An organic compound (X) reacts with Grignard's reagent to give (Y), which on oxidation gives (Z). All of the compounds (X), (Y) and (Z) give positive iodoform test. a) Identify the compounds (X),…

Model Answer: NEB Class 12 Chemistry (2082) – 8 Marks

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Part I: Iodoform Test Series

Isopropyl alcohol (CH₃-CH(OH)-CH₃)Isopropyl alcohol (CH₃-CH(OH)-CH₃)

(Y)(Y)K₂Cr₂O₇/H₂SO₄, heat(Z)(Z)

a) Identification of (X), (Y), and (Z) with Reactions

The given sequence involves a compound (X) that reacts with a Grignard reagent to form (Y), which upon oxidation yields (Z). All three compounds give a positive iodoform test, meaning they must contain the CH₃CO– (acetyl) group or a methyl ketone structure.

Iodoform (CHI₃)Iodoform (CHI₃)

Step-by-Step Reasoning:

1. Positive iodoform test implies the presence of CH₃CO– or CH₃CH(OH)– (secondary alcohol with a methyl group adjacent to –OH).
2. (X) reacts with a Grignard reagent (RMgX) to form (Y), which is a tertiary alcohol (since Grignard reagents add to carbonyls).
3. (Y) on oxidation gives (Z), a ketone (since tertiary alcohols do not oxidize easily, but if (Y) is a secondary alcohol, it would oxidize to a ketone).
4. The only compound that fits this sequence and gives a positive iodoform test is acetaldehyde (ethanal, CH₃CHO) for (X).

Acetaldehyde (CH₃CHO)Acetaldehyde (CH₃CHO)

Reactions:

1. Reaction of (X) with Grignard reagent (CH₃MgBr): $$\mathrm{CH}{}_3\mathrm{CHO}+\mathrm{CH}{}_3\mathrm{MgBr}\stackrel{\mathrm{ether},\mathrm{dry}}{\to }\mathrm{CH}{}_3-\mathrm{CH}(\mathrm{OH})-\mathrm{CH}{}_3(\mathrm{Y})$$
   • (X) = Acetaldehyde (CH₃CHO)
   • (Y) = Isopropyl alcohol (2-propanol, CH₃-CH(OH)-CH₃)

acetaldehydeacetaldehyde+CH₃MgBrether, dry(Y)(Y)

2. Oxidation of (Y) to (Z): $$\mathrm{CH}{}_3-\mathrm{CH}(\mathrm{OH})-\mathrm{CH}{}_3\stackrel{\mathrm{K}{}_2\mathrm{Cr}{}_2\mathrm{O}{}_7/\mathrm{H}{}_2\mathrm{SO}{}_4,\mathrm{heat}}{\to }\mathrm{CH}{}_3-\mathrm{CO}-\mathrm{CH}{}_3(\mathrm{Z})$$
   • (Z) = Acetone (propanone, CH₃COCH₃)

Verification of Iodoform Test:

• (X) = CH₃CHO (contains CH₃CO–)
• (Y) = CH₃-CH(OH)-CH₃ (contains CH₃-CH(OH)–, which can be oxidized to CH₃CO–)
• (Z) = CH₃COCH₃ (contains CH₃CO–)

All three compounds give a positive iodoform test because they contain the CH₃CO– group or can form it upon oxidation.

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b) Why do these compounds give a positive iodoform test?

The iodoform test is positive for compounds containing:

1. Methyl ketones (R-CO-CH₃) – Directly contain CH₃CO–.
2. Secondary alcohols with a CH₃-CH(OH)– group – Can be oxidized to methyl ketones.
3. Ethanol (CH₃CH₂OH) – Oxidized to acetaldehyde, which then gives a positive test.

Mechanism:

• Iodine (I₂) in NaOH oxidizes the methyl ketone or secondary alcohol to a carboxylate ion.
• The triiodomethane (CHI₃, yellow precipitate) forms due to the cleavage of the CH₃CO– group.

Example for (Z) = Acetone (CH₃COCH₃): $$\mathrm{CH}{}_3-\mathrm{CO}-\mathrm{CH}{}_3+3\mathrm{I}{}_2+4\mathrm{NaOH}\to \mathrm{CHI}{}_3(\mathrm{yellow}\mathrm{ppt})+\mathrm{CH}{}_3\mathrm{COONa}+3\mathrm{NaI}+3\mathrm{H}{}_2\mathrm{O}$$

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c) Conversion of (Y) into Pseudonitrol

Pseudonitrol is CH₃NO₂ (nitromethane). To convert (Y) = Isopropyl alcohol (CH₃-CH(OH)-CH₃) into pseudonitrol, we follow these steps:

1. Oxidation of (Y) to Acetone (CH₃COCH₃): $$\mathrm{CH}{}_3-\mathrm{CH}(\mathrm{OH})-\mathrm{CH}{}_3\stackrel{\mathrm{K}{}_2\mathrm{Cr}{}_2\mathrm{O}{}_7/\mathrm{H}{}_2\mathrm{SO}{}_4}{\to }\mathrm{CH}{}_3-\mathrm{CO}-\mathrm{CH}{}_3$$

2. Haloform Reaction on Acetone:

   • Treat acetone with chlorine (Cl₂) in NaOH to form chloroform (CHCl₃) and sodium acetate. $$\mathrm{CH}{}_3-\mathrm{CO}-\mathrm{CH}{}_3+3\mathrm{Cl}{}_2\stackrel{\mathrm{NaOH}}{\to }\mathrm{CHI}{}_3+\mathrm{CH}{}_3\mathrm{COONa}$$ (Note: Here, we use Cl₂ instead of I₂ to get CHCl₃, which can be further converted to CH₃NO₂.)

