Question

an organic compound A reacts with PCL5 and gives alkyl halide B which on reaction with Na metal in presence of dry ether gives hydrocarbon with molecular formula C4H10 identify A and B and write complete chemical reaction​

Answer 1

Explanation:

organic compound A reacts with PCL5 and gives alkyl halide B which on reaction with Na metal in presence of dry ether gives hydrocarbon with molecular formula C4H10 identify A and B and write complete chemical

Answer 2

Answer:

A is  $C_{2}H_{5}OH$

B is   $C_{2}H_{5}Cl$

Complete reaction is-  

$2C_{2}H_{5}OH$ + $PCl_{5}$ →$2C_{2}H_{5}Cl$ →$C_{4}H_{10}$

  A                             B

Explanation:

Given: Final product has molecular formula  $C_{4}H_{10}$.

The given reaction proceeds as-

$A + PCl_{5}$→$B$

$B + Na/ether$→$C_{4}H_{10}$

To find: A and B.

Explanation:

The first reagent used is Phosphorus pentachloride ( $PCl_{5}$). It is a chlorinating agent and when alcohol ( $ROH$) is treated with , then an alkyl halide is produced.

Thus, A is alcohol and B is an alkyl halide (already given in question).

Now, the next reagent is Sodium in dry ether. The reaction between an alkyl halide and sodium in the presence of ether produces higher alkanes, known as the Wurtz reaction.

The Wurtz reaction is –

$2RX + 2Na/ether$→$R-R + 2NaX$

Now the alkane formed is butane ($C_{4}H_{10}$ ), thus the formula of B can be determined as  $C_{2}H_{5}Cl$.

$2C_{2}H_{5}Cl + 2Na/ether$→$C_{2}H_{5}-C_{2}H_{5} + 2NaCl$

Now, the formula of A can be deduced from the formula of B as A is alcohol form of B. Hence, A is ethyl alcohol, $C_{2}H_{5}OH$  .

Hence, the complete reaction can be written as-

$2C_{2}H_{5}OH$ + $PCl_{5}$ →$2C_{2}H_{5}Cl$ →$C_{4}H_{10}$

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