In bromination of Propyne, with Bromine 1,1,2,2-tetrabromopropane is obtained in 27% yield. The amount of 1,1,2,2-tetrabromopropane obtained from 1 g of Bromine in this reaction is _____ $$\times 10^{-1}$$ g. (Molar Mass: Bromine = 80 g/mol)

We need to find the amount of 1,1,2,2-tetrabromopropane obtained from 1 g of bromine in the bromination of propyne, given that the yield is 27%.

The bromination of propyne ($$CH_3C \equiv CH$$) with two equivalents of $$Br_2$$ gives 1,1,2,2-tetrabromopropane:

$$CH_3C \equiv CH + 2Br_2 \rightarrow CH_3CBr_2CHBr_2$$

We have 1 g of bromine. The molar mass of $$Br_2$$ is $$2 \times 80 = 160$$ g/mol. So the moles of $$Br_2$$ available are:

$$n_{Br_2} = \frac{1}{160} \text{ mol}$$

Since 2 moles of $$Br_2$$ are needed per mole of product, the moles of 1,1,2,2-tetrabromopropane that can be formed (at 100% yield) are:

$$n_{product} = \frac{1}{160 \times 2} = \frac{1}{320} \text{ mol}$$

The molar mass of 1,1,2,2-tetrabromopropane ($$C_3H_6Br_4$$) is $$3(12) + 6(1) + 4(80) = 36 + 6 + 320 = 362$$ g/mol.

The theoretical mass of product is:

$$m_{theoretical} = \frac{362}{320} = 1.13125 \text{ g}$$

With a yield of 27%, the actual mass obtained is:

$$m_{actual} = 1.13125 \times 0.27 = 0.3054 \text{ g}$$

Expressing this as $$\times 10^{-1}$$ g: $$m_{actual} = 3.054 \times 10^{-1} \approx 3 \times 10^{-1}$$ g.

Hence, the correct answer is 3.