Home
Class 12
CHEMISTRY
Arrange the following compounds in decre...

Arrange the following compounds in decreasing order of reactivity for EAR :-
(I) `CH_(2)=CH_(2)` (II) `CH_(3)-CH=CH_(2)`
(III) `Ph_(2)C=CH-CH_(3)` (IV) `CH_(2)=CH-CHO`

A

`IV gt I gt II gt III`

B

`III gt II gt I gt IV`

C

`II gt III gt I gt IV`

D

`II gt III gt IV gt I`

Text Solution

AI Generated Solution

The correct Answer is:
To determine the order of reactivity of the given compounds for Electrophilic Addition Reactions (EAR), we need to analyze each compound based on the stability of the carbocation formed during the reaction. The more stable the carbocation, the more reactive the compound will be towards EAR. ### Step-by-Step Solution: 1. **Identify the Compounds**: - (I) `CH2=CH2` (Ethylene) - (II) `CH3-CH=CH2` (Propene) - (III) `Ph2C=CH-CH3` (Diphenylpropene) - (IV) `CH2=CH-CHO` (Acrylaldehyde) 2. **Analyze the Reactivity**: - **Compound (I) `CH2=CH2`**: - Forms a primary carbocation when reacting with an electrophile. - Stability: Least stable due to no resonance and being a primary carbocation. - **Compound (II) `CH3-CH=CH2`**: - Forms a secondary carbocation when reacting with an electrophile. - Stability: More stable than primary carbocation due to being secondary. - **Compound (III) `Ph2C=CH-CH3`**: - Forms a tertiary carbocation when reacting with an electrophile. - Stability: Most stable due to resonance stabilization from the two phenyl groups. - **Compound (IV) `CH2=CH-CHO`**: - Forms a secondary carbocation when reacting with an electrophile. - Stability: Less stable than the tertiary carbocation but more stable than the primary. 3. **Rank the Stability of Carbocations**: - Tertiary carbocation (III) > Secondary carbocation (II) = Secondary carbocation (IV) > Primary carbocation (I) 4. **Order of Reactivity**: - Based on the stability of the carbocations formed, the order of reactivity for EAR is: - (III) > (II) > (I) > (IV) 5. **Final Arrangement**: - The final order of reactivity in decreasing order is: - (III) `Ph2C=CH-CH3` > (II) `CH3-CH=CH2` > (I) `CH2=CH2` > (IV) `CH2=CH-CHO` ### Answer: The decreasing order of reactivity for EAR is: **(III) > (II) > (I) > (IV)**
Promotional Banner

Topper's Solved these Questions

  • REACTION MECHANISM (PART-I)

    ALLEN|Exercise MCQ|21 Videos
  • Redox Reaction

    ALLEN|Exercise All Questions|15 Videos

Similar Questions

Explore conceptually related problems

Arrange the following compounds in decreasing order of acidity : CH_(3)CH_(2)OH H_(2)C=CH_(2)

Arrange the following alkenes in decreasing order of stability : CH_(3) - CH = CH - CH_(3), CH_(2) = CH_(2), CH_(3) - CH = CH_(2)

Arrange the following alkenes in decreasing order of stability : CH_(3) - CH = CH - CH_(3), CH_(2) = CH_(2), CH_(3) - CH = CH_(2)

Arrange the following compounds in decreasing order of acidity. (i) ClCH_(2)CH_(2)CH_(2)COOH (ii) CH_(3)CHClCH_(2)COOH (iii) CH_(3)CH_(2)CHClCOOH

Arrange the following compounds in increasing order of S_(N^(1)) reactivity. (a). (I). ClCH_(2)CH=CHCH_(2)CH_(3) , (II). CH_(3)C(Cl)=CHCH_(2)CH_(3) , (III). CH_(3)CH=CHCH_(2)CH_(2)Cl (b). (I). CH_(3)CH_(2)Br , (II). CH_(2)=CHCH(Br)CH_(3) , (III). CH_(2)=CHBr, (IV). CH_(3)CH(Br)CH_(3) (c). (I). (CH_(3))_(3)CBr , (II). (CH_(3))_(2)CHBr , (III). CH_(3)CH_(2)CH_(2)Br ,

Arrange the following in the increasing order of recativity with NH_(3) (I) CH_(2) O (II) CH_(3) CHO (III) CH_(3) - CO - Ch_(3) (IV) (I) CH_(2

Arrange the folliwing compounds in increasing order of SN^(-2) reactivity. a. I.m ClCH_(2)CH = CHCH_(2)CH_(3) II. CH_(3)C(Cl) = CHCH_(2) CH_(3) III. CH_(3)CH = CHCH_(2)CH_(2)Cl IV. CH_(3)CH = CHCH_(2)(Cl) CH_(3) b. I. CH_(3)CH_(2)Br II. CH_(2) = CHCH(Br) CH_(3) III. CH_(2) = CHBr IV. CH_(3)CH (Br) CH_(3) C. I. (CH_(3))_(3)CCl , II. C_(6)H_(5)C(CH_(3))_(2)Cl III. (CH_(3))_(2)CHCl , IV. CH_(3)CH_(2)CH_(2)Cl

Arrange the following halides in the decreasing order of SN^1 reactivity : (I) CH_3CH_2CH_2Cl, (II) CH_2=CHCH(Cl)CH_3 (III) CH_3CH_2CH(Cl)CH_3