Answer:
Dehydrohalogenation of the given
alkyl halide can, in principle, yield two alkenes \[(I)\] and \[(II)\]
But according to Saytzeff rule,
more highly substituted alkene, i.e., \[(I)\] being more stable is the major product
of dehydrohalogenation. Therefore, in the above reaction, alkene \[(I)\] along
with a small amount of alkene \[(II)\] is produced.
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