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The decreasing order of the stability of...

The decreasing order of the stability of the ions `:`
`underset((I))(CH_(3)=overset(+)(C)H-CH_(3))" "underset((II))(CH_(3)-overset(+)(C)H-OCH_(3))" "underset((III))(CH_(3)-overset(+)(C)H-COCH_(3)^(-))`

A

`IgtIIgtIII`

B

`IIIgtIIgtI`

C

`IIgtIIIgtI`

D

`IIgtIgtIII`

Text Solution

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The correct Answer is:
To determine the decreasing order of stability of the given ions, we need to analyze the effects of substituents on the stability of the carbocations formed in each case. The three ions are: 1. (I) \( \text{CH}_3-\text{C}^+-\text{CH}_3 \) 2. (II) \( \text{CH}_3-\text{C}^+-\text{OCH}_3 \) 3. (III) \( \text{CH}_3-\text{C}^+-\text{COCH}_3^- \) ### Step 1: Analyze Ion (I) For ion (I), we have a carbocation with two methyl groups attached. Methyl groups are electron-donating through the +I (inductive) effect. This stabilizes the positive charge on the carbon atom. **Stability Contribution**: +I effect from two methyl groups. ### Step 2: Analyze Ion (II) For ion (II), we have a carbocation with a methoxy group (\(-OCH_3\)). The methoxy group has a +M (mesomeric) effect, which is stronger than the +I effect. It can donate electron density through resonance, stabilizing the positive charge significantly. **Stability Contribution**: +M effect from the methoxy group. ### Step 3: Analyze Ion (III) For ion (III), we have a carbocation with a carbonyl group (\(-COCH_3\)). The carbonyl group is electron-withdrawing due to its -I (inductive) and -M (mesomeric) effects. This will destabilize the positive charge on the carbocation. **Stability Contribution**: -I and -M effects from the carbonyl group. ### Step 4: Compare the Stability Now we can compare the stability of the three ions based on the effects analyzed: - Ion (II) is the most stable due to the strong +M effect from the methoxy group. - Ion (I) is next stable due to the +I effect from the two methyl groups. - Ion (III) is the least stable due to the destabilizing -I and -M effects from the carbonyl group. ### Conclusion The decreasing order of stability of the ions is: **(II) > (I) > (III)** ---

To determine the decreasing order of stability of the given ions, we need to analyze the effects of substituents on the stability of the carbocations formed in each case. The three ions are: 1. (I) \( \text{CH}_3-\text{C}^+-\text{CH}_3 \) 2. (II) \( \text{CH}_3-\text{C}^+-\text{OCH}_3 \) 3. (III) \( \text{CH}_3-\text{C}^+-\text{COCH}_3^- \) ### Step 1: Analyze Ion (I) For ion (I), we have a carbocation with two methyl groups attached. Methyl groups are electron-donating through the +I (inductive) effect. This stabilizes the positive charge on the carbon atom. ...
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