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Carbocation

Carbocation & Carbanions | Chemistry for JEE Main & Advanced

Carbocation & Carbanions | Chemistry for JEE Main & Advanced

Properties of Carbocation :

1. It is positively charged species.

2. It has a sextet of electrons i.e. diamagnetic.

3. It is formed by heterolysis.

4. It is generally formed due to polar solvent.

Structure :

(sp2), Triangular planar

Stability :

Its stability can be determined with the help of Inductive effect, Hyperconjugation and Resonance effect.

Stability of Carbocation:

Carbocation & Carbanions | Chemistry for JEE Main & Advanced

Carbocation & Carbanions | Chemistry for JEE Main & Advanced Carbocation & Carbanions | Chemistry for JEE Main & Advanced

Carbocation & Carbanions | Chemistry for JEE Main & Advanced

The stability of carbocation can also be determined by Hyperconjugation (no bond Resonance).

Carbocation & Carbanions | Chemistry for JEE Main & Advanced

Carbocation & Carbanions | Chemistry for JEE Main & Advanced

Carbocation & Carbanions | Chemistry for JEE Main & Advanced

9 α C - H bond                6αC - H bond             3α C - H bond

Allylic Carbocation:

Carbocation & Carbanions | Chemistry for JEE Main & Advanced

Carbocation & Carbanions | Chemistry for JEE Main & Advanced

Benzylic Carbocation:

Carbocation & Carbanions | Chemistry for JEE Main & Advanced

Carbocation & Carbanions | Chemistry for JEE Main & Advanced

Q.1. Compare the stability of the following carbocation:

(a) Carbocation & Carbanions | Chemistry for JEE Main & Advanced (b) Carbocation & Carbanions | Chemistry for JEE Main & Advanced (c) Carbocation & Carbanions | Chemistry for JEE Main & Advanced

→ more s character

→ more electronegativity

→ +ve charge on more electronegative element is symbol of unstability.

a > b > c


Q.2. Compare the stability of the following compounds:

(a) Carbocation & Carbanions | Chemistry for JEE Main & Advanced (b) Carbocation & Carbanions | Chemistry for JEE Main & Advanced (c) Carbocation & Carbanions | Chemistry for JEE Main & Advanced (d) Carbocation & Carbanions | Chemistry for JEE Main & Advanced

Sol. d > c > b > a

 Being most electron attracting group it decreases the e- density from positively charged C-atom and decreases the charge density and makes the carbocation less stable.


Q.3. Compare the stability of the following carbocation :

(a) Carbocation & Carbanions | Chemistry for JEE Main & Advanced (b) Carbocation & Carbanions | Chemistry for JEE Main & Advanced (c) Carbocation & Carbanions | Chemistry for JEE Main & Advanced (d) Carbocation & Carbanions | Chemistry for JEE Main & Advanced

Sol. Due to greater size of Iodine, its L.P. will not be available for coordinate bond. Therefore L.P. would not stabilize carbocation.

In case of F due to its small size its lone pair can be easily coordinated to making it most stable.

a > b > c > d (Stability)

* By coordination the carbocation completes its octet and structure having complete octet of its atom is supposed to be most stable.

Carbocation & Carbanions | Chemistry for JEE Main & Advanced

(Each atom has its full octet)

* Carbocation & Carbanions | Chemistry for JEE Main & Advanced

Note : In Resonating Structure of at least one C gets sixtet of e-s and hence less stable than coordinated compound.


Q.4. Compare the stabilities of the following carbocation:

(a) Carbocation & Carbanions | Chemistry for JEE Main & Advanced (b) Carbocation & Carbanions | Chemistry for JEE Main & Advanced (c) Carbocation & Carbanions | Chemistry for JEE Main & Advanced

Sol. N, O, F belongs to same period

→ In period Electronegativity of the atom is deciding factor

→ F being most electronegative, holds its e- pair very firmly.

→ Its L.P. will not be easily available for coordination.

