JEE Challenger
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ChemMedMain 2026 • 6th Apr - Shift 2

IUPAC Name of Aldol Condensation Product

'xx' is the product which is obtained by the hydrolysis of prop-1-yne in the presence of mercuric sulphate under dilute acidic medium at 333 K333\text{ K}. 'yy' is the product which is obtained by the reaction of ethane nitrile with methyl magnesium bromide in dry ether followed by hydrolysis. IUPAC name of product obtained from 'xx' and 'yy' in the presence of barium hydroxide followed by heating is:

Options

A

2-Methylpent-4-en-3-one

B

4-Methylpent-3-en-2-one

Correct
C

4-Methylpent-1-ene

D

2-Methylpent-3-one

Topics & Concepts

Aldehydes, Ketones and Carboxylic AcidsAldehydes and Ketones

Step-by-Step Solution

To find the IUPAC name of the final product, we need to determine the chemical structures of 'xx' and 'yy' through their respective chemical reactions and then perform the final reaction:

Step 1: Identification of 'xx'

Hydration of prop-1-yne (CH3C≡CH\text{CH}_3\text{C}\equiv\text{CH}) in the presence of dilute H2SO4\text{H}_2\text{SO}_4 and HgSO4\text{HgSO}_4 at 333 K333\text{ K} proceeds via Markovnikov addition of water to form an enol intermediate, which undergoes tautomerization:

CH3−C≡CH+H2O→333 KHgSO4/dil. H2SO4[CH3−C(OH)=CH2]⇌CH3−C(=O)−CH3\text{CH}_3-\text{C}\equiv\text{CH} + \text{H}_2\text{O} \xrightarrow[333\text{ K}]{\text{HgSO}_4/\text{dil. H}_2\text{SO}_4} \left[ \text{CH}_3-\text{C(OH)}=\text{CH}_2 \right] \rightleftharpoons \text{CH}_3-\text{C}(=\text{O})-\text{CH}_3

Thus, 'xx' is propan-2-one (acetone).

Step 2: Identification of 'yy'

Reaction of ethanenitrile (CH3CN\text{CH}_3\text{CN}) with a Grignard reagent, methyl magnesium bromide (CH3MgBr\text{CH}_3\text{MgBr}), in dry ether forms an imine salt intermediate, which on acidic hydrolysis yields a ketone:

CH3−C≡N+CH3MgBr→dry etherCH3−C(CH3)=NMgBr\text{CH}_3-\text{C}\equiv\text{N} + \text{CH}_3\text{MgBr} \xrightarrow{\text{dry ether}} \text{CH}_3-\text{C}(\text{CH}_3)=\text{NMgBr}

CH3−C(CH3)=NMgBr→H3O+CH3−C(=O)−CH3+NH3+Mg(OH)Br\text{CH}_3-\text{C}(\text{CH}_3)=\text{NMgBr} \xrightarrow{\text{H}_3\text{O}^+} \text{CH}_3-\text{C}(=\text{O})-\text{CH}_3 + \text{NH}_3 + \text{Mg(OH)Br}

Thus, 'yy' is also propan-2-one (acetone).

Step 3: Reaction of 'xx' and 'yy' in the presence of Ba(OH)2\text{Ba(OH)}_2 followed by heating

Since both 'xx' and 'yy' are acetone (CH3COCH3\text{CH}_3\text{COCH}_3), the reaction with a mild base like barium hydroxide (Ba(OH)2\text{Ba(OH)}_2) followed by heating represents an aldol condensation:

  1. Aldol Addition: CH3−CO−CH3+CH3−CO−CH3→Ba(OH)2CH3−C∣OH(CH3)−CH2−CO−CH3(4-hydroxy-4-methylpentan-2-one)\text{CH}_3-\text{CO}-\text{CH}_3 + \text{CH}_3-\text{CO}-\text{CH}_3 \xrightarrow{\text{Ba(OH)}_2} \text{CH}_3-\underset{\substack{| \\ \text{OH}}}{\text{C}}(\text{CH}_3)-\text{CH}_2-\text{CO}-\text{CH}_3 \quad (\text{4-hydroxy-4-methylpentan-2-one})

  2. Dehydration on Heating (Δ\Delta): CH3−C∣OH(CH3)−CH2−CO−CH3→Δ(CH3)2C=CH−CO−CH3+H2O\text{CH}_3-\underset{\substack{| \\ \text{OH}}}{\text{C}}(\text{CH}_3)-\text{CH}_2-\text{CO}-\text{CH}_3 \xrightarrow{\Delta} (\text{CH}_3)_2\text{C}=\text{CH}-\text{CO}-\text{CH}_3 + \text{H}_2\text{O}

Step 4: IUPAC Naming of the Condensation Product

The structural formula of the product is: C5H3−C4(CH3)=C3H−C2(=O)−C1H3\overset{5}{\text{C}}\text{H}_3-\overset{4}{\text{C}}(\text{CH}_3)=\overset{3}{\text{C}}\text{H}-\overset{2}{\text{C}}(=\text{O})-\overset{1}{\text{C}}\text{H}_3

  • Principal carbon chain: 5 carbon atoms (pent-\text{pent-})
  • Principal functional group: Ketone at C-2\text{C-2} (-2-one\text{-2-one})
  • Unsaturation: Double bond between C-3\text{C-3} and C-4\text{C-4} (-3-en-\text{-3-en-})
  • Substituent: Methyl group at C-4\text{C-4} (4-methyl\text{4-methyl})

Combining these, the IUPAC name is 4-Methylpent-3-en-2-one (commonly known as mesityl oxide).

Correct Option: B

IUPAC Name of Aldol Condensation Product | Chemistry PYQ Solution - JEE Challenger