Predict all the alkenes that would be formed by dehydrohalogenation of the following halides with sodium ethoxide in ethanol and identify the major alkene:
(i)1-Bromo-1 methylcyclohexane
(ii)2-Chloro-2-methyl butane
(iii)2,2,3-Trimethyl-3-bromopentane
(i)
1−bromo−1−methylcyclohexane In the given compound,all β-hydrogen atoms are equivalent.Thus, dehydrohalogenation of this compound gives only one alkene.

(ii)
In the given compound, there are two different sets of equivalent β-hydrogen atoms labelled as a and b.Thus, dehydrohalogenation of the compound yields two alkenes.

Saytzeff’s rule implies that in dehydrohalogenation reactions,the alkene having a greater number of alkyl groups attached to a doubly bonded carbon atoms is preferably produced.Therefore,alkene(I) i.e.,2-methylbut-2-ene is the major product in this reaction.
(iii)
2,2,3-Trimethyl-3-bromopentane In the given compound,there are two different sets of equivalent β-hydrogen atoms labeled as a and b. Thus, dehydrohalogenation of the compound yields two alkenes.According to Saytzeff’s rule,in dehydrohalogenation reactions,the alkene having a greater number of alkyl groups attached to the doubly bonded carbon atom is preferably formed.

Hence, alkene (I)i.e.,3,4,4-trimethylpent-2-ene is the major product in this reaction.
A compound (A) with molecular formula $C_4H_9I$ which is a primary alkyl halide, reacts with alcoholic KOH to give compound (B). Compound (B) reacts with HI to give (C) which is an isomer of (A). When (A) reacts with Na metal in the presence of dry ether, it gives a compound (D), C8H18, which is different from the compound formed when n-butyl iodide reacts with sodium. Write the structures of A, (B), (C) and (D) when (A) reacts with alcoholic KOH.