IIT-JEE 2010 · Paper 2 · Chemistry Q14

← IIT-JEE 2010 · Paper 2

ChemistrySingle correct+3 / -1Question 14 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 compound S is

3-Formyl-4-methylpentanenitrile: (CH₃)₂CH-CH(CHO)-CH₂-CN.

3-Cyano-2,2-dimethylpropanal: NC-CH₂-C(CH₃)₂-CHO.

HOCH₂-C(CH₃)₂-CH(OH)-CN (the cyanohydrin of 3-hydroxy-2,2-dimethylpropanal).

Answer and worked solution

Official answer: D

S is the cyanohydrin of R: HCN adds to the CHO group,
$\ce{HOCH2C(CH3)2CHO + HCN -> HOCH2C(CH3)2CH(OH)CN}.$
Acid hydrolysis gives the 2,4-dihydroxy acid, which closes to the five-membered lactone on heating.

(D)

Topics:Nucleophilic addition Acid derivatives

← Previous question Next question →

IIT-JEE 2010 · Paper 2, Chemistry question 14