Cl 0 1 0.5 8 Na 0 0 1 22 Cl 1 0 0.5 9 Na 1 1 0 23 Cl 0.5 0 1 10 Na 0 1 1 24 Cl 0 0.5 1 11 Na 1 0 1 25 Cl 1 0.5 1 12 Na 0.5 0.5 1 26 Cl 1 1 0.5 13 Na 1 0.5 0.5 27 Cl 0.5 1 1
4. Select Infinite lattice in the Build box. Click OK to close the Crystal Shape dialog box. The sodium chloride lattice appears in the window. However, the chloride ions will appear smaller than the sodium ions. You can fix this in the following way:
❏ Select Edit | Preferences | Periodic Table Settings... Select the Charge tab. Highlight Cl in the box to the right. In the left list the van der Waals radius and charge appears for a Cl ion. Click Change, and a small dialog box opens. Change the radius to 1.67Å. Click OK. Do the same for Na ion (1.16 Å). Click OK to return to the workspace.
5. View the NaCl lattice from different perspectives by rotating the lattice. Note that the Na atoms form a face-centered cubic array with the Cl atoms in the positions of octahedral holes. In actuality, the Cl atoms also form a face- centered cubic array with the Na atoms in the positions of octahedral holes.
View an Extended Lattice
6. Go to the Crystal Shape window.
7. Edit the lattice boundaries so they range from 0 to 2 in the a, b and c directions.
❏ Click on Infinite lattice, and click OK to close the dialog box Select View | Crystal Boundaries and select Show Unit Cell and Show Lattice Boundaries (this option is turned on by default).
8. There should now be eight unit cells within the lattice with one of the unit cells outlined. Dr. Stuart Langley [email protected] 29
9. View the lattice from different angles. Choose one or more orientations to copy and paste into your word processing document. Explain how the figure demonstrates that both the Cl and Na ions form ccp arrays.
Report Guidelines
The aim of the report is to demonstrate an understanding of close packing in atomic and ionic lattices. Select orientations of the lattice, or portion of the lattice, which clearly illustrate close packed layers, octahedral holes and tetrahedral holes. If, for purposes of clarity, you wish to eliminate part of a lattice, simply delete selected atoms or ions. Alternatively, you may select only the atoms you wish to emphasize. It may be useful to paste a structure into a graphics document to add arrows, lines and labels for explanatory purposes.
Dr. Stuart Langley [email protected] 30
Instructor Notes
Typical Results The following figures have been pasted from an editor document for the aluminum lattice. In some cases additional graphics were added by first pasting into a chemical graphics document and adding lines and labels before pasting into the word processing document.
Al unit cell Al atoms at faces Octahedral hole
Three different views of octahedral hole (in center).
Tetrahedral hole in center of cluster (scale=1)
Dr. Stuart Langley [email protected] 31
NaCl lattice viewed along alternating close-packed layers of + Na and Cl- ions (8 unit cells, scale=1)
Dr. Stuart Langley [email protected] 32
Overview
In this exercise you will learn how to locate elements of symmetry in molecules and how to demonstrate the results of symmetry operations.
Recommended Exercises Complete Exercises I, II, III, and IV before beginning this experiment.
Background
Molecular symmetry is of great value in understanding molecular bonding and spectroscopy. The symmetry of a molecule is expressed as a collection of symmetry operations or elements of symmetry. A molecule with a particular set of symmetry elements is said to belong to a particular point group. Molecular orbitals and vibrations within molecules also have symmetry properties. Symmetry operations are manipulations of a molecule which result in indistinguishable representations of the molecule; that is, a molecule looks no different before and following a symmetry operation. The operation may exchange two identical atoms, or it may leave one or more atoms unmoved. The symmetry operations are as follows: Identity (E): The identity is a trivial symmetry operation which involves doing nothing to the molecule. It is mentioned here only for the sake of completeness. All molecules possess an identity and, in some molecules, the identity is the only element of symmetry.
