CHEM 331: Chapter 1/2: Structures (Atoms, Molecules, Bonding) 1. What are the respective hybridizations of the atoms numbered 1 to 4 in this compound?
1: three sigma bonds and a lone pair = sp3 2: three sigma bonds = sp2
3: four sigma bonds = sp3 4: two sigma bonds = sp 2. What hybrid orbitals are used to form the sigma bond between C-1 and C-2, respectively, in the structure shown?
sp2 and sp 3. In which of these structures would the sulfur atom be assigned a formal charge of +1?
Charge = 0 Charge = +1 Charge = +2 Charge = 0 Chapter 3: Alkanes 4. There are four different functional groups in the molecule below. Identify the types of functional groups.
N CH2
H
HCO
NH CH2 C N
1 2 3 4
CH3 CH C
CH3
C N
1 2
CH3 S
O
CH3 CH3 S
O
O CH3 S
O
O
Cl CH3 S Cl
H2N O NH2
O
O
Oamine
ester
amide
ether
5. The heat of combustion of neopentane is -3250 kJ/mol; that of pentane is -3269 kJ/mol. Which statement best describes the relative stability of these two compounds? Neopentane is 2,2-dimethylpropane and is a constitutional isomer of pentane.
a. Pentane and neopentane are isomers and therefore exhibit the same stability b. Pentane is more stable than neopentane by 19 kJ/mol. c. Neopentane is more stable than pentane by 19 kJ/mol. d. Neopentane and pentane are not isomers so no statement can be made regarding
relative stability. 6. Write the IUPAC name for the following:
2,2,8-trimethylnonane 7. Which Newman projection represents the most stable conformation of 3-methylpentane (CH3CH2CH(CH3)CH2CH3) when viewed down the 2-3 carbon-carbon bond?
8. Which compound will have three peaks of equal height and three valleys of equal depth in a diagram of potential energy vs. angle of rotation for one complete rotation around the C-C bond?
The second compound will have the energy diagram where the “hills” and “valleys” are all equal in energy because the second carbon has the same three substituents (all –Br’s) and every staggered and every eclipsed conformation will be the same.
CH3 C CH3CH3
CH2 CH2 CH2 CH2 CH2
CH CH3CH3
H
HCH3
CH2CH3CH3
H
CH3
HH
CH2CH3CH3
H
H
CH3H
CH2CH3CH3
H
H
CH3CH3
CH2CH3CH3
H
wrong compoundsmaller Gauche two Gauche larger Gauche
H C
Cl
H
C H
Cl
HCl C
H
H
C Br
Br
BrH C
Cl
Cl
C H
Cl
ClH C
H
Cl
C H
Cl
Br
Chapter 4: Cycloalkanes 9. What factor is responsible for a greater heat of combustion per CH2 for cyclopropane than the heat of combustion per CH2 for cyclohexane?
a. Cyclohexane has a different hydrogen-to-carbon atom ratio than cyclopropane. (the ratio is the same)
b. Cyclohexane is a strained ring relative to cyclopropane (cyclohexane is the least strained ring)
c. Cyclopropane is a strained ring relative to cyclohexane d. Cyclohexane has more carbon atoms than cyclopropane. (question asks “per CH2,
therefore not based on how many carbons there are) 10. What is the IUPAC name for this compound?
a. ortho-bromocyclohexanol b. cis-2-bromocyclohexanol – both are “up” c. endo-2-bromocyclohexanol d. trans-2-bromocylohexanol
11. Which is the most stable conformation for cis-1-bromo-3-methylcyclohexane? diequatorial and both up or both down
Chapter 5: Stereochemistry 12. What is the correct stereochemical description of the relationship between this pair of compounds (identical, constitutional isomers, enantiomers, diastereomers)? Draw them in their 2-dimensional state:
CH3
H
H
BrH
CH3
Br
HCH3
H
Br
HH
CH3
H
Br
H
HOHO
H
HO
H
H
HO
HO
HO HO
OHand rotate CW:
HO
HO
OH
Br
13. Which statement about these Fischer Projections is correct?
a. I and II are enantiomers. – no – the top chiral center is the same b. II and IV are identical – no – both chiral centers are mirror images c. II is a meso compound – no – no plane of symmetry d. I and II are diastereomers – yes – the top chiral carbon is the same, and the
bottom one is a mirror image. 14. Which of these molecules are meso isomers?
