CEM 351, Fall 2010 - Michigan State University€¦ · CEM 351, Fall 2010 Midterm Exam 1 Friday, October 1, 2010 1:50 – 2:40 p.m. Room 138, Chemistry Name (print) Wright N. Swers

CEM 351, Fall 2010 Midterm Exam 1

Friday, October 1, 2010 1:50 – 2:40 p.m.

Room 138, Chemistry Name (print) Wright N. Swers Signature Student # Section Number (2 pts extra credit)

Grade this question? YES NO! . . 1.(20 pts.) . . 2.(20 pts.) . . 3.(20 pts.) . . 4.(20 pts.) . . 5.(20 pts.) . . 6.(20 pts.) TOTAL (100 pts.)

Score Note: Answer any 5 questions for a total score of 100 pts. Be sure to look over all the questions first before beginning the exam, and indicate which five questions are to be graded by checking the corresponding box. Recitation Day and Time Instructors: Room 1. T 9:10-10:00 a.m. Jason Lam 85 2. M 10:20-11:10 a.m. Carmin Burrell 281 3. T 10:20-11:10 a.m. Carmin Burrell 85 4. Th 10:20-11:10 a.m. Arvind Jaganathan 287 5. T 10:20-11:10 a.m. Jason Lam 283 6. T 11:30-12:20 p.m. Jason Lam 283 7. Th 11:30 - 12:20 Arvind Jaganathan 85 8. M 4:10-5:00 p.m. Carmin Burrell 110 9. Th 4:10-5:00 p.m. Arvind Jaganathan 183

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1. (20 pts) (a) (12 pts) Draw structures to represent the following IUPAC names

(i) 2-isopropoxypropane

O (ii) cis-4-methylcyclohexanol

OH

(iii) 2,2,3,3-tetramethylbutane

or

(b) (2 pts) Which of the compounds in (a) has no dipole? Please circle your answer

(i) (ii) (iii)

(c) (4 pts) Are any of the groups on the cyclohexane in (ii) axial? Please explain with a drawing and a few words; here is a model to work from but you must make your own drawing in the box provided.

Explanation:

(d) (2 pts) How many resonances would you expect in the 1H NMR of (i)? of (iii)? Explanation: (i) Two resonances: (a) a doublet [at ca. 1.1] and (b) a septet [at ca. 3.5 ppm], with the a:b integral ratio of 6:1 (= 12:2). The 6 equivalent Hʼs in the isopropyl groupʼs CH3 groups split the one H on the CH between them into a septet, while they are split into doublets.

(iii) One singlet [at about 0.8 ppm]. All six of those CH3 groups are equivalent. Note: though rough chemical shifts are given above, they werenʼt required for the answer.

The cis related groups are on the same face of the ring [imagine it flat, as in the answer to (ii)]. Here the –CH3 and –OH groups point “up”; the Hʼs on carbons 1 and 4 then show the contrasting “down” (other face) position.

OH

H3CH

H14 23

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2. (20 pts) (a) (3 pts each) Form acceptable Lewis structures for the following uncharged compounds by adding the necessary nonbonding electrons and multiple bonds:

:NC

O

N:H

HUrea

H C N:

Hydrogen cyanide

H

H

(b) (3 pts each) Assign formal charges as necessary to the atoms in the following Lewis structures.

C C

Cl

H1-Chloroethyl cation Nitromethane

(a racing fuel)

C N

O

O

H

H

H

H

H

H

(c) (3 pts) Which of the following alcohols would have the lowest pKa (greatest acid strength)? Explain your answer in terms of the reaction that defines Ka. CH3CH2OH CF3CH2OH CH3OH pKaʼs: 15.9 12.4 15.5

For dissociation of acid HA (below), K is the ordinary equilibrium constant, but since the pKa scale is defined in H2O, Ka leaves out the essentially constant H2O water concentration [H2O]. The more dissociated the acid, the higher the H3O+ and A– concentrations. Since H3O+ is common to all acids, we ask which A– can best accept the negative charge left on it when HA dissociates. CF3CH2OH has three atoms of F, the most electronegative element, that “pull” electrons—negative charge—toward themselves, leaving a more positive carbon next to –OH. This in turn attracts the negative charge in CF3CH2O–, a stabilization mechanism that CH3CH2O– and CH3O– donʼt share.

