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JEE Main Haloalkanes, Alcohols and Ethers — practice questions

Explore 20 free Haloalkanes, Alcohols and Ethers questions in JEE Main Chemistry. Open each question or exercise to read its content, reference answers and available explanation.

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Haloalkanes, Alcohols and Ethers questions and exercises with answers

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For a galvanic cell, the standard cell EMF is given by:Read the answer and worked solution →Oxidation state of Cu in [Cu(NH₃)₄]²⁺:Read the answer and worked solution →Carbonyl carbon hybridization:Read the answer and worked solution →Phenol is more acidic than ethanol because:Read the answer and worked solution →The Williamson ether synthesis works best with:Read the answer and worked solution →NEET · Chemistry · SN1 or SN2Read the answer and worked solution →Fix the wrong word on halide reactionsRead the answer and worked solution →Sort each clue into SN1 or SN2Read the answer and worked solution →Match each named reactionRead the answer and worked solution →Sort the SN1 and SN2 traitsRead the answer and worked solution →Swipe by the carbon holding XRead the answer and worked solution →Spot the false claimRead the answer and worked solution →Pick the bigger halide valueRead the answer and worked solution →Alcohol, phenol or ether?Read the answer and worked solution →True or false: alcohols and phenolsRead the answer and worked solution →NEET · Chemistry · SN1 or SN2Read the answer and worked solution →True or false: alcohols and phenolsRead the answer and worked solution →The Williamson ether synthesis works best with:Read the answer and worked solution →Alcohol, phenol or ether?Read the answer and worked solution →Phenol is more acidic than ethanol because:Read the answer and worked solution →

Worked practice questions

For a galvanic cell, the standard cell EMF is given by:

  1. $E^{\circ}_{\text{cell}} = E^{\circ}_{\text{cathode}} + E^{\circ}_{\text{anode}}$
  2. $E^{\circ}_{\text{cell}} = E^{\circ}_{\text{anode}} - E^{\circ}_{\text{cathode}}$
  3. $E^{\circ}_{\text{cell}} = E^{\circ}_{\text{cathode}} - E^{\circ}_{\text{anode}}$
  4. $E^{\circ}_{\text{cell}} = -2 E^{\circ}_{\text{anode}}$
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Correct answer: C. $E^{\circ}_{\text{cell}} = E^{\circ}_{\text{cathode}} - E^{\circ}_{\text{anode}}$

$E^{\circ}_{\text{cell}} = E^{\circ}_{\text{cathode}} - E^{\circ}_{\text{anode}}$, where both potentials are listed as standard *reduction* potentials. For a spontaneous galvanic cell, $E^{\circ}_{\text{cell}} > 0$, which means the cathode has the higher reduction potential. This is just an algebraic restatement of "reduction happens at the cathode".

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Oxidation state of Cu in [Cu(NH₃)₄]²⁺:

  1. 0
  2. +1
  3. +3
  4. +2 (charge of complex - 4 × 0 (NH₃) = +2)
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Correct answer: D. +2 (charge of complex - 4 × 0 (NH₃) = +2)

NH₃ neutral. Charge on complex +2 = ox of Cu. Cu²⁺ is d⁹ (Jahn-Teller distortion). Deep blue color due to d-d transition.

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Carbonyl carbon hybridization:

  1. sp³
  2. sp³d
  3. sp² (planar, 120° bond angles)
  4. sp
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Correct answer: C. sp² (planar, 120° bond angles)

C=O is sp² (3 σ bonds + 1 π in C=O). Trigonal planar around C. Important for nucleophilic attack from above/below plane.

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