Facebook: UCI ORganic Chemistry Peer Tutoring King 51 Fall 2017

Transcription

1 1 Organic Chemistry 51A Professor King Final Exam Review Session December 7 th, 2017 Will Cabanela (rcabanel@uci.edu) Amanda Pinski (apinski@uci.edu) Zachary Valley (zvalley@uci.edu) Facebook: UCI ORganic Chemistry Peer Tutoring King 51 Fall Below are the IR spectra for the chemical compounds of C4H8O. a. Determine what functional groups are described by the spectra b. Match the spectra with its corresponding chemical structure

2 2

3 3

4 4 2. Below is a scary looking molecule and acetic acid! Answer the following questions! a. Predict the product(s) for the above nucleophilic substitution reaction, showing proper stereochemistry. Provide a mechanism for the product(s) formation below. b. How many transition states would this reaction have? 3

5 5 c. Write the rate law equation for this reaction. Is this a bimolecular or a unimolecular reaction? K[Substrate] Unimolecular 3. Below is another scary molecule and N,N-dimethylformamide (DMF). What nucleophile is needed to produce the following product? 4. Answer the following questions about the featured molecule. a. Draw all resonance structures.

6 6 Note: the nitrogen cannot donate its electrons: the octet rule would be broken for carbon if this was done. b. Draw the resonance hybrid. c. Indicate the major contributor to the resonance hybrid. The minor contributor to resonance is the top right in part A because electronegative atoms, both with negative charges, withdraw electron density away from the carbocation, destabilizing it. There are also three charges. The bottom left is likely to be more stabilized because the secondary carbocation is farther away from EWG s. Ranking 5. Rank the following molecules in order of strongest to weakest acid. Strongest to weakest: A>C>D>E>B Greater pka indicates a weaker acid and a stronger conjugate base.

7 7 Oxygen is more electronegative than nitrogen, causing electron density to be withdrawn from the acidic hydrogen and the acid to be more acidic (lower pka). Therefore, the oxygencontaining compounds will be of lower pka. Nitro groups (page 46 in Dr. King s notes) are extremely electron withdrawing. Iodine is also electronegative, but only weakly so. Since A has only one carbon between the hydroxy and nitro groups as compared to the two carbons in C, it is more acidic and has a lower pka. C and D both have the same number of carbons between the hydroxy and electronegative groups, but since iodine is significantly less electronegative than a nitro group, C has a lower pka. Thus far, we have in terms of increasing pka: A<C<D Alkyl group donate electron density to nitrogen. Although both amines are primary, B has a significantly larger EDG (4 carbons nearby) compared to E (1). The greater abiltity to donate electrons to the nitrogen in B causes the acidic hydrogen to be less acidic and the amine to have a greater pka. Therefore E<B. Overall: A<C<D<E<B 6. Rank the following molecules in order of increasing reactivity with hydroxide ion in a substitution reaction. D<B<A<C<E Hydroxide ion is a strong nucleophile that will undergo SN2 the fastest with methyl alkyl halides followed by primary and secondary alkyl halides. Leaving group also matters: iodide is a better leaving group than chloride because its larger volume allows it to better delocalize the negative charge. Beta branching of primary alkyl halides, as in D, dramatically reduces the rate of SN2 reactions to a degree comparable to that of a tertiary alkyl halide (see page 141 of Dr. King s notes). D will be the slowest since there are no tertiary alkyl halides. The two alkyl chlorides are secondary (B) and primary (A) and the two alkyl iodides are methyl (E) and primary (C). Since iodide is the better leaving group, C and E will react faster than either A or B. E will react faster than C. Thus far, E>C. B will react slower than A because it is secondary. Thus, A>B. Therefore: D<B<A<C<E

8 8 7. Rank the following molecules in order of increasing carbocation stability. A<E<D<B<C Nitro groups are extremely electron withdrawing (see page 46 in Dr. King s notes). It is only one carbon away from the carbocation and its EWG effects will make the carbocation more positively charged, destabilizing it. Primary resonance stabilized carbocations, as in B are more stable than secondary (D) carbocations but less stable than tertiary carbocations (C). Note: E is NOT a primary resonance-stabilized carbocation! There is a carbon between the methoxy and carbocation! Thus, A is least stabilized, C is most stabilized, followed by B, D and E. 8. For single molecules shown (a and b), identify as chiral, achiral, or neither. More than one answer may apply. For pairs of molecules (c and d), identify as enantiomers, diastereomers, identical or neither. a. Meso, achiral b. Achiral (no stereocenters) c. Identical (flip the molecule over 180 degrees: they are the same) d. Diastereomers (stereochemistry is changed at the first carbon, i.e carbon with fluorine)

9 9 9. Stereochemistry: observe the following compound. a. Name the compound. 2-chloro-3-iodobutane (atoms with higher atomic numbers have priority) b. Indicate absolute configuration, R or S, at any stereocenters.

10 10 c. Draw the Newman projection for the staggered conformation shown above looking down the third and second carbons. d. Draw the Newman projection of the highest energy conformation using. Highest energy conformation occurs when torsional and steric strain occurs. Two methyl groups eclipsing and the two halogens eclipsing represent the highest strain possibility. Atom or Group Van der Waal s Radius (pm) H 120 CH3 200 Cl 175 I 198