Alcohols, Phenols and Ethers Practice Questions - DPP for NEET

 Page 1


1. (a) 3-Methyl pentanol-3 will be dehydrated most readily since it
produces tertiary carbonium ion as intermediate.
 
2. (d) 2-Phenylethanol, , is a 1º alcohol which can be
prepared from C
6
H
5
MgBr by treating with ethylene oxide (note
that HCHO will introduce only one carbon atom, i.e. it will give
C
6
H
5
CH
2
OH and not C
6
H
5
CH
2
CH
2
OH).
;
3. (b) Methyl vinyl ether under anhydrous condition at room temperature
undergoes addition reaction.
4. (b) Tertiary alcohols react fastest with conc. HCl and anhydrous ZnCl
2
(lucas reagent) as its mechanism proceeds through the formation of
stable tertiary carbocation.
Mechanism
Page 2


1. (a) 3-Methyl pentanol-3 will be dehydrated most readily since it
produces tertiary carbonium ion as intermediate.
 
2. (d) 2-Phenylethanol, , is a 1º alcohol which can be
prepared from C
6
H
5
MgBr by treating with ethylene oxide (note
that HCHO will introduce only one carbon atom, i.e. it will give
C
6
H
5
CH
2
OH and not C
6
H
5
CH
2
CH
2
OH).
;
3. (b) Methyl vinyl ether under anhydrous condition at room temperature
undergoes addition reaction.
4. (b) Tertiary alcohols react fastest with conc. HCl and anhydrous ZnCl
2
(lucas reagent) as its mechanism proceeds through the formation of
stable tertiary carbocation.
Mechanism
Step 1 :    
Step 2 :   
Step 3 :   
5. (d) The two components should be (CH
3
)
3
CONa + (CH
3
)
3
CBr.
However, tert-alkyl halides tend to undergo elimination reaction
rather than substitution leading to the formation of an alkene,
Me
2
C = CH
2
6. (d)
7. (a) MnO
2
 being a mild oxidising agent stops the oxidation of –
CH
2
OH group at aldehyde stage.
8. (b) The tertiary alkyl halide undergo elimination reaction to give
alkenes
 + NaOC
2
H
5 
--? 
9. (d)
Electron releasing groups (–CH
3
, –OCH
3
, –NCH
3
 etc) intensify the
negative charge of phenoxide ion, i.e., destablises it hence
Page 3


1. (a) 3-Methyl pentanol-3 will be dehydrated most readily since it
produces tertiary carbonium ion as intermediate.
 
2. (d) 2-Phenylethanol, , is a 1º alcohol which can be
prepared from C
6
H
5
MgBr by treating with ethylene oxide (note
that HCHO will introduce only one carbon atom, i.e. it will give
C
6
H
5
CH
2
OH and not C
6
H
5
CH
2
CH
2
OH).
;
3. (b) Methyl vinyl ether under anhydrous condition at room temperature
undergoes addition reaction.
4. (b) Tertiary alcohols react fastest with conc. HCl and anhydrous ZnCl
2
(lucas reagent) as its mechanism proceeds through the formation of
stable tertiary carbocation.
Mechanism
Step 1 :    
Step 2 :   
Step 3 :   
5. (d) The two components should be (CH
3
)
3
CONa + (CH
3
)
3
CBr.
However, tert-alkyl halides tend to undergo elimination reaction
rather than substitution leading to the formation of an alkene,
Me
2
C = CH
2
6. (d)
7. (a) MnO
2
 being a mild oxidising agent stops the oxidation of –
CH
2
OH group at aldehyde stage.
8. (b) The tertiary alkyl halide undergo elimination reaction to give
alkenes
 + NaOC
2
H
5 
--? 
9. (d)
Electron releasing groups (–CH
3
, –OCH
3
, –NCH
3
 etc) intensify the
negative charge of phenoxide ion, i.e., destablises it hence
decrease ionization of parent phenol. Therefore decreases acidity
while electron donating groups (–NO
2
, –COOH, –CHO etc.)
increases acidity.
10. (b) Number of active hydrogen in a compound corresponds to the
number of moles of CH
4
 evolved per mole of the compound.
 Cl
+ 
11. (a) + CO
2
 
+ CH
3
COOH
12. (b)
13. (c)
Page 4


1. (a) 3-Methyl pentanol-3 will be dehydrated most readily since it
produces tertiary carbonium ion as intermediate.
 
