CBSE Class 12 Chemistry 2022 Term 2 Delhi Set 1

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Question : 11 of 12
 
Marks: +1, -0
(a) Complete the following:
(i) CH3CN──────▸2.H2O1,AlH(i−Bu)2⋅A′──────▸H+H2N−OH' B′
(ii) Write IUPAC name of the following compound:
(iii)Write a chemical test to distinguish between the following compounds: Phenol and Benzoic acid
OR
(b) Convert the following:
(i) Benzoic acid to Benzaldehyde
(ii) Propan-1-ol to 2-Bromopropanoic acid
(iii) Acetaldehyde to But-2-enal
Solution:
(a) (i)
CH3C≡N+ DIBAL-H +H2O→CH3CHO +NH2OH+H+→CH3CH=N−OH +H2O
A-Acetaldehyde
B-Acetaloxide
(ii) The IUPAC name is 3-bromobenzaldehyde.
(iii) Phenol and Benzoic Acid can be distinguished by iron chloride (FeCl3) test. As phenol gives violet colouration with neutral FeCl3 solution while benzoic acid gives buff coloured precipitate of ferric benzoate.
6C6H5OH+FeCl3→[Fe(OC6H5)6]3−+3H+
Violet complex +3HCl
3C6H5COOH+FeCl3→(C6H5COO)3Fe+3HCl Violet complex +3HCl
3C6H5COOHBenzoicacid+FeCl3→(C6H5COO)3FeBuffcolouredppt. +3HCl
OR
(b) (i) Benzoic acid to benzaldehyde - Firstly benzoic acid is converted into benzoyl chloride by adding sulphonyl chloride and then by Rosenmund reduction reaction using Pd−BaSO4 converted into benzaldehyde.

(ii) Propan-1-ol to 2-bromo propanoic acid -Firstly propanol is converted into propanoic acid by oxidizing in presence of alk. KMnO4 and then by using Red Phosphorous/ NaOH and Bromine gas is further converted into 2-Bromo-propanoic acid
CH3−CH2−CH2−OHPropanol──────▸ alkaline KMnO4
CH3−CH2−C∥O−OHPropanoicacid+Br2──────▸ red P NaOH CH3−CH|Br−COOH2−Bromopropanoicacid
(iii) Ethanal or Acetaldehyde into But-2-en-1-al Aldol condensation (in presence of dil NaOH ) followed by dehydration.
2CH3CHOEthanal─────▸ dil NaOHCH3−CH|OH−CH2−CHOAcetaldehyde ────▸−H2OΔCH3−CH=CH−CHOBut−2−en−1−al
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