Question
The IUPAC name of the following compound is:
- 4-Bromo-3-cyanophenol
- 2-Bromo-5-hydroxybenzonitrile
- 2-Cyano-4-hydroxybromobenzene
- 6-Bromo-3-hydroxybenzonitrile
The correct answer is: 2-Bromo-5-hydroxybenzonitrile
Related Questions to study
If the line x + y = k is a normal to the parabola y2 = 4x,then the value of k will be-
the equation of tangent and normal in terms of the slope of the parabola are used to find the equations for any given slope.
If the line x + y = k is a normal to the parabola y2 = 4x,then the value of k will be-
the equation of tangent and normal in terms of the slope of the parabola are used to find the equations for any given slope.
The equation of the normal to the parabola x2 = 8y whose slope is 1/m is
equations of normal and tangent in terms of the slope are used to find the normal and tangent for any given slope.
The equation of the normal to the parabola x2 = 8y whose slope is 1/m is
equations of normal and tangent in terms of the slope are used to find the normal and tangent for any given slope.
The equation of the normal to the parabola y2 + 12x = 0 at the upper end of it's latus rectum is
the normal and tangent are always perpendicular to each other at any given point. Therefore, the product of slopes of tangent and normal is equal to -1.
The equation of the normal to the parabola y2 + 12x = 0 at the upper end of it's latus rectum is
the normal and tangent are always perpendicular to each other at any given point. Therefore, the product of slopes of tangent and normal is equal to -1.
The equation of the normal having slope m of the parabola y2 = x + a is
parametric form of slope of parabola = y+tx = 2at + at^3
The equation of the normal having slope m of the parabola y2 = x + a is
parametric form of slope of parabola = y+tx = 2at + at^3
Find the equation of normal to the curve 2y = 3–x2 at the point (1, 1)
equation of straight line :
(y-y1)=m(x-x1)
Find the equation of normal to the curve 2y = 3–x2 at the point (1, 1)
equation of straight line :
(y-y1)=m(x-x1)
The equations of the normal at the ends of the latus rectum of the parabola y2 = 4ax are given by-
the slope of normal = -1/ slope of tangent.
Equation of both normals can be generalized by multiplying both the equations.
The equations of the normal at the ends of the latus rectum of the parabola y2 = 4ax are given by-
the slope of normal = -1/ slope of tangent.
Equation of both normals can be generalized by multiplying both the equations.
The line 
x + my + n = 0 is a normal to the parabola y2 = 4 ax if-
equation of normal of a parabola is given by:
y = -tx + 2at + at3
The line x + my + n = 0 is a normal to the parabola y2 = 4 ax if-
equation of normal of a parabola is given by:
y = -tx + 2at + at3
The equation of the normal to the parabola y2 = 16 x with slope – 1/4 is-
equation of normal of a parabola is given by:
y = -tx + 2at + at3
equation of normal in slope form: y = mx – 2am – am3
The equation of the normal to the parabola y2 = 16 x with slope – 1/4 is-
equation of normal of a parabola is given by:
y = -tx + 2at + at3
equation of normal in slope form: y = mx – 2am – am3
The equation of normal on point (2,4) to the parabola y2 = 8x is-
we can also use equation of normal in slope form: y = mx – 2am – am3
The equation of normal on point (2,4) to the parabola y2 = 8x is-
we can also use equation of normal in slope form: y = mx – 2am – am3
Area of triangle formed by the tangents at three points t1, t2 and t3 of the parabola y2 = 4ax is-
area under a triangle is given by the determinant of the vertex points of the triangle.
Area of triangle formed by the tangents at three points t1, t2 and t3 of the parabola y2 = 4ax is-
area under a triangle is given by the determinant of the vertex points of the triangle.
The equation of the common tangents to the parabolas y2 = 4x and x2 = 32y is-
the equation of tangent to the parabola y2= 4ax is y = mx + a/m
The equation of the common tangents to the parabolas y2 = 4x and x2 = 32y is-
the equation of tangent to the parabola y2= 4ax is y = mx + a/m
The equation of tangent to the parabola x2 = y at one extremity of latus rectum in the first quadrant is
equation of tangent : T=0 of
X2= 4ay is given by
xx1=(y+y1)/2
The equation of tangent to the parabola x2 = y at one extremity of latus rectum in the first quadrant is
equation of tangent : T=0 of
X2= 4ay is given by
xx1=(y+y1)/2
The point of contact of the line 2x – y + 2 = 0 with the parabola y2 = 16 x is-
for a line to be a tangent of a parabola, the value of a/m should be equal to the value of c.
The point of contact of the line 2x – y + 2 = 0 with the parabola y2 = 16 x is-
for a line to be a tangent of a parabola, the value of a/m should be equal to the value of c.
The point on the curve y2 = x the tangent at which makes an angle of 45º with x-axis will be given by
the point of contact of a tangent with the curve is given by (x,y) =(a/m2, 2a/m). this is obtained by solving the equation of tangent with the equation of curve.
The point on the curve y2 = x the tangent at which makes an angle of 45º with x-axis will be given by
the point of contact of a tangent with the curve is given by (x,y) =(a/m2, 2a/m). this is obtained by solving the equation of tangent with the equation of curve.
The equation of the tangent to the parabola y = 2 + 4x –4x2 with slope –4 is
equation of a straight line : y-y1=m(x-x1)
m is given, calculate x1 and y1 by differentiating the equation of parabola and equating it with the slope.
The equation of the tangent to the parabola y = 2 + 4x –4x2 with slope –4 is
equation of a straight line : y-y1=m(x-x1)
m is given, calculate x1 and y1 by differentiating the equation of parabola and equating it with the slope.
