The isomeric deuterated bromide with molecular formula $C _4 H _8 DBr$ having two chiral carbon atoms is
- 2 - Bromo - 1 - deuterobutane
- 2 - Bromo - 2 - deuterobutane
- 2 - Bromo - 3 - deuterobutane
- 2 - Bromo-1 - deutero - 2 - methylpropane
The Correct Option is C
Approach Solution - 1
The correct answer is (C) : 2 - Bromo - 3 - deuterobutane
Approach Solution -2
Top JEE Main Chemistry Questions
What will be the equilibrium constant of the given reaction carried out in a \(5 \,L\) vessel and having equilibrium amounts of \(A_2\) and \(A\) as \(0.5\) mole and \(2 \times 10^{-6}\) mole respectively?
The reaction : \(A_2 \rightleftharpoons 2A\)- At equilibrium the concentrations of and in a sealed vessel at . What will be for the reaction
- Find out the major products from the following reactions
Cobalt chloride when dissolved in water forms pink colored complex $X$ which has octahedral geometry. This solution on treating with cone $HCl$ forms deep blue complex, $\underline{Y}$ which has a $\underline{Z}$ geometry $X, Y$ and $Z$, respectively, are
- The correct order of atomic radii is:
Top JEE Main Chemical Reactions of Haloalkanes and Haloarenes Questions
- $\textbf{A and B formed in the following reactions are:}$ \[ \text{CrO}_2\text{Cl}_2 + 4\text{NaOH} \rightarrow \text{A} + 2\text{NaCl} + 2\text{H}_2\text{O} \] \[ \text{A} + 2\text{HCl} + 2\text{H}_2\text{O}_2 \rightarrow \text{B} + 3\text{H}_2\text{O} \]
Statement I: High concentration of strong nucleophilic reagent with secondary alkyl halides which do not have bulky substituents will follow \(S_N2\) mechanism.
Statement II: A secondary alkyl halide when treated with a large excess of ethanol follows \(S_N1\) mechanism.
In the light of the above statements, choose the most appropriate from the questions given below:
- Find out sum of the coefficients of all the species involved in the balanced equation: \(2MnO_4+I^-\xrightarrow[{Medium}]{Alkaline}\)
- $\textbf{The orange colour of } \text{K}_2\text{Cr}_2\text{O}_7 \textbf{ and purple colour of KMnO}_4 \textbf{ is due to}$
The major products ' A ' and ' B ', respectively, are
Top JEE Main Questions
- Two masses $m$ and $\frac{m}{2}$ are connected at the two ends of a massless rigid rod of length $l$. The rod is suspended by a thin wire of torsional constant $k$ at the centre of mass of the rod-mass system(see figure). Because of torsional constant $k$, the restoring torque is $\tau =k\theta$ for angular displacement $\theta$. If the rod is rota ted by $\theta_0$ and released, the tension in it when it passes through its mean position will be:
What will be the equilibrium constant of the given reaction carried out in a \(5 \,L\) vessel and having equilibrium amounts of \(A_2\) and \(A\) as \(0.5\) mole and \(2 \times 10^{-6}\) mole respectively?
The reaction : \(A_2 \rightleftharpoons 2A\)- The value of is
- The number of terms of an is even. The sum of the odd terms is and of the even terms is . The last term exceeds the first by . Then the number of terms in the series is ______.
A black body is at a temperature of 2880 K. The energy of radiation emitted by this body with wavelength between 499 nm and 500 nm is U1, between 999 nm and 1000 nm is U2 and between 1499 nm and 1500 nm is U3. The Wien's constant, b = 2.88×106 nm-K. Then,
Concepts Used:
Preparation of Haloalkanes
- Preparation by Alcohols: By using alcohol, we can very easily prepare Haloalkane as when R-OH is reacted with a suitable reagent it will form R-X. The following reagents will help in the formation of the reactions:
- Halogen Acids (HX)
- Thionyl chloride (SOCl2)
- Phosphorous halides (PX5 or PX3)
- Preparation by Free Radical Halogenation: The formation of alkyl bromides and alkyl chloride is very much possible with the help of free radical halogenation, but as these radicals are highly reactive and non-specific in nature, they can result in the formation of a mixture of products. Though, it's not a preferred method for the preparation of haloalkanes. For example, free radical chlorination can result in the formation of a number of haloalkanes, which in turn makes it difficult to set apart a single product.
- Preparation by Alkenes: An electrophilic addition reaction can be used to transform an alkene into a haloalkane as alkene will react with HX to form R-X.
Read More: Haloalkanes and Haloarenes