How will you obtain propan-2-ol from grignard’s reagent ? (Give balanced chemical equations).

Propan −2−ol is a secondary alcohol with three carbon atoms. Hence, along with CH 3 MgBr we should use an aldehyde having two carbon atoms, that is, ethanal and follow it up with hydrolysis.

Identify the compounds A, B, C and D in the given reaction:

Hydration of acetylene (followed by keto enol tautomerism) gives acetaldehdye. This is followed by oxidation to form acetic acid. The reaction of acetic acid and calcium hydroxide produces calcium acetate which on dry distillation gives acetone and calcium carbonate. CH ≡ CH → Hg + 2 / H 2 SO 4 H 2 O CH 3 CHO → K 2 Cr 2 O 7 + H 2 SO 4 O CH 3 COOH → Ca OH 2 CH 3 COO 2 Ca → dry distillation ∆ CH 3 COCH 3 A – CH 3 CHO , Acetaldehyde B – CH 3 COOH , Acetic acid C – ( CH 3 COO ) 2 Ca , Calcium acetate D – CH 3 COCH 3 , Acetone

A ketone A ( C 4 H 8 O ) which undergoes Iodoform reaction gives compound B on reduction. B on heating with conc. H 2 SO 4 at 443 K gives a compound C which forms ozonide D. D on hydrolysis with Zn dust gives only E. Identify the compounds A to E. Write the Iodoform reaction with compound A.

A ketone A ( C 4 H 8 O ) is butan - 2 - one which undergoes Iodoform reaction gives compound B, butan - 2 - ol on reduction. B on heating with conc. H 2 SO 4 at 443 K gives a compound C, but - 2 - ene which forms ozonide D. D on hydrolysis with Zn dust gives only E, acetaldehyde. The reaction is given below, The iodoform reaction of compound A is given as, .

Identify the compounds A, B, C and D.

In this reaction sequence, benzoic acid is treated with thionyl chloride produces benzoyl chloride which on ammonolysis form benzamide. The benzamide undergoes bromination in presence of potassium hydroxide which forms aniline. The aniline produce benzene diazonium chloride when reacting with sodamide and hydrochloric acid. Thus, the compound A is benzoyl chloride, B is benzamide and C will be aniline. Then, the compound D is benzene diazonium chloride. The reaction is given below, C 6 H 5 COOH → SOCl 2 C 6 H 5 COCl A → NH 3 C 6 H 5 CONH 2 B → Br 2 / KOH C 6 H 5 NH 2 C → 0 - 5 ° C NaNO 2 + HCl C 6 H 5 N 2 + Cl - D .

Explain why acetic acid is a weaker acid as compared to formic acid.

Acetic acid has electron donating methyl group attached to - COOH group that shows positive inductive effect and increases the electron density on the carbon atom of - COOH group, which reduces the acidity of acid. But, in case of formic acid, HCOOH no such electron donating group is present hence will have higher bond polarity of - C = O bond which in turn shows greater acidity than acetic acid.

Explain why acetaldehyde undergoes aldol condensation, but formaldehyde does not.

Aldol condensation: Aldehydes and ketones having at least one α - hydrogen undergo a reaction in the presence of dilute alkali as catalyst to form β - hydroxy aldehydes or β - hydroxy ketones, respectively. This is known as Aldol reaction. If we observe the structures of acetaldehyde and formaldehyde it is cleared that only acetaldehyde contains α - hydrogens but formaldehyde does not. Because of this reason formaldehyde does not undergo Aldol condensation.

Write chemical equations to illustrate the name reaction HVZ reaction.

Carboxylic acids having an α - hydrogen are halogenated at the α - position on treatment with chlorine or bromine in the presence of small amount of red phosphorus to give α - halocarboxylic acids. The reaction is known as Hell-Volhard-Zelinsky reaction. CH 3 - CH 2 - COOH → Br 2 / P CH 3 − CHBr − COOH → Br 2 / P CH 3 − CBr 2 − COOH → Br 2 / P No further reaction . In the above equation the reaction has been stopped in the last step due to the non-availability of α - hydrogen in the last step's substrate.