Organic Chemistry I (CHE ) Examination I S ep t ember 2 9, KEY Name (PRINT LEGIBLY):

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1 Organic hemistry I (E ) Examination I S ep t ember 2 9, KEY Name (PRINT LEGIBLY): Please provide clear and concise answers to all of the following questions. Use equations and/or drawings to support your answers where appropriate. Please answer questions in the area provided, the back of an exam page or on the clearly labeled spare page. Problem Score 1. (a-d) /28 2. (a-b) /16 3. (a-b) /12 4. (a-b) /16 5. (a-d) /16 6. (a-b) /12 Total /100 PLEA S E observe the following. 1) Write LARGE and LEGIBLY. If I can t read what you write, I can t give you full or partial credit. 2) Read each question carefully before answering. If you don t answer the question I posed, you will not receive credit. 1

2 1a. (5 pts) The σ-bond framework (atom to atom connectivity) of trimethylamine N-oxide is shown below. Fill in the remainder of electrons to draw the best resonance structure for trimethylamine N-oxide. Indicate all unshared pairs of electrons and any formal charges. O - N + 1b. (5 pts) The chemical formula of methacrylonitrile is 4 5 N and its σ-bond framework (atom to atom connectivity) is shown below. Fill in the remainder of electrons to draw the bes t resonance structure for methacrylonitrile. Indicate all unshared pairs of electrons and any formal charges. N 2

3 1c. (6 pts) In the structure you drew above, what are the hybridizations of: 1. the N atom sp 2. the atom directly bonded to N sp 3. the atom bonded to three other atoms sp 2 1d. Draw the best resonance structure for each of the following compounds. Use the curved arrow formalism to show how you got from the structure given to the one you drew. Briefly explain why the resonance structure you drew is better (you will receive only partial-credit for the correct structure without an explanation). (i) (6 pts) O - O - The negative charge is on the more electronegative oxygen atom. [Moreover, aromatic character is restored to the benzene ring. (This last part is not required for full credit). 3

4 (ii) (6 pts) N N the structure on the right is better because all of the atoms in it have an octet of electrons (hence contains no electron deficient atoms) and the structure has no formal charges (fewer than the structure of the left). 2a. (6 pts) ompute the degrees of unsaturation in 2 5 N; show your work. For x y N v O w X z, # rings or π bonds (degree of unsaturation) = [ 2x + 2 (y + z v) ] / 2 degree of unsaturation = [2x2 + 2 (5-1)]/2 = 1 4

5 2b. (10 pts) Draw structures of two isomers whose molecular formula is 2 5 N. N N 2 3. Draw structures of the following compounds: (i) (6 pts) 4-isobutyl-2,5-dimethylheptane (ii) (6 pts) (Z)-3-methyl-2-octene 5

6 4a. (6 pts) Rank the following molecules (A-D) in order of increasing stability. A B D < A < D < B 4b. (10 pts) Briefly explain your answer for the most stable structure and the least stable structure. B is most stable because it contains the most substituted carbon-carbon double bond. The more substituted the carbon-carbon double bond is, the more σ bonds that can hyperconjugate with the π bond and the lower in energy the molecule is. Alternately, you could say the more substituted the carbon-carbon double bond is, the more relatively lowenergy (strong) sp 2 -sp 3 bonds are present. The more strong bonds present, the more stable the molecule. Alkynes require linear geometry since sp hybrid orbitals are pointed 180 from one another. ence is least stable because the angle around the carbon-carbon triple bond in the ring is severely less than 180, leading to high ring strain. The linear nature of the alkyne makes it very difficult to make cycloalkynes. yclooctyne is the smallest cyclic alkyne isolable at room temperature. 6

7 5.(4 pts each, 16 pts total) For each pair of compounds, indicate the relationship of the first compound to the second by circling the correct answer. (a) l 3 is a skeletal isomer of is a configurational diastereomer of is a conformational diastereomer of is an enantiomer of is a resonance structure of is the same compound as has none of the above relationships to l 3 (b) 3 I I is a skeletal isomer of is a configurational diastereomer of is a conformational diastereomer of is an enantiomer of is a resonance structure of is the same compound as has none of the above relationships to I I 3 (c) l is a skeletal isomer of is a configurational diastereomer of is a conformational diastereomer of is an enantiomer of is a resonance structure of is the same compound as has none of the above relationships to l 7

8 (d) is a skeletal isomer of is a configurational diastereomer of is a conformational diastereomer of is an enantiomer of is a resonance structure of is the same compound as has none of the above relationships to 6a. (6 pts) Which of the following two reactions (1 or 2) should have the greates t change? In other words, which reaction should be more thermodynamically favorable? Explain why. (1) ( 3 ) 3 ( 3 ) 3 ( 3 ) 3 ( 3 ) (2) 3 Relief of steric strain (due to adjacent bulky t-butyl groups) would make the transformation in reaction 1 more favorable (hence greater change). The structures in reaction 2 are identical (hence no enthalpy change is expected). 8

9 6b. (6 pts) Arrange the following compounds (A-) in order of increasing acidity. Briefly explain why you ordered the compounds as you did 2 A B A < < B Alkanes, alkenes, and alkynes contain atoms that are sp 3 -, sp 2 -, and sp-hybridized, respectively. The more s character in an orbital, the more stable the orbital is, and hence the more an electron in it is stabilized. Thus in similarly substituted series of compounds, the alkyne will be more acidic than the alkene, which will be more acidic than the alkane. 9