2. (d) 2-Phenylethanol, , is a 1º alcohol which can be
prepared from C
6
H
5
MgBr by treating with ethylene oxide (note
that HCHO will introduce only one carbon atom, i.e. it will give
C
6
H
5
CH
2
OH and not C
6
H
5
CH
2
CH
2
OH).
;
3. (b) Methyl vinyl ether under anhydrous condition at room temperature
undergoes addition reaction.
4. (b) Tertiary alcohols react fastest with conc. HCl and anhydrous ZnCl
2
(lucas reagent) as its mechanism proceeds through the formation of
stable tertiary carbocation.
Mechanism
Step 1 :    
Step 2 :   
Step 3 :   
5. (d) The two components should be (CH
3
)
3
CONa + (CH
3
)
3
CBr.
However, tert-alkyl halides tend to undergo elimination reaction
rather than substitution leading to the formation of an alkene,
Me
2
C = CH
2
6. (d)
7. (a) MnO
2
 being a mild oxidising agent stops the oxidation of –
CH
2
OH group at aldehyde stage.
8. (b) The tertiary alkyl halide undergo elimination reaction to give
alkenes
 + NaOC
2
H
5 
--? 
9. (d)
Electron releasing groups (–CH
3
, –OCH
3
, –NCH
3
 etc) intensify the
negative charge of phenoxide ion, i.e., destablises it hence
decrease ionization of parent phenol. Therefore decreases acidity
while electron donating groups (–NO
2
, –COOH, –CHO etc.)
increases acidity.
10. (b) Number of active hydrogen in a compound corresponds to the
number of moles of CH
4
 evolved per mole of the compound.
 Cl
+ 
11. (a) + CO
2
 
+ CH
3
COOH
12. (b)
13. (c)
14. (b) The commercial alcohol is made unfit for drinking by mixing in it
some copper sulphate (to give it colour) and pyridine (a foul
smelling liquid). It is known as denaturation of alcohol.
15. (b)  
16. (a)  
17. (c) Electron withdrawing group stabilises the benzene ring due to
delocalisation of charge.
–CH
3
 and –CH
2
OH are electron donating group and hence
decrease the stability of benzene ring –OCH
3
 is weaker electron
withdrawing group than –COCH
3
. Hence –COCH
3
 group more
stabilize the phenoxide ion at p-position.
18. (d) This method is suitable for the preparation of a wide variety of
unsymmetrical ethers. The nucleophilic substitution of halides with
alkoxide leads to desired product.
19. (c) Due to presence of methyl alcohol in liquor.
20. (d)  2
21. (4) P/Cl
2
, SOCl
2
, PCl
3
 and PCl
5
.
22. (5) (I), (III), (IV), (V) and (VI) will show iodoform test.
Page 5


1. (a) 3-Methyl pentanol-3 will be dehydrated most readily since it
produces tertiary carbonium ion as intermediate.
 
2. (d) 2-Phenylethanol, , is a 1º alcohol which can be
prepared from C
6
H
5
MgBr by treating with ethylene oxide (note
that HCHO will introduce only one carbon atom, i.e. it will give
C
6
H
5
CH
2
OH and not C
6
H
5
CH
2
CH
2
OH).
;
3. (b) Methyl vinyl ether under anhydrous condition at room temperature
undergoes addition reaction.
4. (b) Tertiary alcohols react fastest with conc. HCl and anhydrous ZnCl
2
(lucas reagent) as its mechanism proceeds through the formation of
stable tertiary carbocation.
Mechanism
Step 1 :    
Step 2 :   
Step 3 :   
5. (d) The two components should be (CH
3
)
3
CONa + (CH
3
)
3
CBr.
However, tert-alkyl halides tend to undergo elimination reaction
rather than substitution leading to the formation of an alkene,
Me
2
C = CH
2
6. (d)
7. (a) MnO
2
 being a mild oxidising agent stops the oxidation of –
CH
2
OH group at aldehyde stage.
8. (b) The tertiary alkyl halide undergo elimination reaction to give
alkenes
 + NaOC
2
H
5 
--? 
9. (d)
Electron releasing groups (–CH
3
, –OCH
3
, –NCH
3
 etc) intensify the
negative charge of phenoxide ion, i.e., destablises it hence
decrease ionization of parent phenol. Therefore decreases acidity
while electron donating groups (–NO
2
, –COOH, –CHO etc.)
increases acidity.
10. (b) Number of active hydrogen in a compound corresponds to the
number of moles of CH
4
 evolved per mole of the compound.
 Cl
+ 
11. (a) + CO
2
 
+ CH
3
COOH
12. (b)
13. (c)
14. (b) The commercial alcohol is made unfit for drinking by mixing in it
some copper sulphate (to give it colour) and pyridine (a foul
smelling liquid). It is known as denaturation of alcohol.
15. (b)  
16. (a)  
17. (c) Electron withdrawing group stabilises the benzene ring due to
delocalisation of charge.
–CH
3
 and –CH
2
OH are electron donating group and hence
decrease the stability of benzene ring –OCH
3
 is weaker electron
withdrawing group than –COCH
3
. Hence –COCH
3
 group more
stabilize the phenoxide ion at p-position.
18. (d) This method is suitable for the preparation of a wide variety of
unsymmetrical ethers. The nucleophilic substitution of halides with
alkoxide leads to desired product.
19. (c) Due to presence of methyl alcohol in liquor.
20. (d)  2
21. (4) P/Cl
2
, SOCl
2
, PCl
3
 and PCl
5
.
22. (5) (I), (III), (IV), (V) and (VI) will show iodoform test.
 
23. (6) All OH, SH and terminal alkyne will react with Grignard reagent
to produce CH
4
.
24. (3) For each   pair one HIO
4
 will be consumed.
25. (6)