Alcohol, Ethers & Phenol

ALCOHOL, ETHERS & PHENOL

Solution to Subjective Problems

LEVEL – I

1. HC ≡ C – CH2 – OH will be a stronger acid. According to the conjugate acid-base theory, if the conjugate base of an acid is weak, the corresponding acid is stronger.

HC ≡ C – CH2 – OH ⎯→ HC ≡ C – CH2 – O– + H+

Due to the –I effect of the sp hybridized carbon atom, the electron density flowing towards the – O– atom will be less and hence less will be the intensification of the charge. The base will be comparatively stable and hence weak.

 

2.
3.
4.
5.                     A =				B =		C =
6.

The mechanism involved in the reaction is

7. H2O ionises phenol into the most reactive phenoxide anion and hence trisubstitution occurs. Moreover H2O can stabilize the arenium ion intermediate formed as well as the electrophile Br+. Hence the rate of the reaction is fast and hence tri-substitutiion results. But in CS2, a non-polar solvent, the above said ionisation and stabilization is absent. Hence the rate of the reaction is slow and hence mono-substitution results.

8.
9.

 

10. a)
b)
c)

LEVEL – II

 

1.	A =			B =
	C =
2.	A =			B =
	C =			D =
	E =
3.

The negative charge on the ortho position can resonate to the para position also and thus we get the para isomer.

4.
5. a)
b)
6.	A =				B =
	C =				D =

7. Atomic weight Atomic ratio

C = 76.6 = 76.6/12 = 6.38 ≈6

H = 6.38 = 6.38/1 = 6.38 ≈ 6

O = 100 – (76.6 + 6.38) = 17.02/16 = 1.06 ≈ 1

Emperical formula = C6H6O

Emperical formula weight = 94

Molecular weight = V.d. × 2 = 47 × 2 = 94

So, molecular formula = C6H6O

Since (A) gives characteristic colour with aqueous FeCl3. So it must be a phenolic compound and the formula of (A) may be

The chemical reactions are as follows

8. A chromate ester is formed in the first step:

The Cr(IV) and Cr(VI) species react tor form 2Cr(V), which is turn also oxidizes alcohols giving Cr(III) will its characteristics colour.

9. We can start with oxiranes of the central part 

10.	A =			B =
	C =
	D =
	E =
11.

(nucleophile attacks the less substituted carbon in base-catalysed reaction)

 

12.	A =		B =
	C =		D =
	E =		F =
	G =

 

13.	Since alcohols are not obtained as products migration of H is faster than migration of R.
14. a)		D+ E (allylic, not vinylic Br reacts ⎯→ A
b)
c)

Although the ether is symmetrical, dehydration of PhCH(OH)CH3 cannot be used because it would readily undergo intermolecular dehydration to give the stable conjugated alkene, PhCH = CH2.

15. a) Add Acid Cr2O72– (orange). The 1° n-butyl alcohol is oxidized; its solution changes colour to green Cr(III). The 3° t-butyl alcohol is unchanged. Alternatively, when Lucas reagent (HCl+ZnCl2) is added, the 3° ROH quickly reacts to form the insoluble t-butyl chloride that appears as a second (lower) layer or a cloudiness. The 1° ROH does  not react and remains dissolved  in the reagent. 
16. b) Add I2 in OH– until the I2 colour  persists. A pale yellow precipitate of CHI3 appears, indicating that ethyl alcohol is oxidized. n-Propyl alcohol does not have the —CH(OH)CH3 goup and is not oxidized. 
17. c) Add Br2 in CCl4; as the Br2 adds to the C = C of the colorless allyl alcohol, its orange colour  disappears. The orange colour  persists in the unreactive n-propyl alcohol. 
18. d) Add  acid Cr2O72–. It ozidizes the alcohol, and the colour changes to green. The ether is unreactive. Alternately, if the two compounds are absolutely dry, add a small piece of Na (caution, use hood and wear goggles!) to each. H2 is released from the alcohol; the ether does not react. 

 

LEVEL – III

 

1.	A =		B =
	C =		D =
	E =		F =

 

2.	A =		B =
	C =		D =
	E =		F =

 

3.	A =					B =
	C =					D =
	E =
4. a)
b)
c)
d)
e)
5. a)		A =			Cl – CH2 – CH2 – OH		B =
		C =					D =
		E =			CHI3

 

b)	A =	H – H	B =	CH3 – – CH2 – CH2OH
	C =	CH3 – CH = CH – CH2OH	D =	CH3 – CH = CH – CHO
	E =	CH3 – CH = CH – CO2H

 

6. a)
b)
c)
d)
e)

 

7.	A =		B =
	C =		D =
	E =		F =

 

8. a)
b)

 

9. a) i)
ii)
iii)
b) i)
ii)
iii)
c) i)
ii)
d) i)
ii)

 

10. a)

 

b)

 

c)
d)
e)
11.		A =		CH3I	B =	PCC
		C =			D =

12. A is an ester because its acidic solution releases CO2 with NaHCO3 and it is sweet smelling.

R has to be H. So the structure of ‘A’ should be H – – OC3H7 (C3H7 may be isopropyl or n-propyl)

13.

14. The reaction is as follows.

The mechanism is

15.

(A) is the normal product