Question

\[1-Bromo-3-chloro\text{ }cyclobutane\] when treated with 2-equivalents of Na in presence of ether, which of the following products will be formed?
(A)
(B)
(C)
(D)

Answer 1

Hint: When two equivalent sodium metals react with two alkyl halides in the presence of dry ether condition, this reaction is known as the Wurtz reaction (coupling reaction). In this reaction, sodium metal is used so that it can donate its electron to halogen and contribute to the formation of a radical. In this question, in place of two alkyl halides, \[1-Bromo-3-chloro\text{ }cyclobutane\]is given. In this case, an internal Wurtz reaction will take place.

Complete Step by Step Solution:
In the first step, \[N{{a}_{2}}\]break into two equivalent sodium metal, and in the second step, the bromine halogen of \[1-Bromo-3-chloro\text{ }cyclobutane\], makes a bond with one sodium metal leaving a radical of cyclobutane behind such as

In the third step, the second equivalent of sodium metal give an electron to the radical compound to form a sodium salt of \[1-Bromo-3-chloro\text{ }cyclobutane\]. Now in the last step, the negative charge on the ring attracted by carbon bonded to chlorine atom as chlorine is a good leaving group such as


The resultant compound formed is a bicycle compound (two cycle formed), named bicycle\[\left[ 1.1.0 \right]\]butane.
Thus, the correct option is D.

Note: Both Chlorine and Bromine are good leaving groups (halogens are good leaving groups because of high electronegativity and stability). But the size of bromine halogen is large as compared to chlorine so due to this, it can easily leave the compound as compared to chlorine. This is the reason why bromine is attacked first by sodium.