1. During estimation of nitrogen present in an organic compound b… - Vidyadip

Question

1. During estimation of nitrogen present in an organic compound by Kjeldahl's method, the ammonia evolved from 0.5 g of the compound in Kjeldahl's estimation of nitrogen, neutralized 10 mL of $1 M H _2 SO _4$. Find out the percentage of nitrogen in the compound.
2. Explain why $\left( CH _3\right)_3 \stackrel{+}{ C }$ is more stable than $CH _3 \stackrel{+}{ C } H _2$ and $\stackrel{+}{ C } H _3$ is the least stable cation.

✓

Answer

$\begin{aligned} & \text { 1. } 1 M \text { of } 10 mL H _2 SO _4=1 M \text { of } 20 mL NH _3 \\ & 100 mL \text { of } 1 M \text { ammonia contains nitrogen }=14 g \\ & 20 mL \text { of } 1 M \text { ammonia will contain nitrogen }=\frac{14 \times 20}{1000} g \\ & \therefore \text { Percentage of nitrogen }=\frac{14 \times 20 \times 100}{1000 \times 0.5}=56.0 \%\end{aligned}$
2. Hyperconjugation interaction in $\left( CH _3\right)_3 \stackrel{+}{ C }$ is greater than in $CH _3 \stackrel{+}{ C } H _2$ as the $\left( CH _3\right)_3 \stackrel{+}{ C }$ has nine C -H bonds. In $\stackrel{+}{ C } H _3$, vacant p orbital is perpendicular to the plane in which $C - H$ bonds lie; hence cannot overlap with it. Thus, $\stackrel{+}{ C } H _3$ lacks hyperconjugative stability. Therefore, $\left( CH _3\right)_3 \stackrel{+}{ C }$ is more stable than $CH _3 \stackrel{+}{ C } H _2$ and $\stackrel{+}{ C } H _3$ is the least stable cation.

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