Alpha Substitution andCondensations of Carbonyl

compounds

Alpha SubstitutionReplacement of a hydrogen on the carbon

adjacent to the carbonyl, C=O.

enolate ion

Condensation withAldehyde or Ketone

Enolate ion attacks a C=O and the alkoxide is protonated. The net result is addition.

C

O

C

_C

O

C

O

C

C

O_

ROHC

O

C

C

OH

Condensation with Esters

Loss of alkoxide ion results in nucleophilic acyl substitution.

Acidity of -Hydrogens

Much more acidic than alkane or alkene (pKa > 40) or alkyne (pKa = 25).Less acidic than water (pKa = 15.7) or alcohol (pike = 16-19).pKa for -H of aldehyde or ketone ~20.In the presence of hydroxide or alkoxide ions, only a small amount of enolate ion is present at equilibrium.

Haloform Reaction

• Methyl ketones replace all three H’s with halogen.

• The trihalo ketone then reacts with hydroxide ion to give carboxylic acid.

Iodoform,yellow ppt.

C

O

CH3excess I2

OH-

C

O

CI3OH

-C

O

OH

CI3-

C

O

O-

HCI3

Positive Iodoformfor Alcohols

If the iodine oxidizes the alcohol to a methyl ketone, the alcohol will give a positive iodoform test.

Aldol Condensation

• Enolate ion adds to C=O of aldehyde or ketone.

• Product is a -hydroxy aldehyde or ketone.

• Aldol may lose water to form C=C.

Mechanism for Aldol Condensation

Dehydration of Aldol

Creates a new C=C bond.

CO

H3CC H

H

CH3C

CH3

OH

H+ or OH

-

heat CO

H3CC

H

CH3C

CH3

Crossed AldolCondensations

• Two different carbonyl compounds.

• Only one should have an alpha H.

Aldol Cyclizations• 1,4-diketone forms cyclopentenone.

• 1,5-diketone forms cyclohexenone.

Planning Aldol SynthesesPlanning Aldol Syntheses

Claisen CondensationTwo esters combine to form a -keto ester.

CH3 O C

O

CH R

CH3OC

O

CH2R

CH3 O C

O

CH R

CH3OC

O

CH2R

enolate ionpKa = 24

CH3 O C

O

CH Rbase

CH3 O C

O

CH2 R

_

CH3 O C

O

C C

R

CH2R

O

pKa = 11

_OCH3

CH3 O C

O

CH C

R

CH2R

O

Crossed Claisen

• Two different esters can be used, but one ester should have no hydrogens.

• Useful esters are benzoates, formates, carbonates, and oxalates.

• Ketones (pKa = 20) may also react with an ester to form a -diketone.

-Dicarbonyl Compounds

• More acidic than alcohols.

• Easily deprotonated by alkoxide ions and alkylated or acylated.

• At the end of the synthesis, hydrolysis removes one of the carboxyl groups.

CH3CH2O C

O

CH2 C

O

OCH2CH3

malonic ester, pKa = 13

CH3 C

O

CH2 C

O

OCH2CH3

acetoacetic ester, pKa =11

Malonic Ester Synthesis• Deprotonate, then alkylate with good

SN2 substrate. (May do twice.)

• Decarboxylation then produces a mono- or di-substituted acetic acid.

Acetoacetic Acid SynthesisProduct is mono- or di-substituted ketone.

Chemical tests

There a number of chemical tests that could be used to determine the presence of organic compounds in a sample.

Alkenes and alkynes

Both class of compounds will decolorized bromine water (purple) to give colorless solution.

Alcohols

Alcohols do not decolorised bromine water. This properth can be used to distinguish alcohols from alkenes and alkynes.

Alcohols can be oxidised by chromic anhydride, CrO3, in aqueous sulfuric acid, the clear orange solution turns blue-green and then opaques within seconds.

Tertiary alcohols do not give this test.

ROH + HCrO4- Opaque, blue-green

1o or 2oClearorange

Alcohols containing the structural features below will give positive iodoform test.

This sort of alcohol will give yellow precipitate (CH3I)when treated with iodine and sodium hydroxide (sodium hypoiodite, NaOI)

The reaction involves oxidation, halogenation and cleavage

R C CH3

H

OH

R C CH3

H

OH

R C CH3

O

R C CI3

O

+ NaOH

+ NaOH

+ NaOH

R C CH3

O

R C CI3

O

+ NaI + H2O

+ 3NaOH

RCOO- Na+ + CHI3

Yellowprecipitate

As would be expected, compounds with left structurewill also gives positive test. Hence, ketone having these features will also give positive results

R C CH3

O

Tollen’s test is used to determine the presence of aldehydes

Tollen’s reagent contains silver ammonia ion, Ag(NK3)2+. Oxiidation of aldehydes is accompanied by reduction of silver ion to free silver (in the form of mirror)

RCHO + Ag(NH3)2+

RCOO- + Ag

Silvermirror

Aldehydes and ketones will react with 2,4-diphenylhydrazine to form insoluble yellow to red solid

To differentiate primary, seconondary and tertiary amines, we can use Hinsberg test.This is done by reacting the amines with benzenesulfonyl chloride and excess of potassium hydroxide.A primary amine yields a clear solution, upon acidification and insoluble material separates.A secondary amine yields an insoluble compound and unaffected by acid.A tertiary amine yields an insoluble compound which dissolves upon acidification.

RNH2 + C6H5SO2ClOH- KOH

C6H5SO2NR- K+

H+

[C6H5SO2NHR]

C6H5SO2NHR

Clearsolution

Insoluble

R2NH + C6H5SO2Cl C6H5SO2NR2

KOH or H+

no reactionOH-

Insoluble

R3N + C6H5SO2Cl R3N R3NH+ Cl-

Insoluble Clear solution

OH-HCl