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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) CH3CN2.H2O1,AlH(iBu)2AH+H2NOH\mathrm{CH}_3 \mathrm{CN} \xrightarrow[2 . \mathrm{H}_2 \mathrm{O}]{1, \mathrm{AlH}(i-\mathrm{Bu})_2} \cdot \mathrm{A}' \xrightarrow[\mathrm{H}^+]{\mathrm{H}_2 \mathrm{N}-\mathrm{OH}}' B\mathrm{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)
  CH3CN+ DIBAL-H +H2OCH3CHO\; \mathrm{CH}_3 \mathrm{C} \equiv \mathrm{N} + \text{ DIBAL-H } + \mathrm{H}_2 \mathrm{O} \rightarrow \mathrm{CH}_3 \mathrm{CHO} NH2OH+H+CH3CH=NOH\mathrm{NH}_2 \mathrm{OH} + \mathrm{H}^+ \rightarrow \mathrm{CH}_3 \mathrm{CH} = \mathrm{N} - \mathrm{OH} H2O\mathrm{H}_2 \mathrm{O}
A-Acetaldehyde
B-Acetaloxide
(ii) The IUPAC name is 3-bromobenzaldehyde.
(iii) Phenol and Benzoic Acid can be distinguished by iron chloride (FeCl3)(\mathrm{FeCl}_3) test. As phenol gives violet colouration with neutral FeCl3\mathrm{FeCl}_3 solution while benzoic acid gives buff coloured precipitate of ferric benzoate.
  6C6H5OH+FeCl3[Fe(OC6H5)6]3+3H+\;6\mathrm{C}_6\mathrm{H}_5\mathrm{OH} + \mathrm{FeCl}_3 \rightarrow \left[ \mathrm{Fe}(\mathrm{OC}_6\mathrm{H}_5)_6 \right]^{3-} + 3\mathrm{H}^+
    Violet complex  +3HCl\;\; \text{Violet complex}\; + 3\mathrm{HCl}
  3C6H5COOH+FeCl3(C6H5COO)3Fe+3HCl\;3\mathrm{C}_6\mathrm{H}_5\mathrm{COOH} + \mathrm{FeCl}_3 \rightarrow (\mathrm{C}_6\mathrm{H}_5\mathrm{COO})_3\mathrm{Fe} + 3\mathrm{HCl} Violet complex +3HCl+ 3\mathrm{HCl}
3C6H5COOHBenzoic acid+FeCl3(C6H5COO)3FeBuff coloured ppt.+3HCl\underset{\text{Benzoic acid}}{3\mathrm{C}_6\mathrm{H}_5\mathrm{COOH}} + \mathrm{FeCl}_3 \rightarrow \underset{\text{Buff coloured ppt.}}{(\mathrm{C}_6\mathrm{H}_5\mathrm{COO})_3\mathrm{Fe}} + 3\mathrm{HCl} +3HCl+ 3\mathrm{HCl}
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 PdBaSO4\mathrm{Pd}-\mathrm{BaSO}_4 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\mathrm{KMnO}_4 and then by using Red Phosphorous/ NaOH\mathrm{NaOH} and Bromine gas is further converted into 2-Bromo-propanoic acid
CH3CH2CH2OHPropanolKMnO4alkaline\underset{\text{Propanol}}{\mathrm{CH}_3-\mathrm{CH}_2-\mathrm{CH}_2-\mathrm{OH}} \xrightarrow[\mathrm{KMnO}_4]{\text{alkaline}}
CH3CH2COOHPropanoic acid+Br2NaOHred P\underset{\text{Propanoic acid}}{\mathrm{CH}_3-\mathrm{CH}_2-\overset{\mathrm{O}}{\overset{\parallel}{\mathrm{C}}}-\mathrm{OH}} + \mathrm{Br}_2 \xrightarrow[\mathrm{NaOH}]{\text{red P}} CH3CHBrCOOH2-Bromopropanoic acid\underset{\text{2-Bromopropanoic acid}}{\mathrm{CH}_3-\underset{\mathrm{Br}}{\underset{\vert}{\mathrm{CH}}}-\mathrm{COOH}}
(iii) Ethanal or Acetaldehyde into But-2-en-1-al Aldol condensation (in presence of dil NaOH\mathrm{NaOH} ) followed by dehydration.
2CH3CHOEthanaldil NaOHCH3CHOHCH2CHOAcetaldehyde\underset{\text{Ethanal}}{2\mathrm{CH}_3\mathrm{CHO}} \xrightarrow{\text{dil NaOH}} \underset{\text{Acetaldehyde}}{\mathrm{CH}_3-\overset{\mathrm{OH}}{\overset{\vert}{\mathrm{CH}}}-\mathrm{CH}_2-\mathrm{CHO}} H2OΔCH3CH=CHCHOBut-2-en-1-al\xrightarrow[\mathrm{-H_2O}]{\Delta} \underset{\text{But-2-en-1-al}}{\mathrm{CH}_3-\mathrm{CH}=\mathrm{CH}-\mathrm{CHO}}
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