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CBSE Class 12 Chemistry 2017 Delhi Set 1 Solved Paper

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Question : 26 of 26
Marks: +1, -0
(a) Write the product (s) in the following reactions:
(ii) CH3−CH∣CH3−O−CH2−CH3\mathrm{CH}_3 - \overset{\mathrm{CH}_3}{\overset{\vert}{\mathrm{CH}}} - \mathrm{O} - \mathrm{CH}_2 - \mathrm{CH}_3 →HI\xrightarrow{\mathrm{HI}} ? ?
(iii) CH3−CH=CH−CH2−OH\mathrm{CH}_3 - \mathrm{CH} = \mathrm{CH} - \mathrm{CH}_2 - \mathrm{OH} →PCC\xrightarrow{\mathrm{PCC}} ?
(b) Give simple chemical tests to distinguish between the following pairs of compounds:
(i) Ethanol and phenol
(ii) Propanol and 2-methylpropan-2-ol
OR
(a) Write the formula of reagents used in the following reactions :
(i) Bromination of phenol to 2, 4, 6-tribromophenol
(ii) Hydroboration of propene and then oxidation to propanol.
(b) Arrange the following compound groups in the increasing order of their property indicated:
(i) p-nitrophenol, ethanol, phenol (acidic character)
(ii) Propanol, Propane, Propanal (boiling point)
(c) Write the mechanism (using curved arrow notation) of the following reaction :
CH3−CH2−O+H2→CH3CH2OH\mathrm{CH}_3 - \mathrm{CH}_2 - \mathrm{O}^{+} \mathrm{H}_2 \xrightarrow{\mathrm{CH}_3\mathrm{CH}_2\mathrm{OH}} CH3−CH2−OH∣+−CH2\mathrm{CH}_3 - \mathrm{CH}_2 - \mathrm{O}^{ \overset{+}{\underset{\vert}{\mathrm{H}}} } - \mathrm{CH}_2 −CH3+H2O-\mathrm{CH}_3 + \mathrm{H}_2\mathrm{O}
Solution:  
(a) (i) Acetylation of salicylic acid:
Hence, the products are aspirin and acetic acid.
(ii) CH3−CH∣CH3−O−CH2−CH3 ⟶HI\mathrm{CH}_3 - \overset{\mathrm{CH}_3}{\overset{\vert}{\mathrm{CH}}} - \mathrm{O} - \mathrm{CH}_2 - \mathrm{CH}_3 \,\overset{\mathrm{HI}}{\longrightarrow} CH3−CH∣CH3−OHIsopropyl alcohol+CH3−CH2−IIodoethane\underset{\text{Isopropyl alcohol}}{\mathrm{CH}_3 - \overset{\mathrm{CH}_3}{\overset{\vert}{\mathrm{CH}}} - \mathrm{OH}} + \underset{\text{Iodoethane}}{\mathrm{CH}_3 - \mathrm{CH}_2 - \mathrm{I}}
Hence, the products are Isopropyl alcohol and iodoethane.
(iii) CH3−CH=CH−CH2−OH⟶PCCCH3−CH=CH−CHOBut-2-enal\mathrm{CH}_3 - \mathrm{CH} = \mathrm{CH} - \mathrm{CH}_2 - \mathrm{OH} \overset{\mathrm{PCC}}{\longrightarrow} \underset{\text{But-2-enal}}{\mathrm{CH}_3 - \mathrm{CH} = \mathrm{CH} - \mathrm{CHO}}
Hence, the product is But-2-enal.
(b) (i) Distinguish between ethanol and phenol:
 Test  Ethanol  Phenol
  Coupling reaction  Negative test  Positive test
Reaction will be :
 Iodoform test  Positive test
Reaction will be :
  Negative test
(ii) Distinction between propanol and 2-methylpropan-2-ol
By Lucas test
CH3−CH2−CH2−OHPropanol→ZnC2Conc. HCl\underset{\text{Propanol}}{\mathrm{CH}_3 - \mathrm{CH}_2 - \mathrm{CH}_2 - \mathrm{OH}} \xrightarrow[\mathrm{ZnC}_2]{\text{Conc. HCl}} Turbidity appears after heating
CH3−C∣CH3∣OH−CH32-Methylpropan-2-ol\underset{\text{2-Methylpropan-2-ol}}{\mathrm{CH}_3 - \underset{\underset{\mathrm{OH}}{\vert}}{\overset{\mathrm{CH}_3}{\overset{\vert}{\mathrm{C}}}} - \mathrm{CH}_3} Turbidity appears very quickly
OR
(a) (i) Br2/H2O\mathrm{Br}_2 / \mathrm{H}_2\mathrm{O}
(ii) (BH3)2,H2O2/OH−(\mathrm{BH}_3)_2, \mathrm{H}_2\mathrm{O}_2 / \mathrm{OH}^{-}
CH3−CH=CH2  (a)  BH3/THF  /  (b)  H2O2/OH\mathrm{CH}_3 - \mathrm{CH} = \mathrm{CH}_2 \; \text{(a)} \; \mathrm{BH}_3 / \mathrm{THF} \; / \; \text{(b)} \; \mathrm{H}_2\mathrm{O}_2 / \mathrm{OH} CH3−CH2−CH2−OH\mathrm{CH}_3 - \mathrm{CH}_2 - \mathrm{CH}_2 - \mathrm{OH}
(b) (i) p-nitrophenol >> phenol >> ethanol
(ii) propanol >> propanal >> propane
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