Arrange the following in increasing order of acidity considering 'H' atom on ∗ carbon atoms. HC≡˙CH,˙CH2=2CH2CICH2˙CH3,FCH2˙CH3,˙CH3−CH3
A
HC≡CH<CH2=CH2<CH3−CH3<CICH2CH3<FCH2CH3
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B
CH2=CH2<HC≡CH<CH3−CH3<CICH2CH3<FCH2CH3
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C
CH3−CH3<CH2=CH2<HC≡CH<ClCH2CH3<FCH2CH3
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D
CH3−CH3<CICH2CH3<FCH2CH3<CH2=CH2<HC≡CH
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Solution
The correct option is CCH3−CH3<CH2=CH2<HC≡CH<ClCH2CH3<FCH2CH3 For more acidity, the negative charge on carbon, after the proton is removed, should be more stable. For this, a group showing −I effect is favourable. Hence, F−CH2CH3 will have highest acidity followed by Cl−CH2CH3. Considering sp,sp2 & sp3 hybridization of carbon, order of electronegativity is sp>sp2>sp3. Hence this is order of the negative charge being stable on the carbon.