When Mg_2C_3 is reacted with water,a compound | vedclass

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(D) The reaction of $Mg_2C_3$ with water produces propyne $(CH_3-C \equiv CH)$,which is compound $X$.$Mg_2C_3 + 4H_2O \rightarrow 2Mg(OH)_2 + CH_3-C \

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$12$

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(D) The reaction of $Mg_2C_3$ with water produces propyne $(CH_3-C \equiv CH)$,which is compound $X$.$Mg_2C_3 + 4H_2O \rightarrow 2Mg(OH)_2 + CH_3-C \equiv CH$Propyne reacts with $Na$ to form sodium propynide $(CH_3-C \equiv C^-Na^+)$,which then reacts with ethyl bromide $(C_2H_5Br)$ via an $S_N2$ mechanism to form pent$-2-$yne $(CH_3-C \equiv C-CH_2-CH_3)$,which is compound $Y$.The structure of pent$-2-$yne is $CH_3-C \equiv C-CH_2-CH_3$.Counting the sigma bonds:- $C-H$ bonds: $3$ (in $CH_3$) $+ 2$ (in $CH_2$) $+ 3$ (in $CH_3$) $= 8$ sigma bonds.- $C-C$ bonds: $1$ ($C-C$ single) $+ 1$ ($C-C$ triple) $+ 1$ ($C-C$ single) $= 3$ sigma bonds.Total sigma bonds $= 8 + 3 = 11$. Wait,let's re-count: $CH_3-C \equiv C-CH_2-CH_3$ has $3$ $(C-H)$ $+ 1$ $(C-C)$ $+ 1$ ($C-C$ sigma from triple) $+ 1$ $(C-C)$ $+ 2$ $(C-H)$ $+ 3$ $(C-H)$ $= 11$ sigma bonds. Checking the options,$12$ is the closest if we consider the structure $CH_3-CH_2-C \equiv C-CH_3$ (pentyne). Let's re-evaluate: $CH_3-CH_2-C \equiv C-CH_3$ has $3+2+1+1+3 = 10$ $C-H$ bonds and $4$ $C-C$ bonds,total $14$. Actually,$CH_3-C \equiv C-CH_2-CH_3$ has $3+2+3 = 8$ $C-H$ bonds and $4$ $C-C$ bonds (including one from triple bond),total $12$ sigma bonds.

When $Mg_2C_3$ is reacted with water,a compound $X$ is formed which on heating with $Na$ followed by treatment with $C_2H_5Br$ gives another compound $Y$. Calculate the number of sigma bonds in $Y$.

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