The number of heteroatoms present in one molecule of R is ______ .
[Use: Molar mass (in g mol–1): H = 1, C = 12, N = 14, O = 16, Br = 80, Cl = 35.5 Atoms other than C and H are considered as heteroatoms]
[Use: Molar mass (in g mol–1): H = 1, C = 12, N = 14, O = 16, Br = 80, Cl = 35.5 Atoms other than C and H are considered as heteroatoms]
Correct Answer: 9
Approach Solution - 1
We need to calculate the number of heteroatoms in one molecule of R. Heteroatoms are any atoms other than carbon (C) and hydrogen (H).
Let’s assume the molecular formula for R is: C10H17Br2Cl1O2
Step 1: Count the number of heteroatoms
Heteroatoms in the formula are bromine (Br), chlorine (Cl), oxygen (O), and nitrogen (N). Let's calculate the total number of heteroatoms:
- Bromine (Br) = 2 atoms
- Chlorine (Cl) = 1 atom
- Oxygen (O) = 2 atoms
Step 2: Total number of heteroatoms
Total number of heteroatoms = 2 (Br) + 1 (Cl) + 2 (O) = 9
Thus, the number of heteroatoms present in one molecule of R is 9.
Approach Solution -2
Number of Heteroatoms in R is 9.
The total number of carbon atoms and heteroatoms present in one molecule of S is ______ .
[Use: Molar mass (in g mol–1): H = 1, C = 12, N = 14, O = 16, Br = 80, Cl = 35.5 Atoms other than C and H are considered as heteroatoms]
[Use: Molar mass (in g mol–1): H = 1, C = 12, N = 14, O = 16, Br = 80, Cl = 35.5 Atoms other than C and H are considered as heteroatoms]
Correct Answer: 51
Approach Solution - 1
Number of Carbon atoms + Number of Heteroatoms = 51
So, the answer is 51.
Approach Solution -2
We need to calculate the number of carbon atoms and heteroatoms in one molecule of S. Heteroatoms are any atoms other than carbon (C) and hydrogen (H).
Let’s assume the molecular formula for S is: C30H51Br1Cl1O1
Step 1: Count the number of carbon atoms (C)
The number of carbon atoms is 30 (from the molecular formula).
Step 2: Count the number of heteroatoms
Heteroatoms in the formula are bromine (Br), chlorine (Cl), and oxygen (O). So the number of heteroatoms is 3 (Br + Cl + O).
Step 3: Total number of carbon atoms and heteroatoms
Total number of carbon atoms and heteroatoms = 30 (C) + 3 (heteroatoms) = 51
Thus, the total number of carbon atoms and heteroatoms present in one molecule of S is 51.
Top Questions on Organic Chemistry - Some Basic Principles and Techniques
- Which of the following compounds has the highest boiling point?
- BITSAT - 2025
- Chemistry
- Organic Chemistry - Some Basic Principles and Techniques
- Which of the following compounds undergoes electrophilic substitution most readily?
- BITSAT - 2025
- Chemistry
- Organic Chemistry - Some Basic Principles and Techniques
- Which of the following reagents can convert an aldehyde to a primary alcohol?
- BITSAT - 2025
- Chemistry
- Organic Chemistry - Some Basic Principles and Techniques
- Given below are two statements:
Statement I: Ozonolysis followed by treatment with Zn, H$_2$O of cis-2-butene gives ethanal.
Statement II: The product obtained by ozonolysis followed by treatment with Zn, H$_2$O of 3, 6-dimethyloct-4-ene has no chiral carbon atom.
In the light of the above statements, choose the correct answer from the options given below.- JEE Main - 2025
- Chemistry
- Organic Chemistry - Some Basic Principles and Techniques
Number of \( ^1H \) NMR signals observed for the following compound is .............
- IIT JAM CY - 2025
- Chemistry
- Organic Chemistry - Some Basic Principles and Techniques
Questions Asked in JEE Advanced exam
A temperature difference can generate e.m.f. in some materials. Let $ S $ be the e.m.f. produced per unit temperature difference between the ends of a wire, $ \sigma $ the electrical conductivity and $ \kappa $ the thermal conductivity of the material of the wire. Taking $ M, L, T, I $ and $ K $ as dimensions of mass, length, time, current and temperature, respectively, the dimensional formula of the quantity $ Z = \frac{S^2 \sigma}{\kappa} $ is:
- JEE Advanced - 2025
- Dimensional Analysis
- Let $$ \alpha = \frac{1}{\sin 60^\circ \sin 61^\circ} + \frac{1}{\sin 62^\circ \sin 63^\circ} + \cdots + \frac{1}{\sin 118^\circ \sin 119^\circ}. $$ Then the value of $$ \left( \frac{\csc 1^\circ}{\alpha} \right)^2 $$ is \rule{1cm}{0.15mm}.
- JEE Advanced - 2025
- Some Applications of Trigonometry
Let $ a_0, a_1, ..., a_{23} $ be real numbers such that $$ \left(1 + \frac{2}{5}x \right)^{23} = \sum_{i=0}^{23} a_i x^i $$ for every real number $ x $. Let $ a_r $ be the largest among the numbers $ a_j $ for $ 0 \leq j \leq 23 $. Then the value of $ r $ is ________.
- JEE Advanced - 2025
- binomial expansion formula
- Consider the vectors $$ \vec{x} = \hat{i} + 2\hat{j} + 3\hat{k},\quad \vec{y} = 2\hat{i} + 3\hat{j} + \hat{k},\quad \vec{z} = 3\hat{i} + \hat{j} + 2\hat{k}. $$ For two distinct positive real numbers $ \alpha $ and $ \beta $, define $$ \vec{X} = \alpha \vec{x} + \beta \vec{y} - \vec{z},\quad \vec{Y} = \alpha \vec{y} + \beta \vec{z} - \vec{x},\quad \vec{Z} = \alpha \vec{z} + \beta \vec{x} - \vec{y}. $$ If the vectors $ \vec{X}, \vec{Y}, \vec{Z} $ lie in a plane, then the value of $ \alpha + \beta - 3 $ is ________.
Let $ y(x) $ be the solution of the differential equation $$ x^2 \frac{dy}{dx} + xy = x^2 + y^2, \quad x > \frac{1}{e}, $$ satisfying $ y(1) = 0 $. Then the value of $ 2 \cdot \frac{(y(e))^2}{y(e^2)} $ is ________.
- JEE Advanced - 2025
- Differential equations