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Chemistry-

General

Easy

Question

The major product formed when 2-bromobutane is treated with alcoholic KOH is :

2-butanol
trans-2-butene
1-butanol
1-butene

The correct answer is: trans-2-butene

Related Questions to study

Name the reagent used to bring about the following transformation: but-2-ene to ethanal:

Name the reagent used to bring about the following transformation: but-2-ene to ethanal:

Chemistry-General

In the following reaction, the product 'R' is : $CaC subscript 2 not stretchy rightwards arrow with straight H subscript 2 straight O on top straight P fraction numerator text hot iron end text over denominator text tube end text end fraction straight Q fraction numerator CH subscript 3 Cl over denominator AlCl subscript 3 end fraction not stretchy rightwards arrow straight R$

In the following reaction, the product 'R' is : $CaC subscript 2 not stretchy rightwards arrow with straight H subscript 2 straight O on top straight P fraction numerator text hot iron end text over denominator text tube end text end fraction straight Q fraction numerator CH subscript 3 Cl over denominator AlCl subscript 3 end fraction not stretchy rightwards arrow straight R$

Chemistry-General

Among the given· compounds, one which can be distinguished by AgNO₃ is:

Among the given· compounds, one which can be distinguished by AgNO₃ is:

Chemistry-General

parallel

The alkene that will give the same product with HBr in the absence as well as in the presence of peroxide is':

The alkene that will give the same product with HBr in the absence as well as in the presence of peroxide is':

Chemistry-General

The 'decreasing order of acidic character among
I) ethane
II) ethene
III) ethyne
IV) propyne
Is:

The 'decreasing order of acidic character among
I) ethane
II) ethene
III) ethyne
IV) propyne
Is:

Chemistry-General

One mole of a symmetrical alkene on ozonolysis gives two moles of an aldehyde having a molecular mass of 44u. The alkene is

One mole of a symmetrical alkene on ozonolysis gives two moles of an aldehyde having a molecular mass of 44u. The alkene is

Chemistry-General

parallel

Identify the product (Y) in the following reaction sequence:

Identify the product (Y) in the following reaction sequence:

Chemistry-General

The radius of the circle passing through the centre or the in-circle of $triangle A B C$ and through the end points of BC is given by

The radius of the circle passing through the centre or the in-circle of $triangle A B C$ and through the end points of BC is given by

CH₃COOH is reacted with $H C identical to C H$ in presence of Hg²⁺ , the product is:

CH₃COOH is reacted with $H C identical to C H$ in presence of Hg²⁺ , the product is:

Chemistry-General

parallel

If $fraction numerator 1 over denominator a end fraction comma fraction numerator 1 over denominator b end fraction comma fraction numerator 1 over denominator c end fraction$ are in A.P., where a, b, c are sides of the triangle ABC, then $s i n to the power of 2 end exponent invisible function application fraction numerator A over denominator 2 end fraction comma s i n to the power of 2 end exponent invisible function application fraction numerator B over denominator 2 end fraction comma s i n to the power of 2 end exponent invisible function application fraction numerator C over denominator 2 end fraction$ are in

If $fraction numerator 1 over denominator a end fraction comma fraction numerator 1 over denominator b end fraction comma fraction numerator 1 over denominator c end fraction$ are in A.P., where a, b, c are sides of the triangle ABC, then $s i n to the power of 2 end exponent invisible function application fraction numerator A over denominator 2 end fraction comma s i n to the power of 2 end exponent invisible function application fraction numerator B over denominator 2 end fraction comma s i n to the power of 2 end exponent invisible function application fraction numerator C over denominator 2 end fraction$ are in

Statement1: If a line intersects a hyperbola at (2, 6) and (4, 4) and one of the asymptotes at (1, 2) then the centre of the hyperbola is (1, 2) because
Statement2: Mid point of the chord intercepted by hyperbola is same as midpoint of the chord intercepted between
asymptotes

Statement1: If a line intersects a hyperbola at (2, 6) and (4, 4) and one of the asymptotes at (1, 2) then the centre of the hyperbola is (1, 2) because
Statement2: Mid point of the chord intercepted by hyperbola is same as midpoint of the chord intercepted between
asymptotes

lf $not stretchy integral subscript blank superscript blank left parenthesis blank s i n blank 2 x plus blank c o s blank 2 x right parenthesis d x equals fraction numerator 1 over denominator square root of 2 end fraction blank s i n blank left parenthesis 2 x minus c right parenthesis plus 0$ , then the value of a and c is

lf $not stretchy integral subscript blank superscript blank left parenthesis blank s i n blank 2 x plus blank c o s blank 2 x right parenthesis d x equals fraction numerator 1 over denominator square root of 2 end fraction blank s i n blank left parenthesis 2 x minus c right parenthesis plus 0$ , then the value of a and c is

parallel

lf $integral subscript blank superscript blank fraction numerator left parenthesis x squared minus 1 right parenthesis over denominator left parenthesis x to the power of 4 plus 3 x squared plus 1 right parenthesis t a n to the power of negative 1 end exponent left parenthesis fraction numerator x squared plus right square bracket over denominator chi end fraction right parenthesis end fraction d x equals k blank l o g blank vertical line t a n to the power of negative 1 end exponent fraction numerator x squared plus 1 over denominator chi end fraction vertical line plus c comma$ then $k$ is equal to

lf $integral subscript blank superscript blank fraction numerator left parenthesis x squared minus 1 right parenthesis over denominator left parenthesis x to the power of 4 plus 3 x squared plus 1 right parenthesis t a n to the power of negative 1 end exponent left parenthesis fraction numerator x squared plus right square bracket over denominator chi end fraction right parenthesis end fraction d x equals k blank l o g blank vertical line t a n to the power of negative 1 end exponent fraction numerator x squared plus 1 over denominator chi end fraction vertical line plus c comma$ then $k$ is equal to

lf $integral subscript blank superscript blank fraction numerator d x over denominator chi square root of 1 minus x cubed end root end fraction equals a blank l o g blank vertical line fraction numerator square root of 1 minus x cubed end root minus 1 over denominator square root of 1 minus x cubed end root plus 1 end fraction vertical line plus c$ , then

lf $integral subscript blank superscript blank fraction numerator d x over denominator chi square root of 1 minus x cubed end root end fraction equals a blank l o g blank vertical line fraction numerator square root of 1 minus x cubed end root minus 1 over denominator square root of 1 minus x cubed end root plus 1 end fraction vertical line plus c$ , then

The gradient of the curve y=f(x) at P is slope of the tangent line at P i.e.,

PM = length of the tangent
PR = length of the normal to the curve y=f(x)
Area of triangle formed by the positive x-axis and the normal and the tangent to $x to the power of 2 end exponent plus y to the power of 2 end exponent equals 4$ at $left parenthesis 1 comma square root of 3 right parenthesis$ is

The gradient of the curve y=f(x) at P is slope of the tangent line at P i.e.,

PM = length of the tangent
PR = length of the normal to the curve y=f(x)
Area of triangle formed by the positive x-axis and the normal and the tangent to $x to the power of 2 end exponent plus y to the power of 2 end exponent equals 4$ at $left parenthesis 1 comma square root of 3 right parenthesis$ is

parallel

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