chem118AF05Midterm1key - Chem 118A, Fall 2005, Midterm 1 key

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Unformatted text preview: Chem 118A, Fall 2005, Midterm 1 key _______________________________ name (1) (10 points total) The molecule shown below is used by some pathogenic fungi to regulate their sexual reproduction. (a) Please draw in any missing lone pairs on the structure below. sp3 tetrahedral ketone O HO 1° alcohol * OH 3° alcohol sp2 trigonal planar * OH 1° alcohol (b) In the structure shown in part (a), please identify each oxygen atom as part of an alcohol (also note whether it is primary, secondary or tertiary), an ether, a ketone, an aldehyde, an ester, or a carboxylic acid. (c) Please identify the hybridization and geometry of the two atoms marked with *'s in part (a). (2) (15 points total) Please rank the following 3 compounds in order of increasing acidity (1=least acidic, 3 = most acidic) and explain your reasoning clearly and concisely. HO Cl Cl Cl Cl OH Br Br OH (2) least acidic (1) Cl Cl most acidic (3) The lefthand compound has more chlorine atoms than the righthand compound. More Cls will provide more inductive stabilization of the conjugate base. In all three, the OH group will be deprotonated. The lefthand compound has two chlorine atoms while the righthand compound has two bromine atoms. Cl is more electronegative than Br and therefore provides greater inductive stabilization of the conjugate base. Chem 118A, Fall 2005, Midterm 1 key _______________________________ name (3) (15 points total) Consider the structure shown below with molecular formula C4H7ON. (a) Please draw in any electrons necessary to make a valid Lewis structure. (b) Please draw one additional valid resonance structure for this molecule. (c) Please list the merits and faults of each of your two structures. • You only needed to draw 2... • Arrows are shown for interconverting each structure with the next in a clockwise direction (you did not need to show the arrows for this question). O H H H H H C C C N C H merits: no formal charges, lots of covalent bonds, complete octets on C, N and O H faults: really none to speak of O H H H H H C C C N C H H H H H H C C O C N C H H H merits: lots of covalent bonds, complete octets on C, N and O faults: separation of charge, positive charge on electronegative N merits: negative charge on electronegative O faults: less covalent bonds, separation of charge, incomplete octet on C Chem 118A, Fall 2005, Midterm 1 key _______________________________ name (4) (10 points total) Please draw a clear and completely labeled molecular orbital interaction diagram that shows how the indicated C–C s-bond of 1,3-butadiene is constructed from orbitals on its carbon atoms. To receive full credit you must also explain clearly why the formation of such a bond is favorable, making reference to your diagram. H 1,3-butadiene H C C C H C H H H H R E node R H antibonding H R sp2-orbital R H sp2-orbital The orbital interaction shown (s-bond formation) allows these two electrons to occupy a lower energy orbital (the bonding orbital shown) than they would if the interaction did not occur. H R bonding R H Chem 118A, Fall 2005, Midterm 1 (5) (15 points total) key _______________________________ name (a) Please draw all of the different constitutional isomers for C6H13Br that can form in the following monobromination reaction. (Note that duplicate structures will cancel each other out). Br2 hn C6H13Br Br Br (b) Please draw the structure of the most stable carbon-centered radical formed in this reaction. (c) Would you expect the reaction to be more or less selective if Cl2 was used instead of Br2? Please explain your answer clearly and concisely. The reaction should be less selective with Cl2 than with Br2. H–Cl bonds have larger bond dissoication energies than do H–Br bonds, meaning that H–Cl bonds are stronger. The Hammond Postulate (actually the Bell-Evans-Polanyi Principle) suggests that stronger bonds will be formed more quickly; i.e. Cl• radicals are more reactive than Br• radicals in hydrogen atom abstraction reactions. More reactive goes along with less selective. Chem 118A, Fall 2005, Midterm 1 ______________key______________ ___ name (6) (20 points total) Below are drawn two particular conformations of propane. (a) Please draw correct Newman projections for each, looking down the indicated C–C bond, in the boxes provided. A H H C CH3 H C H H B H H CH3 C C H H H Newman projection for A: Newman projection for B: H H3C H H H H H H H CH3 H H (b) For each conformation, decide if it is "staggered" or "eclipsed". A: staggered B: eclipsed (c) For each conformation, decide if it is a minimum or a transition state structure. A: minimum B: transition structure (d) State which conformation is lower in energy and explain clearly and concisely why this is so. Conformer A is lower in energy because none of the groups attached to the indicated C–C bond are eclipsed—they are all staggered (i.e. they are as far apart as possible). In conformer B all of the groups are eclipsed and therefore "bump into" each other slightly. Chem 118A, Fall 2005, Midterm 1 ______________key______________ ___ name (7) (15 points total) Consider the reaction shown below. H C H H H H H C C H H H H H C H H H C C H H (a) Please add appropriate curved arrows to the reactant structures shown above to indicate the flow of electrons that leads to the products. (b) Please draw a picture of the transition state for this reaction showing partial bonds and partial charges as appropriate. H C H H H C H CH HH ...
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This note was uploaded on 02/07/2011 for the course CHE 118A taught by Professor Patten during the Fall '08 term at UC Davis.

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