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Imidazole is more basic than pyridine? Why?

Imidazole and its derivatives form an interesting and important class of heterocyclic aromatic amines.
The basic strength of imidazole is approximately 100 times more basic than pyridine. Because protonation of imidazole yields an ion that is stabilized by the electron delocalization represented in the resonance structures given as below:
As seen in above figure the electrostatic potential map of the conjugate acid of imidazole (imidazolium ion) is consistent with the resonance description that shows both nitrogens as equivalent.
Related Questions:

Biological Important of Imidazole and structure:

Imidazole and its derivatives form an interesting and important class of heterocyclic aromatic amines.
An imidazole ring is a structural unit in two biologically important compounds, histidine and histamine. 
Histidine is one of the amino acid building blocks of proteins and is directly involved in key proton-transfer processes. The drop in blood pressure associated with shock is a result of the formation of histamine, which stimulates the dilation of blood vessels.

Related Questions:

Pyridine is almost 1 million times less basic than piperidine? Why?

We know that if nitrogen is part of an aromatic ring, however, its basicity decreases markedly.  So in pyridine nitrogen is part of an aromatic ring hence its basicity is much more less than piperidine. The difference between the two lies in the fact that the nitrogen lone pair occupies an sp3- hybridized orbital in piperidine versus an sp2-hybridized one in pyridine. As we have noted on several occasions, electrons in orbitals with more s character are more strongly held than those with less s character. For this reason, nitrogen holds on to its unshared pair more strongly in pyridine than in piperidine and is less basic.

Cyclohexylamine amine is the stronger base than Aniline? Why?

We can say that aniline is a weaker base than cyclohexylamine because the electron pair on nitrogen of aniline is strongly held by virtue of being delocalized into the π system of the aromatic ring. The unshared pair in cyclohexylamine is localized on nitrogen, less strongly held, and therefore “more available” in an acid–base reaction.

Related Questions:

Tetrahydroquinoline amine is the stronger base than Tetrahydroisoquinoline? Why?

We can say that Tetrahydroisoquinoline  is a weaker base than Tetrahydroquinoline because the electron pair on nitrogen of Tetrahydroisoquinoline is strongly held by virtue of being delocalized into the π system of the aromatic ring. The unshared pair in Tetrahydroquinoline is localized on nitrogen, less strongly held, and therefore “more available” in an acid–base reaction.

Related Questions:

Phenyl group is known to extract negative inductive effect, but each phenyl ring in biphenyl is more reactive than benzene towards Electrophilic substation. Why?

It is because, in biphenyl, one of the phenyl groups acts as donor and the other as electron acceptor. This increases electron density on benzene ring and facilitates Electrophilic attack at ortho and para position.

Are all the five bonds of PCl5 equivalent? Justify your answer.

PCl5 has Sp3 hybridisation and trigonal bipiramidal geometry. PCl5 has three equivalent equatorial and two equivalent axial P – Cl bonds. However, due to greater bond pair – bond pair repulsions, the axial P – Cl bonds are longer and hence different from the three equatorial bonds

Halides of Nitrogen Family:

Silianol (SiH3OH) is more acidic than methanol (CH3OH) why?

We know that acidic strength of an acid also depend upon stability of its conjugate base. So silianol is more acidic than methanol (H3C-OH) because conjugate base silianol (H3Si-OH) stablised by dispersal of negative charge in H3Si-O- ion by 2pπ—3dπ back bonding

Hence methanol is less acidic than silianol.
For more detail on Back bonding click here

Related Questions:

Is all the C-C bond length in fullerene is equal ?

Trisilyl amine, N(SiH3)3 is planar whereas trimethyl amines N(CH3)3 is pyramidal. Explain why?.

In N(SiH3)3, N attains sp2 hydridisation and the lone pair of N is involved in dpi-ppi  back bonding by getting itself delocalized on to empty 3d – orbitals of silicon.

But in N(CH3)3, N is sp3 hybridised in which three of the hybrid orbitals are used in forming s  - bonds with NH3 groups, while the lone pair is present in the fourth hybrid orbital. Thus the molecule is pyramidal.
For More deatils click here Back Bond theory:

What is the molecular formula of Borax ?