Axis of Rotation (Cn): An n-fold axis of rotation is a rotation about an axis by 360°/n. Often, but not always, rotation axes correspond to bond axes. For example, SF6 possesses a C4 axis which passes through opposing S-F bonds.
Rotation by 90° about the axis rotates each fluorine atom perpendicular to the axis into a position previously occupied by another fluorine atom.
Dr. Stuart Langley [email protected] 33
SF also possesses 3 C 6 3 axes of rotation which do not pass through bonds. In the following representation, the C 3 axis is perpendicular to the plane of the paper an passes through the sulfur atom.
Dr. Stuart Langley [email protected] 34
Rotate a molecule 1. Center the molecule in the window by selecting the entire molecule and pressing ctrl + f on the keyboard. It may resize the molecule in the process of centering it. If the molecule is not centered in the window, it will wobble as the rotation commands are executed.
2. Rotate the molecule until the rotation axis is perpendicular to the screen. You should now be looking directly down the rotation axis, which should be centered in the window.
3. Press r on the keyboard to enter rotation mode.
4. Press Z to rotate in the clockwise direction and z to rotate counter-clockwise. Rotate around the x and y axes by pressing x, X, y or Y. Rotation occurs by 5° each time. Thus nine operations result in a 45° rotation, 12 - 60°, 18 - 90°, 24 - 120°, etc. You may rotate continuously by pressing the Space bar. To halt continuous rotation, press the Space bar again.
Plane of Symmetry or Reflection (σ): A mirror plane of symmetry is an operation in which reflection of an atom from one side of the plane to the other results in an indistinguishable representation of the molecule. Atoms which lie within the plan are unmoved. Examples of mirror planes are given below.
Reflect through a mirror plane
1. Orient the molecule so that the plane is parallel to the right and left sides of the window (perpendicular to the window or screen). 2. Select Action | Selection | Mirror.
Improper Rotation or rotation-reflection (Sn): An improper rotation is the result of two operations in sequence; rotation by 360°/n, followed by reflection through a mirror plane perpendicular to the axis of rotation. An example of an improper axis of rotation is shown below.
Dr. Stuart Langley [email protected] 35
Execute and demonstrate an improper rotation 1. Follow the procedures for rotation and reflection in sequence. 2. To demonstrate the example above, the following steps would be executed:
I II
II III
III III
The results of the successive operations of rotation and reflection are shown below.
Inversion (i): An inversion through the center of a molecule can usually be accomplished by a rotation by 180° followed by reflection through a plane (Mirror
operation) perpendicular to the rotation. Note that this is identical to an S 2 as in the
Dr. Stuart Langley [email protected] 36
example below (rotation by 180° about the C-C axis followed by reflection through a plane perpendicular to the C-C axis).
For a planar molecule, an inversion is simply accomplished via rotation by 180° about an axis perpendicular to the plane since subsequent reflection through the plane leaves the molecule unchanged.
Modeling Section
Methane has several elements of symmetry, some of which are challenging to
visualize. Build a model of CH4 and view the atoms by number and the bonds as cylinders. Center the molecule in the window by selecting only the carbon atom and pressing Control + f. For each symmetry element, you should have one document to show the original orientation and one to show the result of the operation. Orient the molecule so as to make apparent the symmetry element (e.g. looking down a rotation axis). Adjust the size of the window so that you can simultaneously view documents showing the original atom placements and those following a symmetry operation. Once you have completed the operation, copy the structures and paste them into a word processing document.
Report Guidelines
Label (e.g. C3), and give the number of each type of symmetry element in the molecule to accompany the examples generated using SCIGRESS.
Dr. Stuart Langley [email protected] 37
Instructor Notes
This exercise could be adapted to numerous other molecules and polyatomic ions.
Typical Results
There are 4 types of symmetry elements in the T d point group, 4 C3 rotation axes, 6
σd mirror planes, 3 C2 axes, and 4 S4 improper axes.
