Description: Unit 7 Homework: Please complete the EVEN problems in Chapters 9 and 10 of your Second Language text, and the following problems from Mendelset.
The 1H and 13C NMR spectra of a compound with chemical formula C10H14O are shown below. The compound's IR spectrum shows a broad peak at 3,300 cm-1. Determine the structure of this compound.
Let's go through the steps you should take to solve any NMR structure elucidation problem.
1.Are there any hints?
Yes. A broad IR peak around 3,300 cm-1 tells you this compound contains an alcohol.
2. How many IHD are there?
(indices of hydrogen deficiency are also called degrees of unsaturation or double bond equivalents, depending on the textbook.)
The formula is C10H14O. That's the same as C10H14 (oxygens don't change IHD count). A fully saturated compound has formula CnH2n+2, so a C10 molecule should have 22 hydrogens (2 x 10 + 2). The difference between C10H22 and C10H14 is 8 hydrogens, which corresponds to 4 IHD.
3.Draw some C10H14O structures with 4 IHD and eliminate, learn, repeat.
This isn't shown on the image below, but if you don't have an idea of the structure of the molecule, just start drawing structures! Never look at a blank page- just start drawing structures with the correct number of IHD (in this case, 4 IHD).
The peaks around ~7 ppm on the 1H NMR tell you its probably aromatic (also, benzene has 4 IHD) so that's a good place to start. Draw a few benzene candidate structures with formula C10H14O, but then eliminate structures that don't fit the data. How? I like to go through this check list:
Eliminate by number of signals: Do the candidate structures you drew give the proper number of 1H and 13C NMR signals? If not, eliminate!
Eliminate by multiplicity: Do your structures have the correct 1H NMR multiplicities? If not, eliminate!
Eliminate by integration: Do your structures have the correct 1H NMR integrations?If not, eliminate!
Eliminate by chemical shifts: Would the structures you drew have chemical shifts that are about the same as the chemical shifts in the 1H spectrum given in the problem?
That's the order in which I usually eliminate candidate structures- fastest method (determining the number of expected NMR signals) to slowest method (predicting chemical shifts).
Some things we know so far about this molecule:
Form the IR we know It contains an alcohol.
The gigantic singlet with integration of 9 screams tert-butyl
The total integration of the aromatic region (~7 ppm) is 4, which means this molecule is disubstituted benzene. They're both doublets with integration of 2 which points to a para-substitution pattern.
As you elminate incorrect structures you will learn what fits the data, and be able to draw better candidate structures. Then you can repeat this process. Once you have a structure that passes these four problems, you probably have the correct structure.
Notice how we didn't even really need the 13C NMR to answer this problem. The 1H NMR is usually enough.
I can't stress how important it is to just draw something! Even if you have no idea what the answer might be, don't even leave a blank page for an NMR problem. Just starting drawing out structures with the proper formula and IHD count!
MendelSet practice problem # 662 submitted by Matt on July 17, 2011.
The 1H and 13C NMR spectra of a compound with chemical formula C4H6O2 are shown below. The compound's IR spectrum shows a sharp peak at 1,700 cm-1. Determine the structure of this compound.
For a detailed explanation of the general strategy for solving NMR structure elucidation problems, see problem 662.
Here are the steps I would take to solve this problem:
1.Are there any any hints?
Yes. The sharp IR peak at 1,700 cm-1 tells you this molecule contains a carbonyl (C=O).
2.How many IHD are there? (also known as DBE or degrees of unsaturation)
C4H6O2 is the same as C4H6 (oxygens can be ignored) which if fully saturated would be C4H10 (from CnH2n+2).
So this compound is missing 4 H's (C4H10 - C4H6), which corresponds to 2 IHD.
3. Draw some C4H6O2 structures with 2 IHD and eliminate, learn, repeat.
Some things we know so far:
From the IR we know this molecule must have a carbonyl group. This "uses" 1C, 1O, and 1 IHD so we have 3C, 1O and 1 IHD remaining to work with).
What is the the remaining 1 IHD? It's probably not another carbonyl since the 13C NMR shows only one carbonyl peak (~170 ppm). It's probably not an alkene since we don't see any vinyl hydrogen peaks on the 1H NMR (~5-6 ppm). That leaves a ring.
Another clue that this molecule contains a ring is that we only see CH2's in the 1H NMR spectrum (all peaks have an integration of 2).
The carbonyl peak on the 13C NMR spectrum is around ~170 ppm, which means it's an oxidation state III carbonyl (ester) rather than an oxidation state II carbonyl (aldehyde or ketone).
Draw a few structures based on these clues and you'll arrive at the correct answer.
MendelSet practice problem # 663 submitted by Matt on July 17, 2011.
Using curved arrows, draw the mechanism for the SN2 reaction below.
