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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 – methylbutane(iii) 2,2,3 – Trimethyl – 3 – bromopentane.

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(i).In 1-Bromo-1-methylcyclohexane, the β-hydrogen atoms on either side of the Br atoms are equal therefore, only product i.e., 1 methylcyclohexene is formed.

(ii).2-Chloro-2-methylbutane has two different sets of equivalent β-hydrogen atoms and hence, in principle can give two alkanes (I and II). But according to Saytzeff’s rule, more highly substituted alkene which, being more stable, is the major product i.e 2-Methylbut-2-ene is the major product.

(iii)3-Bromo-2,2,3-trimethylpentane has two different sets of β-hydrogen atoms and hence, in principle, can give two alkenes. But according to Saytzeff’s rule, more highly substituted alkene which, being more stable, is the major product i.e 3,4,4-Trimethylpent-2-ene is the major product.
Final answer:
(i).In 1-Bromo-1-methylcyclohexane, the β-hydrogen atoms on either side of the Br atoms are equal therefore, only product i.e., 1 methylcyclohexene is formed.

(ii).2-Chloro-2-methylbutane has two different sets of equivalent β-hydrogen atoms and hence, in principle can give two alkanes (I and II). But according to Saytzeff’s rule, more highly substituted alkene which, being more stable, is the major product i.e 2-Methylbut-2-ene is the major product.

(iii)3-Bromo-2,2,3-trimethylpentane has two different sets of β-hydrogen atoms and hence, in principle, can give two alkenes. But according to Saytzeff’s rule, more highly substituted alkene which, being more stable, is the major product i.e 3,4,4-Trimethylpent-2-ene is the major product.

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