Dr. Stuart Langley [email protected] 38
Overview
In this exercise you will use SCIGRESS to compute the infrared spectra of small inorganic molecules. Group theoretical methods will be used to predict the number and symmetries of molecular vibrations possessed by a molecule.
Recommended Exercises Complete Exercises I, II, III, IV, and VI before beginning this experiment.
Background
One of the applications of symmetry in chemistry is the prediction of vibrations which will be active in infrared (IR) and/or Raman spectroscopy. Such predictions can be NOTE confirmed by using MOPAC to generate IR spectra and to examine the vibrations A basic understanding of responsible for the IR peaks. As an example, the symmetries of IR active vibrations in the chemical applications a water molecule are worked out below. The same method is applicable to other for symmetry and group molecules. The water molecule belongs to the C2v point group for which the character theory is necessary. It is advisable to review the table is as follows: relevant material in an inorganic chemistry text or other reference such as F.A. Cotton, Chemical Applications of Group Theory, 3rd ed., Wiley, New York, 1990.
Dr. Stuart Langley [email protected] 39
1. Determine each atom’s contribution to the character of the transformation matrix under each symmetry operation. If an atom moves, it contributes 0 to the character. If an atom remains stationary, determine the contribution of each coordinate. If a coordinate is unchanged, it contributes 1; if it is reversed in sign, it contributes -1; if it is something in between, it makes a fractional contribution. a. The consequences of each operation type are summarized as follows: E: All atoms remain and all coordinates are unchanged. Each coordinate contributes 1 to the character of the reducible representation. Thus, for water, the character for the E element is 9.
C2: The hydrogens move and thus contribute 0. Oxygen remains stationary and the z coordinate remains unchanged, contributing 1. However, x and y are both reversed in sign and thus each contributes -1. The character is thus 1-1-1 = -1
v(xz) (plane of molecule): All atoms remain unchanged. The z and x coordinates are unchanged, each contributing 1. Each y coordinate is reversed, contributing -1. For each atom, the contribution is 1 and the character is 1+1+1 = 3
v(yz): Only O is unchanged. The z and y coordinates contribute 1 each. The x coordinate is reversed and contributes -1. Thus the character is 1+1-1 = 1.
b. The reducible representation is Γ = 9 -1 3 1 2. Determine the irreducible representations of the coordinates by making use of a group property.
3. The reducible representation represents the sum of the motions of all of the coordinates: 3 of which result in translation (x, y, z) and 3 of which result in
rotation (R x, Ry , Rz ) for a nonlinear molecule. Thus, for water, there are 3 vibrations. If the irreducible representation of a vibration is that of a coordinate (x, y, or z), it will be IR active. Subtract the irreducible representations of
translations and rotations. a. Translation (x = B1, y = B2, z = A1)
b. Rotation (Rx = B2, Ry = B1, Rz = A2) c. The vibrations which remain: 3 - 1 = 2 B A1 1 3 - 2 = 1
1 - 1 = 0 B A2 2 2 - 2 = 0 Dr. Stuart Langley [email protected] 40
4. Of the 3 vibrations, 2 are A1 symmetry and 1 is B1 symmetry. A1 is the irreducible representation of the z coordinate and B1 is the irreducible representation of the x coordinate so all three vibrations are IR allowed.
Note that, for a particular point group, Γ will depend upon the number of atoms in the molecule.
If a symmetry operation is a multiple (e.g. 2C3 in C3v) the contribution to the irreducible representation must be multiplied by that multiple in calculating the number of irreducible representations. If a vibration belongs to a degenerate irreducible representation (E or T), then there will be a single vibrational frequency, corresponding to 2 (E) or 3 (T) degenerate transitions. Likewise, when accounting for rotations and vibrations, if 2 rotations, say, belong to an E irreducible representation, that accounts for 1 E irreducible representation.