Arrows in organic chemistry always go from regions of high electron density to regions of low electron density. Most of the time this means arrows start from negative charges and go towards positive charges.
Because bromine is electronegative, the carbon directly bonded to it (also known as the alpha carbon) has a partial positive charge, and can be attacked by a nucleophile such as azide (N3-).
Because this is an SN2 reaction, no carbocation is formed; as the nucleophile attacks the alpha carbon, the leaving group (Br-) leaves.
MendelSet practice problem # 534 submitted by Matt on July 2, 2011.
For the reaction below, draw the structures of the carbocation intermediate and the final product.
The delta (Δ) in the reaciton arrows means that heat is being added to this reaction, which tends to favor elimination over substitution. Also, the reaction is using a non-nucleophilic acid (H2SO4), which tends to favor elimination reactions (H3PO4 is another common reagent for E1 reactions, while HCl or HBr tend to go SN1).
Because this reaction is taking place in acid, a carbocation is likely to form, so this is an E1 reaction. Since water is lost over the course of the reaction, this is a dehydration, which is a type of elimination reaction.
MendelSet practice problem # 348 submitted by Matt on June 7, 2011.
Rank the following anions in order of decreasing stability (1 = most stable)
Charged molecules are generally less table than neutral ones, and each of the molecules below has a negatively charged oxygen.
But not all negative charges are equal; some oxygens are "closer to neutral" than others. How? Because resonance stabilizes charges by sharing electron density over multiple atoms (this is called electron delocalization).
Hydroxide (HO-) doesn't have resonance, so the oxygen has 100% of the negative charge to itself. On the other hand, the sulfonate ester (SO3R-) has three resonance forms, so each oxygen only has ~33% of a negative charge. So the sulfonate ester is the most stable anion.
In general, the more resonance forms a molecule has, the more stable it is.
MendelSet practice problem # 535 submitted by Matt on July 2, 2011.
Rank the following electrophiles in order of decreasing reactivity with NaN3 in DMF. (1 = most reactive)
The only differences among these three molecules is their leaving groups, so whichever has the best leaving group will react fastest with the nucleophile (azide, N3-). Stable molecules make good leaving groups, so the best leaving group of the three is SO3R-, which has has three resonance forms (see problem 535). So the third compound will react the fastest with NaN3.
MendelSet practice problem # 536 submitted by Matt on July 2, 2011.
Indicate the reagents necessary to carry out each transformation.
These are substitution reactions. In each case we're replacing the -OH with either -CN or -I, so the nucleophiles will be NaCN or NaI (SN2 conditions).
Hydroxide (HO-) is a poor leaving group, so the first step in each of these reactions will be to convert the alcohol into a better leaving group. Two good leaving groups are Br- and SO3R-, but which one to use?
Every time an SN2 reation takes place the wedge on the alpha carbon becomes a dash (and vice-versa).
For a), the wedge remains a wedge, so we have to do two SN2 reactions (wedge to dash to wedge again). So we use PBr3 to turn the OH into a Br. (Bromination of an alcohol with PBr3 is an SN2 reaction and so inverts stereochemistry).
For b), the wedge becomes a dash, so we can only do one SN2 reaction. So instead of using PBr3 to make the OH a better leaving group, we use RSO2Cl, which doesn't break the carbon-oxygen bond and so doesn't invert the stereochemistry.
MendelSet practice problem # 537 submitted by Matt on July 2, 2011.
Rank the following electrophiles in order of decreasing reactivity with NaN3 in DMF. (1 = most reactive)
This problem is similar to problem 536. The only difference among these compound is their leaving group, so the compound with the best leaving group will undergo substitution reactions most rapidly.
Good leaving groups are stable. Larger ions tend to be more stable than smaller atoms (due to a smaller charge:size ratio. See problem 288), so when going down the periodic table, stability increases. I- is more stable than Br-, which is more stable than Cl-, etc. So I- is the best leaving group, and 2-iodobutane will react the fastest with a nucleophile.
MendelSet practice problem # 538 submitted by Matt on July 3, 2011.
Rank the following compounds in order of decreasing nucleophilicity. (1 = most nucleophilic)
Nucleophilicity increases with electron density; negatively charged molecules are more nucleophilic than neutral molecules. So HS- is the best nucleophile in this group.
Nucleophilicity also increases with size. This is because the larger an atom, the more polarizable it is. Size increases as we go down the periodic table, so H2S is a better nucleophile than H2O.
MendelSet practice problem # 539 submitted by Matt on July 3, 2011.
Rank the following anions in order of decreasing stability (1 = most stable)
Each of these ions has a charge of -1, and none of them has any resonance forms. So the two things to consider are electronegativity and size. Electronegativity is relevant as we go from left to right across the periodic table. But here we are going up to down on the periodic table, so size is relavant. Larger ions are more stable than smaller ions (due to a smaller charge:size ratio), so I- is the most stable of the group.