(A) Na2[B4O3(OH)4].6H2O
(B) Na2[B4O5(OH)4].6H2O
(C) Na2[B4O5(OH)4].8H2O
(D )Na2[B4O6(OH)2].8H2O

The structure of Borax is Na2[B4O5(OH)4].8H2O.

Hence option (C) is the correct answer for details click on link given below

Four-center two-electron bond (4C-2e Bond): Structure of AlCl3:

(1) Al2Cl6 is neither hypovalent nor hypovalent rather its octet is complete. We will used  MOT here  it cannot act as Lewis acid  due to crowding in spite having vacant d orbital’s   however AlCl3  act as Lewis acid .

(2) Al2Cl6 contains six bonds having two bridge bonds (3c-4e) and four bonds is (2C-2e) bridge bonds are lies perpendicular to plane.
(3) The bond lengths of terminal Al−Cl bonds are shorter (206 pm) while bond length of bridged Al−Cl bonds are longer (221 pm) and also outer bond angle (Cl−Al−Cl) are greater (118) than inner bond angle (Al−Cl−Al).
(4) Maximum six atoms are lies in same plane which four terminal Cl and two aluminium atoms.
(5) AlCl3 in vapor phase or in non-polar solvent is dimeric Al2Cl6 hence sp3 hybridised having 3c-4e bonds.

What is the difference between the structure of AlCl3 and diborane?

The hydrogen bridged dimer B2H6 contains two three centre (3C-2e bond), two electron bonds, A three-centre bond uses two electron to link three atoms, and  four two centre, two electron bond (2C-2e). 3C-2e bridge bond is perpendicular to plane in which 2C-2e bond present.
In contrast the bridge bonding in Al2Cl6 contains two three centre (3C-4e bond), four electron bonds  which can be described in terms of electron pair bond in which a chlorine atom bonded to one aluminium atom act as a Lewis base by donating a lone pair of electrons to the aluminium atom which acts as a Lewis acid.

Chloroform is more acidic than fluoroform why?

We know that acidic strength of the acid also depends upon stability of conjugate bases, so for relative strength of acid, we need to check the relative stabilities of their conjugate bases.

                                                     CF3-, CCl3-, CBr3- CI3-

We are expecting the acidic strength haloform acids as CHF3, CHCl3, CHBr3, CHI3 in decreasing order. Because Fluorine is most electronegative atom so it would be stabilize CF3- more, as electronegativity decreases from F to I the stability of conjugate -ve ion would be but that is not correct the actual order is CHCl3 > CHF3 > CHBr3 > CHI3.

This is because there is effective back bonding in CCl3- and hence the negative charge partially gets stabilised by back donation to the vacant 3d orbitals of Cl. Thus, CHCl3 is a stronger acid than CHF3 and also among them due to 2pπ-3dπ back bonding.

The acidic strengths of the other three haloforms can be compared the inductive effects of their anions. F is very electronegative and hence stabilises the negative charge on the C atom. So, CHF3 is a better acid than CHBr3, and the least acidic is CHI3.

            The overall acidic strength order is, CHCl3 > CHF3 > CHBr3 > CHI3.

Fluorine is more electronegative than chlorine even then, p-flurobenzoic acid is weaker acid than p-chlorobenzoic acid explain ?

Since halogens are more electronegative than carbon and also possesses lone pair electrons , therefore they exert both -I and +R effects . Now in F , the lone pair of electrons are present in 2p-orbitals but in Cl , they are present in 3p-orbitals . Since 2p- orbitals of fluorine and carbon are almost  equal size , therefore , the +R effect is more pronounced in p- flurobenzoic acid than p-chlorobenzoic acid.
Thus in p- flurobenzoic acid +R effect is out weight the -I effect but in p-chlorobenzoic acid, it is the -I effect which is out weight  the +R effect. Hence p- flurobenzoic acid is weaker acid than p-chlorobenzoic acid.

Trifluoroethyl carbocation is less stable than trifluromethyl carbocation why ?

