MS 934 - Carey and Giuliano 8th Ed. CH 19

Submitted by Matt on August 7, 2011.

Textbook and Chapter: Carey and Giuliano 8th Ed. (2010), Chapter 19

Keyword: nucleophilic acyl substitution

Description: Goes over carboxylic acid derivatives and the nucleophilic acyl substitution mechanism. Includes synthesis problems.

Total Problems: 6

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Problem # 726

Show two esters that would yield the two alcohols below after treatment with lithium aluminum hydride.
Solution
Adding LiAlH₄ to an ester yields two alcohols. The carbonyl "side" becomes one alcohol, and the ester "side" becomes the other alcohol.

So there are two esters that would yield these two alcohols after reduction with LiAlH₄.

Problem Details
MendelSet practice problem # 726 submitted by Matt on July 24, 2011.
Subject: Organic Chemistry
Subject Topic(s): Nucleophilic Acyl Substitution (Carboxylic Acid Derivatives)
Question Type: Concept (explain why this is so..)
Keyword(s): nucleophilic acyl substitution
Problem # 727

Show how the ester below can be prepared from propene.
Solution
Esters are prepared from a carboyxlic acid and an alcohol (Fischer esterification).

So the problem becomes, "how can you prepare this ester from an alochol and a carboxylic acid, each containing three carbons? (propene)"

The "pair" of carboxylic acid and alcohol to make this ester is propanoic acid and propanol.

From propene, to make propanol, just do an anti-Markovnikov H₂O addition using borane (BH₃ or B₂H₆) then hydrogen peroxide (H₂O₂).

To make propanoic acid, oxidize propanol to propanoic acid using Jones reagent (aqueous chromium).

You can also convert propanoic acid to its acid chloride using SOCl₂, and then add propanol.

Problem Details
MendelSet practice problem # 727 submitted by Matt on July 24, 2011.
Subject: Organic Chemistry
Subject Topic(s): Nucleophilic Acyl Substitution (Carboxylic Acid Derivatives)
Keyword(s): nucleophilic acyl substitution, Fischer esterification
Problem # 728

The acyl group is a protecting group for amines. Amines can be acylated using acetic anhydride, and deacylated with base.

Propose a mechanism for each reaction.
Solution
These reactions are both nucleophilic acyl substitution under basic conditions. See problem 705 for the mechanism in detail.

Problem Details
MendelSet practice problem # 728 submitted by Matt on July 24, 2011.
Subject: Organic Chemistry
Subject Topic(s): Nucleophilic Acyl Substitution (Carboxylic Acid Derivatives), Amines (Basicity, Hoffman Elimination, Gabriel Amine Synthesis)
Question Type: Mechanism (show a mechanism using curved arrows..)
Keyword(s): nucleophilic acyl substitution
Problem # 729

The ester below was dissolved in a solution of water, a small amount of which was isotopically labeled with O-18, denoted with an asterisk.

After a few hours, some isotopically labeled oxygen was found in the ester. Where was it found in the ester? Can you explain why?
Solution
To answer this problem, you must be familiar with the nucleophilic acyl substitution mechanism. (see problem 725)

Water reacts with an ester to form a carboxylic acid. This is what happens here.

If there is only trace amounts of water, it's possible that the water reacts with the ester to form a carboxylic acid, and then goes back to reform the ester. But, in this process, leaves an isotopically labeled oxygen in the carbonyl position.

Problem Details
MendelSet practice problem # 729 submitted by Matt on July 24, 2011.
Subject: Organic Chemistry
Subject Topic(s): Nucleophilic Acyl Substitution (Carboxylic Acid Derivatives)
Question Type: Mechanism (show a mechanism using curved arrows..)
Keyword(s): nucleophilic acyl substitution, Fischer esterification
Problem # 730

N,N-dimethylformamide (DMF) is shown below. Based on its structure, you might expect to see only one -CH₃ signal in the ¹H NMR spectrum. But instead DMF shows two different -CH₃ signals. Explain.
Solution
DMF appears to have two identical methyl groups. Since these six protons are all equivalent, its ¹H NMR should only show one methyl signal (singlet, 6H).

So why is that the real life the ¹H NMR of DMF shows two methyl signals? (two singlets with integration of 3H).

Because DMF is an amide.Recall that the "real" structure of molecule is the a mixture of its resonance forms. DMF doesn't look like either of the two resonance forms below. In real life, its somewhere in between.

For most carboxylic acid derivatives (such as esters), the resonance form is only a minor contributor and so the real "picture" looks very close to the carbonyl Lewis structure.

But for amides, its resonance form is fairly stable (it's common for nitrogen atoms to be positively charged), and so is a major resonance contributor.

In an amide, the bond between the carbonyl carbon and the nitrogen atom has a high degree of double bond character. (This also explains why it's harder to rotate the C-N "single bond" than you would expect from its Lewis structure- it's sort of like a double bond).

Because the C-N "single bond" is closer to a double bond, the two methyls are not equivalent. One methyl is cis, and the other is trans, and so they show two signals in the ¹H NMR.

Problem Details
MendelSet practice problem # 730 submitted by Matt on July 24, 2011.
Subject: Organic Chemistry
Subject Topic(s): Nuclear Magnetic Resonance Spectroscopy (NMR), Nucleophilic Acyl Substitution (Carboxylic Acid Derivatives), Amino Acids, Peptides, and Proteins
Question Type: Concept (explain why this is so..)
Keyword(s): proton NMR, amides
Problem # 735

Show how each ketone below can be prepared from sodium cyanide and either ethylene or propene.

You may also use methyl Grignard and ethylene oxide.
Solution
Nitriles react with Grignard reagents to form ketones, so the carbonyl carbon on each product must have came from a nitrile.

You also have the restriction that the longest carbon chain building must be three carbons.

a) Working backwards, the left side must have come from a nitrile, so the left side came from a Grignard.

How can we make each starting from propene?

To make the Grignard, prepare propyl bromide from propene via anti-Markovnikov HBr addition (HBr/peroxides), and then add Mg⁰/ether.

To make the nitrile, add sodium cyanide (NaCN) to propyl bromide.

Then add the nitrile and Grignard together, which forms the imine intermediate, which will hydrolyze to the ketone after you add water (H₃O⁺).

b) This problem is harder because you're limited to starting materials with three carbons or less, so we have to make three "cuts" instead of two, like in a).

Working backwards again, prepare propyl Grignard from propene as described in a), and then add ethylene oxide to form 1-pentanol. Form the 6-carbon nitrile by first converting the alcohol to a bromide using PBr₃, and then adding NaCN. Finally, add methyl Grignard followed by H₃O⁺ to form the desired ketone.

Problem Details
MendelSet practice problem # 735 submitted by Matt on July 25, 2011.
Subject: Organic Chemistry
Subject Topic(s): Nitriles
Question Type: Synthesis (show how to prepare X from Y..)
Keyword(s): carboxylic acids, nitriles

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MS 934 - Carey and Giuliano 8th Ed. CH 19

MS 934 - Carey and Giuliano 8th Ed. CH 19

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