Arrange the following carbocations in decreasing order of their stability: ( I ) ( II ) ( III ) ( IV ) ( P ) CH 3 - CH 2 - C ⊕ H 2 ( Q ) CH 3 - CH ⊕ - C 2 H 5 ( R )

In the option (P) , all species are primary carbcation.The stability of carbocation directly proportional to + I Effect (electron donor species or their electron donating power) whereas indirectly proportional - I Effect (Electron withdrawing group or their electron donor capacity). Halogen produce - I Effect whereas alkyl group produces + I Effect. So decreasing order of various carbocations is: propyl carbocation > 2 - bromo- propyl carbocation > 2 - chloro-propyl carbocation > 2 - fluoro-propyl carbocation (as the electronegativity of halogen increases than the magnitude of -I effect increases, so stability of carbocation decreases). In the (Q) step of the question, the stability of various carbocations is- Di-phenyl propyl carbocation ( + I Effect of alkyl group and due to the presence of π & + Charge conjugation with two phenyl rings, positive charge of carbocation delocalises in benzene ring (resonance effect) and nine resonating structures) > phenyl propyl carbocation (+I effect of alkyl group, here the magnitude of resonance effect is lesser due to the presence of one phenyl group) > Tertiary butyl carbocation [ 3 0 - alkyl carbocation & + I Effect of alkyl group] > Secondary butyl carbocation [ 2 0 - alkyl carbocation & + I Effect of alkyl group]. The stability of various alkyl carbocation is - tertiary alkyl carbocation > secondary alkyl carbocation > primary alkyl carbocation. In case of option (R)- Stability order of given carbocation is - C (tertiary carbocation) > B (secondary carbocation) > D (primary carbocation) > A (The stability of carbocation at bridge head carbon is minimum.)

Which ' H ' atom is most acidic in the following compounds. ( I ) CH 3 a - CH b = CH 2 c ( II ) ( III )

To solve this type of question apply following hint - Acidic strength &prop; stability of conjugate base. In case of (I) prop- 1 -ene, the most acidic hydrogen atom is (a) after the removal of H + From (a) carbon, the negative charge of its conjugate base present in the conjugation with double bond ( π & - Charge conjugation) due to this stability of conjugate base increases and acidic character of (a) hydrogen is maximum. In case of option (II) 2 - keto-butanal, the most acidic hydrogen is (b) because it is in between of two carbonyl groups and due to electrometric effect of these group (keto-enol isomerism), it is most acidic. After the removal of H + Ion from (b) carbon. The negative charge of its conjugated base present in the conjugation with two carbonyl double bonds due to this the stability of conjugated base increases and acidic character (b) hydrogen is maximum. In option (III) ethyl benzene, hydrogen (C) is most acidic because after the removal of H + Ion from (C) carbon. The negative charge of its conjugated base present in the conjugation with double bond of benzene ring due to this its delocalisation in benzene ring. So the stability of its conjugated base in increases and hydrogen (C) is most acidic in this compound.