Which of the following is the correct order of reactivity towards sulphonation?

C_{6} D_{6} > C_{6} H_{6} > C_{6} T_{6}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

C_{6} H_{6} > C_{6} D_{6} > C_{6} T_{6}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

C_{6} H_{6} = C_{6} D_{6} = C_{6} T_{6}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

C_{6} T_{6} > C_{6} D_{6} > C_{6} H_{6}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A $C_{6} H_{6} = C_{6} D_{6} = C_{6} T_{6}$
SE reactions (i.e., nitration, halogenation, sulphonation etc.) do not show $1^{o}$ kinetic isotope effect.
Because the $C - H / C - D / C - T$ bond does not breaks in rate determining step.
Hence rate will be equal.
Option C is correct.

Suggest Corrections

Similar questions

C_{6} H_{5} C H_{3} a C_{6} H_{6} b C_{6} D_{6} c C_{6} T_{6} d C_{6} H_{5} B r e C_{6} H_{5} N R_{3}^{+} f C_{6} H_{5} N M e_{2} g

C_{6} H_{5} C H_{3} A

C_{6} H_{6} B

C_{6} D_{6} C

C_{6} T_{6} D

C_{6} H_{5} B r_{3} E

C_{6} H_{5} N R_{3} F

C_{6} H_{5} N M e_{2} G

C_{6} H_{6} ≃ C_{6} D_{6} ≃ C_{6} T_{6}

C - H

C - D

C_{6} H_{5} C H_{3}

C_{6} H_{5} C O O H

C_{6} H_{6}

C_{6} H_{5} N O_{2}

C_{6} H_{5} C H_{3} I, C_{6} H_{5} C O O H I I, C_{6} H_{6} I I I, C_{6} H_{5} N O_{2} I V

Join BYJU'S Learning Program