Question

Find Alkane with least mol. wt (Comp. contain all 1°,2°,3°,4° Carbon) In this question when alkane had 4 times 1° Carbon than 4°C sir said C14H30 is impossible under condition and asked to attach a screenshot/picture of structure and I want to confirm it, so here it is

Answer 1

A $C_{14}H_{30}$ alkane is possible under the condition that $n_{1^\circ C} = 4 n_{4^\circ C}$ and contains all types of carbons (1°, 2°, 3°, 4°). The number of carbons of each type would be $n_1=8, n_2=2, n_3=2, n_4=2$.

Answer 2

Let $n_1, n_2, n_3, n_4$ be the number of 1°, 2°, 3°, and 4° carbons, respectively. For an alkane $C_n H_{2n+2}$, the following relationships hold:

1. Total carbons: $n = n_1 + n_2 + n_3 + n_4$
2. Relationship between carbon types: $n_1 = n_3 + 2n_4 + 2$ For $C_{14}H_{30}$, $n=14$. So, $n_1 + n_2 + n_3 + n_4 = 14$.

The given condition is $n_{1^\circ C} = 4 n_{4^\circ C}$. Substitute this into the second relationship: $4n_4 = n_3 + 2n_4 + 2$ $n_3 = 2n_4 - 2$

The problem states that the compound must contain all types of carbons, meaning $n_1 \ge 1, n_2 \ge 1, n_3 \ge 1, n_4 \ge 1$. From $n_3 = 2n_4 - 2$, for $n_3 \ge 1$, we must have $2n_4 - 2 \ge 1 \implies 2n_4 \ge 3 \implies n_4 \ge 1.5$. Since $n_4$ must be an integer, the minimum value for $n_4$ is 2.

Let's use $n_4=2$: $n_3 = 2(2) - 2 = 2$. (This satisfies $n_3 \ge 1$) $n_1 = 4n_4 = 4(2) = 8$. (This satisfies $n_1 \ge 1$)

Now substitute $n_1=8, n_3=2, n_4=2$ into the total carbon equation: $8 + n_2 + 2 + 2 = 14$ $12 + n_2 = 14$ $n_2 = 2$. (This satisfies $n_2 \ge 1$)

So, for $C_{14}H_{30}$ under the given conditions, the number of each type of carbon must be: $n_1 = 8$ $n_2 = 2$ $n_3 = 2$ $n_4 = 2$

Let's verify the hydrogen count for these values: $H = 3n_1 + 2n_2 + n_3 + 0n_4 = 3(8) + 2(2) + 2 + 0 = 24 + 4 + 2 = 30$. This matches the hydrogen count for $C_{14}H_{30}$. Since a consistent set of non-zero integer values for $n_1, n_2, n_3, n_4$ is derived that satisfies all the given conditions and the general alkane formulas, such an alkane is theoretically possible. The existence of such a set of numbers guarantees the existence of at least one structural isomer.

The statement that $C_{14}H_{30}$ is impossible under the given conditions is incorrect.