MendelSet practice problem # 540 submitted by Matt on July 3, 2011.
Rank the following compounds in order of decreasing reactivity with NaI in acetone. (1 = most reactive)
The only differences among these compounds is the substitution of the alpha carbon (methly, 1º, 2º, or 3º).
This is an SN2 reaction. We know this because NaI is an SN2 reagent- charges give it away; when we see charges (Na+ is positive and I- is negative) it's probably an SN2 reaction. So we want reactants that are less substituted: methyl is more reactive than 1º, which is more reactive than 2º, etc. So methyl bromide reacts the fastest with NaI.
MendelSet practice problem # 541 submitted by Matt on July 3, 2011.
Rank the following compounds in order of decreasing reactivity with water (solvolysis). (1 = most reactive)
This is an SN1 reaction. We know this because none of the reagents have charges (H2O is neutral; if it were HO-, it would probably be SN2 or E2). So a carbocation will be formed in this reaction, and the compound whose carbocation is the most stable will react the fastest. So the 3º bromide will react fastest with water.
MendelSet practice problem # 542 submitted by Matt on July 3, 2011.
For each reaction below, determine whether the primary reaction is SN1, SN2, E1, or E2, and then draw the product.
Note: Me = methyl (CH3)
Predicting SN1/SN2/E1/E2 competition reactions tends to drive students crazy, but it's not so bad once you notice the general pattern:
basic conditions (positive and negative charges) tend to go SN2 or E2 (no carbocation)
neutral or acidic conditions tend to go SN1 or E1 (carbocation is formed).
That's how you determine a SN1/E1 reaction from an SN2/E2 reaction. But how to decide between substitution or elimination? General things to watch for are bulk, nucleophilicity, and heat:
If you see heat (or Δ), the reaction will go elimination.
If you see a big, bulky compound, the reaction will go elmination.
If you see a strong base, the reaction will go elimination. Strong base is anything stronger than RO-.
The exception: if everything is primary, it will probably go SN2.
These rules probably seem confusing, so let's go through these eight examples and see how they apply.
a) NaCN is charged! (Na+ and CN-), so it's SN2 or E2. CN is not a strong base, so it's SN2.
b) KOtBu (potassium tert-butoxide) is charged, so it's SN2 or E2. -OtBu is a strong base, so if anything is more bulky than 1º it will go E2. -OtBu is 3º, so it will definitely go E2 (KOtBu is a classic E2 reagent).
c) NaOMe is charged so E2 or SN2. NaOMe is a strong base, so if anything >1º it will go E2. -OMe is 1º (actually, not even 1º), but the alkyl halide is 2º, so it will go E2.
d) NaOMe is charged so E2 or SN2. NaOMe is a strong base, so if anything >1º it will go E2. But in this case there is no bulk whatsoever- nothing is >1º! NaOMe is 1 and the alkyl halide is also 1º, so it will go SN2.
e) Methanol (MeOH) is neutral so probably E1 or SN1. Methanol is a weak base and there's no bulk, so SN1. In general water and alcohol do a mixture of SN1 and E1 with alkyl halides (mostly SN1).
f) H2SO4 is acidic so probably E1 or SN1. Can't be SN1 though because there is no nucleophile in H2SO4. (HSO4- is a very weak nucleophile). An alcohol with H2SO4 or H3PO4 is a dehydration reaction- E1.
g) H2SO4 is acidic so probably E1 or SN1. In this case we have a nucleophile- Cl-, so it will go SN1.
h) Amines are neutral but they don't so SN1/E1- they tend to go SN2/E2, because they are basic (an amine solution has a basic pH). This amine is really bulky so it will go E2.
MendelSet practice problem # 560 submitted by Matt on July 7, 2011.
Show two ways to prepare the ether below from a combination of an alcohol and an alkyl halide via the Williamson ether synthesis.
Is one way better than the other? Why?
The Williamson ether synthesis takes place in two steps. First an alcohol is deprotonated to form a strong nucleophile (RO-, this step isn't shown in the image below). Then the alkoxide (negative alcohol) attacks an alkyl halide in an SN2 reaction.
So this problem is really asking, which step of conditions is most favorable for an SN2 reaction?
Recall that SN2 reactions compete with E2 reactions. If the nucleophile is too basic, or if there is too much bulk, it will go E2 instead of SN2. (See problem 560 for a full explanation of these competition reactions)
Below, the top combination uses the less substituted (1º) alkyl halide, and so is the best for an SN2 reaction.
The bottom reaction uses a bulkier (2º) alkyl halide, and will probably give a higher percentage of E2 side reaction.
MendelSet practice problem # 703 submitted by Matt on July 21, 2011.