/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 29 Draw a structural formula for th... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 27: Problem 29

Draw a structural formula for the product formed when alanine is treated with the following reagents. (a) Aqueous \(\mathrm{NaOH}\) (b) Aqueous \(\mathrm{HCl}\) (c) \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}, \mathrm{H}_{2} \mathrm{SO}_{4}\) (d) \(\left(\mathrm{CH}_{3} \mathrm{CO}\right)_{2} \mathrm{O}, \mathrm{CH}_{3} \mathrm{COONa}\)

Short Answer

Expert verified
Question: Draw the structural formulas for the products formed when alanine is treated with each of the following reagents: (a) Aqueous NaOH, (b) Aqueous HCl, (c) CH3CH2OH, H2SO4, and (d) (CH3CO)2O, CH3COONa. Answer: (a) CH3CH(NH2)COO- (associated with Na+) (b) CH3CH(NH3+)COOH (associated with Cl-) (c) CH3CH(NH2)COOCH2CH3 (d) CH3CH(NH2)CONHCH3

Step by step solution

01

(Step 1: Understand the Structure of Alanine)

Alanine is an amino acid with the following structure: CH3CH(NH2)COOH. It has an amine group (NH2) and a carboxyl group (COOH) along with a methyl group (CH3) as its side chain. Begin by drawing the structural formula for alanine.
02

(Step 2: Treatment with Aqueous NaOH)

Aqueous NaOH is a strong base that has a tendency to deprotonate acidic protons. In alanine, the acidic protons are in the carboxyl group (COOH). NaOH will remove one hydrogen from the carboxyl group, forming the carboxylate anion and a molecule of water. The product will have the structure: CH3CH(NH2)COO-. A sodium ion from NaOH will be associated with this structure.
03

(Step 3: Treatment with Aqueous HCl)

Aqueous \(\mathrm{HCl}\) is a strong acid that will protonate the amine group of alanine. The lone pair of electrons on the nitrogen atom of the amine group (NH2) will accept a proton from the HCl, forming a positively charged ammonium ion. The product will have the structure: \(\mathrm{CH}_{3} \mathrm{CH}(\mathrm{NH}_{3}^{+})\mathrm{COOH}\). A chloride ion from HCl will be associated with the ammonium group in the structure.
04

(Step 4: Treatment with CH3CH2OH, H2SO4)

\(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}\) (ethanol) and \(\mathrm{H}_{2} \mathrm{SO}_{4}\) (sulfuric acid) will form an esterification reaction between the carboxyl group of alanine and the hydroxyl group of ethanol. Remove the water molecule formed and combine the remaining structures to form the ester product. The product's structure will be: \(\mathrm{CH}_{3} \mathrm{CH}(\mathrm{NH}_{2})\mathrm{COOCH}_{2} \mathrm{CH}_{3}\)
05

(Step 5: Treatment with (CH3CO)2O, CH3COONa)

The reaction of alanine with \(\left(\mathrm{CH}_{3} \mathrm{CO}\right)_{2} \mathrm{O}\) (acetic anhydride) and \(\mathrm{CH}_{3}\mathrm{COONa}\) (sodium acetate) is a non-aqueous amide formation. The carboxyl group of alanine reacts with acetic anhydride to form an amide bond and releases acetic acid as a byproduct. The product's structure will be: \(\mathrm{CH}_{3} \mathrm{CH}(\mathrm{NH}_{2})\mathrm{CONHCH}_{3}\) The product structures formed are: (a) CH3CH(NH2)COO- (associated with Na+) (b) CH3CH(NH3+)COOH (associated with Cl-) (c) CH3CH(NH2)COOCH2CH3 (d) CH3CH(NH2)CONHCH3

