In the Hunsdiecker reaction,

Question

In the Hunsdiecker reaction, (A) Number of carbon atoms decrease (B) Number of carbon atoms increase (C) Number o f carbon atoms remain same (D) none of the above

Tardigrade Admin · Accepted Answer

In the Hunsdiecker reaction → CH3CH2COOAg + Br2 xrightarrowCCl4 CH3 CH2Br + CO2 + AgBr Silver Propionate Ethyl Bromide (No. of C = 3) (No. of C = 2) No. of carbon Atoms decrease Therefore, option ‘a’ is correct.

Q. In the Hunsdiecker reaction,

Number of carbon atoms decrease

Number of carbon atoms increase

Number o f carbon atoms remain same

none of the above

In the Hunsdiecker reaction $\to$
$C H_{3} C H_{2} COO A g + B r_{2} CC l_{4} C H_{3} C H_{2} B r + C O_{2} + A g B r$
Silver Propionate Ethyl Bromide
(No. of C = 3) (No. of C = 2)
No. of carbon Atoms decrease Therefore, option ‘a’ is correct.

Q. In the Hunsdiecker reaction,

Number of carbon atoms decrease

Number of carbon atoms increase

Number o f carbon atoms remain same

none of the above

In the Hunsdiecker reaction → CH3​CH2​COOAg+Br2​CCl4​​CH3​CH2​Br+CO2​+AgBr Silver Propionate Ethyl Bromide (No. of C = 3) (No. of C = 2) No. of carbon Atoms decrease Therefore, option ‘a’ is correct.

In the Hunsdiecker reaction $\to$
$C H_{3} C H_{2} COO A g + B r_{2} CC l_{4} C H_{3} C H_{2} B r + C O_{2} + A g B r$
Silver Propionate Ethyl Bromide
(No. of C = 3) (No. of C = 2)
No. of carbon Atoms decrease Therefore, option ‘a’ is correct.