Class 12th

2, 4-DNP (Brady's Reagent) is used to test carbonyl compound

(a) Na₂CO₃ -> Solvay

(b) Ti -> Van-Arkel

(c) Cl₂ -> Deacon

(d) NaOH -> Castner – kellner

It is an intramolecular aldol condensation

A > B > C > D

Lone pair is localized in (A) while all 3 have delocalized lone pair but they can be compared by 3° > 2° > 1° because methyl group increases the basicity.

Consider the equation of plane,

$P:\left(2x+3y+z+20\right)+\lambda \left(x-3y+5z-8\right)=0$  

$?$  Plane P is perpendicular to 2x + 3y + z + 20 = 0

So,  $4+2\lambda +9-9\lambda +1+5\lambda =0$  

0  $\lambda =7$  

P : 9x – 18y + 36z – 36 = 0

Or P : x – 2y + 4z = 4

If image of

$\left(2,-\frac{1}{2},2\right)$  

In plane P is (a, b, c) then

$\frac{a-2}{1}=\frac{b+\frac{1}{2}}{-2}=\frac{c-2}{4}$  

and  $\left(\frac{a+2}{2}\right)-2\left(\frac{b-\frac{1}{2}}{2}\right)+4\left(\frac{c+2}{2}\right)=4$     

clearly 

$a=\frac{4}{3},b=\frac{5}{6}andc=-\frac{2}{3}$  

So, a : b : c = 8 : 5 : -4

Both (A) and (B) are correct but R is not correct explanation of A. In both oxidation state of metal is +1, also both have similar lattice structure.  

(a) Sucrose -> a-D glucose and b-D fructose

(b) Lactose -> b-D- galactose and b-D glucose

(c) maltose -> a-D glucose and a-D-glucose

$\frac{x-1}{1}=\frac{y-2}{2}=\frac{z-1}{2}=\frac{-2\left(1+4+2-16\right)}{1+2^2+2^2}$

(x, y, z) = (3, 6, 5)

now point Q and line both lies in the plane.

So, equation of plane is

$\left|\begin{array}{ccc}\hfill x\hfill & \hfill y\hfill & \hfill z+1\hfill \\ \hfill 3\hfill & \hfill 6\hfill & \hfill 6\hfill \\ \hfill 1\hfill & \hfill 1\hfill & \hfill 2\hfill \end{array}\right|=0$ a

=> 2x – z = 1

option (B) satisfies.

Let AB   $\equiv x-2y+1=0$

AC $\equiv x-2y+1=0$  

So vertex A = (1, 1)

altitude from B is perpendicular to AC and passing through

orthocentre.

So, BH = x + 2y – 7 = 0

CH = 2x + y – 7 = 0

now solve AB & BH to get B (3, 2) similarly CH and AC to get C (2, 3) so centroid is at (2, 2)

Line $\perp$  to the normal

=>3p + 2q – 1 = 0

$\left(2,-1,-3\right)$ lies in the plane 2p + q = 8

From here p = 15, q = -22

Equation of plane 15x – 22y + z – 5 = 0

Distance from origin = $\left|\frac{5}{\sqrt[]{15^2+{\left(-22\right)}^2+1^2}}\right|=\sqrt[]{\frac{5}{142}}$