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Haloalkanes and Haloarenes question

2024 · 31 Jan · Shift 1 · Q9
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Haloalkanes and Haloarenes question

2024 · 31 Jan · Shift 1 · Q9

JEE MainChemistryHaloalkanes and HaloarenesMCQ+4 / −1
The product (C) in the below mentioned reaction is : CH3−CH2−CH2−Br→ΔKOH(alc) A→HBrB→KOH(aq)ΔC\mathrm{CH}_3-\mathrm{CH}_2-\mathrm{CH}_2-\mathrm{Br} \xrightarrow[\Delta]{\mathrm{KOH}_{(\text {alc) }}} \mathrm{A} \xrightarrow{\mathrm{HBr}} \mathrm{B} \xrightarrow[\mathrm{KOH}_{(\mathrm{aq})}]{\Delta} \mathrm{C}CH3​−CH2​−CH2​−BrKOH(alc) ​Δ​AHBr​BΔKOH(aq)​​C
  1. A
    Propan-1-ol
  2. B
    Propyne
  3. C
    Propan-2-ol
  4. D
    Propene
View written solutionFree

Correct answer: C

  1. First reaction: alcoholic KOH with heat
    Starting compound is 1-bromopropane, CH3−CH2−CH2−Br\mathrm{CH_3-CH_2-CH_2-Br}CH3​−CH2​−CH2​−Br.

    Alcoholic KOH and heat favor elimination (β\betaβ-elimination), so HBr is removed: CH3−CH2−CH2−Br→ΔKOH(alc)CH3−CH=CH2\mathrm{CH_3-CH_2-CH_2-Br} \xrightarrow[\Delta]{KOH(alc)} \mathrm{CH_3-CH=CH_2}CH3​−CH2​−CH2​−BrKOH(alc)Δ​CH3​−CH=CH2​ Thus, A=propeneA = \text{propene}A=propene

  2. Second reaction: addition of HBr
    Propene reacts with HBr by Markovnikov addition (since no peroxide is mentioned).

    CH3−CH=CH2+HBr→CH3−CHBr−CH3\mathrm{CH_3-CH=CH_2 + HBr \rightarrow CH_3-CHBr-CH_3}CH3​−CH=CH2​+HBr→CH3​−CHBr−CH3​

    Thus, B=2-bromopropaneB = \text{2-bromopropane}B=2-bromopropane

  3. Third reaction: aqueous KOH with heat
    Aqueous KOH favors nucleophilic substitution, replacing Br by OH: CH3−CHBr−CH3→ΔKOH(aq)CH3−CHOH−CH3\mathrm{CH_3-CHBr-CH_3 \xrightarrow[\Delta]{KOH(aq)} CH_3-CHOH-CH_3}CH3​−CHBr−CH3​KOH(aq)Δ​CH3​−CHOH−CH3​

    Thus, C=propan-2-olC = \text{propan-2-ol}C=propan-2-ol

  4. Matching with options

    • A: Propan-1-ol →\to→ incorrect
    • B: Propyne →\to→ incorrect
    • C: Propan-2-ol →\to→ correct
    • D: Propene →\to→ incorrect

Therefore, the product CCC is: Propan-2-ol\boxed{\text{Propan-2-ol}}Propan-2-ol​

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