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Aldehydes Ketones and Carboxylic Acids question

2022 · 25 Jul · Shift 2 · Q11
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Aldehydes Ketones and Carboxylic Acids question

2022 · 25 Jul · Shift 2 · Q11

JEE MainChemistryAldehydes Ketones and Carboxylic AcidsMCQ+4 / −1
CH3−CH2−CN⟶EtherCH3MgBrA→H3O+B⟶HClZn−HgCC{H_3} - C{H_2} - CN\mathrel{\mathop{\kern0pt\longrightarrow} \limits_{Ether}^{C{H_3}MgBr}} A\xrightarrow{{H_3}{O^ + }} B\mathrel{\mathop{\kern0pt\longrightarrow} \limits_{HCl}^{Zn - Hg}} CCH3​−CH2​−CNEther⟶CH3​MgBr​AH3​O+​BHCl⟶Zn−Hg​C The correct structure of C is
  1. A
    CH3−CH2−CH2−CH3C{H_3} - C{H_2} - C{H_2} - C{H_3}CH3​−CH2​−CH2​−CH3​
  2. B
    JEE Main 2022 (Online) 25th July Evening Shift Chemistry - Aldehydes, Ketones and Carboxylic Acids Question 98 English Option 2
  3. C
    JEE Main 2022 (Online) 25th July Evening Shift Chemistry - Aldehydes, Ketones and Carboxylic Acids Question 98 English Option 3
  4. D
    CH3−CH2−CH=CH2C{H_3} - C{H_2} - CH = C{H_2}CH3​−CH2​−CH=CH2​
View written solutionFree

Correct answer: A

  1. Identify the starting compound

    The given compound is propionitrile (ethyl cyanide): CH3−CH2−CN\mathrm{CH_3-CH_2-CN}CH3​−CH2​−CN

  2. Reaction with Grignard reagent

    A nitrile reacts with one mole of Grignard reagent to form an imine magnesium salt, which on hydrolysis gives a ketone.

    Here: CH3−CH2−CN→EtherCH3MgBrA\mathrm{CH_3-CH_2-CN \xrightarrow[Ether]{CH_3MgBr} A}CH3​−CH2​−CNCH3​MgBrEther​A

    The methyl group from CH3MgBr\mathrm{CH_3MgBr}CH3​MgBr adds to the carbon of the nitrile group.

    So intermediate AAA is the magnesium imine complex, and on hydrolysis: A→H3O+B\mathrm{A \xrightarrow{H_3O^+} B}AH3​O+​B gives the ketone: B=CH3−CH2−CO−CH3\mathrm{B = CH_3-CH_2-CO-CH_3}B=CH3​−CH2​−CO−CH3​

    This is 2-butanone.

  3. Reduction with Zn-Hg/HCl

    The reagent Zn−Hg/HCl\mathrm{Zn-Hg/HCl}Zn−Hg/HCl is Clemmensen reduction, which reduces the carbonyl group of aldehydes/ketones to a methylene group: R−CO−R′→Zn−Hg/HClR−CH2−R′\mathrm{R-CO-R' \xrightarrow{Zn-Hg/HCl} R-CH_2-R'}R−CO−R′Zn−Hg/HCl​R−CH2​−R′

    Therefore: CH3−CH2−CO−CH3→Zn−Hg/HClCH3−CH2−CH2−CH3\mathrm{CH_3-CH_2-CO-CH_3 \xrightarrow{Zn-Hg/HCl} CH_3-CH_2-CH_2-CH_3}CH3​−CH2​−CO−CH3​Zn−Hg/HCl​CH3​−CH2​−CH2​−CH3​

  4. Final structure of CCC

    C=CH3−CH2−CH2−CH3\mathrm{C = CH_3-CH_2-CH_2-CH_3}C=CH3​−CH2​−CH2​−CH3​

  5. Option check

    • A: CH3−CH2−CH2−CH3\mathrm{CH_3-CH_2-CH_2-CH_3}CH3​−CH2​−CH2​−CH3​ ✅
    • D: CH3−CH2−CH=CH2\mathrm{CH_3-CH_2-CH=CH_2}CH3​−CH2​−CH=CH2​ ❌ not formed in Clemmensen reduction

Hence, the correct answer is Option A.

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