Chemistry Class 12 NCERT Solutions: Chapter 10 Haloalkanes and Haloarenes Part 6

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Q: 10. 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.

Answer:

(i)

Q 10 1-Bromo-1-Methylcyclohexane

Q 10 1-Bromo-1-Methylcyclohexane

Q 10 1-Bromo-1-Methylcyclohexane

1−bromo−1−methylcyclohexane

In the given compound, all β-hydrogen atoms are equivalent. Thus, dehydrohalogenation of this compound gives only one alkene.

Q 10 i 1 Dehydrohalogenation of this Compound Gives only one Alkene

Q 10 I 1 Dehydrohalogenation Gives Only One Alkene

Q 10 i 1 Dehydrohalogenation of this Compound Gives only one Alkene

(ii)

Q 10 ii 1 Two Different Sets of Equivalent β-hydrogen Atoms Labelled as a and b

Q 10 Ii 1 Two Sets of Equivalent Beta-Hydrogen Atoms

Q 10 ii 1 Two Different Sets of Equivalent β-hydrogen Atoms Labelled as a and b

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.

Q 10 ii 2 The Compound Yields Two Alkenes

Q 10 Ii 2 the Compound Yields Two Alkenes

Q 10 ii 2 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)

Q 10 iii 2, 2, 3-Trimethyl-3-bromopentane

Q 10 Iii 2, 2, 3-Trimethyl-3-Bromopentane

Q 10 iii 2, 2, 3-Trimethyl-3-bromopentane

2,2,3-Trimethyl-3-bromopentane

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

Q 10 iv 2 - Ethyl -3, 3 - dimethybut-1-ene (II).

Q 10 Iv 2 - Ethyl -3, 3 - Dimethybut-1-Ene (II).

Q 10 iv 2 - Ethyl -3, 3 - dimethybut-1-ene (II).

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.