(1)The median AD of triangle ABC is bisected at E,BE meets AC in F then AF:AC
(a)1:3 (b)1:2 (c)3:4(d)1:4
(2)CH3-CH2-CO-18O-CH2-CH3+NAOH
which bond is broken???
answer to question 1>
i do not the exact solution.
but since it is an MCQ i would use extreme case analysis.
that is, i would assume the triangle to be a equilateral triangle.this should hold good becuase the given question does not specify anything.
when i calculated in this way i got 1:3
is the answer correct.
answer question 2>
the bond between the oxygens is broken.
in general the alkoxide group always come out of the ester on hydrolysis.
therefore in this case a sodium salt of the acid and the alcohol with the radioactive oxygen are the products.
i do not the exact solution.
but since it is an MCQ i would use extreme case analysis.
that is, i would assume the triangle to be a equilateral triangle.this should hold good becuase the given question does not specify anything.
when i calculated in this way i got 1:3
is the answer correct.
answer question 2>
the bond between the oxygens is broken.
in general the alkoxide group always come out of the ester on hydrolysis.
therefore in this case a sodium salt of the acid and the alcohol with the radioactive oxygen are the products.
Hey thanks Aditya your answer is right.
(1.) Let the co-ordinates of B,C,A be (0,0),(c,0),(a,b) .
As D is the mid-point of BC , co-ordinates of D is (c/2,0)
As E is the mid-point of AD , co-ordinates of E is [(a/2+c/4),b/2] .
Equation of BE is :
=> y=(2bx)/(c+2a) -----------(1)
Equation of AC is :
=> (a-c)*y = b*(x-c) -------------(2)
Solving (1) & (2) :
=> x=(c+2a)/3
=> y=2b/3
Co-ordinates of F are [(c+2a)/3 , 2b/3]
Let AF:FC = m:1
Using section formula :
=> (m*0+b*1)/(m+1) = 2b/3
=> m = 1/2
=> AF:FC = 1:2
=> AF:AC = 1:3
So answer is 1:3
(2)O-O18 bond is broken . - +
The products are : CH3-CH2-COOH & CH3-CH2-O18Na
As D is the mid-point of BC , co-ordinates of D is (c/2,0)
As E is the mid-point of AD , co-ordinates of E is [(a/2+c/4),b/2] .
Equation of BE is :
=> y=(2bx)/(c+2a) -----------(1)
Equation of AC is :
=> (a-c)*y = b*(x-c) -------------(2)
Solving (1) & (2) :
=> x=(c+2a)/3
=> y=2b/3
Co-ordinates of F are [(c+2a)/3 , 2b/3]
Let AF:FC = m:1
Using section formula :
=> (m*0+b*1)/(m+1) = 2b/3
=> m = 1/2
=> AF:FC = 1:2
=> AF:AC = 1:3
So answer is 1:3
(2)O-O18 bond is broken . - +
The products are : CH3-CH2-COOH & CH3-CH2-O18Na