Synthesis of Amides
Trending Questions
Q.
General solution of the differential equation is given by
Q.
In Gabriel synthesis, amine is always an
- aliphatic primary amine
- aliphatic secondary amine
- aromatic primary amine
- aromatic secondary amine
Q. The solution of the equation log(dydx)=ax+by is
- ebyb=eaxa+c
- e−by−b=eaxa+c
- e−bya=eaxa+c
- None of these
Q.
They yield of acid amide in the reaction, RCOCl+NH3→RCONH2 , is maximum when
Acid chloride and ammonia are treated in equimolar ratio
Acid chloride and ammonia are treated in 1 : 2 molar ratio
Acid chloride and ammonia are treated in 2 : 1 molar ratio
All the three gives nearly similar result
Q. If dydx+3cos2xy=1cos2x, x∈(−π3, π3), and y(π4)=43, then y(−π4) equals :
- −43
- 13+e6
- 13+e3
- 13
Q. The family of curves passing through (0, 0) and satisfying the differential equation y′′y′=1 (where y′=dydx and y′′=d2ydx2) is
- y=k
- y=kx
- y=k(ex+1)
- y=k(ex−1)
Q. The solution of the differential equation dydx−y+3xloge(y+3x)+3=0 is
(where c is a constant of integration)
(where c is a constant of integration)
- x−2loge(y+3x)=c
- x−loge(y+3x)=c
- x−12(loge(y+3x))2=c
- y+3x−12(logex)2=c
Q. The solution of dydx=ax+bcy+d represents a parabola if
- a=0, c=0
- a=1, b=2
- a=0, c≠0
- a=1, c=1
Q. A differentiable function f satisfies the equation f(x)=x∫0(f(t)cost−cos(t−x))dt, then the correct option(s) is/are
- f′(0)=−1
- f(x) has a maximum value 1−e
- f(x) has a minimum value 1−e−1
- f′′(π2)=e
Q. Let y=f(x) be a non-negative function defined on [0, 2] satisfying the differential equation y3y′′+1=0. If f′(1)=0 and f(1)=1, then
- the maximum value of f(x) is 1.
- the minimum value of f(x) is 0.
- the solution y=f(x) of the given differential equation represents a semi-circle with centre (1, 0).
- the area bounded by the curve y=f(x) and x−axis is π4 sq. units.
Q. The general solution of dydx=1−3y−3x1+x+y is
(where ′c′ is constant of integration)
(where ′c′ is constant of integration)
- x+3y+2ln|1+x+y|=c
- x+y+2ln|1−x−y|=c
- 3x+y+2ln|1+x+y|=c
- 3x+y+2ln|1−x−y|=c
Q.
The solution of the equation dydx=ex−y+x2 e−y is
ey=ex+x33+c
ey=ex+2x+c
ey=ex+x3+c
y=ex+c
Q. Which of the following differential equations can be solved using variable separable method.
1. (x2+y2)dx−2xydy=0
2. (x2+sinx)dx+(siny)dy=0
1. (x2+y2)dx−2xydy=0
2. (x2+sinx)dx+(siny)dy=0
- Only 1
- Only 2
- Both 1 and 2
- None of the above
Q.
The solution of the differential equation dydx=sec x (sec x+tan x) is
y = sec x + tan x + c.
y = sec x + cot x + c.
y = sec x - tan x + c
None of these
Q. Let f be a differentiable function such that f′(x)=f(x)+∫20f(x)dx, f(0)=4−e23, then f(x) is
- ex−(e2−13)
- ex−(e2−23)
- ex+e−[arg|z−1|]3
- 2ex−e2−1[arg(−|z|)] (where [.] greatest integer & z is a complex number)
Q. Rearrangement of an oxime to an amide in the presence of strong acid is called
Curtius rearrangement
Fries rearrangement
- Beckman rearrangement
- Sandmeyer reaction
Q. If the curve y=y(x) satisfies the differential equation dydx−2xy1+x2=0 and y(0)=1, then y(1) is equal to
- 1
- 12
- 2
- 0
Q. The solution of x3dx+yx2dy√x2+y2=ydx−xdy, y(1)=1 is
- √x2+y2=(√2−1)x
- √x2+y2+yx=1+√2
- √x2+y2+yx2=√2−1
- (x2+y2)2+xy2=√2