Consider the given reaction. The total number of oxygen atoms present per molecule of the product (P) is____.
Consider the given reaction. The total number of oxygen atoms present per molecule of the product (P) is____.
Correct Answer: 1
Approach Solution - 1
The reaction shown is an ozonolysis of 2-butene. The ozone (O3) cleaves the double bond, and the reduction with Zn/H2O results in the formation of aldehydes or ketones. For 2-butene, the ozonolysis reaction can be depicted as:
Reactant: \[\text{CH}_{3}\text{CH}=\text{CH}\text{CH}_{3}\]
Step 1: Ozonolysis cleaves the C=C bond.
Step 2: The products are two molecules of acetaldehyde (\[\text{CH}_{3}\text{CHO}\]).
Product (P): Each acetaldehyde molecule contains one oxygen atom.
| Molecule | Oxygen Atoms |
|---|---|
| Acetaldehyde (\[\text{CH}_{3}\text{CHO}\]) | 1 |
Since 2-butene is symmetrical, only one type of product is formed, which is acetaldehyde. The total number of oxygen atoms per molecule of product (P) is therefore 1.
Approach Solution -2
Upon ozonolysis, the compound CH3–CH=CH–CH3 is cleaved at the double bond, resulting in two molecules of CH3–C=O.
Each resulting molecule contains one oxygen atom.
Hence, the total number of oxygen atoms present per molecule of the product is: 1
So, the correct answer is: 1.
Top Questions on Classification Of Organic Compounds
The correct increasing order of stability of the complexes based on \( \Delta \) value is:
- JEE Main - 2025
- Chemistry
- Classification Of Organic Compounds
Match List-I with List-II: List-I
- JEE Main - 2025
- Chemistry
- Classification Of Organic Compounds
- Which is a heterocyclic compound?
- KEAM - 2025
- Chemistry
- Classification Of Organic Compounds
- The final product A formed in the following multistep reaction sequence is
- JEE Main - 2024
- Chemistry
- Classification Of Organic Compounds
- Given below are two statements: Statement I:
IUPAC name of \(HO–CH_2–(CH_2)_5–CH_2–COCH_3\) is 7-hydroxyheptan-2-one.
Statement II: 2-oxoheptan-7-ol is the correct IUPAC name for above compound.
In the light of the above statements, choose the most appropriate answer from the options given below:- JEE Main - 2024
- Chemistry
- Classification Of Organic Compounds
Questions Asked in JEE Main exam
Let $ P_n = \alpha^n + \beta^n $, $ n \in \mathbb{N} $. If $ P_{10} = 123,\ P_9 = 76,\ P_8 = 47 $ and $ P_1 = 1 $, then the quadratic equation having roots $ \alpha $ and $ \frac{1}{\beta} $ is:
- JEE Main - 2025
- Relations and functions
- Let \( \langle a_n \rangle \) be a sequence such that \( a_0 = 0 \), \( a_1 = \frac{1}{2} \), and \( 2a_{n+2} = 5a_{n+1} - 3a_n \).n= 0,1,2,3.... Then \( \sum_{k=1}^{100} a_k \) is equal to:
- JEE Main - 2025
- Sequences and Series
Two blocks of masses \( m \) and \( M \), \( (M > m) \), are placed on a frictionless table as shown in figure. A massless spring with spring constant \( k \) is attached with the lower block. If the system is slightly displaced and released then \( \mu = \) coefficient of friction between the two blocks.
(A) The time period of small oscillation of the two blocks is \( T = 2\pi \sqrt{\dfrac{(m + M)}{k}} \)
(B) The acceleration of the blocks is \( a = \dfrac{kx}{M + m} \)
(\( x = \) displacement of the blocks from the mean position)
(C) The magnitude of the frictional force on the upper block is \( \dfrac{m\mu |x|}{M + m} \)
(D) The maximum amplitude of the upper block, if it does not slip, is \( \dfrac{\mu (M + m) g}{k} \)
(E) Maximum frictional force can be \( \mu (M + m) g \)Choose the correct answer from the options given below:
- JEE Main - 2025
- Newtons Laws of Motion
Let \( f : \mathbb{R} \to \mathbb{R} \) be a twice differentiable function such that \[ (\sin x \cos y)(f(2x + 2y) - f(2x - 2y)) = (\cos x \sin y)(f(2x + 2y) + f(2x - 2y)), \] for all \( x, y \in \mathbb{R}. \)
If \( f'(0) = \frac{1}{2} \), then the value of \( 24f''\left( \frac{5\pi}{3} \right) \) is:
- JEE Main - 2025
- Differential Calculus
- When sec-butylcyclohexane reacts with bromine in the presence of sunlight, the major product is:
- JEE Main - 2025
- Organic Reactions