'X' is: 
Show Hint
Remember Markovnikov’s rule for electrophilic addition reactions. The most stable carbocation intermediate leads to the major product.
The Correct Option is B
Solution and Explanation
Step 1: Identify the Reactants
The reactants are tetrahydrofuran (THF) and 2-methylpropene. The reaction is catalyzed by HF and takes place under heat.
Step 2: Determine the Reaction Mechanism
This reaction is an electrophilic addition of THF to the alkene. HF protonates the alkene to form a carbocation. The oxygen in THF acts as a nucleophile and attacks the carbocation. Finally, deprotonation occurs to yield the product.
Step 3: Determine the Major Product
The major product is determined by Markovnikov’s rule, which states that the proton adds to the carbon of the double bond with more hydrogens. In this case, the carbocation will form on the more substituted carbon of 2-methylpropene, leading to product (1).
Top Questions on Hydrocarbons
- Hydroxy compound ($\text{A}$) with molecular mass $= 122$ react with excess of acetic anhydride and gives compound ($\text{X}$) with molecular mass $= 290$, then find the no. of hydroxy groups in given compound ($\text{A}$).
- JEE Main - 2026
- Chemistry
- Hydrocarbons
Consider the following reaction sequence.
- JEE Main - 2026
- Chemistry
- Hydrocarbons
Which of the following hydrocarbons reacts easily with MeMgBr to give methane?
- WBJEE - 2026
- Chemistry
- Hydrocarbons
- $\text{C}_3\text{H}_6\text{Cl}_2$ (X) $\xrightarrow[\Delta]{\text{NaNH}_2 (2 \text{eq})}$ (Y) $\xrightarrow[\text{(ii) NaBH}_4]{\text{(i) Hg(OAc)}_2/\text{H}_2\text{O}}$ $\text{CH}_3\text{COCH}_3$. (Y) $\xrightarrow[\Delta]{\text{Fe/tube}}$ (Z).
Statement-I : Y gives yellow ppt. with $\text{NaOH}/\text{I}_2$.
Statement-II : Two types of H-atoms and one aromatic ring is present in Z and ratio of Z and X is 1 : 3.
Choose the correct option.- JEE Main - 2026
- Chemistry
- Hydrocarbons
- Ph-CH=CH$_2 \xrightarrow[\text{(PhCOO)}_2]{\text{HBr}}$ Product.
Correct statement(s) regarding product :
(a) Ph-CH(Br)-CH$_3$ is minor product
(b) Benzene is also form a bi product
(c) Reaction follow free radical mechanism
(d) In absence of peroxide carbocation mechanism is followed- JEE Main - 2026
- Chemistry
- Hydrocarbons
Questions Asked in JEE Main exam
- Let $\vec{a}=2\hat{i}-\hat{j}-\hat{k}$, $\vec{b}=\hat{i}+3\hat{j}-\hat{k}$ and $\vec{c}=2\hat{i}+\hat{j}+3\hat{k}$. Let $\vec{v}$ be the vector in the plane of $\vec{a}$ and $\vec{b}$, such that the length of its projection on the vector $\vec{c}$ is $\dfrac{1}{\sqrt{14}}$. Then $|\vec{v}|$ is equal to
- JEE Main - 2026
- Vector Algebra
A substance 'X' (1.5 g) dissolved in 150 g of a solvent 'Y' (molar mass = 300 g mol$^{-1}$) led to an elevation of the boiling point by 0.5 K. The relative lowering in the vapour pressure of the solvent 'Y' is $____________ \(\times 10^{-2}\). (nearest integer)
[Given : $K_{b}$ of the solvent = 5.0 K kg mol$^{-1}$]
Assume the solution to be dilute and no association or dissociation of X takes place in solution.- JEE Main - 2026
- Solutions
- The wavelength of photon 'A' is 400 nm. The frequency of photon 'B' is \(10^{16}\,\text{s}^{-1}\). The wave number of photon 'C' is \(10^{5}\,\text{cm}^{-1}\). The correct order of energy of these photons is:
- JEE Main - 2026
- General Chemistry
- Given below are two statements: Statement I: The increasing order of boiling point of hydrogen halides is HCl<HBr<HI<HF. Statement II: The increasing order of melting point of hydrogen halides is HCl<HBr<HF<HI. In the light of the above statements, choose the correct answer from the options given below:
- JEE Main - 2026
- General Chemistry
Inductance of a coil with \(10^4\) turns is \(10\,\text{mH}\) and it is connected to a DC source of \(10\,\text{V}\) with internal resistance \(10\,\Omega\). The energy density in the inductor when the current reaches \( \left(\frac{1}{e}\right) \) of its maximum value is \[ \alpha \pi \times \frac{1}{e^2}\ \text{J m}^{-3}. \] The value of \( \alpha \) is _________.
\[ (\mu_0 = 4\pi \times 10^{-7}\ \text{TmA}^{-1}) \]- JEE Main - 2026
- Ray optics and optical instruments
Concepts Used:
Hydrocarbons
Hydrocarbons can be described as organic compounds that consists only hydrogen and carbon atoms. These compounds are of different types and thereby have distinct natures. Hydrocarbons are colorless gases and are known for discharging faint odours. These have been categorized under four major classes named as alkynes, alkanes, alkenes, and aromatic hydrocarbons.
Types of Hydrocarbons
- Saturated hydrocarbons - Saturated hydrocarbons are those compounds where there is a single bond exists between carbon atoms and are saturated with atoms of hydrogen.
- Unsaturated hydrocarbons - Hydrocarbons comprises of at least one double or triple bond between carbon atoms are known as unsaturated hydrocarbons.
- Aliphatic hydrocarbons - The term denotes the hydrocarbons formed as an outcome of the chemical degradation of fats. Aliphatic hydrocarbons are basically chemical compounds.
- Aromatic hydrocarbons - They are distinguished because of the presence of benzene rings in them. They give away distinct types of aroma. These hydrocarbons comprises of only hydrogen and carbon atoms.