Iiiiiiiv major The sum of the molecular mass of gas A - The number of - hydrogens in B - The number of stereoisomers of C - The degree of unsaturation inD is-

(i) PhSO_3H + NaHCO_3 → PhSO_3^ Na^+ + H_2CO_3 H_2CO_3 → H_2O + Mol. wt. = 44CO_2 (ii) Number of α hydrogen in (B) is 10. (iii) Total number of stereoisomers ...

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The sum of the molecular mass of gas (A) + The number of $α -$ hydrogen(s) in (B) + The number of stereoisomers of (C) + The degree of unsaturation in (D) is:

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Solution

(i) $P h S O_{3} H + N a H C O_{3} \to P h S O_{3}^{-} N a^{+} + H_{2} C O_{3}$
$H_{2} C O_{3} \to H_{2} O + C O_{2} Mol. wt. = 44$

(ii) Number of $α -$ hydrogen in (B) is 10.

(iii) Total number of stereoisomers of (c) is 4.

(iv) Degree of unsaturation of (D) is 1.

$A + B + C + D = 59$

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The sum of the molecular mass of gas (A) + The number of α−hydrogen(s) in (B) + The number of stereoisomers of (C) + The degree of unsaturation in (D) is:

(i) PhSO3H+NaHCO3→PhSO−3Na++H2CO3 H2CO3→H2O+CO2Mol. wt. = 44 (ii) Number of α−hydrogen in (B) is 10. (iii) Total number of stereoisomers of (c) is 4. (iv) Degree of unsaturation of (D) is 1. A+B+C+D=59

The sum of the molecular mass of gas (A) + The number of $α -$ hydrogen(s) in (B) + The number of stereoisomers of (C) + The degree of unsaturation in (D) is:

(i) $P h S O_{3} H + N a H C O_{3} \to P h S O_{3}^{-} N a^{+} + H_{2} C O_{3}$
$H_{2} C O_{3} \to H_{2} O + C O_{2} Mol. wt. = 44$

(ii) Number of $α -$ hydrogen in (B) is 10.

(iii) Total number of stereoisomers of (c) is 4.

(iv) Degree of unsaturation of (D) is 1.

$A + B + C + D = 59$