Modeling Section
Simulation of the Infrared Spectrum of H 2O (g) 1. Build a water molecule in the Workspace. 2. Optimize the molecule with MOPAC using PM3 parameters. Append a Force calculation. When you append calculations, MOPAC executes the selected calculations in sequence; in this case, a geometry optimization followed by an IR/vibrational spectrum. ❏ Open a new experiment and select chemical sample, IR transitions, and MO-G PM3 FORCE. The experiment manager automatically does an optimization followed by an IR calculation. Select Run. 3. View the vibrational spectrum. ❏ Activate the molecule window after the calculations are complete. Select Analyze | IR Transitions. An IR spectrum appears in a separate window. Each separate vibration is represented by a peak with a triangle symbol at the minimum. 4. Adjust the x and y axes as needed. For example, the water spectrum should be displayed between 0 and 4000 cm-1 and should have no peaks at frequencies below 1000 cm-1. To check frequencies above 4000 cm-1, double click just below the horizontal axis to open the Axis Attributes/ Axis Parameters dialog box. ❏ Enter a Wavelength Range of 6000 to 1000 cm-1. Click OK to rescale the spectrum. Rescale the spectrum, as desired, to view parts of the spectrum in greater detail. 5. Resolve overlapping peaks as needed. ❏ Click on a curve with the Select tool. A red line appears, defining 6. Click on a peak symbol to view the vibration corresponding to a particular peak. The structure window will rearrange to show the atom motions involved in the vibration. To better view the vibration, click anywhere in the structure window and reorient the structure. 7. Paste the spectrum and vibrations into a word processing document and assign each vibration to an irreducible representation. Dr. Stuart Langley [email protected] 41
N2F2 1. Build a molecule of N2F2 in the Workspace. Optimize and execute a Force calculation as in the case of the water molecule. 2. View the molecule and determine the point group. Obtain the character table for the point group. 3. Use the methodology given in the Background section to determine the number and symmetries of the vibrations of the molecule. Determine which of the vibrations are IR active.
4. View the computed IR spectrum of N2F2. View the vibrations, both IR active and IR inactive, and assign their symmetries (irreducible representations).
Report Guidelines
✎ Paste the spectra and vibrations for water into a word processing document and assign each vibration to an irreducible representation. ✎ Use group theoretical methods to determine the symmetries of the vibrational modes of N2F2 and determine which vibrations are IR active.
✎ Paste the spectra and vibrations for N2F2 into a word processing document and assign each vibration to an irreducible representation.
Dr. Stuart Langley [email protected] 42
Instructor Notes
Typical Results Water
N2F2
Number of each symmetry species:
nAg = 14 [(12)(1) + (0)(1) + (0)(1) + (4)(1)] = 4
nBg = 14 [(12)(1) + (0)(-1) + (0)(1) + (4)(-1)] = 2
nAu = 14 [(12)(1) + (0)(1) + (0)(-1) + (4)(-1)] = 2
nBu = 14 [(12)(1) + (0)(-1) + (0)(-1) + (4)(1)] = 4
Reducible representations of translations (Au, 2Bu) and rotations (Ag, 2Bg).
Remainder are vibrations: 3Ag, 1Au, 2Bu
IR active modes: 1Au, 2Bu
Dr. Stuart Langley [email protected] 43
Further Discussion
An additional system worth studying is SF4. The structure is based on a trigonal bipyrimidal distribution of electron pairs, and since there is one lone pair, there are two structures possible. The two structures do not have the same number of IR active transitions, so that infrared spectroscopy provides an experimental criterion for determining the correct structure. One structure belongs to the C2v point group and prediction of the number of IR active vibrations is
straightforward. The other belongs to C3v and here transformation of the atoms under the C3 operation is more difficult.
S F 4
The structure of SF4 is based upon a trigonal bipyramidal distribution of electrons, one being a lone pair. Thus there are two possible structures one with the lone pair in an axial position and one with the lone pair in an equatorial position.
