SN₁ reaction is a first order nucleophilic substitution, e.g.,

The concentration of nucleophile does not appear in the rate law expression:
Reaction rate = k[ RX]
In a multistep organic reaction, the rate-limiting step is the slowest step. Rate determining step is represented by the following energy level diagram:


A reaction energy level diagram for an reaction. The rate limiting step is spontaneous dissociation of an alkyl halide to give a carbocation intermediate

reaction is:

SN₁ reaction is a first order nucleophilic substitution, e.g.,

The concentration of nucleophile does not appear in the rate law expression:
Reaction rate = k[ RX]
In a multistep organic reaction, the rate-limiting step is the slowest step. Rate determining step is represented by the following energy level diagram:


A reaction energy level diagram for an reaction. The rate limiting step is spontaneous dissociation of an alkyl halide to give a carbocation intermediate

In the graph 3 for reaction, the rate limiting step is the spontaneous dissociation of alkyl halide and is given by:

SN₁ reaction is a first order nucleophilic substitution, e.g.,

The concentration of nucleophile does not appear in the rate law expression:
Reaction rate = k[ RX]
In a multistep organic reaction, the rate-limiting step is the slowest step. Rate determining step is represented by the following energy level diagram:


A reaction energy level diagram for an reaction. The rate limiting step is spontaneous dissociation of an alkyl halide to give a carbocation intermediate

Select the correct statement(s) about the graph 2:

SN₁ reaction is a first order nucleophilic substitution, e.g.,

The concentration of nucleophile does not appear in the rate law expression:
Reaction rate = k[ RX]
In a multistep organic reaction, the rate-limiting step is the slowest step. Rate determining step is represented by the following energy level diagram:


A reaction energy level diagram for an reaction. The rate limiting step is spontaneous dissociation of an alkyl halide to give a carbocation intermediate

In reaction, the hybridization changes in rate determination step. Select the correct change among the following:

The removal of two atoms or groups, one generally hydrogen and the other a leaving group resulting in the formation of unsaturated compound is known as elimination reaction.

In (elimination) reactions, the bond is broken heterolytic ally (in step 1) to form a carbocation (as in reaction) in which ) is lost (rate determining step). The carbocation (in step 2) loses a proton from the -carbon atom by a base (nucleophile) to form an alkene. reaction is favoured in compounds in which the leaving group is at secondary or tertiary position. In formed simultaneously. reactions occur in one step through a transition state.

reactions are most common in haloalkanes (particularly, ) and better the leaving group higher is the reaction. In reactions, both the leaving groups should be antiplanar.
cb (Elimination unimolecular conjugate base) reaction involves the removal of proton by. a conjugate base (step 1) to produce carbanion which loses a leaving group to form an alkene (step 2) and is a slow step.
Neopentyl bromide undergoes dehydrohalogenation to give alkene even though it has no -hydrogen. This is due to:

The removal of two atoms or groups, one generally hydrogen and the other a leaving group resulting in the formation of unsaturated compound is known as elimination reaction.

In (elimination) reactions, the bond is broken heterolytic ally (in step 1) to form a carbocation (as in reaction) in which ) is lost (rate determining step). The carbocation (in step 2) loses a proton from the -carbon atom by a base (nucleophile) to form an alkene. reaction is favoured in compounds in which the leaving group is at secondary or tertiary position. In formed simultaneously. reactions occur in one step through a transition state.

reactions are most common in haloalkanes (particularly, ) and better the leaving group higher is the reaction. In reactions, both the leaving groups should be antiplanar.
cb (Elimination unimolecular conjugate base) reaction involves the removal of proton by. a conjugate base (step 1) to produce carbanion which loses a leaving group to form an alkene (step 2) and is a slow step.
2-Br-omopentane is heated with potassium ethoxide in ethanol. The major product obtained is:

The removal of two atoms or groups, one generally hydrogen and the other a leaving group resulting in the formation of unsaturated compound is known as elimination reaction.

In (elimination) reactions, the bond is broken heterolytic ally (in step 1) to form a carbocation (as in reaction) in which ) is lost (rate determining step). The carbocation (in step 2) loses a proton from the -carbon atom by a base (nucleophile) to form an alkene. reaction is favoured in compounds in which the leaving group is at secondary or tertiary position. In formed simultaneously. reactions occur in one step through a transition state.

reactions are most common in haloalkanes (particularly, ) and better the leaving group higher is the reaction. In reactions, both the leaving groups should be antiplanar.
cb (Elimination unimolecular conjugate base) reaction involves the removal of proton by. a conjugate base (step 1) to produce carbanion which loses a leaving group to form an alkene (step 2) and is a slow step.

This reaction is an example of:

For any real x, the expression cannot exceed

The number of real roots of is

If z₁, z₂, z₃ are complex numbers such that∣z₁∣=∣ z₂ ∣=∣ z₃ ∣=, thus, ∣ z₁ + z₂ + z₃ ∣ is

If the roots of the equation z² + az + b = 0 are purely imaginary, then

The cube roots of unity

If z satisfies ∣ z − 1∣ < ∣ z + 3 ∣, then = 2z + 3 − i satisfies :

A(z₁), B(z₂) and C(z₃) be the vertices of an equilateral triangle in the argand plane such that ∣z₁∣ = ∣z₂∣ = ∣z₃∣. Then which of the following is false ?

If the system of equations x-ky-z=0,kx-y-z=0,x+y-z=0 has a non-zero solution then the possible values of k are

For such questions, we should remember the requirement for non zero solution. We have to be careful when finding the determinant.

If then exists

Whenever we have to find the inverse of the matrix, we should check the determinant of the matrix. The determinant must be non zero for a inverse to exist.