The middle compound has two chiral carbons and a plane of symmetry (sort of diagonally through molecule – rotate that if you need to see it better) 15. Determine if each of the configurations in each of these two molecules is R or S. Assignment of priorities, then manipulate so #4 is in back for each
S R 16. What is the stereochemical classification of (1S, 2S)-1,2-cyclohexanediol and (1R, 2S)-1,2-cyclohexanediol?
a. enantiomers b. diastereomers c. meso compounds d. racemates
If one chiral center is the opposite but the other is the same, then they are diastereomers.
CH3
HHO
HO H
CH3
CH3
HHO
H OH
CH3
CH3
OHH
H OH
CH3
CH3
OHH
HO H
CH3
I II III IV
Cl
Cl
H
Cl
H
Cl H
Cl
Cl
H
CO2H
SH
CH3
CH3CH2
CH2Br
CH2CH3
H OCH3
1
1
2
3
4
2
3
4
17. Which statement is true? a. Compounds with R stereocenters rotate plane-polarized light clockwise. [R/S have
no relationship with (+)/(-)] b. For equal concentrations and equal path lengths, solutions of (+) and (-)
enantiomers rotate plane-polarized light equally, but in opposite directions c. Racemic mixtures can rotate plane-polarized light either clockwise or
counterclockwise. [Racemic mixtures cancel each other out – rotation of 0] d. Meso compounds can rotate plane-polarized light either clockwise or
counterclockwise. [Meso compounds aren’t optically active – rotation of 0] 18. Which of the following molecules is the enantiomer of this molecule?
19. How many stereocenters (chiral centers) are present in this molecule? 3
20. Which of these molecules could have an enantiomer?
meso – no Yes no chiral C’s no chiral C’s 21. Which is capable of exhibiting cis-trans isomerism?
linear – no ring – yes no – Trisubstituted no – not disubstituted
ClCH2
H3C
HBr HBr
ClCH2
CH3
Br
HClCH2
H3C
H
BrClCH2
H3C
CH3 CH3
CH3CHCH3
OH
* **
HO OH
OH
OH
OH
HO
OH
HO
CH3C CCH3
OH
OH(CH3)2C CHCl CH3CH C CHCH3
Br
H
ClCH2
H3C
22. Which molecule shown below is the same stereoisomer as the one in this Fischer projection?
Chapter 7: Intro to Alkenes 23. What is the IUPAC name for this compound? a. (E)-3,5-dimethyl-2-hexene b. (Z)-3,5-dimethyl-2-hexene c. (Z)-1,2-dimethyl-2-isobutylethylene d. (E)-3,5,5-trimethyl-2-pentene Six carbons, double bond between 2 and 3, two extra methyls on 3/5. High priority on left is CH3 and high priority on right is larger carbon group. Opposite = E. 24. What would be the stereochemical classification of the major product of this reaction?
a. R-enantiomer b. S-enantiomer c. Achiral – no chiral carbons d. Racemic
CH3
CH3
Cl H
H Cl
CH3
CH3
H
Cl
Cl
H=
CH3
CH3
H Cl
Cl H
CH3
CH3 H
ClH
Cl
H
ClCH3
ClCH3
H
Cl
HCH3
ClCH3
Hmirror
H3C CH3
Cl ClH H
meso diastereomer
rotate into eclipsed:
H3C Cl
H
H3C Cl
H
meso
rotate into eclipsed:
H3C H
Cl
H3C Cl
H
same
CH3CH C(CH3)2HBr Br
CH3
H CH2CH(CH3)2
CH3
CH3
CH3
Cl H
H Cl
25. Cyclohexene reacts with bromine to yield 1,2-dibromocyclohexane. Molecules of the product would be:
a. a racemic form and, in their most stable conformation, they would have both bromine atoms equatorial.
b. a racemic form and, in their most stable conformation, they would have one bromine atom equatorial and one axial.
c. a meso compound and, in its most stable conformation, it would have both bromine atoms equatorial.
d. A meso compound and, in its most stable conformation, it would have one bromine atom equatorial and one axial.
