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Aldehydes Ketones and Carboxylic Acids question

2017 · Shift 1 · Q15
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  5. /2017 · Shift 1 · Q15

Aldehydes Ketones and Carboxylic Acids question

2017 · Shift 1 · Q15

JEE AdvancedChemistryAldehydes Ketones and Carboxylic AcidsMCQ+3 / −1
Columns 1, 2 and 3 contain starting materials, reaction conditions, and type of reactions, respectively.

Column 1 Column 2 Column 3
(I) Toluene (i) NaOHNaOHNaOH/Br2 (P) Condensation
(II) Acetophenone (ii) Br2/hv (Q) Carboxylation
(III) Benzaldehyde (iii) (CH3CO)2O/
CH3COOK		 (R) Substitution
(IV) Phenol (iv) NaOHNaOHNaOH/CO2 (S) Haloform
The only CORRECT combination that gives two different carboxylic acids is :
  1. A
    (II) (iv) (R)
  2. B
    (IV) (iii) (Q)
  3. C
    (III) (iii) (P)
  4. D
    (I) (i) (S)
View written solutionFree

Correct answer: C

  1. We must check each given combination and see:
    • what reaction actually occurs,
    • whether the reaction type matches,
    • and whether the product formed is two different carboxylic acids.
  1. Option A: (II) (iv) (R)(II)\ (iv)\ (R)(II) (iv) (R)
  • (II)(II)(II) = Acetophenone (C6H5COCH3)\left(C_6H_5COCH_3\right)(C6​H5​COCH3​)
  • (iv)(iv)(iv) = NaOH/CO2\mathrm{NaOH/CO_2}NaOH/CO2​
  • (R)(R)(R) = Substitution

Now, NaOH/CO2\mathrm{NaOH/CO_2}NaOH/CO2​ is the condition for Kolbe-Schmitt carboxylation of phenoxide ion, not for acetophenone.

So this combination is not appropriate.

Also, it does not give two different carboxylic acids.

Hence, A is incorrect.

  1. Option B: (IV) (iii) (Q)(IV)\ (iii)\ (Q)(IV) (iii) (Q)
  • (IV)(IV)(IV) = Phenol
  • (iii)(iii)(iii) = (CH3CO)2O/CH3COOK\mathrm{(CH_3CO)_2O/CH_3COOK}(CH3​CO)2​O/CH3​COOK
  • (Q)(Q)(Q) = Carboxylation

Phenol with acetic anhydride in presence of sodium acetate undergoes acetylation, not carboxylation.

Reaction:

C6H5OH→(CH3COOK)(CH3CO)2OC6H5OCOCH3C_6H_5OH \xrightarrow[(CH_3COOK)]{(CH_3CO)_2O} C_6H_5OCOCH_3C6​H5​OH(CH3​CO)2​O(CH3​COOK)​C6​H5​OCOCH3​

This gives phenyl acetate, not carboxylic acids.

Hence, B is incorrect.

  1. Option C: (III) (iii) (P)(III)\ (iii)\ (P)(III) (iii) (P)
  • (III)(III)(III) = Benzaldehyde
  • (iii)(iii)(iii) = (CH3CO)2O/CH3COOK\mathrm{(CH_3CO)_2O/CH_3COOK}(CH3​CO)2​O/CH3​COOK
  • (P)(P)(P) = Condensation

This is the Perkin reaction.

Benzaldehyde reacts with acetic anhydride in presence of sodium acetate to give cinnamic acid after hydrolysis:

C6H5CHO+(CH3CO)2O→CH3COOKC6H5CH=CHCOOHC_6H_5CHO + (CH_3CO)_2O \xrightarrow{CH_3COOK} C_6H_5CH=CHCOOHC6​H5​CHO+(CH3​CO)2​OCH3​COOK​C6​H5​CH=CHCOOH

This is indeed a condensation reaction.

Now, the question says this combination gives two different carboxylic acids. In Perkin reaction, the product mixture/overall transformation involves formation of:

  • cinnamic acid as the main product,
  • and acetic acid is also formed from acetic anhydride during the process/hydrolysis.

Thus, two different carboxylic acids are involved/obtained:

  • C6H5CH=CHCOOH\mathrm{C_6H_5CH=CHCOOH}C6​H5​CH=CHCOOH (cinnamic acid)
  • CH3COOH\mathrm{CH_3COOH}CH3​COOH (acetic acid)

So this combination is correct.

Hence, C is correct.

  1. Option D: (I) (i) (S)(I)\ (i)\ (S)(I) (i) (S)
  • (I)(I)(I) = Toluene
  • (i)(i)(i) = NaOH/Br2\mathrm{NaOH/Br_2}NaOH/Br2​
  • (S)(S)(S) = Haloform

Haloform reaction is shown by compounds containing:

−COCH3or−CH(OH)CH3-COCH_3 \quad \text{or} \quad -CH(OH)CH_3−COCH3​or−CH(OH)CH3​

Toluene does not undergo haloform reaction with Br2/NaOH\mathrm{Br_2/NaOH}Br2​/NaOH.

Hence, D is incorrect.

  1. Final conclusion

The only correct combination is:

(III) (iii) (P)(III)\ (iii)\ (P)(III) (iii) (P)

which corresponds to Option C.

  1. Comparison with stored answer

Stored correct answer = C

Our derived answer = C

So, they agree.

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