Page No. 231

8.1. What is meant by the following terms? Give an example of the reaction in each case.

(i) Cyanohydrin

(ii) Acetal

(iii) Semicarbazone

(iv) Aldol

(v) Hemiacetal

(vi) Oxime

(vii) Ketal

(viii) Imine

(ix) 2, 4-DNP-derivative

(x) Schiff’s base

Solution

(i) Cyanohydrins are organic compounds having the formula RR′C(OH)CN, where R and R′ can be alkyl or aryl groups.

Aldehydes and ketones react with hydrogen cyanide (HCN) in the presence of excess sodium cyanide (NaCN) as a catalyst to field cyanohydrin. These reactions are known as cyanohydrin reactions.

Cyanohydrins are useful synthetic intermediates.

(ii) Acetals are gem−dialkoxy alkanes in which two alkoxy groups are present on the terminal carbon atom. One bond is connected to an alkyl group while the other is connected to a hydrogen atom.

When aldehydes are treated with two equivalents of a monohydric alcohol in the presence of dry HCl gas, hemiacetals are produced that further react with one more molecule of alcohol to yield acetal.

(iii) Semicarbazones are derivatives of aldehydes and ketones produced by the condensation reaction between a ketone or aldehyde and semicarbazide.

Semicarbazones are useful for identification and characterization of aldehydes and ketones.

(iv) A β-hydroxy aldehyde or ketone is known as an aldol. It is produced by the condensation reaction of two molecules of the same or one molecule each of two different aldehydes or ketones in the presence of a base.

(v) Hemiacetals are α−alkoxyalcohols.

General structure of a hemiacetal

Aldehyde reacts with one molecule of a monohydric alcohol in the presence of dry HCl gas.

(vi) Oximes are a class of organic compounds having the general formula RR′CNOH, where R is an organic side chain and R′ is either hydrogen or an organic side chain. If R′ is H, then it is known as aldoxime and if R′ is an organic side chain, it is known as ketoxime.

(vii) Ketals are gem−dialkoxyalkanes in which two alkoxy groups are present on the same carbon atom within the chain. The other two bonds of the carbon atom are connected to two alkyl groups.

Ketones react with ethylene glycol in the presence of dry HCl gas to give a cyclic product known as ethylene glycol ketals.

(viii) Imines are chemical compounds containing a carbon ­nitrogen double bond.

Imines are produced when aldehydes and ketones react with ammonia and its derivatives.

(ix) 2, 4−dinitrophenylhydragones are 2, 4−DNP−derivatives, which are produced when aldehydes or ketones react with 2, 4−dinitrophenylhydrazine in a weakly acidic medium.

To identify and characterize aldehydes and ketones, 2, 4−DNP derivatives are used.

(x) Schiff’s base (or azomethine) is a chemical compound containing a carbon-nitrogen double bond with the nitrogen atom connected to an aryl or alkyl group-but not hydrogen. They have the general formula R₁R₂C = NR₃. Hence, it is an imine. It is named after a scientist, Hugo Schiff.

Aldehydes and ketones on treatment with primary aliphatic or aromatic amines in the presence of trace of an acid yields a Schiff’s base.

8.2. Name the following compounds according to IUPAC system of nomenclature:

(i) CH₃CH(CH₃)—CH₂ CH₂—CHO

(ii) CH₃CH₂COCH(C₂H₅)CH₂CH₂Cl

(iii) CH₃CH = CHCHO

(iv) CH₃COCH₂COCH₃

(v) CH₃CH(CH₃)CH₂C(CH₃)2COCH₃

(vi) (CH₃)₃CCH₂COOH.

(vii) OHCC₆H₄CHO-p

Solution

(i) 4-methylpentanal

(ii) 6-Chloro-4-ethylhexan-3-one

(iii) Butc-2-en-1-al

(iv) Pentane-2, 4-dione

(v) 3, 3, 5-trimethylhexan-2-one

(vi) 3, 3-dimethylbutanoic acid

(vii) Benzene-1, 4-dicarbaldehyde.

8.3. Draw the structures of the following compounds.

(i) 3-Methylbutanal

(ii) p-Nitropropiophenone

(iii) p-Methylbenzaldehyde

(iv) 4-Methylpent-3-en-2-one

(v) 4-Chloropentan-2-one

(vi) 3-Bromo-4-phenylpentanoic acid

(vii) p, p’-Dihydroxybenzophenone

(viii) Hex-2-en-4-ynoic acid

Solution

(i) 3-Methylbutanal

(ii) p-Nitropropiophenone

(iii) p-Methylbenzaldehyde

(iv) 4-Methylpent-3-en-2-one

(v) 4-Chloropentan-2-one

(vi) 3-Bromo-4-phenylpentanoic acid

(vii) p, p’-Dihydroxybenzophenone

(viii) Hex-2-en-4-ynoic acid

8.4. Write the IUPAC names of the following ketones and aldehydes. Wherever possible, give also common names.

(i) CH₃CO(CH₂)₄CH₃

(ii) CH₃CH₂CH BrCH₂CH(CH₃)CHO

(iii) CH₃(CH₂)₅CHO

(iv) Ph—CH = CH—CHO 

(vi) PhCOPh

Solution

(i) CH₃CO(CH₂)₄CH₃

IUPAC Name: Heptan-2-one

Common Name: Methyl-n-pentyl ketone.

