Suppose the double bonded methylene group of $1$ -bromo- $2$ -propene is labelled with $^{13} C, H_{2}^{13} C = C H - {C H}_{2} B r$ . What product will be formed by methanolysis of this substrate?

H_{2}^{13} C = C H - {C H}_{2} - O {C H}_{3}

and

H_{2} C = C H -^{13} C H - O {C H}_{3}

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

H_{2} C =^{13} C H - {C H}_{2} - O {C H}_{3}

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

H_{2}^{13} C = C H - {C H}_{2} - O {C H}_{3}

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

H_{2} C = C H -^{13} {C H}_{2} - O {C H}_{3}

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

Open in App

Solution

The correct option is A $H_{2}^{13} C = C H - {C H}_{2} - O {C H}_{3}$ and $H_{2} C = C H -^{13} C H - O {C H}_{3}$
Two products are obtained. In the first product, the bromine atom is replaced by the methoxy group. In the second product, the Br atom is replaced by the methoxy group and the double bond is also rearranged.
$^{13} C H_{2} = C H C H_{2} B r {C H_{2} O H - ---- \to - H B r}^{13} C H_{2} = C H C H_{2} O C H_{3} + C H_{2} = C H^{13} C H_{2} O C H_{3}$

Suggest Corrections

Similar questions

C H_{3} - C H_{2} - C H = C H_{2} + C H_{3} O H^{\oplus} C H_{3} - C H_{2} - C | O C H_{3} H - C H_{3}

C H_{3} - C | H O C H_{3} - O | | C - C | H C H_{3} - O - C H_{2} - C H_{3}

I - C H_{3} - * C H - C H_{3}

I I - C H_{3} - * C H - O C H_{3}

I I I - C H_{3} - * C H - C H_{2} - O C H_{3}

C H_{3} - \oplus C H - C H_{3}

C H_{3} - \oplus C H - O C H_{3}

C H_{3} - \oplus C H - C H_{2} - O C H_{3}

C H_{2} - + C H - C H_{3}

C H_{3} - + C H - O C H_{3}

C H_{3} - + C H - C O C H_{3}

Join BYJU'S Learning Program