Sign in
12thPass logo
New chatPYQ LibraryDoubtsRank report
Sign in to see Recents

Your guest activity stays on this device

Sign in to save progress →
Sign in

Alcohols Phenols and Ethers question

2010 · Shift 1 · Q16
Guest · filters and generic practice availableBrowsing as a guest · PYQ filters and generic practice are available. Sign in only for personalised features and saved progress.
  1. PYQ Library
  2. /JEE Advanced
  3. /Chemistry
  4. /Alcohols Phenols and Ethers
  5. /2010 · Shift 1 · Q16

Alcohols Phenols and Ethers question

2010 · Shift 1 · Q16

JEE AdvancedChemistryAlcohols Phenols and EthersMCQ+3 / −1
In the reaction IIT-JEE 2010 Paper 1 Offline Chemistry - Alcohols, Phenols and Ethers Question 12 English the products are :
  1. A
    IIT-JEE 2010 Paper 1 Offline Chemistry - Alcohols, Phenols and Ethers Question 12 English Option 1
  2. B
    IIT-JEE 2010 Paper 1 Offline Chemistry - Alcohols, Phenols and Ethers Question 12 English Option 2
  3. C
    IIT-JEE 2010 Paper 1 Offline Chemistry - Alcohols, Phenols and Ethers Question 12 English Option 3
  4. D
    IIT-JEE 2010 Paper 1 Offline Chemistry - Alcohols, Phenols and Ethers Question 12 English Option 4
View written solutionFree

Correct answer: D

Step-by-step Solution:

  1. Identify Reactants and Reaction Type: The reaction involves an ether, specifically ethoxybenzene (also known as phenetole), reacting with hydroiodic acid (HI). Reactant: C6H5-O-CH2-CH3 Reagent: HI This is a classic example of ether cleavage by a strong acid.

  2. Mechanism of Ether Cleavage with HI: The reaction proceeds via a two-step mechanism:

    • Step 1: Protonation of the Ether Oxygen: The oxygen atom in the ether has lone pairs of electrons, making it a Lewis base. It gets protonated by the strong acid HI to form a protonated ether (an oxonium ion). C6H5−O−CH2−CH3+HI⇌[C6H5−O+(H)−CH2−CH3]+I−C_6H_5-O-CH_2-CH_3 + HI \rightleftharpoons [C_6H_5-\stackrel{+}{O}(H)-CH_2-CH_3] + I^-C6​H5​−O−CH2​−CH3​+HI⇌[C6​H5​−O+​(H)−CH2​−CH3​]+I−

    • Step 2: Nucleophilic Attack by Iodide Ion (I-): The iodide ion is a strong nucleophile. It attacks one of the carbon atoms attached to the positively charged oxygen, leading to the cleavage of a C-O bond. There are two possible sites for attack: a) The sp2sp^2sp2 hybridized carbon of the phenyl group (C6H5-). b) The sp3sp^3sp3 hybridized primary carbon of the ethyl group (-CH2-CH3).

  3. Analysis of the Nucleophilic Attack Sites:

    • Attack on the Phenyl Carbon: The C(aryl)-O bond has a partial double bond character due to resonance between the lone pairs on the oxygen and the benzene ring. This makes the bond stronger and difficult to break. Furthermore, SN2S_N2SN​2 reactions do not occur on sp2sp^2sp2 hybridized carbons of an aromatic ring. Therefore, the iodide ion does not attack the phenyl carbon.

    • Attack on the Ethyl Carbon: The C(alkyl)-O bond is a simple sigma bond. The carbon of the ethyl group is a primary sp3sp^3sp3 hybridized carbon, which is susceptible to SN2S_N2SN​2 attack. The iodide ion (I-) will attack this less sterically hindered carbon.

  4. Formation of Products: The iodide ion attacks the ethyl carbon in an SN2S_N2SN​2 reaction. This breaks the O-CH2CH3 bond. Phenol acts as the leaving group. I−+CH3−CH2‾−O+(H)−C6H5⟶CH3−CH2−I+C6H5−OHI^- + CH_3-\underline{CH_2}-\stackrel{+}{O}(H)-C_6H_5 \longrightarrow CH_3-CH_2-I + C_6H_5-OHI−+CH3​−CH2​​−O+​(H)−C6​H5​⟶CH3​−CH2​−I+C6​H5​−OH The products formed are ethyl iodide (CH3CH2I) and phenol (C6H5OH).

  5. Evaluate the Options:

    • A: C6H5-CH2-CH2-I and H2O - Incorrect. The reactant structure is changed.
    • B: C6H5-CH2-OH and CH3I - Incorrect. The reactant is ethoxybenzene, not a related isomer.
    • C: C6H5-I and CH3-CH2-OH - Incorrect. This would result from cleavage of the strong C(aryl)-O bond, which does not happen.
    • D: C6H5-OH and CH3-CH2-I - Correct. This matches the products derived from the correct reaction mechanism.

Conclusion:

The reaction of ethoxybenzene with HI results in the cleavage of the alkyl-oxygen bond, not the aryl-oxygen bond. This yields phenol and ethyl iodide. Therefore, option D is the correct answer.

PreviousNext

More from Alcohols Phenols and Ethers

  • In the reaction The intermediate(s) is(are) Includes diagram2010 · Multiple correct
  • The compounds P,Q and S were separately subjected to nitration using HNO3​/H2​SO4​ mixture. The major product formed in each case respectively, is :2010 · MCQ
  • A tertiary alcohol H upon acid catalysed dehydration gives a product I. Ozonolysis of I leads to compounds J and K compound J upon reaction with KOH gives benzyl alcohol and a compound L, whereas K on reaction with KOH gives only M.… Includes diagram2008 · MCQ
  • A tertiary alcohol H upon acid catalysed dehydration gives a product I. Ozonolysis of I leads to compounds J and K compound J upon reaction with KOH gives benzyl alcohol and a compound L, whereas K on reaction with KOH gives only M.… Includes diagram2008 · MCQ
  • A tertiary alcohol H upon acid catalysed dehydration gives a product I. Ozonolysis of I leads to compounds J and K compound J upon reaction with KOH gives benzyl alcohol and a compound L, whereas K on reaction with KOH gives only M.… Includes diagram2008 · MCQ
  • The reaction sequence given below is carried out with 16 moles of X. The yield of the major product in each step is given below the product in parentheses. The amount (in grams) of S produced is ​. Use: Atomic mass… Includes diagram2025 · Numerical
  • The correct reaction/reaction sequence that would produce a dicarboxylic acid as the major product is :2025 · MCQ
  • For the reaction sequence given below, the correct statement(s) is(are) Includes diagram2025 · Multiple correct