→ Stability by it will be minimum.

a > b > c

Q.5. Compare the following carbocation in order of their stability.

(a) Carbocation & Carbanions | Chemistry for JEE Main & Advanced (b) Carbocation & Carbanions | Chemistry for JEE Main & Advanced

Sol. If periods of atoms which have to donate their electrons for coordination (for stability) is different then atomic size will be deciding factor. The atom whose size is greater will be unable to make its e- pair available for coordination.

b > a.

Q.6. Compare the stability of the following compounds:

(a) Carbocation & Carbanions | Chemistry for JEE Main & Advanced (b) Carbocation & Carbanions | Chemistry for JEE Main & Advanced (c) Carbocation & Carbanions | Chemistry for JEE Main & Advanced

Sol → more s-character

→ more e.n.

→ attracts e-

→ reduces stability

b > a > c

Carbanion

1. It is a -ve charged species 2. It has octet of electrons. 3. diamagnetic

Structure :


If -ve charge is in Resonance then the hybridisation of carbanion is sp2(Triangular planar shape).

* If -ve charge is not in Resonance then the hybridisation of carbanion is sp3(pyramidal).

Stability :

Its stability can be determined with the help of

(1) Inductive effect

(2) Resonance effect

Carbocation & Carbanions | Chemistry for JEE Main & Advanced

Carbocation & Carbanions | Chemistry for JEE Main & Advanced

Carbocation & Carbanions | Chemistry for JEE Main & Advanced

Q.1.Carbocation & Carbanions | Chemistry for JEE Main & Advanced

a > b (stability)

* Stability of the carbanion is as follows

Carbocation & Carbanions | Chemistry for JEE Main & Advanced


Q.2. Compare the stability of the following carbocation:

(a) Carbocation & Carbanions | Chemistry for JEE Main & Advanced (b) Carbocation & Carbanions | Chemistry for JEE Main & Advanced (c) Carbocation & Carbanions | Chemistry for JEE Main & Advanced

Sol. c > a > b


Q.3. Compare the stability of the following carbanion:

 

(a) Carbocation & Carbanions | Chemistry for JEE Main & Advanced 
(b) Carbocation & Carbanions | Chemistry for JEE Main & Advanced 
(c) Carbocation & Carbanions | Chemistry for JEE Main & Advanced

→ become more stable

Sol. b > a > c


Q.4. Compare the stability of the following carbanion:

(a) Carbocation & Carbanions | Chemistry for JEE Main & Advanced (b) Carbocation & Carbanions | Chemistry for JEE Main & Advanced (c) Carbocation & Carbanions | Chemistry for JEE Main & Advanced

Sol. a > b > c


Q.5. Arrange the following anion in order of their stability:

(a) Cl-, (b) Br- (c) F- (d) I(maximum size)

→ maximum dispersion of -ve charge

→ max stability

Sol. d > b > a > c


Q.6. Compare the stability of the following:

Carbocation & Carbanions | Chemistry for JEE Main & Advanced


Q.7. Compare the acidic strength:

(a) HCl (b) HF (C) HBr (D) HI

Sol. Acidic strength ≈ stability of the anion formed (conjugate base)

As we know, I> Br- > Cl- > F-

→ HI > HBr > HCl > HF


Q.8. Compare the Acidic strength of the following:

(a) NH3 (b) pH3 (c) AsH3 (d) SbH3 (e) BiH3

Sol. Anion formed from these acids are:

Carbocation & Carbanions | Chemistry for JEE Main & Advanced

→ acidic strength e > d > c > b > a


Q.9. Compare the acidic strength of the following compounds:

CH4, NH3, H2O, HF

Sol. The conjugate base of the given acid is as follows:

Carbocation & Carbanions | Chemistry for JEE Main & Advanced

we have already proved that

Carbocation & Carbanions | Chemistry for JEE Main & Advanced > Carbocation & Carbanions | Chemistry for JEE Main & Advanced > Carbocation & Carbanions | Chemistry for JEE Main & Advanced > Carbocation & Carbanions | Chemistry for JEE Main & Advanced (Stability)

→ HF > H2O > NH3 > CH4 (acidic strength)


Q.10. Compare the stability of the following carbanion.

