Chemistry 21b FinalExamination - California Institute of...

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Chemistry 21b Final Examination Out: 12 March 2018 Due: 16 March 2018, 2 pm This is an open book examination, and so you may use the PDF’d versions of Atkins and Friedman, Harris and Bertolucci, and/or Maczek along with the posted Lecture Notes and Problem Sets (with Solutions). Use of a calculator or packages such as Mathematica is also permitted. The exam must be done in one continuous sitting of three hours. I believe you should be able to complete the exam in the time allotted; but if not, please draw a line across where the time limit is reached and continue if you wish to. The TAs will record the grades for points awarded within the time allocated and that after. The problems are worth: 1=20 (10 for a, 10 for b) 2=30 (spread over a-g) 3=50 (10 each for a,b, 15 for c, d) To facilitate the grading of the exam, please begin the solution of each problem on a new sheet, and remember to sign your name on every page of your solutions. Good luck!

Transcript of Chemistry 21b FinalExamination - California Institute of...

Page 1: Chemistry 21b FinalExamination - California Institute of ...web.gps.caltech.edu/~gab/ch21b/docs/Ch21b_Final_March_2018.pdf · Chemistry 21b FinalExamination Out: 12 March 2018 Due:

Chemistry 21b

Final Examination

Out: 12 March 2018Due: 16 March 2018, 2 pm

This is an open book examination, and so you may use the PDF’d versions of Atkinsand Friedman, Harris and Bertolucci, and/or Maczek along with the posted Lecture Notesand Problem Sets (with Solutions). Use of a calculator or packages such as Mathematica isalso permitted. The exam must be done in one continuous sitting of three hours. I believeyou should be able to complete the exam in the time allotted; but if not, please draw a lineacross where the time limit is reached and continue if you wish to. The TAs will recordthe grades for points awarded within the time allocated and that after. The problems areworth:

1=20 (10 for a, 10 for b)2=30 (spread over a-g)3=50 (10 each for a,b, 15 for c, d)

To facilitate the grading of the exam, please begin the solution of each problem

on a new sheet, and remember to sign your name on every page of your solutions.

Good luck!

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1. Use group theoretical arguments to answer the following. You can be brief!

a). Assess the polarization of the 1A2 ←1A1 transition in H2CO and the 1B2u ←

1Ag

transition in ethylene (CH2=CH2), including vibrational coupling if necessary. Is thelatter transition electric dipole allowed?

b). Below the normal modes of vibration of the diborane molecule are shown. What isthe point group for this molecule? From the symmetries listed, which modes are IRactive, which are Raman active, and why is there no overlap between these two lists?

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2. This is a suite of combined IR/NMR spectroscopy problems, with UV/Vis results addedfor some. From the spectra and information below, derive the molecular structure. Explainyour reasoning.

a). The compound whose IR and NMR spectra are shown below has a parent ion peak at102 amu and a strong fragment at 57 amu. There is no UV absorption above 205 nm.

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b). The mass spectrum of this compound shows an intense molecular ion peak at 172 amuand an M +2 peak of approximately the same size. The largest fragment ion appearsat 65 amu. The IR spectrum of this solid compound was obtained by casting a filmon salt plates from a CCl4 solution. What is the structure of this compound?

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c). The local anesthetic benzocaine has the formula C9H11NO2. From the spectra below,determine the structure.

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d). The mass of this compound, a neat liquid, is 86 amu, and it has a near UV absorptionband at 280 nm in 95% ethanol. From the spectra below, determine the structure.

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e). The UV spectrum of this molecule shows no maximum longward of 205 nm, and theIR spectrum was obtained on a neat liquid. What is the structure of this compound?(Hint: The single NMR peak with integral = 3 actually arises from two functionalgroups that participate in hydrogen bonds – hence the peaks are averaged on thetimescale of the NMR measurement and no spin-spin splitting occurs.)

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f). The chemical formula of this molecule is C6H12O2, and in the NMR spectrum belowthe peaks near δ=3.9 and 0.8 are triplets, that near 1.9 is a singlet. The relativeintegrated areas from high to low chemical shift are 2, 3, 2, 2, 3.

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g). Last, but not least, the two NMR peaks below are the only ones seen in this moleculewith chemical formula C5H8O. The integrated areas of the two NMR features areequal.

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3. A selection of statistical thermodynamics questions:

a). From group theoretical arguments, what is the symmetry number σ that must beused in the canonical rotational partition function for: 1 = trans-dibromoethylene(HBrC=CBrH), 2 = Deuterated methane (CH3D), 3 = Sulfur hexafluoride (SF6,all bond lengths equal), 4 = Allene (C3H4, carbon atoms linear), 5 = Phosphorouspentafluoride (PF5, trigonal bipyramidal structure), 6=silane (SiH4), and 7=Pt(CN)4(planar)?

b). For a crystal with a sublimation energy U0 and considered to consist of a 3D suite ofcoupled harmonic oscillators, it can be shown that the approximate partition function(that is, the Q that ignores anharmonicity) can be expressed as

Q =

(

e−hν/2kT

1 − e−hν/kT

)3N

eU0/kT ,

where hν/k ≡ ΘE is a constant characteristic of the crystal (the Einsteintemperature, specifically). Calculate the heat capacity from this simple partitionfunction and show that at high temperatures one obtains the law of Dulong and Petit,namely, that CV → 3Nk.

c). An approximate canonical partition function for a dense gas is of the form

Q(N, V, T ) =1

N !

(

2πmkT

h2

)3N/2

(V −Nb)NeaN2/V kT ,

where a and b are constants that are given in terms of molecular parameters. Derivethe equation of state from this partition function along with the thermodynamicenergy and heat capacity. How do these equations compare to what we learned aboutideal gases in class (think about the dilute limit, or N/V → 0)? And, hopefully theequation of state looks familiar to you. What, qualitatively, is the physical origin ofthe constants a and b?

d). Finally, consider a paramagnetic crystal of spin-1/2 spins, where the interactionsbetween the spins can be neglected. If a magnetic field B is applied, the spin energiestherefore become ±µB, where µ is the magnetic dipole moment of the spins. Whatis the partition function for a mole of such spins, in the presence of a magneticfield? From this, determine the molar entropy of the crystal (the hint here is thatthe derivative of ln[cosh(ax)] is a[tanh(ax)], where 2cosh(ax) = (eax + e−ax) is thehyperbolic cosine). What are the low temperature (that is, T → 0) limits of S in thepresence and absence of a magnetic field, and how can these limits be explained giventhe statistical thermodynamic interpretation of the Third Law? If such a crystal iscooled to a temperature Ti in the presence of a magnetic field Bi, thermally isolated,and the field changed to Bf in a reversible fashion, the entropy remains constant. Inthis event, there is a very simple prediction for the final spin temperature in terms ofTi, Bi and Bf . What is it? This approach is used experimentally to cool paramagneticsolids to very low temperatures.