imine derivatives

8/19/2019 Imine Derivatives

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Just as the reaction of an aldehyde or ketone with a 1° amine forms an imine (Section 21.11) in

the presence of mild acid, so, too, reaction of an aldehyde or ketone with a reagent having the

general structure NH2Z (where Z contains an O or N atom bonded to the NH2 group) forms an

imine derivative. The overall reaction results in replacement of C –– O by C –– NZ.

R'

R' = H or alkyl

mild acidOC

RH2N Z

ZR'

imine derivative

NCR

+ H2O+General reaction

Thus, treatment of an aldehyde or ketone with hydroxylamine (NH 2OH), 2,4-dinitrophenylhy-

drazine [(O2N)2C6H3NHNH2], or semicarbazide (NH2NHCONH2) forms oximes, 2,4-dinitro-

phenylhydrazones, and semicarbazones, respectively.

H mild acid

OCCH3CH2CH2

H

CCH3CH2CH2

H

N

N

H2N OH

OH

oximehydroxylamine

N

C

CH3

+

mild acidOC

CH3

2,4-dinitrophenylhydrazine

+

Examples

CH3CH3NO2H2NNH

O2N

2,4-dinitrophenylhydrazone

NO2

O2N

H NH2

N

Nmild acidO

O

semicarbazide

+ H2NNH

semicarbazone

C

NH2

O

The mechanism for these reactions is exactly the same as the mechanism for imine formation

with a 1° amine. There are two key steps: nucleophilic addition of the NH2Z reagent, and lossof water to form the C –– N of the imine derivative.

B

Supplementary

Topic

Imine Derivatives

B-1


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B-2 Topic B Imine Derivatives

Imine derivatives were once widely used to characterize carbonyl compounds. Reaction of a

liquid aldehyde or ketone with NH2Z generally forms a solid imine derivative as product. Since

the melting points of thousands of these derivatives are known, this experimental value can be

compared to known melting points to help identify the carbonyl compound from which it was

derived. Today, carbonyl compounds are more commonly identied using the spectroscopic

techniques discussed in Chapters 13 and 14.

Problem B.1 What imine derivative is formed when cyclopentanone is treated with each reagent in the presenceof mild acid?

a. H2N OH b. H2NNH C

NH2

O

c. H2N NH2 d. NO2H2NNH

O2N

Problem B.2 Draw the product formed when each carbonyl compound is treated with 2,4-dinitrophenylhydrazine.

a. CHO b.

O

c. CH3CH2CHO d.

O

Problem B.3 What carbonyl compound and NH2Z reagent are used to prepare each imine derivative?

a.

NOH

b. NNH c.

NNH

NH2

O

Problem B.4 Although semicarbazide (H2NNHCONH2 ) contains two NH2 groups, only one reacts to form asemicarbazone. Explain why the two NH2 groups have different reactivities.

Mechanism B.1 Conversion of an Aldehyde or Ketone to an Imine Derivative

Part [1] Nucleophilic addition to the carbonyl group

C+

R'

OC

R

NH2Z nucleophilicattack

protontransfer

[1] [2]R

O –

R'

NH2Z R C

OH

R'

NHZ• Nucleophilic attack of the reagent NH2Z,

followed by proton transfer, forms an

unstable intermediate (Steps [1]–[2]). These

steps result in the addition of H and NHZ to

the carbonyl group.

Part [2] Elimination of H2O to form the imine derivative

C

H

CR'

C

R

NZ

eliminationof H2O

resonance-stabilized cation

iminederivative

++

+[3] [4] [5]

R

OH

R'NHZ

OH2

R

OH2

H2O

R' H R' ZC

R

N

R'

C+

R

NZ

H

NHZ

H2O+

H3O++

• Elimination of H2O forms the imine derivative

in three steps. Protonation of the OH group

in Step [3] forms a good leaving group,

leading to loss of water in Step [4], giving a

resonance-stabilized cation. Loss of a proton

forms the imine derivative in Step [5].


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Topic B Imine Derivatives B-3

Problem B.5 Purpald ( A ) is the trade name for a reagent that reacts with an aldehyde to form a colorlesscompound B, which is air-oxidized to give a magenta or purple product C.

[1]

RCHO

Purpald

A

+ 2 H2O

N

NHNH2HS N

N

NH2

B

N

NHHS N

R

N

HN NH[2]

O2

C

N

NHS N

R

N

N N

a. Draw a stepwise mechanism for the conversion of A to B.

b. Explain why ketones react with Purpald, but do not form a colored product; that is, Reaction [1]

occurs, but Reaction [2] does not.

imine derivatives

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