IIT-JEE 2010 · Paper 2 · Chemistry Q12

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ChemistrySingle correct+3 / -1Question 12 of 57

Passage

Paragraph for Questions 12 to 14

Two aliphatic aldehydes P and Q react in the presence of aqueous $\ce{K2CO3}$ to give compound R, which upon treatment with HCN provides compound S. On acidification and heating, S gives the product shown below:

3-Hydroxy-4,4-dimethyldihydrofuran-2(3H)-one (pantolactone): a five-membered lactone with a gem-dimethyl carbon next to the CH(OH) carbon α to C=O.

The compounds P and Q respectively are

2-Methylpropanal ((CH₃)₂CH-CHO) and ethanal (CH₃CHO).

2-Methylpropanal ((CH₃)₂CH-CHO) and methanal (HCHO).

3-Methylbutanal ((CH₃)₂CH-CH₂-CHO) and ethanal (CH₃CHO).

3-Methylbutanal ((CH₃)₂CH-CH₂-CHO) and methanal (HCHO).

Answer and worked solution

Official answer: B

Work back from the product. Pantolactone, $\ce{HOCH2C(CH3)2CH(OH)COOH}$ lactonised, comes from the cyanohydrin of $\ce{HOCH2C(CH3)2CHO}$ (hydrolysis of CN to COOH, then ring closure with the primary OH).

That aldehyde is the crossed aldol of 2-methylpropanal (the only α-H source, giving the quaternary carbon) with formaldehyde (which supplies the $\ce{CH2OH}$ ):
$\ce{(CH3)2CHCHO + HCHO ->[K2CO3] HOCH2C(CH3)2CHO}.$

(B) 2-methylpropanal and methanal

Topics:Aldol and related condensations

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IIT-JEE 2010 · Paper 2, Chemistry question 12