The removal of two atoms or groups, one generally hydrogen $open parentheses H to the power of plus end exponent close parentheses$ and the other a leaving group $open parentheses L to the power of minus end exponent close parentheses$ resulting in the formation of unsaturated compound is known as elimination reaction.

In $E subscript 1 end subscript$ (elimination) reactions, the $C minus L$ bond is broken heterolytic ally (in step 1) to form a carbocation (as in $S subscript N to the power of 1 end exponent end subscript$ reaction) in which $open parentheses L to the power of minus end exponent close$ ) is lost (rate determining step). The carbocation (in step 2) loses a proton from the $beta$ -carbon atom by a base (nucleophile) to form an alkene. $E subscript 1 end subscript$ reaction is favoured in compounds in which the leaving group is at secondary $open parentheses 2 to the power of ring operator end exponent close parentheses to the power of ´ end exponent$ or tertiary $open parentheses 3 to the power of ring operator end exponent close parentheses$ position. In $E subscript 2 end subscript text end text left parenthesis e l i m i n a t i o n right parenthesis text end text r e a c t i o n s comma text end text t w o text end text s i g m a text end text b o n d s text end text a r e text end text b r o k e n text end text a n d text end text a$ formed simultaneously. $E subscript 2 end subscript$ reactions occur in one step through a transition state.

$E subscript 2 end subscript$ reactions are most common in haloalkanes (particularly, $1 to the power of ring operator end exponent$ ) and better the leaving group higher is the $E subscript 1 end subscript$ reaction. In $E subscript 2 end subscript$ reactions, both the leaving groups should be antiplanar.
$E subscript 1 end subscript$ 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 $beta$ -hydrogen. This is due to:

Chemistry-

The removal of two atoms or groups, one generally hydrogen $open parentheses H to the power of plus end exponent close parentheses$ and the other a leaving group $open parentheses L to the power of minus end exponent close parentheses$ resulting in the formation of unsaturated compound is known as elimination reaction.

In $E subscript 1 end subscript$ (elimination) reactions, the $C minus L$ bond is broken heterolytic ally (in step 1) to form a carbocation (as in $S subscript N to the power of 1 end exponent end subscript$ reaction) in which $open parentheses L to the power of minus end exponent close$ ) is lost (rate determining step). The carbocation (in step 2) loses a proton from the $beta$ -carbon atom by a base (nucleophile) to form an alkene. $E subscript 1 end subscript$ reaction is favoured in compounds in which the leaving group is at secondary $open parentheses 2 to the power of ring operator end exponent close parentheses to the power of ´ end exponent$ or tertiary $open parentheses 3 to the power of ring operator end exponent close parentheses$ position. In $E subscript 2 end subscript text end text left parenthesis e l i m i n a t i o n right parenthesis text end text r e a c t i o n s comma text end text t w o text end text s i g m a text end text b o n d s text end text a r e text end text b r o k e n text end text a n d text end text a$ formed simultaneously. $E subscript 2 end subscript$ reactions occur in one step through a transition state.

$E subscript 2 end subscript$ reactions are most common in haloalkanes (particularly, $1 to the power of ring operator end exponent$ ) and better the leaving group higher is the $E subscript 1 end subscript$ reaction. In $E subscript 2 end subscript$ reactions, both the leaving groups should be antiplanar.
$E subscript 1 end subscript$ 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:

Chemistry-General

For any real x, the expression $2 open parentheses K minus x close parentheses open square brackets x plus square root of x to the power of 2 end exponent plus K to the power of 2 end exponent end root close square brackets$ cannot exceed

The number of real roots of $open parentheses sin invisible function application 2 to the power of x end exponent close parentheses open parentheses cos invisible function application 2 to the power of x end exponent close parentheses equals fraction numerator 1 over denominator 4 end fraction open parentheses 2 to the power of x end exponent plus 2 to the power of negative x end exponent close parentheses$ is

If z₁, z₂, z₃ are complex numbers such that∣z₁∣=∣ z₂ ∣=∣ z₃ ∣= $open vertical bar fraction numerator 1 over denominator z subscript 1 end subscript end fraction plus fraction numerator 1 over denominator z subscript 2 end subscript end fraction plus fraction numerator 1 over denominator z subscript 3 end subscript end fraction close vertical bar equals 1$ , thus, ∣ z₁ + z₂ + z₃ ∣ is

