Chemistry-
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Question

An organic compound A upon reacting with NH3 gives B. On heating B gives C. C in presence of KOH reacts with Br2 to given CH3CH2NH2 . A is :

  1. CH3COOH    
  2. CH3CH2CH2COOH    
  3.    
  4. CH3CH2COOH    

The correct answer is: CH3CH2COOH

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Related Questions to study

General
Maths-

If r, s, t are prime numbers and p, q are the positive integers such that the LCM of p, q is r squared s to the power of 4 t squared, then the number of ordered pair (p, q) is-

Step by step solution :
It is given that L C M space left parenthesis p comma q right parenthesis space equals space r squared s to the power of 4 t squared
That is, at least one of p and q must have r squareds to the power of 4 and t squared in their prime factorizations.
Now, consider the cases for power of r as follows:
Case 1 : p contains  s to the power of 4 then q has  s to the power of K with K = (0,1,2,3)
That is, number of ways=4.
Case 2 : q contains  s to the power of 4 then q has  s to the power of K with K = (0,1,2,3)
That is, number of ways=4.
Case 3 : Both p and q contains s to the power of 4 
Then, number of ways=1.
Therefore, exponent of r may be chosen in 2+2+1=5 ways.
Similarly, exponent of s may be chosen in 4+4+1=9 ways and 
Exponent of s may be chosen in 2+2+1=5 ways
Thus, the total number of ways is:
 5 space cross times 9 space cross times 5 space equals space 225
Hence, the number of the ordered pair (pq) is 225.

If r, s, t are prime numbers and p, q are the positive integers such that the LCM of p, q is r squared s to the power of 4 t squared, then the number of ordered pair (p, q) is-

Maths-General
Step by step solution :
It is given that L C M space left parenthesis p comma q right parenthesis space equals space r squared s to the power of 4 t squared
That is, at least one of p and q must have r squareds to the power of 4 and t squared in their prime factorizations.
Now, consider the cases for power of r as follows:
Case 1 : p contains  s to the power of 4 then q has  s to the power of K with K = (0,1,2,3)
That is, number of ways=4.
Case 2 : q contains  s to the power of 4 then q has  s to the power of K with K = (0,1,2,3)
That is, number of ways=4.
Case 3 : Both p and q contains s to the power of 4 
Then, number of ways=1.
Therefore, exponent of r may be chosen in 2+2+1=5 ways.
Similarly, exponent of s may be chosen in 4+4+1=9 ways and 
Exponent of s may be chosen in 2+2+1=5 ways
Thus, the total number of ways is:
 5 space cross times 9 space cross times 5 space equals space 225
Hence, the number of the ordered pair (pq) is 225.
General
Maths-

A rectangle has sides of (2m – 1) & (2n – 1) units as shown in the figure composed of squares having edge length one unit then no. of rectangles which have odd unit length

We have, number of ways of selecting vertical sides  1+3+5+....+(2m1which is equal to m squared
Similarly, the number of ways of selecting horizontal sides  1+3+5+....+(2n1which is equal to n squared
So, the number of rectangles possible with odd side length =
 number of ways of selecting vertical sides × number of ways of selecting horizontal sides.
Therefore, we get the number of rectangles possible with odd side length = m squared cross timesn squared

Error converting from MathML to accessible text.
.

A rectangle has sides of (2m – 1) & (2n – 1) units as shown in the figure composed of squares having edge length one unit then no. of rectangles which have odd unit length

Maths-General
We have, number of ways of selecting vertical sides  1+3+5+....+(2m1which is equal to m squared
Similarly, the number of ways of selecting horizontal sides  1+3+5+....+(2n1which is equal to n squared
So, the number of rectangles possible with odd side length =
 number of ways of selecting vertical sides × number of ways of selecting horizontal sides.
Therefore, we get the number of rectangles possible with odd side length = m squared cross timesn squared

Error converting from MathML to accessible text.
.
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Type-2: 
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