Explanation
For quantitative estimation of nitrogen, in an organic compound, Kjeldahl and Duma’s methods are used. In Kjeldahl method, the nitrogen method the nitrogen element of organic compound is changed to the ammonia.
(CH3) – C – Cl→(30) tert-butyl chloride
CH3 CH2 CH2 CH2Cl→1-chlorobutane(10)
Cl – CH2 – CH(CH3) CH3→2-methyl-1-chloropropane(10)
CH3 – CH2CH(Cl) CH3→2-chlorobutane(20)
SN1 reaction is most effective in 30 carbon.
When nitro group is present in the benzene nucleus it withdraws electron from o and p positions. Thus electron density at o and p position decreases and m-position becomes positions of compartively higher electron density. So electrophilic attack occurs at m positions.
Order of reactivity is I > II > III chlorobenzene has only one deactivating group. In 2,4-dinitrochlorobenzene has three deactivating group.
Cyclohexyl alcohol.
HO – CH2 – CH = CH2 Vinylcarbinol
Carbocations have three pairs (6) electrons in the valence shell.
Aldehydes and ketones undergo nucleophilic addition reactions
Homolylic fission of C — C bond gives free radicals in which carbon issp2 - hybridized.
Resonance structures are separated by a double headed arrow (↔)
Angle increases progressively sp3(109º-28’), sp2(120º), sp(180º)
Hyperconjugation
Only sp2 - hybridized C-atoms of phenyl ring and CH2 = CH group hasessentially planar geometry. All the rest of the groups have sp3- hybridizedC- atoms and hence have non- planar geometry.
Electrophilic addition .
Greater the electronegativity, higher is the I-effect, i.e, – NR2 < – OR < – F.
In structure (c), N contains ten electrons in the valence shell which is not possible.
The bond order in ethane
(I) is 1, in ethene
(II) it is 2, in acetylene
(III) it is3 while due to the resonance the bond order in benzene
(IV) is 1.5 Sincethe bond length decreases as the bond order increases,
therefore, theactual order is I > IV > II > III.
I has 8 + 2 = 10π electrons and hence is aromatic. II and V have 4,III has 8 while IV has 8 +1 = 9 π-electrons and hence all these speciesare not aromatic.
During catalytic hydrogenation, the hydrogens are transferred from thecatalyst to the same side of the double bond, Evidently, smaller thenumber of R substituents, lesser is the steric hindrance and hence fasteris the rate of hydrogenation. Thus, option (a) with two R groups on thesame side of the molecule is correct.
As electronegativity of the atom decreases (F>O > N > C), its tendency todonate a pair of electrons, i.e., nucleophilicity increases. Thus, CH-3 hasthe highest nucleophilicity.
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