Alcohol and Ether ‌–‌ ‌JEE‌ ‌Advanced‌ ‌Previous‌ ‌Year‌ ‌Questions with Solutions

Alcohol and Ether JEE Advanced Previous Year Questions with Solutions have been made exclusively for JEE aspirants to help them strengthen their understanding of the different aspects of the question paper and boost their preparation for the entrance exam. With a list of questions provided here, students can engage in solving the Alcohol and Ether JEE Advanced previous year questions to revise and gain a better understanding of the different topics given in this chapter. Additionally, solving the questions and going through the detailed solutions will enhance the student’s problems solving skills and at the same time get a better grip on the important concepts.

Alcohol and Ether JEE Advanced Previous Year Questions with Solutions has been created to ensure that aspirants prepare in the right manner. We have included as many questions that we could find from all the previous sessions of JEE‌ ‌Advanced‌‌. The questions and solutions can also be accessed in the form of a PDF and downloaded freely from the link given below.

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JEE Advanced Previous Year Questions on Alcohol and Ether

Question 1. Total number of hydroxyl groups present in a molecule of the major product P is.

Solution: (6)

The total number of hydroxyl groups present in a molecule of the major product P is 6

In step (i), the Lindlar catalyst converts the given alkyne into a cis alkene. Further in step (ii), syn dihydroxylation takes place with the help of dilute KMnO₄.

Question 2. The number of the hydroxyl group(s) in Q is ___

Solution: (4)

∴ The total number of hydroxyl groups in Q is four.

In the first step, the acid-catalyzed dehydration of alcohol takes place to obtain an alkene. In this reaction, secondary carbocation rearranges to tertiary carbocation to form a more stable alkene. The next step is the dihydroxylation of two double bonds which introduces 4 -OH groups in the molecule.

Question 3. The correct combination of names for isomeric alcohols with molecular formula C₄H₁₀O is/are.

A. tert-butanol and 2-methylpropan-2-ol

B. tert-butanol and 1,1-dimethylethan-1-ol

C. n-butanol and butan-1-ol

D. Isobutyl alcohol and 2-methylpropan-1-ol

Solution: (A, C and D)

Isomeric alcohols of C₄H₁₀O are:

OR

Question 4. For the identification of β-naphthol using dye test, it is necessary to use:

A. Dichloromethane solution of β-naphthol

B. Acidic solution of β-naphthol

C. Neutral solution of β-naphthol

D. Alkaline solution of β-naphthol

Solution: (D)

In alkaline solution -OH convert to −O^– which reacts with azo compound i.e, Ph − N ≡ N⁺ to form an azo dye.

D is the correct option.

Question 5. An organic compound (C₈H₁₀O₂) rotates plane-polarized light. It produces pink colour with a neutral FeCl₃ solution. What is the total number of all the possible isomers for this compound?

Solution: (6)

DBE (Double bond equivalent) of C₈H₁₀O₂ is

= Number of carbon atoms – (Number of monovalent atoms) / 2 + 1

= 8 – 10 / 2 + 1 = 4

It gives pink colour with a neutral FeCl₃ solution. It means the phenolic group should be present in the compound.

Note: C* represent chiral carbon. So it will have (d and l) optically active isomers.

Total optically active isomer = 6

Question 6. The number of resonance structures for ‘N’ is ___.

Solution: (9)

Naphthalen-2-ol acts as an acid and reacts with NaOH to form salt sodium naphthalen-2-olate which is compound N. The number of resonance structures for N is 9.

Question 7. The reactivity of compound Z with different halogens under appropriate conditions is given below:

The observed pattern of electrophilic substitution can be explained by:

A. The steric effect on the halogen

B. The steric effect of the tert – butyl group

C. The electronic effect of the phenolic group

D. The electronic effect of the tert – butyl group

Solution: (A, B and C)

This problem includes the concept of the effect of steric and electronic effects on the reactivity of organic compounds.

The steric effect of halogens is as follows

Cl₂ < Br₂ < I₂

The electronic effect of the phenolic group directs the approaching electrophile towards ortho and para positions.

The tertiary butyl group has a large size so it causes a steric effect around the aromatic nucleus.

Hence orientation in electrophilic substitution reaction is decided by –

(A) The steric effect of halogen

(B) The steric effect of the tertiary butyl group

(C) The electromeric effect of phenolic group

Question 8. Total number of isomers, considering both structural and stereoisomers, of cyclic ethers with the molecular formula C₄H₈O is ____.

Solution: (10)

One of the four-membered cyclic structures exists as pair of enantiomers, contributing 2 to the total.

Two of the three-membered cyclic structures exist as pairs of enantiomers, contributing 4 to the total.

Also one of the three-membered structures exhibits geometrical isomerism adding 1 to the total structure.

Question 9. Match the chemical conversions in List-I with appropriate reagents given in List-II.

Solution: (A)

Question 10. The major product(s) of the following reaction is(are):

A. P

B. Q

C. R

D. S

Solution: (B)

If we look at phenol, we can say that the hydroxide group is the ortho and para directing group. Thus, bromination occurs at ortho and para positions to form 2,4,6-tribromophenol.

On the other hand, SO₃ is a good leaving group. The sulphonic acid group is also replaced with bromine. Thus, the product is Q.

Generally, phenols are very reactive towards bromine water, that even substituents like −SO₃H, etc. that are present in o- and p-positions, are replaced by halogens.

Alcohols Phenols and Ethers – Important Topics