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Show the mechanism for formation of a Boc derivative by reaction of an amino acid with di-tert-butyl dicarbonate.

Short Answer

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The amino acid reacts with di-tert-butyl dicarbonate to form a Boc-protected amino acid.

Step by step solution

01

Understanding the Reactants

We begin by identifying the reactants involved in the formation of a Boc derivative. The two reactants are an amino acid and di-tert-butyl dicarbonate (Boc anhydride). An amino acid has an amino group (-NH2) that will react with Boc anhydride to form the Boc-protected compound.
02

Nucleophilic Attack on Boc Anhydride

The amino group of the amino acid acts as a nucleophile and attacks the carbonyl carbon of one of the esters in di-tert-butyl dicarbonate. This forms a tetrahedral intermediate. This step leads to the addition of the amino group to the Boc anhydride.
03

Formation of the Boc-Protected Intermediate

The tetrahedral intermediate collapses, and one of the t-butyl groups of the Boc anhydride leaves as t-butanol (t-BuOH). This step creates an acyl substitution product, where the carbonyl carbon is now bonded to the nitrogen, forming the Boc-protected amino acid.
04

Product Formation and Final Adjustment

The newly formed Boc-protected amino acid may equilibrate any tautomeric forms, but the significant result is the addition of the Boc group to the amino acid. The final product is a Boc-protected amino acid, ready for use in further peptide synthesis or other applications.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Amino Acid Protection
For chemists working with amino acids, protecting the functional groups is a crucial step in many synthetic processes. Amino acids have both an amino group and a carboxyl group, making them susceptible to reactions. In peptide synthesis, we often need to "protect" one group to avoid unwanted reactions while the other is being modified.
- Boc (tert-butoxycarbonyl) protection is a popular method to shield the amino group during synthesis. - The use of Boc is advantageous due to its stability under mildly acidic conditions and its easy removal under acidic conditions.
This way, reactions can be carried out on other parts of the molecule without disturbing the protected amino group.
Nucleophilic Attack
The concept of nucleophilic attack is central to understanding chemical reaction mechanisms. A nucleophile is an atom or molecule with an excess of electrons, making it attracted to positively charged, electron-deficient centers. In our Boc amino acid protection example, the amino group (-NH2) acts as a nucleophile.
- It attacks the electrophilic carbon in the carbonyl group of di-tert-butyl dicarbonate. - This crucial first step leads to the formation of a new covalent bond, setting the stage for subsequent reaction steps. Understanding nucleophilic attack helps demystify how molecules interact and transform during chemical reactions.
Tetrahedral Intermediate
During the process of forming a Boc-protected amino acid, the reaction passes through a tetrahedral intermediate stage. This form is a transient structure that appears right after the nucleophilic attack.
- As the amino group attacks the carbonyl carbon of Boc anhydride, the carbon atom temporarily has four bonds instead of the typical three. - This intermediate is called "tetrahedral" because the central carbon has a geometry resembling a tetrahedron. As short-lived as it is, this intermediate plays a pivotal role in the rearrangement of bonds needed for the final product formation.
Acyl Substitution
Acyl substitution is a type of reaction that occurs after the tetrahedral intermediate is formed. This involves the replacement of one group (leaving group) attached to an acyl carbon with another group. In our context: - The tetrahedral intermediate formed after the nucleophilic attack on Boc anhydride collapses. - One of the t-butyl groups leaves as t-butanol, a process that results in the formation of a carbon-nitrogen bond.
This transformation results in the stabilization of the reaction's final product, the Boc-protected amino acid, ready for further synthetic steps in peptide chain creation.

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Most popular questions from this chapter

Proline has \(\mathrm{pK}_{\mathrm{a} 1}=1.99\) and \(\mathrm{p} K_{\mathrm{a} 2}=10.60 .\) Use the HendersonHasselbalch equation to calculate the ratio of protonated and neutral forms at \(\mathrm{pH}=2.50 .\) Calculate the ratio of neutral and deprotonated forms at \(\mathrm{pH}=9.70\)

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Cysteine is the only amino acid that has \(\mathrm{L}\) stereochemistry but an \(R\) configuration. Make up a structure for another L. amino acid of your own creation that also has an \(R\) configuration.

What alkyl halides would you use to prepare the following \(\alpha\) -amino acids by the amidomalonate method? (a) Leucine (b) Histidine (c) Tryptophan (d) Methionine

Proteins can be cleaved specifically at the amide bond on the carboxyl side of methionine residues by reaction with cyanogen bromide, \(\mathrm{BrC}=\mathrm{N}\). The reaction occurs in several steps: (a) The first step is a nucleophilic substitution reaction of the sulfur on the methionine side chain with BrCN to give a cyanosulfonium ion, \(\left[\mathrm{R}_{2} \mathrm{SCN}\right]^{+} .\) Show the structure of the product, and propose a mechanism for the reaction. (b) The second step is an internal \(S_{N}^{2}\) reaction, with the carbonyl oxygen of the methionine residue displacing the positively charged sulfur leaving group and forming a five-membered ring product. Show the structure of the product and the mechanism of its formation. (c) The third step is a hydrolysis reaction to split the peptide chain. The carboxyl group of the former methionine residue is now part of a lactone (cyclic ester) ring. Show the structure of the lactone product and the mechanism of its formation. (d) The final step is a hydrolysis of the lactone to give the product shown. Show the mechanism of the reaction.

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