Dr. Stuart Langley [email protected] 44
NMR Spectroscopy
Preparation for 6F5Z2006
Dr. Ryan Mewis [email protected] 45
Dr. Ryan Mewis [email protected] 46
Inorganic and organic synthesis - spectroscopy component (6F5Z2006)
Recommended Reading: Foundations of molecular structure determination Duckett, Gilbert and Cockett (OUP, 2nd Ed) Good all round text detailing IR, Raman, UV-vis, NMR, mass spectrometry and X-ray crystallography Nuclear Magnetic Resonance Hore (OUP, 2nd Ed) Introductory primer that covers NMR NMR: The Toolkit. How Pulse sequences work Hore, Jones and Wimperis (OUP, 2nd Ed) More advanced primer that covers pulse sequences
Review Yr 1 Notes: Specifically the Application of 13C NMR spectroscopy in organic chemistry (Dr Birkett). Also refer to Dr Beatriz Macia-Ruiz’s notes in this document on 13C and IR spectroscopy.
Topics that will be covered this year:
A background on the inherent sensitivity issues of NMR and how this links to the NMR experiment What the 1H NMR spectrum provides the spectroscopist with e.g. number of signals, integration ratio, chemical shift of the signals and signal multiplicity J-couplings due to first-order effects and the Karplus plot. Determination of geometric isomerism using J-coupling will also be covered Second order effects Determination of geometric isomerism using J-coupling J-coupling in aromatic ring systems 19F NMR spectroscopy Advanced methods such as NOESY (Nuclear Overhauser Spectroscopy) 2D methods such as COSY (Correlation Spectroscopy) and HMQC (Heteronuclear Multiple-Quantum Correlation) Mass spectrometry – how the mass spectrum is collected Production of fragments and their relative stability Common fragmentation pathways e.g. retro Diels-Adler, McLafferty Rearrangement, retro-ene reaction, tropylium and acylium ions Deducing structures using a suite of analytical data
Dr. Ryan Mewis [email protected] 47
Beatriz Maciá Ruiz
ESSENTIAL 1st YEAR
ORGANIC CHEMISTRY
Preparation for 6F5Z2006
Beatriz Maciá Ruiz
Beatriz Maciá Ruiz
1. NOMENCLATURE
Revise class notes, including nomenclature of (cyclo)alkanes, (cyclo)alkenes, (cyclo)alkynes, aromatic compounds and organic compounds with functional groups.
Beatriz Maciá Ruiz
2. ISOMERISM AND STEREOCHEMISTRY
Revise the concepts in the following diagram:
Chain Isomers
Structural Isomers Position Isomers
Isomers Functional Group Isomers (same molecular formula) Conformational Isomers Stereoisomers E-and Z- Isomers Configurational Isomers Isomers with chiral centres
In particular, revise:
2.1. Stability in substituted cyclohexanes:
Beatriz Maciá Ruiz
2.2. CIP Rules (establish priority between substituents):
1st- Order by atomic number
Cl > F > O > N > C > H
…or in case of isotopes, by mass
2nd- If same priority at first atom: Go to first point of difference
3rd- Multiple bonds: Add double or triple representations of atoms at the respective other end of the multiple bond.
