An organic compound of molecular formula C4H10O does not react with sodium. With excess of HI, it gives only one type of alkyl halide. The compound is

Question

An organic compound of molecular formula C4H10O does not react with sodium. With excess of HI, it gives only one type of alkyl halide. The compound is (A) ethoxyethane (B) 1 - butanol (C) 1 - methoxy propane (D) 2 - methoxypropane

Tardigrade Admin · Accepted Answer

Symmetrical ether (here ethoxyethane) gives only one type of halide with excess of HI. underset textEthoxyethaneC2H5OC2H5+2HI arrow 2C2H5I +H2O

Q. An organic compound of molecular formula $C_{4} H_{10} O$ does not react with sodium. With excess of $H I$ , it gives only one type of alkyl halide. The compound is

ethoxyethane

$1$ - butanol

$1$ - methoxy propane

$2$ - methoxypropane

Symmetrical ether (here ethoxyethane) gives only one type of halide with excess of $H I$ .
$Ethoxyethane C_{2} H_{5} O C_{2} H_{5} + 2 H I \to 2 C_{2} H_{5} I + H_{2} O$

Q. An organic compound of molecular formula C4​H10​O does not react with sodium. With excess of HI, it gives only one type of alkyl halide. The compound is

ethoxyethane

1 - butanol

1 - methoxy propane

2 - methoxypropane

Symmetrical ether (here ethoxyethane) gives only one type of halide with excess of HI. EthoxyethaneC2​H5​OC2​H5​​+2HI→2C2​H5​I+H2​O

Q. An organic compound of molecular formula $C_{4} H_{10} O$ does not react with sodium. With excess of $H I$ , it gives only one type of alkyl halide. The compound is

$1$ - butanol

$1$ - methoxy propane

$2$ - methoxypropane

Symmetrical ether (here ethoxyethane) gives only one type of halide with excess of $H I$ .
$Ethoxyethane C_{2} H_{5} O C_{2} H_{5} + 2 H I \to 2 C_{2} H_{5} I + H_{2} O$