Anisole is treated with HI under two different conditions. C + D {{HI}(g)} {C₆H₅OCH₃ {{conc. HI}} A + B The nature of A to D will be

Correct Answer is : Both A and B as well as both C and D are CH3I{CH₃I}CH3I and C6H5OH{C₆H₅OH}C6H5OH

A and B are {CH₃I} and {C₆H₅OH}, while C and D are {CH₃OH} and {C₆H₅I}

A and B are {CH₃OH} and {C₆H₅I} , while C and D are {CH₃I} and {C₆H₅OH}

Both A and B as well as both C and D are {CH₃I} and {C₆H₅OH}

A and B are {CH₃I} and {C₆H₅OH} , while there is no reaction in the second case

Correct Answer is : Both A and B as well as both C and D are CH3I{CH₃I}CH3I and C6H5OH{C₆H₅OH}C6H5OH

Test Index

VITEEE 2016 Paper

Section: Mathematics

Question : 77 of 125

Marks: +1, -0

\xleftarrow{\mathrm{HI}(g)}

\mathrm{C_6H_5OCH_3}

\xrightarrow{\text{conc. HI}}

The nature of A to D will be

A and B are $\mathrm{CH_3I}$ and $\mathrm{C_6H_5OH}$ , while C and D are $\mathrm{CH_3OH}$ and $\mathrm{C_6H_5I}$
A and B are $\mathrm{CH_3OH}$ and $\mathrm{C_6H_5I}$ , while C and D are $\mathrm{CH_3I}$ and $\mathrm{C_6H_5OH}$
Both A and B as well as both C and D are $\mathrm{CH_3I}$ and $\mathrm{C_6H_5OH}$
A and B are $\mathrm{CH_3I}$ and $\mathrm{C_6H_5OH}$ , while there is no reaction in the second case

Solution:

Although in both cases products are

\mathrm{CH_3I}

and

\mathrm{C_6H_5OH}

; the two reactions follow different mechanism.

\mathrm{C_6H_5-O-CH_3}

\xrightarrow[\mathrm{S_N2}]{\mathrm{HI}(g)}

\mathrm{CH_3I + C_6H_5OH}

\mathrm{C_6H_5-O-CH_3}

\xrightarrow[\mathrm{S_N1}]{\text{conc. HI}}

\mathrm{CH_3 I + C_6H_5 OH}

Remember that during

\mathrm{S_N1}

reaction ,

\mathrm{CH_3^+}

is formed because it is more stable than

\mathrm{C_6H_5^+}

Go to Question:

Prev Question

Next Question