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Deprotonation
In chemistry, deprotonation is the removal of a proton (H鈦�) from a molecule, making it a significant process in acid-base reactions. When alanine, an amino acid, is treated with a strong base like sodium hydroxide (NaOH), deprotonation occurs at the carboxyl group (COOH).
The carboxyl group has a weakly acidic proton that can be taken away by the base:
  • The proton (H鈦�) is removed from the carboxyl group.
  • This loss of a proton forms a carboxylate anion (COO鈦�).
The overall reaction ends with the sodium ion (Na鈦�) associating with the carboxylate, forming CH鈧僀H(NH鈧�)COO鈦籒a鈦�. This ionic product highlights how deprotonation can alter an amino acid, impacting its solubility and reactivity.
Protonation
Protonation is the addition of a proton (H鈦�) to a molecule, typically affecting amine groups in amino acids. When alanine is subjected to an acidic environment using hydrochloric acid (HCl), the amine group undergoes protonation.
The nitrogen in the amine group (NH鈧�) has a lone pair of electrons that can bond with a proton:
  • The lone pair on nitrogen accepts a proton to form NH鈧冣伜.
  • This creates a positive charge on the nitrogen, forming an ammonium ion.
The resulting structure, CH鈧僀H(NH鈧冣伜)COOHCl鈦�, sees a chloride ion associating with the positive ammonium group. Protonation transforms the physical and chemical properties of alanine, potentially affecting its function in biochemical reactions.
Esterification
Esterification is a chemical reaction that forms an ester from an acid and an alcohol. In this process with alanine, ethanol ( CH鈧僀H鈧侽H) in the presence of sulfuric acid ( H鈧係O鈧�) interacts with the carboxyl group of alanine. This reaction involves the combination of two groups after the release of a water molecule:
  • The carboxyl group (COOH) binds with the hydroxyl group (OH) of ethanol.
  • The resulting compound is an ester, CH鈧僀H(NH鈧�)COOCH鈧侰H鈧�.
Esterification forms a covalent bond and is crucial in organic chemistry for creating diverse molecules with functions varying from aroma-producing compounds to biological molecules.
Amide Formation
Amide formation involves creating an amide bond between a carboxyl group and an amine. When alanine reacts with acetic anhydride ( (CH鈧僀O)鈧侽) and sodium acetate (CH鈧僀OONa), an amide bond forms:
  • Acetic anhydride acts as a reagent, facilitating the removal of water from the reactants.
  • The amino group in alanine forms a new bond with a part of the acetic anhydride.
  • This process yields an amide product, CH鈧僀H(NH鈧�)CONHCH鈧�, and acetic acid as a byproduct.
Amide formation is a pivotal reaction in synthesizing proteins and peptides, as amide bonds are foundational links in these biological polymers.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

The isoelectric point of histidine is \(7.64\). Toward which electrode does histidine migrate on paper electrophoresis at pH 7.0?

Draw a structural formula for Lys-Phe-Ala. Label the \(N\)-terminal amino acid and the \(C\)-terminal amino acid. What is the net charge on this tripeptide at \(\mathrm{pH} 6.0\) ?

Examine the \(\alpha\)-helix conformation. Are amino acid side chains arranged all inside the helix, all outside the helix, or randomly?

Why is Arg often referred to as a basic amino acid? Which two other amino acids are also basic amino acids?

Do the following compounds migrate to the cathode or to the anode on electrophoresis at the specified pH? (a) Histidine at \(\mathrm{pH} 6.8\) (b) Lysine at \(\mathrm{pH} 6.8\) (c) Glutamic acid at \(\mathrm{pH} 4.0\) (d) Glutamine at pH \(4.0\) (e) Glu-Ile-Val at pH \(6.0\) (f) Lys-Gln-Tyr at \(p \mathrm{H} 6.0\)
See all solutions

Recommended explanations on Chemistry Textbooks

Chemistry Branches

Read Explanation

Chemical Analysis

Read Explanation

Making Measurements

Read Explanation

Kinetics

Read Explanation

Physical Chemistry

Read Explanation

Ionic and Molecular Compounds

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.