3. Conversion of Chloroform to Pseudonitrol:

   • Chloroform (CHCl₃) reacts with silver nitrite (AgNO₂) to form nitromethane (CH₃NO₂). $$\mathrm{CHCl}{}_3+3\mathrm{AgNO}{}_2\to \mathrm{CH}{}_3\mathrm{NO}{}_2+3\mathrm{AgCl}$$

Alternative Route (Direct from Isopropyl Alcohol):

• Isopropyl alcohol can be dehydrated to propene, then ozonolyzed to acetaldehyde, and finally converted to nitromethane via iodoform reaction followed by nitrite substitution.

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d) Two Methods for the Preparation of (Z) = Acetone (CH₃COCH₃)

1. Dry Distillation of Calcium Acetate: $$(\mathrm{CH}{}_3\mathrm{COO}){}_2\mathrm{Ca}\stackrel{\mathrm{\Delta }}{\to }\mathrm{CH}{}_3-\mathrm{CO}-\mathrm{CH}{}_3+\mathrm{CaCO}{}_3$$

   • Procedure: Heat calcium acetate (from acetic acid + Ca(OH)₂) to get acetone.

2. Oxidation of Isopropyl Alcohol (2-Propanol): $$\mathrm{CH}{}_3-\mathrm{CH}(\mathrm{OH})-\mathrm{CH}{}_3\stackrel{\mathrm{K}{}_2\mathrm{Cr}{}_2\mathrm{O}{}_7/\mathrm{H}{}_2\mathrm{SO}{}_4}{\to }\mathrm{CH}{}_3-\mathrm{CO}-\mathrm{CH}{}_3+\mathrm{H}{}_2\mathrm{O}$$

   • Procedure: Oxidize isopropyl alcohol with acidic potassium dichromate.

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OR

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Part II: Conversion Reactions

a) Ethanol into Chloroform (CHCl₃)

Steps:

1. Oxidation of Ethanol to Acetaldehyde: $$\mathrm{CH}{}_3\mathrm{CH}{}_2\mathrm{OH}\stackrel{\mathrm{K}{}_2\mathrm{Cr}{}_2\mathrm{O}{}_7/\mathrm{H}{}_2\mathrm{SO}{}_4}{\to }\mathrm{CH}{}_3\mathrm{CHO}+\mathrm{H}{}_2\mathrm{O}$$

2. Aldol Condensation (if needed, but direct haloform is better):

   • Haloform Reaction: $$\mathrm{CH}{}_3\mathrm{CHO}+3\mathrm{Cl}{}_2\stackrel{\mathrm{NaOH}}{\to }\mathrm{CHI}{}_3(\mathrm{Chloroform})+\mathrm{CH}{}_3\mathrm{COONa}$$
   • Procedure:
     • Mix ethanol with bleaching powder (Ca(OCl)Cl) in alkaline medium.
     • Chloroform is formed via haloform reaction.

Final Reaction: $$\mathrm{CH}{}_3\mathrm{CH}{}_2\mathrm{OH}\stackrel{\mathrm{K}{}_2\mathrm{Cr}{}_2\mathrm{O}{}_7/\mathrm{H}{}_2\mathrm{SO}{}_4}{\to }\mathrm{CH}{}_3\mathrm{CHO}\stackrel{\mathrm{Cl}{}_2/\mathrm{NaOH}}{\to }\mathrm{CHCl}{}_3$$

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b) Phenol into DDT (Dichlorodiphenyltrichloroethane)

Steps:

1. Reimer-Tiemann Reaction (Formylation of Phenol): $$\mathrm{C}{}_6\mathrm{H}{}_5\mathrm{OH}+\mathrm{CHCl}{}_3\stackrel{\mathrm{NaOH},\mathrm{heat}}{\to }\mathrm{C}{}_6\mathrm{H}{}_4(\mathrm{OH})\mathrm{CHO}(\mathrm{Salicylaldehyde})$$

2. Condensation with Chloral (Trichloroacetaldehyde): $$\mathrm{C}{}_6\mathrm{H}{}_4(\mathrm{OH})\mathrm{CHO}+\mathrm{CCl}{}_3\mathrm{CHO}\stackrel{\mathrm{conc}\cdot \mathrm{H}{}_2\mathrm{SO}{}_4}{\to }\mathrm{DDT}(\mathrm{C}{}_{14}\mathrm{H}{}_9\mathrm{Cl}{}_5)$$

   • Procedure:
     • Treat phenol with chloral (CCl₃CHO) in the presence of concentrated sulfuric acid.
     • The product is DDT, a pesticide.