Chapter 8: Alkene Reactions 26. What set of reagents would convert the alkene in this alcohol as the major product?
a. H2O, H+ b. BH3 followed by NaOH, H2O2 c. Hg(OAc)2, H2O followed by NaBH4 d. HBr followed by NaOH
Answer (a) would work on paper but the carbocation must form next to a quaternary carbon with lots of possible methyl groups that can do methide shifts! Carbocations rearrange! Answer (b) makes the incorrect regiochemistry Answer (d) would do SN2 and E2 with NaOH, so two products form. 27. Draw the product of the following reaction: BD3 is just like BH3 but its easier to see the placement of the new D:
Br2
Br
Brdown
up
BrBr
equals
1. BD3
CH3
2. NaOH, H2O2CH3DOH
CH3 C
CH3
CH3
CH CH2 CH3 C
CH3
CH3
CH CH3
OH
28. What intermediate is involved in the reaction shown?
29. The addition of bromine, Br2, to the alkene shown will give
a. one meso compound b. one pair of enantiomers c. two meso compounds d. four stereoisomers
Achiral plus achiral = not optically active (and with a methy on the left and ethyl on the right, it cannot be a meso compound) Chapter 9: Alkynes and Reactions 30. What is the major product isolated from the reaction shown? Markovnikov = Methyl ketone!
31. What reagent(s) can be used to reduce a triple bond to a double bond having Z (cis) geometry (as shown in the figure)? a. Na, NH3 b. NaBH4 c. H2, Pd/CaCO3/Pb d. H2, Pd/C Chapter 10: Radicals 32. For which free radical can the most resonance forms be written that show the delocalization of the radical? a. b. c. d.
Cl C Cl Cl C Cl Cl C Cl Cl C Cl
dichlorocarbene
CH3CH2CCH3
OCH3CH2CH2CH CH3CH2CHCH2OH CH3CH2CHCH3
O OH OH
CH2CH2 CH
no resonance no resonance 4 resonance forms SEVEN resonance forms!!
CHCl3, KOH Cl
Cl
Br2
H2O, HgSO4CH3CH2 C C HH2SO4
R' RR' R
H H
33. Which of the indicated sites would most readily undergo hydrogen atom abstraction to generate a radical? #2
34. What is the expected stereochemistry of the organic product from this reaction?
a. (S)-isomer only b. (R)-isomer only c. equal amounts of (R)-isomer and (S)-isomer – can’t control stereochemistry in
radical reaction d. unequal amounts of (R)-isomer and (S)-isomer
35. Which radical is the most stable?
Chapter 11: Substitutions and Eliminations 36. The reaction of benzyl bromide (C6H5CH2Br) with azide ion (N3
⊖) is shown in the box. When the benzyl bromide concentration is constant and the azide concentration doubles, the reaction rate is observed to increase by a factor of two. When the azide concentration is held constant and the benzyl bromide concentration doubled, the rate of the reaction doubles. What is the correct rate law for this reaction?
a. Rate = k [C6H5CH2Br]2 [N3⊖]2
b. Rate = k [C6H5CH2Br]4 [N3⊖]2
c. Rate = k [C6H5CH2Br]2 [N3⊖]
d. Rate = k [C6H5CH2Br] [N3⊖] – SN2 reaction Rate = k[nuc][RX]
CH3
H3C CH3
4
321
1º
3º1º
3º allylic
CH3
CH2
CH3
CH3
CH3
CH3
CH3
CH3
3º allylic2º allylic3º1º
C6H5CH2Br + N3 C6H5CH2N3 + Br
Cl2, lightCH3
H
CH3
Cl + HCl
37. What is the structure of the reagent that would yield the two products shown?
SN1 and E1 reaction with neak nucleophile/base – need alkyl halide where SN1 occurs:
38. What is the expected reaction pathway (E2, E1, SN2, SN1) for the following reaction?
1º alkyl halide with strong nucleophile = SN2 39. What is the configuration of the substitution product formed from the reaction of KOH with (R)-1-chloro-1-deuteriobutane (CH3CH2CH2CHDCl)?