(ii) CH₃CH₂CHBrCH₂CH(CH₃)CHO

IUPAC Name: 4-Bromo-2-methylhexanal

Common Name: (γ-Bromo-α-methyl-caproaldehyde)

(iii) CH₃(CH₂)₅CHO

IUPAC Name: Heptanal

Common name: n-heptyl aldehyde

(iv) Ph – CH = CH – CHO

IUPAC Name: 3-phenylprop-2-enal

Common name: β-Pheynolacrolein

IUPAC Name: Cyclopentanecarbaldehyde

Common name: Cyclopentanecarboxaldehyde

(vi) PhCOPh

IUPAC Name: Diphenylmethanone

Common Name: Benzophenone.

8.5. Draw structures of the following derivatives:

(i) The 2,4-dinitrophenylhydrazone of benzaldehyde

(ii) Cyclopropanone oxime

(iii)Acetaldehydedimethylacetal

(iv) The semicarbazone of cyclobutanone

(v) The ethylene ketal of hexan-3-one

(vi) The methyl hemiacetal of formaldehyde

Solution

(i) The 2,4-dinitrophenylhydrazone of benzaldehyde

(ii) Cyclopropanone oxime

(iii) Acetaldehydedimethylacetal

(iv) The semicarbazone of cyclobutanone

(v) The ethylene ketal of hexan-3-one

(vi) The methyl hemiacetal of formaldehyde

8.6. Predict the products formed when cyclohexanecarbaldehyde reacts with following reagents :

(i) PhMgBr and then H₃O⁺

(ii) Tollen’s reagent

(iii) Semicarbazide and weak acid

(iv) Excess ethanol and acid

(v) Zinc amalgam and dilute hydrochloric acid

Solution

(i) When cyclohexanecarbaldehyde reacts PhMgBr and then H₃O⁺, the final product is Cyclohexylphenylcarbinol/Cyclohyxylphenyl methanol.

(ii) When cyclohexanecarbaldehyde reacts Tollen’s reagent, the final product is Cyclohexanecarboxylate ion, silver, ammonium and water.

(iii) When cyclohexanecarbaldehyde reacts Semicarbazide and weak acid, the final product is Cyclohexanecarbaldehyde semicarbazone.

(iv) When cyclohexanecarbaldehyde reacts excess ethanol and acid, the final product is Cyclohexanecarbaldehyde diethyl acetal.

(v) When cyclohexanecarbaldehyde reacts Zinc amalgam and dilute hydrochloric acid, the final product is Methylcyclohexane.

8.7. Which of the following compounds would undergo aldol condensation, which the Cannizzaro reaction and which neither? Write the structures of the expected products of aldol condensation and Cannizzaro reaction.

(i) Methanal

(ii) 2-Methylpentanal

(iii) Benzaldehyde

(iv) Benzophenone

(v) Cyclohexanone

(vi) 1-Phenylpropanone

(vii) Phenylacetaldehyde

(viii) Butan-1-ol

(ix) 2,2-Dimethylbutanal

Solution

Aldehydes and ketones having at least one α-hydrogen atom undergo aldol condensation. The compounds
(ii) 2-methylpentanal, (v) cyclohexanone, (vi) 1-phenylpropanone, and (vii) phenylacetaldehyde contain one or more α-hydrogen atoms. Therefore, these undergo aldol condensation.

Aldehydes having no α-hydrogen atoms undergoes Cannizzaro reactions. The compounds (i) Methanal, (iii) Benzaldehyde, and (ix) 2, 2-dimethylbutanal do not have any α-hydrogen atom. Therefore, these undergo Cannizzaro reactions.

Compound (iv) Benzophenone is ketone having no α-hydrogen atom and compound (viii) Butan-1-ol is an alcohol. Hence, these compounds do not undergo either aldol condensation or Cannizzaro reaction.

Aldol condensation:

Cannizzaro reaction:

8.8. How will you convert ethanal into the following compounds?

(i) Butane-1,3-diol

(ii) But-2-enal

(iii) But-2-enoic acid

Solution

(i) On treatment with dilute alkali, ethanal produces 3-hydroxybutanal, which on reduction gives butane-1, 3-diol.

(ii) On treatment with dilute alkali, ethanal gives 3-hydroxybutanal which on heating produces but-2-enal.

(iii) When treated with Tollen’s reagent, But-2-enal produced in the above reaction produces but-2-enoic acid.