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Chapter 20: Q20P (page 1038)

Predict the major products formed when the following amines undergo exhaustive methylation, treatment with Ag2O, and heating.

  1. Hexan-2-amine
  2. 2-methylpiperidine
  3. N-ethylpiperidine

5.

6.

Short Answer

Expert verified

(a)

(b)

(c)

(d)

(e)

(f)

Step by step solution

01

Step-1. Explanation of part (a):

The process of producing tertiary amines and alkenes from quaternary ammonium is called Hofmann elimination. This process can also be referred to as exhaustive methylation. The Hofmann rule states that the major alkene product is the least substituted and least stable product when it comes to asymmetrical amines. Methyl iodide is used in excess because it has no beta hydrogens and therefore cannot compete in an elimination reaction. Silver oxide ion is used for deprotonation of water to form a hydroxide ion.

In part (a), Hofmann exhaustive methylation of the given reactant produces the following product as shown.

Formation of the product

02

Step-2. Explanation of part (b):

The process of producing tertiary amines and alkenes from quaternary ammonium is called Hofmann elimination. This process can also be referred to as exhaustive methylation. The Hofmann rule states that the major alkene product is the least substituted and least stable product when it comes to asymmetrical amines. Methyl iodide is used in excess because it has no beta hydrogens and therefore cannot compete in an elimination reaction. Silver oxide ion is used for deprotonation of water to form a hydroxide ion.

In part (b), Hofmann exhaustive methylation of the given reactant produces the following product as shown.

Formation of the product

03

Step-3. Explanation of part (c):

The process of producing tertiary amines and alkenes from quaternary ammonium is called Hofmann elimination. This process can also be referred to as exhaustive methylation. The Hofmann rule states that the major alkene product is the least substituted and least stable product when it comes to asymmetrical amines. Methyl iodide is used in excess because it has no beta hydrogens and therefore cannot compete in an elimination reaction. Silver oxide ion is used for deprotonation of water to form a hydroxide ion.

In part (c), Hofmann exhaustive methylation of the given reactant produces the following product as shown.

Formation of the product

04

Step-4. Explanation of part (d):

The process of producing tertiary amines and alkenes from quaternary ammonium is called Hofmann elimination. This process can also be referred to as exhaustive methylation. The Hofmann rule states that the major alkene product is the least substituted and least stable product when it comes to asymmetrical amines. Methyl iodide is used in excess because it has no beta hydrogens and therefore cannot compete in an elimination reaction. Silver oxide ion is used for deprotonation of water to form a hydroxide ion.

In part (d), Hofmann exhaustive methylation of the given reactant produces the following product as shown.

Formation of the product

05

Step-5. Explanation of part (e):

The process of producing tertiary amines and alkenes from quaternary ammonium is called Hofmann elimination. This process can also be referred to as exhaustive methylation. The Hofmann rule states that the major alkene product is the least substituted and least stable product when it comes to asymmetrical amines. Methyl iodide is used in excess because it has no beta hydrogens and therefore cannot compete in an elimination reaction. Silver oxide ion is used for deprotonation of water to form a hydroxide ion.

In part (e), Hofmann exhaustive methylation of the given reactant produces the following product as shown.

Formation of the product

06

Step-6. Explanation of part (f):

The process of producing tertiary amines and alkenes from quaternary ammonium is called Hofmann elimination. This process can also be referred to as exhaustive methylation. The Hofmann rule states that the major alkene product is the least substituted and least stable product when it comes to asymmetrical amines. Methyl iodide is used in excess because it has no beta hydrogens and therefore cannot compete in an elimination reaction. Silver oxide ion is used for deprotonation of water to form a hydroxide ion.

In part (f), Hofmann exhaustive methylation of the given reactant produces the following product as shown.

Formation of the product

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Most popular questions from this chapter

Show how you would use an acid chloride as an intermediate to synthesize(a) N-phenylbenzamide (PhCONHPh)from benzoic acid and aniline.(b) phenyl propionate(CH3CH2COOPh) from propionic acid and phenol.

Glutathione (GSH) is a tripeptide that serves as a mild reducing agent to detoxify peroxides and maintain the cysteine residues of hemoglobin and other red blood cell proteins in the reduced state. Complete hydrolysis of glutathione gives Gly, Glu, and Cys. Treatment of glutathione with carboxypeptidase gives glycine as the first free amino acid released. Treatment of glutathione with 2,4-dinitrofluorobenzene (Sanger reagent, Problem 24-21, page 1282), followed by complete hydrolysis, gives the 2,4-dinitrophenyl derivative of glutamic acid. Treatment of glutathione with phenyl isothiocyanate does not give a recognizable phenylthiohydantoin, however.

(a) Propose a structure for glutathione consistent with this information. Why would glutathione fail to give a normal product from Edman degradation, even though it gives a normal product from the Sanger reagent followed by hydrolysis?

(b) Oxidation of glutathione forms glutathione disulfide(GSSG). Propose a structure for glutathione disulfide, and write a balanced equation for the reaction of glutathione with hydrogen peroxide.

Question: Rank the compounds in each set in order of increasing acid strength.

(a)CH3CH2COOH,CH3CHBrCOOH,CH3CBr2COOH

(b) CH3CH2CH2CHBrCOOH,CH3CH2CHBrCH2COOH,CH3CHBrCH2CH2COOH

(c)

,CH3CH2COOH,

Show how you would synthesize the following compounds, starting with acetylene and any compounds containing no more than four carbon atoms.

(a)hex-1-yne

(b)hex-2-yne

(c)cis-hex-2-ene

(d)trans-hex-2-ene

(e)1,1-dibromohexane

(f)2,2-dibromohexane

(g)pentanal, CH3CH2CH2CH2CHO

(h)pentan-2-one, CH3-CO-CH 2CH2 CH3

(i) (+/)-3,4-dibromohexane

(j)meso-butan-2,3-diol

(k)2-methylhex-3-yn-2-ol

Show how you would use extractions with a separatory funnel to separate a mixture of the following compounds: benzoic acid, phenol, benzyl alcohol, aniline.
See all solutions

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