Question

Which statement(s) is/are WRONG?
A) Acetylene is insoluble in conc. ${H_2}S{O_4}$ due to the non-formation of vinyl carbocation$({H_3}C - \mathop C\limits^ \oplus {H_2})(HSO_4^ - )$
B) Ethylene is soluble in conc. ${H_2}S{O_4}$ due to the formation of vinyl carbocation $({H_3}C - \mathop C\limits^ \oplus {H_2})(HSO_4^ - )$
C) But-2-yne dissolves in conc. ${H_2}S{O_4}$ due to the formation of vinyl carbocation $(Me - \mathop C\limits^ \oplus = CH - Me)(HSO_4^ - )$
D) More the s-character in the positively charged C, the more stable is the carbocation and more likely is its formation.

Answer 1

Solubility follows the rule “like dissolves like”. If a molecule is polar, it will dissolve in a polar solvent like concentrated ${H_2}S{O_4}$. That is, ${H^ + }$from ${H_2}S{O_4}$will attack the carbon atom to form carbocation only if there is some degree of polarity. Further, s-character increase means a lesser number of electron donating groups on the positively charged carbon.

Complete step-by-step answer:
Hydrocarbons (saturated or unsaturated) are insoluble in water as they are non-polar in nature. Water is a polar solvent and dissolves polar solutes. So, you can make unsaturated hydrocarbons water soluble, by protonation of them with strong acid such as sulphuric acid.

Statement A is correct because acetylene is a highly symmetrical and non-polar molecule and does not get affected by conc. ${H_2}S{O_4}$. In fact, conc. ${H_2}S{O_4}$is used as a drying agent in the preparation of acetylene.

Statement B is also correct because ethylene is a slightly polar molecule and ${H^ + }$from conc. ${H_2}S{O_4}$attacks one of the doubly bonded carbon-atoms to form a carbocation which forms a bond of ionic nature with and hence, is soluble in conc. ${H_2}S{O_4}$.

Statement C is also a correct statement as methyl groups may be unsymmetrically distributed around doubly-bonded carbon which makes ${H^ + }$ from conc. ${H_2}S{O_4}$ to attack it and form a relatively stable carbocation whose combination with $HSO_4^ -$ makes But-2-yne soluble in conc. ${H_2}S{O_4}$

Statement D is a wrong statement because with increasing s-character, the stability of carbocation decreases. This happens because as the s-character increases, unsaturation increases, making electron donating alkyl groups less and less. Thus, the positive charge if acquired by such a carbon remains intensified on it and does not distribute or spread itself as happens in $s{p^3}$carbon. So, as the s-character increases, carbocation stability decreases because of intensification of positive charge.Thus option D is the wrong statement while all the above three statements are correct.

Hence, the correct answer is option ‘D’.

Note: London forces or weak van der Waals forces do not impact the dissolution process of a compound in conc. ${H_2}S{O_4}$ as these forces are negligible as compared to strong polar forces that exist between ${H_2}S{O_4}$molecules. Further, the s-character increase, stabilizes the carbanion and not carbocation.