(d) (5 pts) Which of the following compounds has a dipole moment (circle Yes or No): CH3NH2 Yes No

CO2 Yes No

CH3OCH3 Yes No (remember, dimethyl ether is bent at O)

(CH3)2C=C(CH3)2 Yes No

BF3 Yes No

HA + H2O H3O+ + A–K

K =[H3O+][A–][H2O][HA] but Ka =

[H3O+][A–][HA]

(water omitted)

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3. (20 pts) Consider the four compounds A-D:

Cl Cl

Cl

Cl

A

C

B

D

1-chloroethyl

(a) (3 pts) Which one of A-D would you expect to show the 1-chloroethyl cation (seen in problem 2b, at left) as a major fragment in its mass spectrum (circle the letter): A B C D (b) (6 pts) Fill in the three Newman projection blanks below to represent the three staggered conformations you would see viewing down the bond in compound A shown by the arrow as the sp3 carbon at the other end underwent rotation. Be sure to pay close attention to the wedged/dashed bond notation as used above.

H3C

H

H

H Cl

CH2CH3

H

H

CH3

H Cl

CH2CH3

H

CH3

H

H Cl

CH2CH3

CH3

betw

een

CH2C

H 3 an

d Cl

(c) (1 pt) (1 pt) Circle your drawing of the highest energy conformation. (d) (8 pts) Name the four compounds according to IUPAC conventions:

A 3-chloropentane B 2-chloro-3-methylbutane C 1-chloro-3-methylbutane D 2-chloro-2-methylpentane

(e) (2 pts) One of the compounds A-D has a different formula than the other three. Which is it, and what is its formula (circle the letter, fill in blank)? A B C D Formula: C6H13Cl

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4. (20 pts) Structural analysis by spectroscopy (a) (3 pts each) For each pair of compounds in the grid below, propose a spectroscopic method to tell them apart, and explain what you would look for with the suggested method.

OHO

OO O

O

Cl Cl

A

B

C

D

Compounds Method & Explanation (Many possible)

O

IR: Detects O-H stretch and only one C-O stretch in the left hand compound, just two C-O stretches in right. 1H NMR: Left: three singlets in 9:2:1 ratio; Right: two (9:3). MS: Left: huge 31 peak (CH2OH+) from tBu loss; Right: mainly CH3 losses.

IR: Left: Only CH2 stretches and bends; Right: Strong C=O stretch 1H NMR: Left: A broad singlet; Right: two triplets in 1:1 ratio; 13C NMR: Left: 1 peak; Right: 3 peaks, one (C=O) downfield MS: Left: Strong molecular ion; others at Mmol -(n x 14); Right: Weak molecular ion; large -28 peak

IR: Left: Aldehyde C-H stretch doublet; Right: CH3 bends1H NMR: Left: doublet:triplet in 2:1 ratio; Right: sharp singlet ca. 2.5 ppm MS: Left: strong 29 peak (CHO+); Right: weak molecular ion; huge CH3CO (43) loss.

1H NMR: Left: four resonances: singlet, triplet, multiplet, triplet with integrals 6:2:2:3; Right: three resonances: doublet, septet, singlet (6:1:6). 13C NMR: Left: 5 resonances, one (quaternary) weak; Right: 4 resonances, one weak

(b) (8 pts) Below are 1H NMR and IR spectra for ethyl acetate, a solvent in laquers and fingernail polish remover. Please identify key features that confirm this structure—i.e., assign the 1H NMR peaks to sites in the molecule, explain their splitting patterns, and predict their respective integrals. Similarly, identify at least three features of the IR spectrum that confirm the structure.

