Organic Chemistry

Problem # 661

The proton NMR of cyclohexane gives only one peak when the NMR is run at room temperature.

But when the temperature is lowered to -100 ºC the proton NMR spectrum shows two peaks. Explain.

Every proton on cyclohexane appears identical, but remember than cyclohexane is usually in a chair form, so there are really two types of protons: axial and equatorial.

In theory each of these two protons should give its own NMR peak. But at room temperature the molecule is undergoing chair flip so rapidly the two peaks converge into one.

But at low temperature the rate of chair flip slows down enough that two distinct peaks emerge.

Problem Details

MendelSet practice problem # 661 submitted by Matt on July 17, 2011.

Subject: Organic Chemistry

Subject Topic(s): Nuclear Magnetic Resonance Spectroscopy (NMR)

Question Type: Concept (explain why this is so..)

Keyword(s): cyclohexane chair forms, proton NMR

Problem # 660

Using your knowledge of ¹H NMR, predict the NMR spectrum for the compound below. (draw out the spectrum you would expect to see). Be sure to include:

peak integrations
peak multiplicities
chemical shifts (approximate)

Solution

There are four types of protons (hydrogens) in the compound below, so we would expect to see four peaks (A, B, C, and D).

A is not adjacent to any other types of protons, so its multiplicity is 1, also called singlet (s). ( n=0, so n + 1 = 0 + 1 = 1). A also is a CH3, so it will have an integration of 3. Finally, A's proton are on a carbon adjacent to an oxygen, so its chemical shit will be about ~4 ppm.

B is a CH2 so its integration is 2. It's adjacent to a 2 hydrogens (a CH₂) so its multiplicty is 3, also called a triplet (t). (n=2, so n + 1 = 3). B is also adjacent to oxygen, so it will be close to ~4 ppm, but a little further downfield than A because it's more substituted than A.

C is also a CH2 so its integration is 2. It's also adjacent to 2 hydrogens so it's a triplet (n+1 rule). C's hydrogens are on a carbon adjacent to a carbonyl (C=O), so its chemical shift will be around ~2 ppm.

D is an aldehyde proton, which is always a singlet with near ~10 ppm. It's only one proton, so its integration is 1.

Note: notice that the aldehyde proton doesn't couple to another other protons! So even through C is surrounded by a CH₂ on its left and an aldehyde proton on its right, when determining multiplicity we only count the CH₂, and not the aldehyde proton.

Problem Details

MendelSet practice problem # 660 submitted by Matt on July 17, 2011.

Subject: Organic Chemistry

Subject Topic(s): Nuclear Magnetic Resonance Spectroscopy (NMR)

Question Type: Concept (explain why this is so..)

Keyword(s): proton NMR

Problem # 659

The mass spec of 4-nonanone shows peaks at m/z = 58, 71, 86, 99.

Using curved arrows or hooks, show how each of these fragments can form via alpha cleavage or the McLafferty rearrangement. (and draw the structure of the indicated species in the appropriate box).

Solution

Note that I drew the McLafferty rearrangements using arrows (2 electrons moving at once). Some textbooks use hooks instead, but the results are the same.

In most ungraduate organic chemistry courses, being able to draw an alpha cleavage is much more important than a being able to draw a McLafferty rearrangement (which tends to only show up on bonus problems).

Problem Details

MendelSet practice problem # 659 submitted by Matt on July 17, 2011.

Subject: Organic Chemistry

Subject Topic(s): Mass Spectrometry (MS)

Question Type: Mechanism (show a mechanism using curved arrows..)

Keyword(s): mass spec, alpha cleavage, McLafferty rearrangement

Problem # 658

The mass spec of methyl ethyl ether shows peaks at m/z = 45 and 59.

Use hooks to show the alpha cleavages that result in these two fragments.

Solution

MS ionization will knock off an electron from the heteroatom (atom that's not C or H), in this case, the oxygen, leaving behind a positively charge compound. This is the molecular ion (M⁺).

Oxygen usually undergoes homolytic cleavage- the bond splits and each atom gets one electron. Since only one electorn is involed, we use hooks instead of the usual curved arrows.

If your professor is into mass spec cleavage mechanisms, I suggest you practice this mechanism, called an alpha cleavage (see image below).

Problem Details

MendelSet practice problem # 658 submitted by Matt on July 17, 2011.

Subject: Organic Chemistry

Subject Topic(s): Mass Spectrometry (MS)

Question Type: Mechanism (show a mechanism using curved arrows..)

Keyword(s): mass spec, homolytic cleavage, alpha cleavage

Problem # 657

The mass spec of chlorocyclohexane shows a peak at m/z = 83.

