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