(a) Give the products expected when acetic formic anhydride reacts with

(i) aniline and

(ii) benzyl alcohol

(b) Propose mechanisms for these reactions.

Short Answer

Expert verified

(a)

(b)

Step by step solution

01

(a) Role of aniline in reaction one

In this reaction, lone pair of electron of the nitrogen atom of aniline acts as a nucleophile, and it attacks the more electron deficient centre of the carbonyl carbon atom, i.e. aldehyde carbonyl

02

The reaction of acetic formic anhydride with aniline

When acetic formic anhydride react with aniline then there is the formation of N-phenyl formamide and acetic acid as a side product,In this reaction nucleophilic substitution reaction take place.

03

Role of benzyl alcohol in reaction two

In this reaction, lone pair of electron of oxygen atom of benzyl alcohol acts as a nucleophile and it attack to the more electron deficient centre of carbonyl carbon atom i.e. aldehyde carbonyl carbon.

04

The reaction of acetic formic anhydride with benzyl alcohol

When acetic formic anhydride reacts with aniline, then there is the formation of ester and acetic acid as a side product. Inthis reaction, a nucleophilic substitution reaction takes place.

05

(b) Mechanism for reaction with acetic formic anhydride and aniline

In the first step of this reaction, lone pair of nitrogen atom attacks as a nucleophile and it attack the less hindered carbonyl carbon and there is a formation of intermediate product, In the second step negative charge on the oxygen atom, form a double bond with carbon and there is elimination of acetate ion.

In the last step acetate ion abstract the proton and neutralize the nitrogen atom and there is a formation of final product.

06

The mechanism for reaction with acetic formic anhydride and benzyl alcohol

In the first step of this reaction, lone pair of oxygen atom attacks as a nucleophile and attack the less hindered carbonyl carbon, and there is a formation of an intermediate product. In the second step, the negative charge on the oxygen atom forms a double bond with carbon, and the acetate ion is eliminated.

In the last step, the acetate ion abstracts the proton and neutralizes the oxygen atom, and there is a formation of the final product.

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Most popular questions from this chapter

Show how you would convert glycine to following derivatives. Show the structure of the product in each case.

(a) Glycine isopropyl ester

(b) N-benzoylglycine

(c) N-benzyloxycarbonylglycine

(d) Tert-butyloxycarbonylglycine

Which of the following compounds are chiral? Draw each compound in its most symmetric conformation, star any asymmetric carbon atoms, and draw any mirror planes. Label any meso compounds. You may use Fischer projections if you prefer.

(a) meso-2,3-dibromo-2,3-dichlorobutane

(b) -2,3-dibromo-2,3-dichlorobutane

(c) (2R,3S)-2-bromo-3-chlorobutane

(d) (2R,3S)-2,3-dibromobutane

(e) (R,R)-2,3-dibromobutane

(f)

(g)

(h)

Question: Show how you would synthesize the following carboxylic acids, using the indicated starting materials.

(a) hex-3-ynepropanoicacid

(b) cis-cyclodecenedecanedioicacid

(c ) iodobenzenephenylaceticacid

(d)pentan-3-ol2-ethylbutanoicacid

(e)o-xylenephthalicacid

(f) allylbromidebut-3-enoicacid

Question: A carboxylic acid has two oxygen atoms, each with two nonbonding pairs of electrons.

  1. Draw the resonance forms of a carboxylic acid that is protonated on the hydroxy oxygen atom.
  2. Compare the resonance forms with those given previously for an acid protonated on the carbonyl oxygen atom.
  3. Explain why the carbonyl oxygen atom of a carboxylic acid is more basic than the hydroxy oxygen.

Question:

  1. Why do most long-chain fatty acids show a large peak in the mass spectrum at m/z 60?
  2. Use equations to explain the prominent peaks at m/z 74 and m/z 87 in the mass spectrum of 2-methylpentanoic acid.
  3. Why doesn’t the mass spectrum of 2-methylpentanoic acid show a large peak at m/z 60?

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