Final Reaction: $$\mathrm{C}{}_6\mathrm{H}{}_5\mathrm{OH}\stackrel{\mathrm{CHCl}{}_3/\mathrm{NaOH}}{\to }\mathrm{C}{}_6\mathrm{H}{}_4(\mathrm{OH})\mathrm{CHO}\stackrel{\mathrm{CCl}{}_3\mathrm{CHO}/\mathrm{H}{}_2\mathrm{SO}{}_4}{\to }\mathrm{DDT}$$

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c) Ethoxyethane (Diethyl Ether) into Methoxymethane (Dimethyl Ether)

Steps:

1. Cleavage of Ethoxyethane with HI: $$\mathrm{CH}{}_3\mathrm{CH}{}_2-\mathrm{O}-\mathrm{CH}{}_2\mathrm{CH}{}_3+\mathrm{HI}\to \mathrm{CH}{}_3\mathrm{CH}{}_2\mathrm{I}+\mathrm{CH}{}_3\mathrm{CH}{}_2\mathrm{OH}$$

   • Procedure: Heat ethoxyethane with hydriodic acid (HI) to form iodoethane and ethanol.

2. Dehydration of Ethanol to Ethene: $$\mathrm{CH}{}_3\mathrm{CH}{}_2\mathrm{OH}\stackrel{\mathrm{conc}\cdot \mathrm{H}{}_2\mathrm{SO}{}_4,443\mathrm{K}}{\to }\mathrm{CH}{}_2=\mathrm{CH}{}_2+\mathrm{H}{}_2\mathrm{O}$$

3. Hydroboration-Oxidation to Methanol: $$\mathrm{CH}{}_2=\mathrm{CH}{}_2+\mathrm{BH}{}_3\stackrel{\mathrm{THF}}{\to }\mathrm{CH}{}_3\mathrm{CH}{}_2-\mathrm{BH}{}_2\stackrel{\mathrm{H}{}_2\mathrm{O}{}_2/\mathrm{NaOH}}{\to }\mathrm{CH}{}_3\mathrm{CH}{}_2\mathrm{OH}$$ (This is a detour; a better method is direct methylation.)

Better Method (Direct Methylation):

• Treat ethoxyethane with methyl iodide (CH₃I) in the presence of silver oxide (Ag₂O) to replace ethyl groups with methyl: $$\mathrm{CH}{}_3\mathrm{CH}{}_2-\mathrm{O}-\mathrm{CH}{}_2\mathrm{CH}{}_3+2\mathrm{CH}{}_3\mathrm{I}+\mathrm{Ag}{}_2\mathrm{O}\to \mathrm{CH}{}_3-\mathrm{O}-\mathrm{CH}{}_3+2\mathrm{CH}{}_3\mathrm{CH}{}_2\mathrm{I}+2\mathrm{AgI}$$
  • Procedure: Use Williamson ether synthesis with sodium methoxide (CH₃ONa) and methyl iodide.

Final Reaction: $$\mathrm{CH}{}_3\mathrm{CH}{}_2-\mathrm{O}-\mathrm{CH}{}_2\mathrm{CH}{}_3\stackrel{\mathrm{CH}{}_3\mathrm{I}/\mathrm{NaOCH}{}_3}{\to }\mathrm{CH}{}_3-\mathrm{O}-\mathrm{CH}{}_3$$

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d) Aniline into Chlorobenzene

Steps:

1. Diazotization of Aniline: $$\mathrm{C}{}_6\mathrm{H}{}_5\mathrm{NH}{}_2+\mathrm{NaNO}{}_2+\mathrm{HCl}\stackrel{0-5°\mathrm{C}}{\to }\mathrm{C}{}_6\mathrm{H}{}_5\mathrm{N}{}_2{}^{+}\mathrm{Cl}{}^{-}(\mathrm{Benzenediazonium}\mathrm{chloride})+2\mathrm{H}{}_2\mathrm{O}$$

2. Sandmeyer Reaction (Replacement of N₂⁺ with Cl⁻): $$\mathrm{C}{}_6\mathrm{H}{}_5\mathrm{N}{}_2{}^{+}\mathrm{Cl}{}^{-}+\mathrm{CuCl}\stackrel{\mathrm{CuCl}}{\to }\mathrm{C}{}_6\mathrm{H}{}_5\mathrm{Cl}+\mathrm{N}{}_2+\mathrm{CuCl}{}_2$$

   • Procedure:
     • Treat diazotized aniline with copper(I) chloride (CuCl) in HCl.
     • Chlorobenzene is formed with the evolution of nitrogen gas.

Final Reaction: $$\mathrm{C}{}_6\mathrm{H}{}_5\mathrm{NH}{}_2\stackrel{\mathrm{NaNO}{}_2/\mathrm{HCl},0-5°\mathrm{C}}{\to }\mathrm{C}{}_6\mathrm{H}{}_5\mathrm{N}{}_2{}^{+}\mathrm{Cl}{}^{-}\stackrel{\mathrm{CuCl}}{\to }\mathrm{C}{}_6\mathrm{H}{}_5\mathrm{Cl}$$