JEE Advanced 2016 · Paper 2 · Chemistry Q24

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ChemistrySingle correct+3 / -1Question 24 of 54

The major product of the following reaction sequence is

Isobutyrophenone $\xrightarrow[\text{ii) HCHO/}\ce{H+}\text{ (catalytic amount)}]{\text{i) HCHO (excess)/NaOH, heat}}$

Isobutyrophenone (PhCOCH(CH₃)₂) treated with i) HCHO (excess)/NaOH, heat, then ii) HCHO/H⁺ (catalytic amount).

5,5-Dimethyl-4-phenyl-1,3-dioxane: a six-membered O-CH₂-O ring with a phenyl at C4 and gem-dimethyl at C5.

PhCO-C(CH₃)₂-CH₂-O-CH₂-OH: the ketone with a hemiformal-ether side chain.

4-Hydroxy-5,5-dimethyl-4-phenyl-1,3-dioxane: a 1,3-dioxane with OH and phenyl both on C4 and gem-dimethyl at C5.

3-(Hydroxymethyl)phenyl ketone: Ar-CO-C(CH₃)₂-CH₂OH with a CH₂OH group meta on the ring.

Answer and worked solution

Official answer: A

Step i. $\ce{PhCOCH(CH3)2}$ has one α-H. Aldol with HCHO gives $\ce{PhCO-C(CH3)2-CH2OH}$ . Excess HCHO/NaOH then reduces the carbonyl by hydride transfer (crossed Cannizzaro-type, as in the pentaerythritol synthesis), giving the 1,3-diol
$\ce{PhCH(OH)-C(CH3)2-CH2OH}.$

Step ii. HCHO with catalytic acid converts the 1,3-diol to its cyclic formal: 5,5-dimethyl-4-phenyl-1,3-dioxane.

(A)

Topics:Named reactions Aldol and related condensations

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JEE Advanced 2016 · Paper 2, Chemistry question 24