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Compound X on reaction with O 3 _3 3 ā followed by Zn/H 2 O _2O 2 ā O gives formaldehyde and 2-methylpropanal as products. The compound X is:
medium
Hydrocarbons
2022
chemistry
3 ā 3- 3 ā Methylbut ā 1 ā -1- ā 1 ā ene
2 ā 2- 2 ā Methylbut ā 1 ā -1- ā 1 ā ene
Explanation To solve this problem, we need to understand the reaction of compound X with ozone ( O 3 _3 3 ā ) followed by reduction with Zn/H 2 _2 2 ā O. This process is known as ozonolysis, which cleaves double bonds in alkenes to form carbonyl compounds. \\
Given that the products are formaldehyde (
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Methylbut
ene
HCHO
) and 2-methylpropanal (
C
H
O
), we need to determine the structure of compound X.
Step 1: Identify the carbonyl products.
ā¢ Formaldehyde is
HCHO
which indicates that one of the carbon atoms involved in the double bond is a terminal carbon (
CH
group).
ā¢ 2-Methylpropanal is
(CH
)_2CHCHO
which suggests the presence of a branched alkyl group adjacent to the carbonyl group.
Step 2: Analyze the possible structures of compound X.
ā¢ Compound X must have a double bond that, when cleaved, results in the given carbonyl compounds.
Step 3: Consider the options.
Option 1: 3-Methylbut-1-ene
ā¢ Structure: CH
=C(CH
)CH
CH
ā¢ Ozonolysis of this compound would cleave the double bond between the terminal CH
and the adjacent carbon, forming formaldehyde (
HCHO
) and 2-methylpropanal (
(CH
)_2CHCHO
).} \\
Option 2: 2-Methylbut-1-ene
ā¢ Structure: CH
=CHCH(CH
)CH
ā¢ Ozonolysis would not produce formaldehyde, as the double bond is not terminal.
Option 3: 2-Methylbut-2-ene
ā¢ Structure: (CH
)_2C=CHCH
ā¢ Ozonolysis would not produce formaldehyde, as the double bond is internal.
Option 4: Pent-2-ene
ā¢ Structure: CH
CH=CHCH
CH
ā¢ Ozonolysis would not produce formaldehyde, as the double bond is internal.
Conclusion: The only compound that can produce both formaldehyde and 2-methylpropanal upon ozonolysis is 3-Methylbut-1-ene.
Therefore, the correct option is Option 1: 3-Methylbut-1-ene.