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JEE Main 20241 FebruaryMorning Shift
Exam
JEE Main
Year
2024
Date
1 February
Shift
Morning Shift
Subject
Chemistry
Chapter
Aldehydes and Ketones

JEE Main 2024 · 1 February · Morning Shift · Chemistry · Aldehydes and Ketones

Match List – I with List –II. Choose the correct answer from...

Match List – I with List –II.

Match List – I with List –II. Choose the correct answer from options given below:

Choose the correct answer from options given below:

Options

  1. AA−(III),(B)−(IV),(C)−(I),(D)−(II)A - \left(III\right), \left(B\right) - \left(IV\right), \left(C\right) - \left(I\right), \left(D\right) - \left(II\right)
  2. BA−(IV),(B)−(II),(C)−(I),(D)−(III)A - \left(IV\right), \left(B\right) - \left(II\right), \left(C\right) - \left(I\right), \left(D\right) - \left(III\right)
  3. CA−(IV),(B)−(II),(C)−(III),(D)−(I)A - \left(IV\right), \left(B\right) - \left(II\right), \left(C\right) - \left(III\right), \left(D\right) - \left(I\right)
  4. DA−(III),(B)−(IV),(C)−(II),(D)−(I)A - \left(III\right), \left(B\right) - \left(IV\right), \left(C\right) - \left(II\right), \left(D\right) - \left(I\right)

Correct answer

B. A−(IV),(B)−(II),(C)−(I),(D)−(III)A - \left(IV\right), \left(B\right) - \left(II\right), \left(C\right) - \left(I\right), \left(D\right) - \left(III\right)

Step-by-step solution

DIBAL-H i.e., Di isobutyl aluminium hydride causes partial reduction of cyanide group along with the ester group and further hydrolysis forms corresponding aldehyde as the product. The reaction is as follows,

CH3(CH2)5COOC2H5→DIBAL-H,H2OCH3(CH2)5CHO{CH}_{3}{\left({CH}_{2}\right)}_{5}{COOC}_{2}{H}_{5}\overset{\text{DIBAL-}H, {H}_{2}O}{\to}{CH}_{3}{\left({CH}_{2}\right)}_{5}CHO

The Clemmensen reduction is a reaction that is used to reduce aldehydes or ketones to alkanes using hydrochloric acid and zinc amalgam.

C6H5COC6H5→Zn(Hg) and conc.HClC6H5CH2C6H5{C}_{6}{H}_{5}{COC}_{6}{H}_{5}\overset{Zn ( Hg ) \text{ and } \text{conc.}HCl}{\to}{C}_{6}{H}_{5}{CH}_{2}{C}_{6}{H}_{5}

A Grignard reagent is an organomagnesium compound which can be described by the chemical formula ‘R-Mg-X’ . Here aldehyde will react with Grignard's reagent to give alcohol.

C6H5CHO→H2OCH3MgBrC6H5CH(OH)CH3{C}_{6}{H}_{5}CHO\mathop{\to}_{{H}_{2}O}^{{CH}_{3}MgBr}{C}_{6}{H}_{5}CH ( OH ) {CH}_{3}

CH3COCH2COOC2H5→NaBH4,H+CH3CH(OH)CH2COOC2H5{CH}_{3}{COCH}_{2}{COOC}_{2}{H}_{5}\overset{{NaBH}_{4}, {H}^{+}}{\to}{CH}_{3}CH ( OH ) {CH}_{2}{COOC}_{2}{H}_{5}

In this reaction, NaBH4,H+{NaBH}_{4}, {H}^{+} is used as a reducing agent, to get CH3CH(OH)CH2COOC2H5{CH}_{3}CH ( OH ) {CH}_{2}{COOC}_{2}{H}_{5} as a product.

Hence option B is the answer.

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