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Both of them are tertiary carbocations. But the one on the left is a bridge head carbocation which is very unstable due to angle strain( the angle shd be 120 but here it is very low)
The lone pair on the nitrogen gets involved in cojugation in the case of pyrolle. On the other hand the lone pair on nitrogen has no part in the conjugation in the case of pyridine.
The reason is that the sp2 nitrogen is more electronegative than the sp2 carbon which has greater electronegativity than the sp3 nitrogen.
Instead of saying that the coagulating capacity is more for the ion with the highest charge we can also say that more the concentration of a particular charged ion then more is its coagulating power.
For example, for a negatively charged colloid, K3(PO4) > K2(SO4) > KCl in coagulating power
By the definition of an external indicator it is easy to say that they change colour when it is subjected to temperature changes.
6CH3-CH=CH2 + B2H6 ---> 2B(CH3-CH2-CH2)3 + CH3COOH ---> CH3-CH2-CH3
There's already a discussion of this in the above link in goiit
N doesnt have d-orbitals
If u draw the diagram then u will realise that r+R,r+R,2R form a right angled triangle
The efficiency of the direct heat engine=1-Qc/Qh
Qc/Qh=Tc/Th(For carnot engines)
For refrigerators the efficiency is the coefficient of perfromance: COP=Qc/(Qh-Qc)
It will be zero. The two axial dipole moments cancel. On the plane the equatorial bonds are all 120 degrees apart. So they cancel.
are u talking about cine substitutions then u can have a look in this link:http://en.wikipedia.org/wiki/Meta-_(chemistry)
u need to realise what is happening in the rate determining step. First right the mechanism of the organic reaction. If u see one nucleophile(electron rich entity) replacing another then it's nucleophilic substitution. If the nucleophile adds on to the reactant then its addition. Eg:- Alkyl halide reacts with aq.KOH. Here OH^- replaces the X- so its substitution. But when some nucleophile like CN- adds on to alkene then its addition.
Same for a electrophile(electron wanting entities).
I prefer converting the 3-D structures to fischer projection and then assigning the configuration.
That will prevent any silly mistakes
see this mate.
smidt reaction of an aldehyde will end in a cyanide.
see this: http://www.organic-chemistry.org/namedreactions/schmidt-reaction.shtm
This can be done in 3 steps.
1.Hydrolysis to acetylene
2.Acid catalysed hydrolyis(in presence of Hg^2+) to get acetaldehyde
3. Smidt reaction to get ethane nitrile
actually it follows an addition-elimination mechanism.
The energy difference should be there because then only the electron jumps to that lower and most stable energy level and thus induces a stability in the molecule.
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