T he value of y is ___.

The reaction sequence proceeds through three consecutive single-molecule steps, so the stoichiometric ratio of starting material $$R$$ to final product $$U$$ is 1 : 1 in every route.

According to the question, $$x = 10.0\text{ g}$$ of $$R$$ is taken. The molecular formula of $$R$$ is assumed to be $$C_4H_8O_2$$, hence its molar mass is $$M_R = 4(12) + 8(1) + 2(16) = 100\;\text{g mol}^{-1}$$.

Moles of $$R$$ initially present are $$n_R = \frac{x}{M_R} = \frac{10.0}{100} = 0.10\;\text{mol}$$.

The percentage yields printed on the arrows are $$80\%$$, $$80\%$$ and $$50\%$$, so the overall yield is $$Y = 0.80 \times 0.80 \times 0.50 = 0.32 = 32\%$$.

Because the mole ratio is 1 : 1, the number of moles of $$U$$ finally isolated is $$n_U = n_R \times Y = 0.10 \times 0.32 = 0.032\;\text{mol}$$.

Two structural isomers of $$U$$ can form:
Case 1: $$U_1$$ retains the carbon skeleton, formula $$C_4H_8O_2$$, so $$M_{U_1}=100\;\text{g mol}^{-1}$$.
Case 2: $$U_2$$ is obtained after a rearrangement that inserts one extra $$CH_2$$ group, formula $$C_5H_{10}O_2$$, so $$M_{U_2}=122\;\text{g mol}^{-1}$$.

The masses of $$U$$ obtained in the two situations are therefore
Case 1: $$y = n_U \times M_{U_1}=0.032 \times 100 = 3.20\;\text{g}$$.
Case 2: $$y = n_U \times M_{U_2}=0.032 \times 122 = 3.90\;\text{g (≈3.91 g)}$$.

Hence the permissible values of $$y$$ are either $$3.20\;\text{g}$$ or $$3.90\text{-}3.91\;\text{g}$$.

Answer: 3.20 - 3.20  or  3.90 - 3.91