In case of trifluroethylcarbocation the highly electronegative fluorine atom , withdraws shared pair electrons between C-F bonds towards itself to great extent and intensifies positive charge on carbon. Greater the intensification of charge more is the instability.theirfore this carbocation is destabilised.
In case of trifluroethylcarbocation carbocation, the unshared paired of electrons in P - Orbital of each of the fluorine can be shifted into vacant Orbital of carbon atom of carbocation via P-P overlap.

It's leads to dispersal of charge and provide stability to the carbonation.

Stability of Heteroatomic carbocations:

The carbocaton contaning hetero atoms adjacent to positive carbon of cation such as oxygen, nitrogen and sulphur etc which are more electronegative than carbon, you might expect that they would by definition be electron withdrawing groups that destabilize carbocations.
But there is opposite effect, if the oxygen, nitrogen or sulphur atom is present at adjacent to carbocation, the overall effect is carbocation stabilization.
This is due to the fact that although these heteroatoms are electron withdrawing groups by induction, they are electron donating groups by resonance, and it is this resonance effect which is more powerful.
ILLUSTRATIVE EXAMPLE(1): Give the correct Stability of given carbocation .
SOLUTION: Sulphur containing carbocation is more stable because lone pair of sulphur atom show more resonating effect due to least electronegativity as compared to nitrogen as well as oxygen atom similarly  nitrogen show more resonance than oxygen atom
Conceptual Facts:
(1) If number of conjugation increases stability of carbocation increases ...
(2) We known that on increasing conjugation stability increases. their is exception in option (A) their is three nitrogen may involving in resonance but  actually not, 
 Because positive carbon does not involved in resonance due to bridge head carbon. we known according to Bredt,s rule bridge carbon cannot for duble bond.
Hence correct stability order is  (B) > (C) >> (A): 
(3)  Similarly 

Hence correct stability order is  (B) > (C) >> (A): 

ILLUSTRATIVE EXAMPLE(3): Give the correct Stability of given carbocation.
SOLUTION: (1):  (C) > (B) > (A)     and    (2):   (C) > (B) > (A)  

ILLUSTRATIVE EXAMPLE(4): Give the correct Stability of given carbocation.
SOLUTION: (1):  (D) > (B) > (C) > (A)     and    (2):  (D) > (B) > (C) > (A)  
Other important examples:


Dancing resonance or Sigma Resonance:

Dancing resonance is a special stability mechanism which increases stability of carbocations attached directly to the three membered rings. For example Cyclopropylmethyl Carbocation.
In cyclopropane all the carbon is sp3 hybridized and the bond angle for the same should be 109 degree 28′ but the actual bond angle is 60 due to which angle strain develops. So in order to minimize the strain p orbital bents due to which it acquires partial sigma and partial pi bond character which behave like pi bond. And resonance take place between sigma and vacant p-orbitals hence called P-orbitals overlapping or sigma resonance.

We know by Drago’s rule bond angle is directly proportional to the s-character while inversely proportional to p-character.  
Dancing resonance is a hypothetical phenomenon which is reduces strain of the ring, hence the carbocation is more stable.  CH2+ has a vacant p orbital and a very effective overlapping takes place between p-orbital and electron density of cyclopropane, due to this its stability is very high. There is a conjugation between the sigma bond and positive charge.
The exceptional stability of cyclopropane methyl cation can be explained by the concept of dancing resonance concept. The stability of additional cyclopropyl group , is result of more conjugation between the bent orbital of cyclopropyl ring and cationic carbon.
The most stable carbocation known till date in organic chemistry is explain by Dancing resonance.

ILLUSTRATIVE EXAMPLE: Why is the Tropylium carbocation less stable than the tricyclopropylmethyl carbocation?

Why is the Tropylium carbocation less stable than the tricyclopropylmethyl carbocation?

Tropylium is highly stable due to conjugated system, that being resonance stabilized and the number of canonical forms of tropylium is more. But it is less stable than tricyclopropylmethyl carbocation because tricyclopropylmethyl carbocation undergoes a strong  stability mechanism factor that is sigma resonance or dancing resonance  (sigma-tropic rearrangement ).

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