Answer:
Option A is the Answer.
Step-by-step explanation:
Treatment of alkyne with Bromine in the presence of
produces a trans product that is, a alkenyl bromide.
![CH_3 \ CH_2 \ CH_2 \ CH_2 \ CH_2 - C = CH \frac {(CCl_4)}{(Br_2 )} > CH_3 \ CH_2 \ CH_2\ CH_2 \CH_2\ -CBr = CHBr](https://img.qammunity.org/2020/formulas/chemistry/college/qf5yuybfab82bf65s36d7ji8wn1i3yl4b5.png)
E-1, 2-dibromoheptene
It is a trans intermediate on further halogenations produces a tetrabromopentane.
It is an electrophilic expansion response in which the triple bond breaks to turn into a twofold bond and creates a dibromoalkene (E-1,2 dibromoheptene), and on further expansion, the twofold bond turns into a solitary bond and delivers a tetrabromoalkane.