Q: For DEPT spectra (sect. 13.6), I see where CH, CH2, and CH3 show up on the NMRs - but I don't understand how you can differentiate between two CH3's or two CH2's for a compound. Could you explain this?
A: You do this by chemical shifts. For the example in Fig 13.11 (pg. 489), one CH2 is allylic (next to a double bond) while the other is between two sp3 carbons. The CH2 next to the double bond will be downfield relative to the other one. The same holds for the three CH3's. Two are next to a double bond and should be downfield relative the the other -CH3 which is attached to an sp3 carbon. You can not easily differentiate between the two allylic -CH3's.
Q: What is the difference between a protic and an aprotic solvent? How can you tell them apart?
A: Protic solvents have a acidic hydrogen. For the most part these will be O-H bonds such as those in water and alcohol solvents. Aprotic solvents do not have an acidic hydrogen; i.e., there is no O-H bond.
Q: For the allylic bromination of alkenes, the book sometimes writes to use NBS with hv, hv and CCL4 or just CCL4 in the reactions. Do you need to use both hv and CCL4 or one or the other?
A: CCl4 is the solvent for the reaction. It is not too important to know. However, hn and NBS should always be a reagent for allylic bromination.
Q: I have a question about problem 10.3. Why is the product formed from the abstraction of the "a" hydrogens achiral? Isn't the carbon with the halogen also bound to 2 hydrogen atoms, which would make it achiral? I'm confused.
A: You are not confused, there is a typo in the solutions manuel. Chlorination at any of the primary carbons (a and e) will yield achiral products a will chorination at site e.
Q: On page 23
of the text, on the last paragraph at the very bottom of the page,
it says that C-H bonds are 106 pm and C-C bonds are 120 pm. And
then says, "...making the triple bond in acetylene the shortest
and strongest of any carbon-carbon bond." So 120 pm is shorter
than 106 pm? I'm wondering if this is a typo. Please let me know.
A: No. 106 pm is the length of the C-H bond when the C is sp hybridized. The 120 pm is the bond length of the C-C triple bond when both C's are sp hybridized.
The C-C bond length of a C-C double bond when both carbons are sp2 hybridized is 133 pm and the C-C single bond (sp3 hybridization) is 154 pm.
Q: For the oxidative
cleavage of alkenes to carbonyl compounds, some of the examples
the original compound is split into two separate fragments, either
a ketone or an aldehyde depending on how substituted a carbon
is, while for other other alkenes the compound doesnít
break up the carbon chain. How can we tell which will happen?
(one example of this in on page 6 of the last hand-out,
the slide on the bottom of the page)
A: When the alkene is part of a ring (that is the sp2 hybridized carbons of the alkene are both carbons within the ring), then they are tethered together by the other ring carbons. After oxidative cleavage of the alkene, the carbonyls are still tethered together by the other ring atoms. Thus the carbonyls are part of the the same compound.
Q: On the second
and third exams, will the cumulative material be tested specifically
(such as more minor details) or will it mostly be used as basis
for the questions on new material?
A: The second exam will specially be on material covered since the first exam. The material is inherently cumulative. While I may not specially ask a question on material from Chapter 1-4, it is assumed that you know that material and some question on exam 2 may be based on prior knowledge of Chapter 1-4.
Q: Why do cycloalkenes only exist as the cis stereoisomer and not as the trans stereoisomer?
A: The distance between the trans position of an alkene is further away than the cis positions. For common ring sizes such as a cyclohexene, the carbon chain is not long enough to span this distance between the trans positions. You can have trans cycloalkene in larger rings. If you try to draw or make a model of trans cyclohexene this will become immediately clear.
Q: I am unclear on the terminology disubstituted and trisubstituted.
A: Di-, tri- and tetrasubstitued alkenes means how many non-hydrogen groups are on the two alkene carbons. The there are two substituents on the alkenes, then it is disubstituted, etc.
Q: I am having a hard time following the (E) and (Z)explanation for the last three diagrams on page 200. (The three that you are asked to work through, below the text examples in red and green)
A: to determine E or Z for the three examples at the bottom of pg. 200, look at each carbon of the alkene. Look at the atoms attached to the the alkene carbons. For the first one (E)-3-methyl-1,3-pentadiene, one alkene carbon has an -H and a -CH3. The atom with the highest atomic number gets the high priority, so 6 gets priority over 1 (atomic #'s for C and H). On the other alkene carbon the is a vinyl group (-CH=CH2) and a -CH3. You have to convert the double bonds to the equivalent number of single bonds, See the top example on that page. The first atom attached to the alkene carbon are both C's, so you go out the the next highest priority atoms. For the -CH3 groups it is H (atomic number 1) and for the vinyl group it is a C (atomic # 6), so the vinyl group gets priority. The high priority groups are on opposite side, so the alkene is E.
Q: How is the chlorination of methane a self-sustaining process when methane is not produced during the reaction? Is it assumed that there is an excess of methane? I was looking at the propagation steps on page 156 of McMurry, and I see that chloromethane is produced, but not methane.
A: Methane is a reactant and once the methane is consumed the reaction is over. The mechanism is self-sustaining or self-propagating until all the methane is consumed. What we mean by this, is that the reactive intermediate in the reaction, the chlorine radical which is initially generated in the initiation step, is regenerated in the the propagation step of the reaction. So once the reaction is initiated, it should be self-sustaining until the methane is all consumed (i.e., it does not, in theory, need to be initiated again).
Q: Do we need to know the structures of the functional groups listed in table 3.1 on page 76 for the exam?
A: You need to be able to identify the functional groups. We went over the ones you need to be able to identify in class, which is most of the list. The sulfur containing ones are less important and you don't need to commit them to memory.
Q: In a chemical reaction in which an alkane is being used as a reactant, how can we distinguish whether or not it is a cycloalkane? This is under the assumption that it makes a difference in the reaction , which it may very well not.
A: As we will see later, a cyclic substrate can have different chemical reactivity than a non-cyclic. These example will be with other functional groups present. There are relatively few reactions of alkanes and cycloalkanes and there is little difference in reactivity for these reactions. So, I guess the answer to your question is that it is not going to make much of a difference at this point. You can tell if a reactant is an alkane versus a cycloalkane from its chemical formula. An alkane has a general formula of C(n)H(2n+2) where a cycloalkane has the formula of C(n)H(2n).