What is the decreasing order of stability of given carbocations?

3 > 2 > 1 > 4 > 5

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3 > 2 > 5 > 4 > 1

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3 > 2 > 4 > 5 > 1

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3 > 2 > 5 > 1 > 4

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Solution

The correct option is D $3 > 2 > 5 > 4 > 1$
Stability order of carbocation is $a r o m a t i c = r e s o n a n c e < 3^{o} < 2^{o} < 1^{o}$ . Hence the correct order is $3 > 2 > 5 > 4 > 1$
3= $3^{o}$ ,
2= $2^{o}$ with $6 α H$ :
5 = $2^{o}$ with $4 α H$ ,
4= $2^{o}$ with $2 α H$ ,
1= $2^{o}$ with $2 α H$ and less -I effect.

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Consider the following carbocations

(1) $C H_{3} - \oplus C H_{2}$ (2) $C H_{2} = \oplus C H$ (3) $C H_{2} = C H - \oplus C H_{2}$ (4) $C_{6} H_{5} - \oplus C H_{2}$ stability of these carbocations in decreasing order is

Find the mode from the given data.

1, 2, 3, 4, 1, 2, 3, 4, 1, 2, 3, 4, 1, 2, 3, 4, 1, 2 , 3, 4, 1, 2, 3, 4, 1 ,2 , 3, 4, 5, 5, 1, 5, 5, 5, 3, 5, 1, 2, 4, 5, 6, 1, 2, 4, 2, 1

Q. The decreasing order of energy for the electron represented by the following sets of quantum number is
(1)

n = 4, l = 0, m = 0, s = + \frac{1}{2}

(2)

n = 3, l = 2, m = + 1, s = - \frac{1}{2}

(3)

n = 3, l = 2, m = - 1, s = + \frac{1}{2}

(4)

n = 3, l = 1, m = 0, s = + \frac{1}{2}

(5)

n = 5, l = 2, m = + 1, s = - \frac{1}{2}

Question 4(a) :

Look at the figures and write ‘<’ or ‘>’ between the given pairs of fractions:

$\frac{1}{6} □ \frac{1}{3}$

Gopi spun the spinner 30 times and got the following results:

2, 1, 4, 2, 4, 5, 1, 4, 1, 5, 3, 1, 4, 2, 3, 1, 5, 4, 4, 2, 3, 3, 5, 4, 2, 2, 5, 3, 1, 4

What is the frequency of 4?

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What is the decreasing order of stability of given carbocations?

The correct option is D 3>2>5>4>1Stability order of carbocation is aromatic=resonance<3o<2o<1o. Hence the correct order is 3>2>5>4>13= 3o, 2= 2o with 6αH : 5 = 2o with 4αH, 4= 2o with 2αH, 1= 2o with 2αH and less -I effect.