Arrange the following in the increasing order of the property given as indicted a.Arrange the following in the increasing order of the property given as indicted a. 2-Bromobutanoic acid, 3-bromobutanoic acid, 2-methypropanoic acid, butanoic acid (Acid strength) c. CH3CHO, CH3CH2OH, CH3OCH3, CH3CH2CH3 (Boiling point)

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Solution

1. 2-Bromobutanoic acid > 3-Bromobutanoic acid > Butanoic acid > 2-Methylpropanoic acid.
2. This is because -I effect of Bromine is more active in 2nd position as compared to 3 and +I effect of methyl group hinders the release of H⁺.
3. CH₃CH₂OH > CH₃CHO > CH₃OCH₃ > CH₃CH₂CH₃.
4. This is because,the effect of H-bonding is maximum in alcohol,then aldehyde,ether and no H-bonding in alkane. Due to this aggregation of molecules is less thus,they are low boiling.

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