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 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 $A equals open square brackets table row cell fraction numerator 1 over denominator 2 end fraction open parentheses e to the power of i x end exponent plus e to the power of negative i x end exponent close parentheses end cell cell fraction numerator 1 over denominator 2 end fraction open parentheses e to the power of i x end exponent minus e to the power of negative i x end exponent close parentheses end cell row cell fraction numerator 1 over denominator 2 end fraction open parentheses e to the power of i x end exponent minus e to the power of negative i x end exponent close parentheses end cell cell fraction numerator 1 over denominator 2 end fraction open parentheses e to the power of i x end exponent plus e to the power of negative i x end exponent close parentheses end cell end table close square brackets$ then $A to the power of negative 1 end exponent$ 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.

If $A equals open square brackets table row cell fraction numerator 1 over denominator 2 end fraction open parentheses e to the power of i x end exponent plus e to the power of negative i x end exponent close parentheses end cell cell fraction numerator 1 over denominator 2 end fraction open parentheses e to the power of i x end exponent minus e to the power of negative i x end exponent close parentheses end cell row cell fraction numerator 1 over denominator 2 end fraction open parentheses e to the power of i x end exponent minus e to the power of negative i x end exponent close parentheses end cell cell fraction numerator 1 over denominator 2 end fraction open parentheses e to the power of i x end exponent plus e to the power of negative i x end exponent close parentheses end cell end table close square brackets$ then $A to the power of negative 1 end exponent$ 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.

If $l subscript 1 end subscript superscript 2 end superscript plus m subscript 1 end subscript superscript 2 end superscript plus n subscript 1 end subscript superscript 2 end superscript equals 1 comma$ etc., and $l subscript 1 end subscript l subscript 2 end subscript plus m subscript 1 end subscript m subscript 2 end subscript plus n subscript 1 end subscript n subscript 2 end subscript equals 0 comma$ etc. and $capital delta equals open vertical bar table row cell l subscript 1 end subscript end cell cell m subscript 1 end subscript end cell cell n subscript 1 end subscript end cell row cell l subscript 2 end subscript end cell cell m subscript 2 end subscript end cell cell n subscript 2 end subscript end cell row cell l subscript 3 end subscript end cell cell m subscript 3 end subscript end cell cell n subscript 3 end subscript end cell end table close vertical bar$ then

If $A equals open square brackets table row alpha 0 row 1 1 end table close square brackets$ and $B equals open square brackets table row 1 0 row 3 1 end table close square brackets comma$ then value of $alpha$ for which $A to the power of 2 end exponent equals B$ is

For such questions, we should know how to multiply to matrices. When there is equal sign between two matrices, the elements of both the matrix should be equal.

If $A equals open square brackets table row alpha 0 row 1 1 end table close square brackets$ and $B equals open square brackets table row 1 0 row 3 1 end table close square brackets comma$ then value of $alpha$ for which $A to the power of 2 end exponent equals B$ is

For such questions, we should know how to multiply to matrices. When there is equal sign between two matrices, the elements of both the matrix should be equal.

For the primitive integral equation $y d x plus y to the power of 2 end exponent d y equals x d y comma x element of R comma y greater than 0 comma y equals y open parentheses x close parentheses comma y open parentheses 1 close parentheses equals 1 comma$ then $y open parentheses negative 3 close parentheses$ is

For such questions, we should know different method of differentiation and integration.

For the primitive integral equation $y d x plus y to the power of 2 end exponent d y equals x d y comma x element of R comma y greater than 0 comma y equals y open parentheses x close parentheses comma y open parentheses 1 close parentheses equals 1 comma$ then $y open parentheses negative 3 close parentheses$ is

For such questions, we should know different method of differentiation and integration.

The differential equation of all circles which pass through the origin and whose centre lies on y-axis is

For such questions, we should know the equation of cricle with its centre at a point other than origin.

The differential equation of all circles which pass through the origin and whose centre lies on y-axis is