Question: Benzene is passed through \[Na\] in \[N{H_3}\] solution then the major product was passed through oz...

Question

Benzene is passed through $Na$ in $N{H_3}$ solution then the major product was passed through ozone and $Zn - {H_2}O$ solution then the product of it was passed through $Zn - Hg$ in $Conc.HCl$ solution to give a product. What and how is the product formed?

Answer 1

Solution

Benzene is an aromatic compound and when reacted with sodium in ammonia undergo reduction to form $1,4$ cyclohexadiene. When this compound passes through ozone and $Zn - {H_2}O$ solution forms an aldehyde compound. When aldehyde is passed through $Zn - Hg$ in $Conc.HCl$ solution forms propane.

Complete answer:
Given that benzene is passed through $Na$ in $N{H_3}$ solution. The reaction is named as Birch reduction. Aromatic compound when passed through $Na$ in $N{H_3}$ solution undergoes reduction to form a $1,4$ cyclohexadiene.
The chemical reaction involved will be as follows-

When this $1,4$ cyclohexadiene is passed through ozone and $Zn - {H_2}O$ solution the double bond will converts into carbonyl group forms two molecules of dialdehyde compound named as $1,3 - propane - di - al$ . Two moles of dialdehyde compound form in this reaction.
The chemical reaction involved will be as follows-

Later, $1,3 - propane - di - al$ passed through $Zn - Hg$ in $Conc.HCl$ solution undergoes clemmensen reduction to form alkanes. Carbonyl group converts into alkane in clemmensen reduction. Two molecules of $1,3 - propane - di - al$ convert into two molecules of propane.
The chemical reaction involved will be as follows:

Thus, Benzene is passed through $Na$ in $N{H_3}$ solution then the major product was passed through ozone and $Zn - {H_2}O$ solution then the product of it was passed through $Zn - Hg$ in $Conc.HCl$ solution to give a product. The product is propane.

Note:
Birch reduction can occur in two types based on the group present on the aromatic ring. When there is a presence of an electron releasing group then the reduction takes place at $2,5$ positions. When there is a presence of an electron withdrawing group then the reduction takes place at $1,4$ positions.