Use curved arrows to show the heterolytic cleavage that accounts for this fragment.

Solution

Heteroatoms (atoms that are not C or H) are always the most likely atoms to lose an electron during mass spec, and this will be no exception. Mass spec will blow off an electron from the chlorine, leaving behind the molecular ion (M⁺). But the molecular ion will then fragment.

The problem told us it was a heterolytic cleavage, in which both electrons form the bond go towards the positive charge. Because two electrons are involved, we use curved arrows as usual.

This results in a neutral chlorine radical and a carbocation with formula C₆H₁₁, for a total mass of 83 (6 x 12 + 11 x 1). Note that the chlorine radical doesn't give an MS peak because it is neutral.

Problem Details

MendelSet practice problem # 657 submitted by Matt on July 17, 2011.

Subject: Organic Chemistry

Subject Topic(s): Mass Spectrometry (MS)

Question Type: Mechanism (show a mechanism using curved arrows..)

Keyword(s): mass spec, heterolytic cleavage

Problem # 616

Let's go through another way to make benzyne.

First, let's form a Grignard reagent. Then, let's elminate to form benzyne.

Solution

Halogens get more reactive towards Grignard formation as we go down the periodic table, so Mg⁰ will form a Grignard with Br before F.

Grignards behave like negatively charged carbon atoms- very unstable. So even though fluorine is a poor leaving group, the negative carbon will cause it to eliminate to form benzyne.

Problem Details

MendelSet practice problem # 616 submitted by Matt on July 10, 2011.

Subject: Organic Chemistry

Subject Topic(s): Nucleophilic Aromatic Substitution and Benzyne

Question Type: Mechanism (show a mechanism using curved arrows..)

Keyword(s): benzyne, Grignard

Problem # 615

Let's go through a benzyne reaction (also called elimination-addition).

In the reaction below, the strong base (NaNH₂) will form a benzyne intermediate, which when forms either ortho nitroaniline or meta nitroaniline.

Used curved arrows to show the formation of each intermediate and the final products.

Solution

First the ^-NH₂ acts as a base to eliminate a leaving group and form Benzyne. Then the ^-NH₂ acts as a nucleophile and attacks the benzyne, and the "leaving group" is the triple bond.

Problem Details

MendelSet practice problem # 615 submitted by Matt on July 10, 2011.

Subject: Organic Chemistry

Subject Topic(s): Nucleophilic Aromatic Substitution and Benzyne

Question Type: Mechanism (show a mechanism using curved arrows..)

Keyword(s): benzyne

Problem # 612

Draw a mechanism for the nucleophilic aromatic substitution (S_NAr) reaction below. Show all resonance forms of the intermediate.

Solution

For S_NAr to work you need to have electron withdrawing groups (EWG) either ortho or para to the leaving group.

This molecule has two leaving groups (chlorines), but only one chlorine has EWG ortho/para to it. So that's the carbon where the nucleophile (NH₃) will attack.

Why do EWG need to be ortho/para? Because that's the only way for the negative charge to be stabilized by resonance. Try it- if you have the NH₃ attack the other carbon with a chlorine, you will not be able to draw a resonance form that places the negative charge on one of the EWG.

Problem Details

MendelSet practice problem # 612 submitted by Matt on July 10, 2011.

Subject: Organic Chemistry

Subject Topic(s): Nucleophilic Aromatic Substitution and Benzyne

Question Type: Mechanism (show a mechanism using curved arrows..)

Keyword(s): SNAr

Problem # 611

Draw in the arrows to show the electron flow and resonance forms in the nucleophilic aromatic substitution reaction below.

Note: Depending on the textbook, nucleophilic aromatic substitution is referred to as NAS, S_NAr, or addition-elimination.

Solution

SNAr is sort of like SN2, except the leaving group doesn't leave right away; a tetrahedral intermediate is formed first.

The trick with SNAr (or NAS, addition-elimination, etc.) is to draw resonance forms that stabilizes the negative charge that forms. That's why EWG increase the rate of S_NAr (they stabilize negative charges).

Problem Details

MendelSet practice problem # 611 submitted by Matt on July 10, 2011.

Subject: Organic Chemistry

Subject Topic(s): Nucleophilic Aromatic Substitution and Benzyne

Question Type: Mechanism (show a mechanism using curved arrows..)

Keyword(s): SNAr

Problem # 596

Indicate the eletrophile formed by each set of reagents/conditions below.

Solution

Note that some textbooks don't draw EAS electrophiles with formal positive charges.

Some textbooks instead draw electrophiles with partial positive charges and a leaving group. It's just a convention, and doesn't affect the product of an EAS reaction. See the problem 595 for an explanation of the difference.