Mechanism (show a mechanism using curved arrows..)

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 # 595

Phenol can be prepared from benzene and hydrogen peroxide in the presence of a really strong acid. Propose a mechanism for this reaction.

Solution

The electrophile we need for this EAS reaction is HO⁺. We can generate it by protonating one of the oxygens in peroxide, which causes it to act as a leaving group.

On the left side of this image I make a note about the structure of the electrophile. Some textbooks use electrophiles with formal positive charges, such as HO⁺. Other textbooks don't use formal positive charges, and instead use partial positive charges with a leaving group, such as HO-OH₂⁺. Using HO⁺as the electrophile is analogous to an S_N1 reaction, while using HO-OH₂⁺. is analogous to an S_N2 reaction. When you draw mechanisms, you should use whichever convention your textbook uses. Both lead to the same product.

Problem Details

MendelSet practice problem # 595 submitted by Matt on July 9, 2011.

Subject: Organic Chemistry

Subject Topic(s): Electrophilic Aromatic Substitution

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

Keyword(s): EAS mechanism

Problem # 587

Use curved arrows to draw a mechanism for the generic electrophilic aromatic substitution (EAS) reaction below.

Solution

Notice that the electrophile takes the place of the hydrogen.

Problem Details

MendelSet practice problem # 587 submitted by Matt on July 9, 2011.

Subject: Organic Chemistry

Subject Topic(s): Electrophilic Aromatic Substitution

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

Keyword(s): EAS mechanism

Problem # 581

Draw all products for the two reactions below.

The allylic alkene gives two products- the 1,2 product, and the 1,4 product.

However, the benzylic alkene only gives the one product (analogous to the 1,2 product), instead of multiple products (like the 1,4 product, 1,6 product, and 1,8 product). Why is this the case?

Solution

The allylic carbocation has two resonance forms, and so two positions where a nucleophile can attack, yielding two different products (1,2 and 1,4).

The benzylic carbocation has four resonance forms, and so in theory, four positions where a nucleophile can attack. But all but one of these resonance forms yield a product that is not aromatic, and so do not form.

Aromatic compounds are very stable; aromatic starting materials tend to only form products that are also aromatic.