2.3. Naming Z/E isomers (cis (same side)/trans (opposite side) double bonds, respectively)
2.3 Naming enantiomers: R/S nomenclature
Beatriz Maciá Ruiz
3. REACTION INTERMEDIATES. Concepts to revise:
3.1. Electrophiles (species that accept e-)
3.2. Nucleophiles (species with e- to share)
3.3. Inductive effect (the most electronegative atom atracts the e- Polar bond)
3.4. Mesomeric effect (to delocalize e- through bonds) Example:
3.5. Carbocations
3.6. Carbanions
Beatriz Maciá Ruiz
3.7. Radicals
3.8. Language in organic chemistry
4. REACTIONS IN ORGANIC CHEMISTRY 4.1. Radical halogenation
Beatriz Maciá Ruiz
4.2. Nucleophilic substitution (SN1 and SN2)
General picture:
Beatriz Maciá Ruiz
4.3. Elimination reactions (the nucleophile acts as a base)
4.4. Additions to double bonds (hydrogenation, bromination, hydration and adition of HX)
Beatriz Maciá Ruiz
General picture:
4.5. Aromatic electrophilic substitution (SE Ar) (halogenation, nitration, sulfonation, alkylation and acylation)
General picture:
Mechanism:
Beatriz Maciá Ruiz
Beatriz Maciá Ruiz
4.6. Nucleophilic addition to carbonyl compounds
Beatriz Maciá Ruiz
Beatriz Maciá Ruiz
4.7. Nucleophilic substitution to carbonyl compounds
General picture:
Reactivity trend:
General mechanism:
Beatriz Maciá Ruiz
5. SPECTROSCOPY: Molecular response to radiative stimulus ΔE = hν ν = c/λ
Beatriz Maciá Ruiz
IR - Generalities:
* Streching requires more en\ergy than Bending - Streching vibration has higher than Bending vibrations * To vibrate a short (and stronger) bond requires more energy than to vibrate a long (and weaker) bond - Streching C=C has higher than streching C-C * Higher for the vibration of bonds with lighter atoms (Hooke’s law) - Streching C-H has higher than streching C-D
Beatriz Maciá Ruiz
13C NMR Spectroscopy - chemical shift () vs. absorption of energy (resonance) as the y-axis
Chemical shift for 13C NMR ( = 0 – 250 ppm) Coupling to H is removed by “broad band” irradiation of all Hs Each carbon atom in a specific carbon environment will produce one singlet in a 13C NMR spectrum.
Hybridisation and type of Approximate chemical shift range (ppm) carbon atom 0 – 80 ppm (moves downfield (to higher values with 3 sp the change from CH3 to CH2 to CH to C)
sp 80 – 100 ppm
110 -140 ppm (alkene carbon signals tend to be more 2 sp alkene upfield (lower values) than aromatic carbon signals)
sp2 aromatic 115 – 150 ppm
190 – 210 ppm sp2 C=O Carbonyl carbon signals are often the most downfield signal in the 13C NMR spectrum
Beatriz Maciá Ruiz
6. WHERE YOUR 2nd YEAR STARTS: Second year lectures in the unit 6F5Z2006 (Core Concepts 1; Organic and Inorganic Chemistry) will focus on the carbonyl group with some emphasis on the chemistry of enols/enolates. Many aldehydes/ketones can exist as two forms – a keto and an enol. Consider acetone:
O OH
Keto Enol form form The formation of an enol/enolate arises from deprotonation of the proton:
O H H H O O O O H H H H B H H H H H Enolate Enol Mechanism of Formation of enols/enolates
This proton can be removed by base to give an anion that is resonance stabilised – we can delocalise the anion over additional atoms so it is more stable than expected. This phenomenon is also observed elsewhere, such as in nitro- and cyano-substituted compounds
Beatriz Maciá Ruiz
Resonance Stabilisation of anions to carbonyls, nitro groups and cyano groups.
Thus, we can think of the proton of aldehydes/ketones/esters as being acidic
Acidity of alpha protons to a carbonyl:
Lower pKa means more acidic H H abstraction is easier when the carbanion generated (ENOLATE) is more stable . As a result protons to carbonyl carbons are far more acidic than those on a simple
alkane such as butane (pKa~50) where no stabilisation can occur.
Enolates as nucleophiles:
Enols/enolates are far more than a curiosity in organic chemistry. They are very useful nucleophiles. In first year lectures you saw how simple alkenes can act as nucleophiles and can undergo addition reactions with protons and bromine for instance and the mechanism is very similar (See section 4.4 of these notes). Enolates can be considerd as alkenes which are more powerful nucleophiles owing to the presence of the O- group and can react with a wide range of electrophiles with regeneration of the carbonyl group:
O O E E R1 R1 R2 R2 General mechanism for reaction of enolates with electrophiles
Enolates, being powerful nucleophiles can also react with carbon-based electrophiles such as haloalkanes and are thus very useful species for forming carbon-carbon bonds – one of the most important transformations in organic chemistry.
