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Aldehydes Ketones and Carboxylic Acids question

2003 · Shift 0 · Q17
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Aldehydes Ketones and Carboxylic Acids question

2003 · Shift 0 · Q17

JEE MainChemistryAldehydes Ketones and Carboxylic AcidsMCQ+4 / −1
When CH2CH_2CH2​ = CHCHCH - COOHCOOHCOOH is reduced with LiAlH4LiAlH_4LiAlH4​, the compound obtained will be
  1. A
    CH2CH_2CH2​ = CHCHCH - CH2OHCH_2OHCH2​OH
  2. B
    CH3CH_3CH3​ - CH2CH_2CH2​ - CH2OHCH_2OHCH2​OH
  3. C
    CH3CH_3CH3​ - CH2CH_2CH2​ - CHOCHOCHO
  4. D
    CH3CH_3CH3​ - CH2CH_2CH2​ - COOHCOOHCOOH
View written solutionFree

Correct answer: A

  1. Identify the substrate

    The given compound is acrylic acid: CH2=CH−COOHCH_2=CH-COOHCH2​=CH−COOH

  2. Recall the action of LiAlH4LiAlH_4LiAlH4​

    Lithium aluminium hydride is a strong reducing agent.

    • It reduces carboxylic acids (−COOH)(-COOH)(−COOH) to primary alcohols (−CH2OH)(-CH_2OH)(−CH2​OH).
    • It generally does not reduce an isolated C=C bond in this context.

  3. Apply the reduction

    So, in CH2=CH−COOHCH_2=CH-COOHCH2​=CH−COOH the −COOH-COOH−COOH group becomes −CH2OH-CH_2OH−CH2​OH while the double bond remains unchanged.

    Therefore: CH2=CH−COOH→LiAlH4CH2=CH−CH2OHCH_2=CH-COOH \xrightarrow{LiAlH_4} CH_2=CH-CH_2OHCH2​=CH−COOHLiAlH4​​CH2​=CH−CH2​OH

  4. Match with the options

    • A: CH2=CH−CH2OHCH_2=CH-CH_2OHCH2​=CH−CH2​OH ✅
    • B: CH3−CH2−CH2OHCH_3-CH_2-CH_2OHCH3​−CH2​−CH2​OH ❌ double bond also reduced, which LiAlH4LiAlH_4LiAlH4​ does not do here
    • C: CH3−CH2−CHOCH_3-CH_2-CHOCH3​−CH2​−CHO ❌ not the reduction product of a carboxylic acid with LiAlH4LiAlH_4LiAlH4​
    • D: CH3−CH2−COOHCH_3-CH_2-COOHCH3​−CH2​−COOH ❌ unrelated hydrogenation product

  5. Final answer

    The compound obtained is: CH2=CH−CH2OH\boxed{CH_2=CH-CH_2OH}CH2​=CH−CH2​OH​ Hence, Option A is correct.

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