a. (S)-1-deuterio-1-butanol b. (R)-1-deuterio-1-butanol c. meso-1-deuterio-1-butanol3 d. a racemic mixture of (a) and (b)
40. Put the nucleophiles (CH3OH, CH3O⊖, CH3NH2) in order, fastest to slowest, for reaction with propyl bromide (CH3CH2CH2Br + Nuc → CH3CH2CH2Nuc). Strongest nucleophile has anion (CH3O⊖). The other two nucleophiles are neutral. Which is a stronger nucleophile – oxygen or nitrogen? Which one is more willing to give up its electrons? Nitrogen. Why? Its less electronegative so more willing to share its electrons. Order? CH3OH CH3O⊖, CH3NH2, CH3OH 41. Which alkyl halide would you expect to undergo SN1 hydrolysis most rapidly? a. (CH3)3CI b. (CH3)3CBr c. (CH3)3CCl d. (CH3)3CF best leaving group is iodide
O OCH3
C6H5
O C6H5+? CH3OH
O C6H5 O C6H5O Br
C6H5
O C6H5
Br
CH2CH2Br CH3CH2C C
Cl
HD
R1º
KOH
SN2 OH
DH
S
42. Which anion is the most nucleophilic towards methyl iodide in an SN2 reaction?
43. What is the major elimination product from the reaction shown?
Need to ring flip first:
44. What is the product of the reaction shown?
Rotate left chiral carbon to put H in antiperiplanar position:
45. What would be the possible product(s) of the dehydrohalogenation of trans-1-bromo-2-methylcyclohexane?
CH3CH2CH2 OCH3CH2CH2 S O CH3CH2 C
O
O
KOtBu
H
CH3
H
Br
D
H
D
H
Br
HCH3
H DH
CH3
H
E2 with trans-diaxial hydrogen/halide
H3CBr
H
H
D
C6H5
KOtBu
HBr
CH3
PhH
DC6H5= Ph (phenyl)
H3C
D
Ph
H
KOHBr H
CH3H
Br
HCH3
H
H
H
CH3
H
CH3
H
46. Why would concentrated hydrobromic acid be an inappropriate catalyst for the dehydration of alcohols?
a. HBr is too weakly acidic to protonate the alcohol. Not true! Used to convert 3º alcohols to 3º bromides
b. The conjugate base, Br⊖, is a good nucleophile and it would attack the carbocation to form an alkyl bromide.
c. HBr is strongly acidic, so the water molecule would not be a good leaving group after protonation of the alcohol. False!
d. HBr would be more like to promote rearrangement of the carbocation intermediate. No more than any other acid involved in a reaction that makes a carbocation.
47. Which carbocation is not likely to undergo rearrangement? Already most stable:
48. Which alcohol is dehydrated fastest in concentrated H2SO4? 3º alcohols do E1 fastest
49. Consider the following typical nucleophilic substitution reaction. The effect of doubling the volume of solvent would be to multiply the reaction rate by a factor of: a. ¼ b. ½ c. 2 d. 4 Tricky question – if you double the solvent volume, you cut the concentration of both the nucleophile and the alkyl halide in half. If you cut them BOTH in half, you cut the rate to a quarter of its original rate.
CH3CH2CCH3
OH
CH3
CH3CH2CHCH2OHCH3
CH3CHCHCH3
CH3
OHCH3CHCH2CH2OH
CH3
CH3Br + KOH CH3OH + KBr
Chapter 18: Ethers 50. Which set of products is expected from the reaction shown? a. b. c. d.
51. What is the product of the reaction shown?
52. Which reaction would product phenyl propyl ether? a. b.
c. d.
Answer (b) has no leaving group, (c) cannot do SN2 on sp2 carbon and (d) has no nucleophile. 53. Which of the following molecules shown below is the product of the reaction of (E)-1-phenylpropene with meta-chloroperbenzoic acid? Epoxide – maintain stereochemistry!
Br OH Br
BrCH2CH3
+ +BrCH2CH3 HOCH2CH3
+ No Reaction
O CH3
NaOCH3H3CO CH3
OH
O NaCH3CH2CH2Br
BrCH3CH2CH2ONa
BrCH3CH2CH2Br
CO2H CH3
HO
CH3
H HO
H
H CH3O
HBrO CH2CH3heat
O NaCH3CH2CH2OH
CH3mCPBA