JEE Advanced (Subjective Type Questions): Alcohols, Phenols & Ethers- 2 | Chapter-wise Tests for JEE Main & Advanced PDF Download

13. 3,3-Dimethylbutan-2-ol loses a molecule of water in the presence of concentrated sulphuric acid to give tetramethylethylene as a major product. Suggest a suitable mechanism. (1996 - 2 Marks)

Solution : 
The steps involved in the suggested mechanism are as follows.
(a) The protonation of hydroxyl group.

(b) The removal of H₂O to form a secondary (2^o) carbonium ion

(c) The conversion of 2^o carbonium to the more stable 3^o carbonium ion by the shift of CH³ group

(d) The removal of H⁺ to form a double bond

14. A compound D (C₈H₁₀O) upon treatment with alkaline solution of iodine gives a yellow precipitate. The filtrate on acidification gives a white solid E (C₇H₆O₂). Write the structures of D and E and explain the formation of E.       (1996 - 2 Marks)

Solution :

NOTE :
The reaction of D (C₈H₁₀O ) with alkaline solution of iodine is an iodoform reaction. This reaction is possible if the compound D has 

The high carbon content in D indicates that D is an aromatic compound containing a benzene ring. To account for the given formula, the compound D may be C₆H₅CH(OH)CH₃   

The given reactions are

15. An optically active alcohol A (C₆H₁₀O) absorbs two moles of hydrogen per mole of A upon catalytic hydrogenation and gives a product B. The compound B is resistant to oxidation by CrO₃ and does not show any optical activity. Deduce the structures of A and B. (1996 - 2 Marks)

Solution :
 TIPS/Formulae :
(a) Since (B) is resistant to oxidation, it must be ter-alcohol.
(b) Since (B) is optically inactive, it must have at least two  alkyl groups similar.

Thus the five carbon atoms can be adjusted into three alkyl groups (of which two are similar) either as –CH₃, –CH₃, and –C₃H₇, or as –C₂H₅, –C₂H₅ and –CH₃,
Thus the possible structure of alcohol (B) is either

Hence the corresponding compound (A) is either

However, the compound (A) is optically active, so (A) and hence also (B) should have right side structure.

16. Predict the structures of the intermediates/products in the following reaction sequence :                                        (1996 - 2 Marks)

Solution : 

17. 2, 2-Dimethyloxirane can be cleaved by acid (H⁺). Write mechanism. (1997 - 2 Marks)

Solution : 

TIPS/Formulae :
The oxirane ring is cleaved via S_N2 mechanism

18. Which of the following is the correct method for synthesising methyl-t-butyl ether and why?
 (i) (CH₃)₃CBr + NaOMe →
 (ii) CH₃Br + NaO-t-Bu →                                                                 (1997 - 2 Marks)

Solution : 

The method given in (ii) is the correct method for the formation of ether because method (i) leads alkene as the main product.
NOTE :
3° alkyl halides are easily dehydrohalogenated by base.

19. Write the intermediate steps for each of the following reaction.

Solution :

20. Explain briefly the formation of the products giving the structures of the intermediates. (1999 - 3 Marks)

Solution : 

(i) NOTE :
Since the large propenyl group is attached to the carbon atom bearing the hydroxyl group, so the reaction is likely to occur via S_N1 mechanism.

NOTE :
In the intermediate carbocation, Ia carbon bearing positive charge has CH3 group which decreases the positive charge and hence prevents cyclisation of the
compound.

21. A biologically active compound, bombykol (C₁₆H₃₀O) is obtained from a natural source. The structure of the compound is determined by the following reactions.
 (2002 - 5 Marks)

(a) On hydrogenation, bombykol gives a compound A, C₁₆H₃₄O, which reacts with acetic anhydride to give an ester;
 (b) Bombykol also reacts with acetic anhydride to give another ester, which on oxidative ozonolysis (O₃/H₂O₂) gives a mixture of butanoic acid, oxalic acid and 10-
 acetoxydecanoic acid.
 Determine the number of double bonds in bombykol. Write the structures of compound A and bombykol. How many geometrical isomers are possible for bombykol?

Solution : 

TIPS/Formulae :
Let us summarise the given facts.

(i) Hydrogenation of bombykol (C₁₆H₃₀O) to C₁₆H₃₄O (A) indicates the presence of two double bonds in bombykol.
(ii) Reaction of A with acetic anhydride to form ester indicates the presence of an alcoholic group in A and hence also in bombykol.
(iii) Products of oxidative ozonolysis of bombykol ester suggests the structure of bombykol.

The structure of the bombykol ester suggests that bombykol has the following structure :
CH₃CH₂CH₂CH = CH – CH = CH.(CH₂)₈.CH₂OH (Bombykol) and the structure of A is
CH₃CH₂CH₂CH₂CH₂CH₂CH₂(CH₂)₈.CH₂OH or C₁₆H₃₃OH.
Four geometrical isomers are possible for the above bombykol structure (as it has two double bonds).

22. An organic compound (P) of molecular formula C₅H₁₀O is treated with dil. H₂SO₄ to give two compounds (Q) and (R) both of which respond iodoform test. The rate of reaction of (P) with dil. H₂SO₄ is 10¹⁰ faster than the reaction of ethylene with dil. H₂SO₄. Identify the organic compounds, (P), (Q) and (R) and explain the extra reactivity of (P).               (2004 - 4 Marks)

Solution : 

(i) Molecular formula of P, C₅H₁₀O indicates 1º of unsaturation. So it should have double bond.
(ii) Acidic hydrolysis of P to Q and R, both of which responds iodoform test, indicates that Q and R should have following structure.
CH₃CH₂OH, (CH₃)₂CHOH, CH₃CHO or CH₃COR
The only possible linkage that can explain such hydrolysis is ether. Hence P should have following type of structure.
C₂ – Component – O – C₃ – Component
Further either the C₂ – or the C₃ – component should have double bond, thus the possible structure for P should be either of the following two structures which explains all the
given reactions. 

Extra reactivity of P toward dil. H₂SO₄ than ethylene is due to formation of highly stable carbocation

23. Identify (X) and (Y) in the following reaction sequence. (2005 - 2 Marks)

Solution :