Question

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Answer 1

Explanation:

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Answer 2

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(i) When chloroform reacts with aq. KOH, the chlorines on CHCl3 are successively replaced by -OH groups from KOH via. nucleophilic reaction (SN2). In theory, first it forms CHCl2(OH), then CHCl(OH2), and then CH(OH)3, while eliminating KCl with each step. Since Cl- is an excellent leaving group, CHCl(OH2) will spontaneously release one molecule of H2O and produce HC(=O)Cl or formyl chloride, which again will be hydrolyzed to produce HCOOK or Potassium Formate.

1 mole of Chloroform will react with 4 moles of KOH to produce 1 mole of HCOOK (potassium formate), 3 moles of KCl and 2 moles of H2O.

1 CHCl3 + 4 KOH —-> HCOOK + 3 KCl + 2 H2O.

This is if you allow the reaction to go to completion. If the reaction conditions are not severe enough, you can isolate some of the intermediates as well.

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(ii)  When iodoform is heated with silver powder, both react to give acetylene or ethyne and silver chloride as a product. Silver is a reducing agent that reduces the chloroform into ethyne.

2CHI3 + 6Ag -----> CH = CH (acetylene) + 6AgI

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(iii) You get the iodoform reaction. The products from acetone ((CH3)2C=O) are acetic acid (CH3COOH) and iodoform (CHI3).

(CH3)2C=O + I + NaOH → CH3COOH + CHI3

This is a general reaction for alpha-methyl ketones and was long used as a qualitative test for this structure as CHI3 is a solid at room temp, insoluble in the reaction medium, and yellow in color. So, the formation of CHI3 is easily observable and diagnostic.

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(iv) When n-butyl chloride is treated with alcoholic KOH,dehydrohalogenation takes place to give but-1-ene.

This reaction is a dehydrohalogenation reaction.

CH₃CH₂CH₂CH₂-Cl+KOH—>CH₂=CH-CH₂CH₃ +KCl+H₂O

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(v) In this reaction, the analyte is heated with alcoholic potassium hydroxide and chloroform. If a primary amine is present, the isocyanide (carbylamine) is formed, as indicated by a foul odor. The carbylamine test does not give a positive reaction with secondary and tertiary amines.

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(vi) When phenol treated with chloroform in the presence of aqueous sodium hydroxide(base) at 340 K temperature followed by hydrolysis yields 2-hydroxybenzaldehyde also called as salicylaldehyde and reaction called as Reimer-Tiemann reaction.

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(vii) Chlorobenzene does not undergo hydrolysis under normal conditions. However, it undergoes hydrolysis when heated in an aqueous sodium hydroxide solution at a temperature of 623 K and a pressure of 300 atm to form phenol.

C6H5Cl + NaOH ----->>C6H5ONa+HCl------->> C6H5OH

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