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The correct order for acid strength of compounds ${\text{CH}} \equiv {\text{CH}}$, ${\text{C}}{{\text{H}}_3}{\text{ - C}} \equiv {\text{CH}}$ and ${\text{C}}{{\text{H}}_2}{\text{ = C}}{{\text{H}}_2}$is as follows:
(A)${\text{CH}} \equiv {\text{CH > C}}{{\text{H}}_{\text{2}}}{\text{ = C}}{{\text{H}}_{\text{2}}}{\text{ > C}}{{\text{H}}_3}{\text{ - C = CH}}$
(B)${\text{CH}} \equiv {\text{CH > C}}{{\text{H}}_3}{\text{ - C}} \equiv {\text{CH > C}}{{\text{H}}_2}{\text{ - C}}{{\text{H}}_2}$
(C)${\text{C}}{{\text{H}}_3}{\text{ - C}} \equiv {\text{CH > C}}{{\text{H}}_2}{\text{ = C}}{{\text{H}}_2}{\text{ > CH}} \equiv {\text{CH}}$
(D)${\text{C}}{{\text{H}}_3}{\text{ - C}} \equiv {\text{CH > CH}} \equiv {\text{CH > C}}{{\text{H}}_2}{\text{ = C}}{{\text{H}}_2}$

Last updated date: 13th Jun 2024
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Hint: The given compounds are three hydrocarbons alkane, alkene, and alkyne with saturated and unsaturated carbon atoms respectively. The s-character of these compounds depends upon their saturation and hybridization and acidic character more for more s-character.

Complete Step-by-step solution:
Let us understand how s-character affects the acidity of the given organic hydrocarbon. If the s-character is more in the bond between carbon and hydrogen atoms in the given compound, the s-orbital being aligned along the axis of the C-H bond, the electron cloud is more attracted towards the nucleus of the central atom (carbon) and thus the hydrogen atom becomes more available to be donated to lewis bases.
We know that in alkanes the bonds are single bonds formed by ${\text{s}}{{\text{p}}^{\text{3}}}$ hybridization, in alkenes the the bonds are double bonds formed by ${\text{s}}{{\text{p}}^{\text{2}}}$ hybridization and alkynes the bonds are triple bonds formed by sp hybridization.
In the above three compounds ${\text{CH}} \equiv {\text{CH}}$ has triple bonds so it is sp hybridized with 50% s-character, ${\text{C}}{{\text{H}}_3}{\text{ - C}} \equiv {\text{C}}{{\text{H}}_3}$ is alkyne with triple bond and the hybridization sp with 50% S-character and ${\text{C}}{{\text{H}}_2}{\text{ = C}}{{\text{H}}_2}$ has double bond with sp hybridization and its s-character is 33.3%. With the increase in s-character the acidic strength increases. So the correct order of acidic strength for above compounds follows alkyne > alkene > alkane.

The answer for the question is option (B) ${\text{CH}} \equiv {\text{CH}} > {\text{C}}{{\text{H}}_3}{\text{ - C}} \equiv {\text{CH > C}}{{\text{H}}_2}{\text{ = C}}{{\text{H}}_2}$

NOTE: In the compound ${\text{C}}{{\text{H}}_3}{\text{ - C}} \equiv {\text{CH}}$ there are two types of bonds, the triple and the single bond but as the priority in naming the hydrocarbon goes for triple bond. So the compound is alkyne but it is not alkane.