O

O

012345PPM

1H NMR

abc

a (3H)

b (3H)

(CH3 group,no C-H'snext door)

c (2H) (CH3 group next to CH2)

(CH2 next to CH3)

CH stretches

C=O stretch

OC-O-CH3 stretches

IR

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5. (20 pts) More on structural language and analyses: consider the following compounds

O

ClCH3H

CH3Cl H(careful!)

(i) (ii) (iii) (iv)

O

(a) (6 pts) Identify the number of sp3, sp2, and sp carbons in compound (i):

sp3 0 1 2 3 4 5 6

sp2 0 1 2 3 4 5 6

sp 0 1 2 3 4 5 6 (b) (2 pts) Identify the number of secondary carbons in compound (i):

0 1 2 3 4 5 6 7 8 (c) (2 pts) What is the hybridization of the oxygens in (iii)?

sp3 sp2 sp

(d) (4 pts) Identify among the compounds above the one that should show only one resonance in both 13C and 1H NMR spectra. (i) (ii) (iii) (iv) (e) (6 pts) Write the chemical formula--e.g. C5H8Cl2 for (iv)—for compounds (i)-(iii). (i): C6H12 (ii): C8H18 (iii): C4H8O2

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6. (20 pts) (a) (4 pts) Draw the 3D (i.e. wedges and dashes) structure of a molecule of formula C2H4Br2 that has no dipole in its lowest energy conformation.

Br

Br

H

H

H

H

(b) (6 pts) Now draw the 3D representation of the C2H4Br2 isomer that has two peaks in the 1H NMR spectrum—a doublet (1:1) and a quartet (1:3:3:1)—and in its Mass Spectrum, it has a 1:2:1 triplet for the parent ion, with the sub peaks separated by 2 mass units. Explain your choice of structure and how it gives rise to the above spectroscopic behavior. What will be the parent ion masses, and where will the MS show a 1:1 doublet?

H

H

Br

H

Br

H

1H NMR: CH3 1:1 doublet split by single H next door

MS: Mass is C2H4 = 28 + 2Br's, thus it Mmol can be 28 + 79 79 81 81 (Bromine isotopes) +79 +81 +79 +81 158 or 160 or 160 or 162 = 186, 188 (2 ways), 190

1H NMR: CH 1:3:3:1quartet split by 3 Hs next door

(c) (10 pts) In Chapters 1 and 2 we saw that bonding comes from lowering electron pairs’energies. But discussing mass spectrometry, we learned about knocking the least tightly held (i.e. highest energy) electrons out of molecules to make positive ions. Now let’s look at the following 8-valence electron systems: Ne H-F H2O NH3 CH4 (all have a full octet like Neon). The cost of pulling out an electron, known as the ionization energy (IE) or ionization potential (IP) seems to follow electronegativity trends until…we get to CH4! Can you explain this seemingly surprising fact? Yes you can! [Units of IE are electron volts (eV); one eV = 23.06 kcal/mol, so these numbers are big compared, say, to the H-H bond strength in H2 (104.2 kcal/mol = 4.52 eV) or the C-H bond strength in methane (105.1 kcal/mol = 4.55 eV).]. Hint: consider where each valence electron pair resides; how does the valence electron set of CH4 differ from the others?

Ne 21.6 eV HF 16.0 H2O 12.6 NH3 10.0 CH4 12.6 (just like H2O’s value!)

All but CH4 have lone pair (nonbonding) electrons. CH4 has only bonding electron pairs, and since bonding lowers the electrons' energies by sharing them between atoms, those require higher energy to pull out. Compare the IEs of the atoms, where in each case we're pulling an electron from a p orbital: Ne: 21.6; F: 17.4; O: 13.6; N: 14.5; C: 11.3; B: 8.3 Here the "hiccup" is between N and O; Why?

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