Organic Compounds Containing Oxygen: Chemistry | JEE Main
Find out the major product for the following reaction.
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Hint 1 of 3
What type of reaction is initiated when an alkyl halide is heated in the presence of potassium tert-butoxide (t-BuOK)?
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Step-by-step solutionView
Correct answer
Sterically hindered potassium tert-butoxide preferentially abstracts the least hindered beta-proton, yielding the less substituted Hofmann alkene as the major product.
Option analysis
Why each option works or fails
A ·
Assuming that elimination always favors the more substituted, thermodynamically stable Zaitsev alkene regardless of base sterics. Check the steric bulk of the base: bulky bases like tert-butoxide encounter severe steric clash at internal positions and selectively abstract from the less hindered, accessible primary carbon.
B ·
Believing that elimination occurs preferentially towards an internal carbon with specific E/Z stereochemistry while ignoring the steric hindrance of the base. Identify the base as bulky (t-BuOK), which directs deprotonation to the methyl group instead of the internal CH2/CH positions.
C ·
Selecting an internal alkene isomer formed by abstraction of a sterically congested proton. Recall that a bulky base gives kinetic control leading to Hofmann regiochemistry (the terminal alkene), not internal regioisomers.
D ·
None. This is the correct major product. Potassium tert-butoxide is a sterically hindered base that preferentially abstracts the most accessible proton from the least hindered methyl group, yielding the Hofmann alkene as the major product.
Reviewed route
Solution
StepWorking
01identify
The reaction is an acid-catalyzed dehydration of a secondary alcohol (4-methylpentan-2-ol) using concentrated H2SO4 and heat (E1 mechanism involving carbocation formation, rearrangement, and elimination).
02mechanism
Protonation of the -OH group by H2SO4 followed by loss of H2O yields a secondary carbocation: (CH3)2CH-CH2-CH(+)-CH3.
03mechanism
A 1,2-hydride shift from the adjacent tertiary carbon to the secondary carbocation generates a more stable tertiary carbocation: (CH3)2C(+)-CH2-CH2-CH3.
04mechanism
Elimination of a β-proton from -CH2- gives the most substituted, most stable alkene according to Zaitsev's rule, yielding 2-methylpent-2-ene: (CH3)2C=CH-CH2-CH3.
05product
The major product is 2-methylpent-2-ene, matching Option (3).
✓verify
2-methylpent-2-ene has a trisubstituted double bond (7 hyperconjugative α-hydrogens), making it thermodynamically more stable than 4-methylpent-2-ene (disubstituted) or 4-methylpent-1-ene / 2-methylpent-1-ene.
Hints that build this answer step by step
What type of reaction is initiated when an alkyl halide is heated in the presence of potassium tert-butoxide (t-BuOK)?
E2 elimination promoted by a strong, sterically hindered base
How does the steric bulk of potassium tert-butoxide influence the regioselectivity of the E2 elimination?
It favors abstraction of the less hindered primary β-hydrogen, giving the Hofmann product.
Which double bond position corresponds to deprotonation at the least hindered β-position?
The terminal alkene formed by deprotonation of the adjacent methyl group
Why does a 1,2-hydride shift occur instead of immediate elimination?
Carbocation rearrangement occurs rapidly when it can convert a secondary carbocation into a more stable tertiary carbocation prior to the elimination step.
