What will be the major product when 2-methylbutane undergoes bromination?
2-bromo-2-methyl butane.
What is the bromination of methane?
A Free Radical Substitution Reaction The organic product is bromomethane. One of the hydrogen atoms in the methane has been replaced by a bromine atom, so this is a substitution reaction. However, the reaction doesn’t stop there, and all the hydrogens in the methane can in turn be replaced by bromine atoms.
What will be the products of the bromination of methane?
When a mixture of methane and bromine is exposed to ultraviolet light – typically sunlight – a substitution reaction occurs and the organic product is bromomethane. That means that you could get any of bromomethane, dibromomethane, tribromomethane or tetrabromomethane.
What is the name of the product obtained after 2-methylbutane bromination reaction at exposure to UV light?
2-bromo-2-methylbutane
The relative rates of attack at the different types of H atom are 3°:2°:1° = 1640:82:1. So the major product is C, 2-bromo-2-methylbutane.
What is the organic product when 2-methylbutane is allowed to react with Br2 in the presence of light or heat?
2- methylbutane on reacting with bromine in the presence of sunlight gives mainly. 1 – bromo-2- methylbutane.
Which is the main major product for the Monochlorination of 2-methylbutane?
2-chloro-2-methylpropane
The reaction occurs as: So, instead of forming 2-chloro-2-methylpropane as the major product, it forms 1-chloro-2-methylpropane as a major product because of the presence of 9 primary hydrogens. Hence, the option (c)- 1-chloro-2-methylpropane as the major product is the correct answer.
What is bromination give example?
The bromination of benzene is an example of an electrophilic aromatic substitution reaction. Then, a proton is removed from the intermediate to form a substituted benzene ring.
How does bromine react with methane?
Alkanes undergo a substitution reaction with halogens in the presence of light. For instance, in ultraviolet light , methane reacts with halogen molecules such as chlorine and bromine. This reaction is a substitution reaction because one of the hydrogen atoms from the methane is replaced by a bromine atom.
What is the common name of 2-methylbutane?
Isopentane
Isopentane, also called methylbutane or 2-methylbutane, is a branched-chain saturated hydrocarbon (an alkane) with five carbon atoms, with formula C5H12 or CH(CH3)2(C2H5). Isopentane is an extremely volatile and extremely flammable liquid at room temperature and pressure.
What is the major product of chlorination of 2 Methylbutane?
In 2-Methylpropane there are 9 primary hydrogens present and tertiary hydrogen is present. The reaction occurs as: So, instead of forming 2-chloro-2-methylpropane as the major product, it forms 1-chloro-2-methylpropane as a major product because of the presence of 9 primary hydrogens.
Which is more effective in the bromination of 2-methylpropane?
Thus, in the radical bromination of 2-methylpropane, the tertiary alkyl radical is much more effectively generated by the bromine radical’s attack than the primary alykl radical. However, chlorination may be a relevant reaction, when the alkane contains only one type of C-H bond, such as in cyclohexane (only secondary).
What is the major product obtained in the bromination of?
The major product obtained in the bromination of (CH₃)₂CHCH₂CH₃ is (CH₃)₂CBrCH₂CH₃. The free-radical bromination of this alkane produces four products: 1-bromo-3-methylbutane (A) 2-bromo-3-methylbutane (B) 2-bromo-2-methylbutane (C) 1-bromo-2-methylbutane(D) But bromine is highly regioselective in where it attacks.
Why is 2-chloro-2-methylbutane preferred in the transition state?
Therefore, the stability of the product radical has less influence on the activation energy of the reaction. Thus, according to the Hammond postulate, the transition state is more reactant-like. The statistical factor is responsible for this. Thus, 2-chloro-2-methylbutane is indeed slightly preferred – as you have expected.
How to calculate the relative amounts of bromine?
But bromine is highly regioselective in where it attacks. The relative rates of attack at the different types of H atom are 3°:2°:1° = 1640:82:1. To calculate the relative amounts of each product, we multiply the numbers of each type of atom by the relative reactivity.