Beatriz Maciá Ruiz
O O O 1 R1 R 1 2 R2 R R R2 H H H H 3 X R R3 B X - Br, I New Bond Reaction of enolates with haloalkanes
Enolates also react with other aldehydes/ketones (which have an electrophilic carbonyl carbon) in a more complex process known as the aldol reaction.
New Bond R'' R'' O R''' O R''' OH H B- R R R R'' R''' R H2O R R' R' R' R' R' O O O O O
General Mechanism of the Aldol Reaction
Next term: we will look at these process in detail and their uses in synthesising useful organic molecules.
We will also look at other reactions associated with carbonyl groups:
Conjugate additions to -unsaturated carbonyls. The Michael Addition. The Robinson Annulation. Oxidations and Reductions in Organic Chemistry.
In addition we will be discussing how to plan a synthesis of a target molecule in a logical fashion applying what is known as Retrosynthetic Analysis.
If you would like to read ahead over the summer here are a few references:
For an introduction see: “Chemistry3”, Ch 23,p1054-1057, p1074-p1090.
For good coverage at second year level see:
“Organic Chemistry”, P.Y. Bruice, 5th Ed, Ch 18, p850-891, Ch 19, p908-920.
“Organic Chemistry With Biological Applications”, John McMurry, 2nd Ed, Ch 14, p 574 and p 588- p592, Ch 17, p695-733.
For more advanced, but rigorous, coverage see:
“Organic Chemistry”, Clayden et al; 1st Ed, Ch 6, p139-141, Ch 10, p227-241, Ch 21, p523-538, Ch26 p667-671, p676-680, Ch 27, p689-698, Ch 29, p749-753, p760-762.
For good coverage of retrosynthetic analysis read Ch 30, p771-801.
Beatriz Maciá Ruiz
Preparation for 6F5Z2009 “Biological Processes & Pharmaceutical Chemistry”
Dr. Kirsty Shaw [email protected] 69
Dr. Kirsty Shaw [email protected] 70
Biological Processes and Pharmaceutical Chemistry (6F5Z2009)
Recommended Reading: Campbell Biology, Reece et al. (9th Edition), Pearson Publishing
Review Yr 1 Notes: Industrial Chemistry (6F4Z2006) - Specifically the Pharmaceutical Chemistry topics (Dr. Caprio & Dr. Birkett)
Content: Element 1 will cover some basic biological principles so it is important, especially if you have not studied Biology for a while, that you revise DNA structure, DNA replication, transcription and translation as this will provide you with a good basis for the taught course.
DNA Structure: DNA Replication:
Transcription & Translation: Further Reading: The following topics will be covered in the course: How mutations occur and how these can lead to disease states or be exploited as drug targets. The design of pharmaceutical drugs to inhibit DNA replication, transcription or translation. Aspects of genetic engineering, including gene cloning for commercial insulin production, transgenics and gene therapy. Key features and life cycles of bacteria, viruses and parasites. A study of drug targets and the production of antibacterial, antiviral and antiparasitic drugs, including issues around the development of drug resistance.
Dr. Kirsty Shaw [email protected] 71
Biological Processes and Pharmaceutical Chemistry (6F5Z2009)
Recommended Reading: Biochemistry, Campbell/Farrell, (6th Edition), Thomson publishing Review Yr 1 Notes: Industrial Chemistry (6F4Z2006) - Specifically the Pharmaceutical Chemistry topics (Dr. Caprio & Dr. Birkett).
Topics that will be covered include:
Krebs cycle & metabolism (Krebs cycle not expected to learn by heart, but type of metabolic processes involved) AcCoA acylation reactions in the body (Organic Chemistry within metabolism)
Electron transport chain conversion NADH/FADH2 into ATP (energy) Biomimetic receptors (Molecularly Imprinted Polymers) as sensors, which is a preview for what is included in the 6F6Z2007 Advanced Pharmaceutical Chemistry unit. Polymeric receptors for drug delivery in body and use as therapeutics.
Dr Marloes Peeters [email protected] 72