(a) Carbocation & Carbanions | Chemistry for JEE Main & Advanced 
(b) Carbocation & Carbanions | Chemistry for JEE Main & Advanced
 (c) Carbocation & Carbanions | Chemistry for JEE Main & Advanced 
(d) Carbocation & Carbanions | Chemistry for JEE Main & Advanced

Sol. d > c > b > a

* M or -M is not distance dependent


Q.11. Compare the stability of the following carbocation

(a) Carbocation & Carbanions | Chemistry for JEE Main & Advanced 
 (b) Carbocation & Carbanions | Chemistry for JEE Main & Advanced 
 (c) Carbocation & Carbanions | Chemistry for JEE Main & Advanced 
 (d) Carbocation & Carbanions | Chemistry for JEE Main & Advanced

Sol. a > b > c > d


Q.12. Compare the stability of the following carbocation.

 

(a) Carbocation & Carbanions | Chemistry for JEE Main & Advanced 
(b) Carbocation & Carbanions | Chemistry for JEE Main & Advanced 
(c) Carbocation & Carbanions | Chemistry for JEE Main & Advanced 
(d) Carbocation & Carbanions | Chemistry for JEE Main & Advanced

Sol.  M (OH) > M (OCH3)

b > c > d > a


Q.13. Compare the stability of the following carbocation

(a) Carbocation & Carbanions | Chemistry for JEE Main & Advanced 
(b) Carbocation & Carbanions | Chemistry for JEE Main & Advanced 
(c) Carbocation & Carbanions | Chemistry for JEE Main & Advanced

Sol. c > a > b


Q.14. Compare the order of dehydration of the following alcohols :

(a) Carbocation & Carbanions | Chemistry for JEE Main & Advanced 
(b) Carbocation & Carbanions | Chemistry for JEE Main & Advanced
 (c) C - C - C - OH

Sol. After the formation of carbocation

Carbocation & Carbanions | Chemistry for JEE Main & Advanced Carbocation & Carbanions | Chemistry for JEE Main & Advanced, Carbocation & Carbanions | Chemistry for JEE Main & Advanced
 

Since 3° carbocation is most stable therefore it will show the greatest tendency to lose water as after the loss of water it comes in a stable form.

The document Carbocation & Carbanions | Chemistry for JEE Main & Advanced is a part of the JEE Course Chemistry for JEE Main & Advanced.
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FAQs on Carbocation & Carbanions - Chemistry for JEE Main & Advanced

1. What is a carbocation?
Ans. A carbocation is a positively charged carbon atom that has only three bonds and an empty p orbital. It is a reactive intermediate in organic chemistry and is formed by the loss of a pair of electrons from a carbon atom.
2. How is a carbocation formed?
Ans. Carbocations can be formed through various methods, such as the departure of a leaving group in a nucleophilic substitution reaction or the protonation of a neutral molecule. They can also be generated by the cleavage of a covalent bond or through rearrangement reactions.
3. What are the properties of carbocations?
Ans. Carbocations are highly reactive species due to the presence of an electron-deficient carbon atom. They tend to undergo reactions to stabilize themselves, such as rearrangements or interactions with nucleophiles. Carbocations also show resonance stabilization when possible, and their stability follows the order: tertiary > secondary > primary.
4. What is a carbanion?
Ans. A carbanion is a negatively charged carbon atom that has gained an extra pair of electrons. It is a reactive intermediate in organic chemistry and is formed through the donation of a pair of electrons to a carbon atom.
5. How are carbanions formed?
Ans. Carbanions can be formed by various methods, such as the deprotonation of a carbon atom in the presence of a strong base, the addition of a nucleophile to a carbonyl compound, or the cleavage of a covalent bond. They are also involved in many important organic reactions, such as nucleophilic substitutions and elimination reactions.
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