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Aldehydes, Ketones and Carboxylic Acids · Question 17

JEE Main 2026 (Online) 23rd January Morning ShiftMCQMedium📈 Topic in 19% papers+4−1
1

[figure] The correct sequence of reagents for the above conversion of X to Y is :

  1. (i) NaOH(aq), (ii) Jones reagent, (iii) H3O+
  2. (i) NaOEt, (ii) B2H6/H2O2, (iii) Jones reagent
  3. (i) Jones reagent, (ii) NaOEt, (iii) Hot KMnO4/KOH
  4. (i) B2H6/H2O2, (ii) NaOEt, (iii) Jones reagent

Answer

(2) (i) NaOEt, (ii) B2H6/H2O2, (iii) Jones reagent

Solution

X is 1-bromo-1-phenylethane; in Y the oxygen function is on the terminal carbon, so an anti-Markovnikov route is needed:

[figure]

(i) NaOEt (strong base): dehydrobromination gives styrene, C6H5CH=CH2

(ii) B2H6/H2O2 (hydroboration–oxidation): anti-Markovnikov addition of water gives 2-phenylethanol, C6H5CH2CH2OH

(iii) Jones reagent (CrO3/H2SO4): oxidises the 1° alcohol to phenylacetic acid, C6H5CH2COOH (Y)

Option (1) puts –OH on the benzylic carbon (Jones reagent then gives a ketone), in option (3) Jones reagent cannot act on the alkyl halide, and in option (4) B2H6 cannot act on an alkyl halide.

Answer: (2)

JEE Main 2026 (Online) 23rd January Morning Shift

Keep practising — Aldehydes, Ketones and Carboxylic Acids

2

Which of the following statements are TRUE about Haloform reaction? Choose the correct answer from the options given below :

  1. A, B & C Only
  2. A, C & E Only
  3. B, D & E Only
  4. A & C Only
JEE Main 2026 (Online) 23rd January Evening Shift
3

Iodoform test can differentiate between Choose the correct answer from the options given below :

  1. A, B & E Only
  2. A & D Only
  3. A & E Only
  4. B, C & E Only
JEE Main 2026 (Online) 23rd January Evening Shift
4

Consider the following reaction of benzene. [figure] In compound (Q), the percentage of oxygen is %. (Nearest integer)

JEE Main 2026 (Online) 23rd January Evening Shift
5

Match the LIST-I with LIST-II Choose the correct answer from the options given below :

  1. A-IV, B-III, C-II, D-I
  2. A-III, B-I, C-IV, D-II
  3. A-II, B-I, C-IV, D-III
  4. A-III, B-IV, C-I, D-II
JEE Main 2026 (Online) 22nd January Morning Shift
6

The compound A, C8H8O2 reacts with acetophenone to form a single product via cross-Aldol condensation. The compound A on reaction with conc. NaOH forms a substituted benzyl alcohol as one of the two products. The compound A is :

  1. 2-hydroxy acetophenone
  2. 4-hydroxy benzaldehyde
  3. 4-methyl benzoic acid
  4. 4-methoxy benzaldehyde
JEE Main 2026 (Online) 22nd January Evening Shift
7

An organic compound "P" of molecular formula C6H12O3 gives positive iodoform test but negative Tollens' test. When "P" is treated with dilute acid, it produces "Q". "Q" gives positive Tollens' test and also iodoform test. The structure of "P" is :

  1. H- O \|C-CH2-CH2- OCH3 |CH-OCH3
  2. CH3- O \|C- OCH3 |CH- OCH3 |CH2
  3. CH3- O \|C-CH2- OCH3 |CH-OCH3
  4. CH3- O \|C- OCH3 | OCH3 |C-CH3
JEE Main 2026 (Online) 21st January Morning Shift
8

Match List - I with List - II. Choose the correct answer from the options given below :

  1. A-II, B-III, C-IV, D-I
  2. A-IV, B-III, C-I, D-II
  3. A-IV, B-I, C-II, D-III
  4. A-II, B-IV, C-I, D-III
JEE Main 2026 (Online) 21st January Evening Shift
9

A [(ii) H3O+](i) NaOH B [(ii) H2SO4, ](i) EtOH C 'A' shows positive Lassaigne's test for N and its molar mass is 121. 'B' gives effervescence with aq NaHCO3. 'C' gives fruity smell. Identify A, B and C from the following.

  1. [figure]
  2. [figure]
  3. [figure]
  4. [figure]
JEE Main 2025 (Online) 8th April Evening Shift
10

When [figure] undergoes intramolecular aldol condensation, the major product formed is :

  1. [figure]
  2. [figure]
  3. [figure]
  4. [figure]
JEE Main 2025 (Online) 8th April Evening Shift
11

Which of the following compounds is least likely to give effervescence of CO2 in presence of aq. NaHCO3?

  1. Ph- NH3 Cl
  2. [figure]
  3. [figure]
  4. [figure]
JEE Main 2025 (Online) 7th April Morning Shift

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