JEE Main 2015 · 10 Apr · Shift 1 · Chemistry Q54

← JEE Main 2015 · 10 Apr · Shift 1

ChemistrySingle correct+4 / -1Question 54 of 90

A compound A with molecular formula $\ce{C10H13Cl}$ gives a white precipitate on adding silver nitrate solution. A on reacting with alcoholic KOH gives compound B as the main product. B on ozonolysis gives C and D. C gives Cannizaro reaction but not aldol condensation. D gives aldol condensation but not Cannizaro reaction. A is :

\ce{C6H5-CH2-C(CH3)2-Cl}

\ce{C6H5-CH2-CH2-CH(Cl)-CH3}

\ce{C6H5-CH2-CH2-CH2-CH2-Cl}

o-Propylbenzyl chloride (a benzene ring with

\ce{-CH2CH2CH3}

and, on the adjacent carbon,

\ce{-CH2Cl}

)

Benzene ring with CH₂-CH₂-CH₃ and, ortho to it, CH₂-Cl.

Answer and worked solution

Official answer: A

A gives AgCl readily, so it is a reactive (benzylic or tertiary) chloride. C, which gives Cannizzaro but not aldol, has no α-H (e.g. $\ce{C6H5CHO}$ ). D, which gives aldol but not Cannizzaro, has α-H (e.g. acetone).

So B is $\ce{C6H5CH=C(CH3)2}$ , which gives $\ce{C6H5CHO}$ and $\ce{(CH3)2CO}$ on ozonolysis. Then A is $\ce{C6H5CH2C(CH3)2Cl}$ , option (1).

Topics:Elimination reactions Alkenes Aldol and related condensations

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JEE Main 2015 · 10 Apr · Shift 1, Chemistry question 54