Organic Compounds Containing Oxygen: Chemistry | JEE Main
Find out the major products from the following reaction
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Hint 1 of 2
What reaction pathway dominates when treating the given alkyl halide with a strong base under heating?
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Step-by-step solutionView
Correct answer
The reaction proceeds via an E2 mechanism where the base abstracts a proton from the adjacent carbon that yields the more substituted, thermodynamically stable alkene (Zaitsev's rule).
Option analysis
Why each option works or fails
A ·
Believing that the reagent causes substitution or an alternative functionalization without elimination, or mistaking the regiochemical outcome of the base. Identify the conditions as promoting an elimination (E2) pathway rather than substitution, and predict double bond formation accordingly.
B ·
None. This is the correct product. Correctly applies E2 elimination governed by Zaitsev's rule using an unhindered base to afford the most substituted alkene.
C ·
Assuming that elimination gives the less substituted (Hofmann) alkene instead of the more substituted (Zaitsev) alkene. When using a non-bulky base, the elimination preferentially abstracts the β-hydrogen that produces the more stable, more substituted alkene.
D ·
Assuming that a standard elimination base acts in an unexpected non-regioselective manner or shifts the double bond to an unfavorable position. Check the available β-hydrogens and verify which elimination path provides the most thermodynamically stable double bond.
Reviewed route
Solution
StepWorking
01identify
The reaction sequence has two steps. First, 2-phenylpropan-2-ol undergoes acid-catalyzed dehydration using concentrated H2SO4 and heat (E1 elimination). Next, reductive ozonolysis (O3, Zn/H2O) cleaves the resulting alkene.
02mechanism
Step 1: Protonation of the tertiary -OH group of 2-phenylpropan-2-ol by H2SO4 yields an alkyloxonium ion. This ion loses water to form a resonance-stabilized benzylic tertiary carbocation: Ph-C+(CH3)2. Loss of a β-proton (H+) from one of the methyl groups gives α-methylstyrene [2-phenylprop-1-ene, Ph-C(CH3)=CH2].
03mechanism
Step 2: Ozonolysis of α-methylstyrene (Ph-C(CH3)=CH2) cleaves the double bond. Treatment with O3 followed by reductive workup with Zn/H2O oxidatively cleaves the C=C bond. This reaction forms carbonyl compounds. The Ph-C(CH3)= fragment becomes acetophenone (Ph-CO-CH3). The =CH2 fragment becomes formaldehyde (HCHO).
04product
The major products are acetophenone (Ph-CO-CH3) and formaldehyde (HCHO).
✓verify
Under reductive conditions (Zn/H2O), aldehydes are preserved and not further oxidized to carboxylic acids. Thus HCHO remains formaldehyde, matching Option (1).
Hints that build this answer step by step
What reaction pathway dominates when treating the given alkyl halide with a strong base under heating?
Bimolecular elimination (E2)
Which regiochemical rule governs the major product in this E2 elimination using an unhindered base?
Zaitsev's rule, yielding the more substituted and more stable alkene.
