Chemistry 14C Spring 2016 Final Exam Part B Solutions Page 1

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1 hemistry 14 Spring 2016 Final Exam Part B Solutions Page 1 Statistics: igh score, average and low score will be posted on the course web site after exam grading is complete. A note about exam keys: The answers presented here may be significantly longer than expected from a student taking the exam. An exam key serves not only to reveal what was expected, but to instruct you as well. To see the projected course grade cutoffs, consult the grading scale on the hem 14 course web page. 1. Barbituric acid has eight sp 2 atoms (all oxygen atoms, all nitrogen atoms, and three carbon atoms) and eight lone pairs (two on each oxygen atom and one on each nitrogen atom). 2. The least electronegative element in barbituric acid is. E = 3.5, E = 3.0, E = 2.5, and E = Any other contributors violate at least one resonance contributor preference rule, and are therefore less significant δ δ δ δ+ δ+ 2 2 δ onjugation provides stabilization to a molecule by providing electron delocalization. r (better):... a longer pi electron wavelength. ther answers may be possible. Planarity is a result of conjugation, and not a cause of stability. 7. Aromaticity is an important feature of molecular structure because aromaticity causes stability (or planarity or electron delocalization or...). ther answers may be possible. 8. Yes to all parts. 9. The structure shown is D-ribose. The on the D/L carbon points to the right. 10. The cyclic form of ribose is more likely to be a furanose. 11. Including the stereoisomer shown, there are eight stereoisomers of the ribose structure. There are three stereocenters, and no meso possibilities. 2 3 = 8 total possible stereoisomers.

2 hemistry 14 Spring 2016 Final Exam Part B Solutions Page nly one enantiomer is possible. It is formed by inverting all of the stereocenters onfigurational isomers (stereoisomers) and diastereomers. 14. ompared to ribose, glucose has more carbon atoms. Ribose is a pentose whereas glucose is a hexose. 15. A hydrogen bond donor requires a hydrogen atom with a large δ + (or bonded to a high E atom), whereas all hydrogen bond acceptor atoms have a lone pair. 16. Water: Middle. Urea: Least. arbon dioxide: Most. arbon dioxide is a gas a room temperature; the others are not. Urea has more hydrogen bonding, dipole-dipole, and London force attractive interactions than does water. 17. Many answers are possible. Argon and methane are perhaps the most obvious. 18. The pk a of melamine's bond is less than the pk a of aniline's bond because melamine is a stronger acid than aniline. 19. For the equilibrium shown K eq is less than 1 because an acid-base equilibrium favors the weakest acid (pk a barbituric acid > pk a 3 + ). The pk a values of water and barbiturate anion are irrelevant because these molecules are not acids in this reaction (they are bases). 20. Many answers are possible. For example we could replace a hydrogen atom with a fluorine atom and make the molecule more acidic via an inductive effect. F 21. Solubility (soluble in nonpolar solvents and insoluble in polar solvents) and origin (biological origin). 22. Prostaglandin Alanine is the only acceptable answer. ther one-carbon side chains are either acidic (cysteine) or hydrophilic (serine). 25. ne of the DA base pairs occurs between nucleobases G and, and it involves three hydrogen bonds. r: ne of the DA base pairs occurs between nucleobases A and T, and it involves two hydrogen bonds. 26. In DA, phosphorus is bonded to.

3 hemistry 14 Spring 2016 Final Exam Part B Solutions Page Mass spectrum: m/z = 141 (M): Molecular mass (lowest mass isotopes) = 141. dd number of nitrogen atoms. m/z = 142 (M+1): 9.36%/1.1% = or 9. m/z = 143 (M+2): < 4% so no S, l, or Br. Formula ( 8 ): ( 8 ) = 45 amu for oxygen, nitrogen, and hydrogen. xygens itrogens = Formula omments = Rejected: Violates -rule = Reasonable = Rejected: More than three signals in 1 -MR spectrum Formula ( 9 ): ( 9 ) = 33 amu for oxygen, nitrogen, and hydrogen. xygens itrogens = Formula omments = Rejected: Poor fit with 1 -MR integration = Rejected: More than three signals in 1 -MR spectrum DBE: 8 - (15/2) + (1/2) + 1 = 2 Two rings, two pi bonds, or one ring plus one pi bond. IR: Zone 1: Alcohol : Absent - no strong peak. Amine/amide : Absent - no peak. Terminal alkyne : Absent - no peak; no in zone 3. Zone 2: Aryl/vinyl : Absent - no peak > 3000 cm -1. Alkyl : Present. Aldehyde : Absent - no peak ~ 2700 cm -1 ; 13 -MR = signal is a singlet. arboxylic acid : Absent - not broad enough; not enough oxygens. Zone 3: Alkyne : Absent - no peak; not enough DBE for triple bond plus =. itrile : Absent - no peak; not enough DBE for triple bond plus =. Zone 4: =: 1680 cm -1. ot conjugated (no benzene ring or = in zone 5). nly an amide fits. annot be an alkene, because the 13 -MR signal at ppm would not be accounted for. Zone 5: Benzene ring: Absent - not enough DBE; no 13 -MR signals ppm. Alkene =: Maybe; only one 13 -MR signal in ppm range means that if this is an alkene, it must be symmetrical.

4 hemistry 14 Spring 2016 Final Exam Part B Solutions Page 4 1 -MR: hemical shift Splitting Integral # Implications 3.47 ppm singlet x or 3 x 2.18 ppm triplet in in ppm pentet in in 2 * 2 x in 2 2 x in 2 x in x in 2 x in 2 () ppm singlet x or 3 x 2 or 6 x 1.47 ppm triplet in in 2 2 x in 2 2 x in Totals (2 x ) + 2 = 6 15 * Rejected because the implications do not also contain a in MR: ppm singlet agrees with amide from IR. There are seven 13 -MR signals and eight carbons in the formula, so the molecule contains some symmetry (probably the two methyl groups at 1.52 ppm in the 1 - MR). Atom check: 8 15 (from mass spectrum) (from 1 -MR) - (amide from IR) =. This carbon is not accounted for in the 1 -MR so it is not bonded to any hydrogen atoms. It is probably the 76.0 ppm singlet in the 13 -MR. DBE check: alculated 2 from formula - 1 (=) = 1 left over. Appears to be a ring. Pieces: 2 x (no ) 2 in in in = (3 o amide) Assembly: According to the 1 -MR splitting, the three 2 groups form a group. 2 x (no ) = (3 o amide) The 2 x can only be attached to the (no ) or the amide nitrogen. Their chemical shift (1.52 ppm) is low for, so let's guess they are attached to the (no ). ( ) 2 (no ) = (3 o amide) The other cannot be attached to the ( ) because this is not equivalent to the other two. This leaves it to be attached to the amide on the carbonyl side or on the nitrogen side. The methyl's 1 -MR chemical shift (3.47 ppm) is more consistent with 3 (typically 2 3 ppm) than with 3 = (typically ppm) ( ) 2 (no ) = (3 o amide)

5 hemistry 14 Spring 2016 Final Exam Part B Solutions Page 5 The remaining pieces can be assembled two ways: 3 3 Either answer was accepted for full credit. ommon (but incorrect) answers: 3 ot at 3.47 ppm lose, but not acceptable: The amide methyl group does not have a 1 -MR chemical shift of 3.47 ppm. Equivalent ( ) 2 ot acceptable( and not close): Two of the cyclobutane methylene groups are equivalent; this molecule has only four signals in its 1 -MR spectrum.