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Ethanol is prepared industrially by:
  1. hydration of ethylene
  2. fermentation of sugars
  3. both of the above
  4. none of these
Solution (C)
Hydration of alkenes:
Description: 50495.png
Description: 47216.png
Fermentation of sugars:
Description: 47230.png
Description: 47238.png


The alcohol that produces turbidity immediately with Description: 47247.png conc. HCl at room temperature:
  1. 1-hydroxybutane
  2. 2-hydroxybutane
  3. 2-hydroxy-2-methylpropane
  4. 1-hydroxy-2-methylpropane
Solution (C)
Description: 47258.png
Description: 47266.png; 3° reacts immediately
Description: 47274.png
Description: 47283.png; 2° reacts after 5 min
Description: 47291.png
Description: 47302.png; 1° reacts only on heating.


The alcohol which easily reacts with conc. HCl is:
  1. Description: 47314.png
  2. Description: 47323.png
  3. Description: 47332.png
  4. Description: 47342.png
Solution (B)
Tertiary alcohol readily reacts with halogen acid:
Description: 47349.png
The presence of three alkyl groups increases electron density on 3° carbon atom. Hence –OH group is easily removed. After the removal of –OH group, 3° carbonium ion is formed, which is most stable.


At low temperature, phenol reacts with Description: 47356.png in CS2 to form:
  1. m-bromophenol
  2. o- and p-bromophenol
  3. p-bromophenol
  4. 2,4,6-tribromophenol
Solution (B)
Description: 50505.png
In the presence of non-polar solvent (CS2), the ionization of phenol is suppressed. The ring is slightly activated and hence monosubstitution occurs. On the other hand, with Br2 water, phenol forms 2,4,6-tribromophenol.
Description: 50544.png
In aqueous solution, phenol ionizes to give phenoxide ion. Due to the presence of negative charge on oxygen, the benzene ring is highly activated and hence trisubstituted product is obtained.


An unknown compound (A) C8H10O3 on acetylation with CH3COCl/Py forms acetyl derivative of (A) whose molecular weight is 280. (A) on treatment with CH2N2 gives methyl ether of (B) having molecular weight 182. The number of phenolic hydroxyls and alcoholic hydroxyls in the compound (A) will respectively be:
  1. 1, 3
  2. 3, 1
  3. 2, 1
  4. 1, 2
Solution (C)
Number of hydroxyl group by acetylation method
Description: 47369.png Description: 47382.png
Number of phenolic hydroxyl
Description: 47395.png Description: 47402.png
Hence number of alcoholic –OH = 3 – 2 = 1.

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