Question

The reaction sequence given below is carried out with 12 moles of X. The yield of the major product in each step is given below the product in parenthesis. The amount (in grams) of S produced is: [Atomic masses: C = 12 u, H = 1 u, Cl = 35.5 u, Br = 80 u]

$C_xH_y \xrightarrow[Fe \ tube]{Red \ Hot} P \xrightarrow[AlCl_3]{C_2H_5Cl} Q \xrightarrow[CCl_4]{NBS} R \xrightarrow[\Delta]{alc.KOH} S$ (X) (100%) (50%) (50%) (50%)

Answer 1

52

Answer 2

The problem involves a sequence of organic reactions, and we need to calculate the final mass of the product S, considering the yield at each step.

Step 1: Identify X and P

The first reaction is $C_xH_y \xrightarrow[Fe \ tube]{Red \ Hot} P$. This reaction is characteristic of the trimerization of ethyne (acetylene) to form benzene.
Therefore, X is ethyne ($C_2H_2$) and P is benzene ($C_6H_6$).
The balanced equation is:

$3 C_2H_2 \xrightarrow[Red \ Hot \ Fe \ tube]{} C_6H_6$

Given initial moles of X ($C_2H_2$) = 12 moles.
The yield for P is 100%.
From stoichiometry, 3 moles of $C_2H_2$ produce 1 mole of $C_6H_6$.
Moles of P produced = $(12 \text{ moles } C_2H_2) \times \frac{1 \text{ mole } C_6H_6}{3 \text{ moles } C_2H_2} \times 100\% = 4 \text{ moles } C_6H_6$.

Step 2: Identify Q

The second reaction is $P \xrightarrow[AlCl_3]{C_2H_5Cl} Q$. This is a Friedel-Crafts alkylation reaction where benzene (P) reacts with chloroethane ($C_2H_5Cl$) in the presence of anhydrous $AlCl_3$.
The product Q is ethylbenzene.

$C_6H_6 + C_2H_5Cl \xrightarrow{AlCl_3} C_6H_5C_2H_5 + HCl$

Q = Ethylbenzene ($C_6H_5C_2H_5$)

The yield for Q is 50%.
Moles of Q produced = $(4 \text{ moles } C_6H_6) \times 50\% = 2 \text{ moles } C_6H_5C_2H_5$.

Step 3: Identify R

The third reaction is $Q \xrightarrow[CCl_4]{NBS} R$. NBS (N-bromosuccinimide) in $CCl_4$ is a reagent used for benzylic bromination. Ethylbenzene (Q) has a benzylic hydrogen, which will be substituted by bromine.
The product R is (1-bromoethyl)benzene.

$C_6H_5CH_2CH_3 \xrightarrow{NBS/CCl_4} C_6H_5CH(Br)CH_3 + \text{succinimide}$

R = (1-bromoethyl)benzene ($C_6H_5CH(Br)CH_3$)

The yield for R is 50%.
Moles of R produced = $(2 \text{ moles } C_6H_5C_2H_5) \times 50\% = 1 \text{ mole } C_6H_5CH(Br)CH_3$.

Step 4: Identify S

The fourth reaction is $R \xrightarrow[\Delta]{alc.KOH} S$. Alcoholic KOH is a strong base used for dehydrohalogenation (elimination reaction). (1-bromoethyl)benzene (R) will undergo elimination to form an alkene.
The product S is styrene (phenylethene).

$C_6H_5CH(Br)CH_3 \xrightarrow{alc.KOH, \Delta} C_6H_5CH=CH_2 + HBr$

S = Styrene ($C_6H_5CH=CH_2$)

The yield for S is 50%.
Moles of S produced = $(1 \text{ mole } C_6H_5CH(Br)CH_3) \times 50\% = 0.5 \text{ moles } C_6H_5CH=CH_2$.

Step 5: Calculate the mass of S

The molecular formula of S (styrene) is $C_8H_8$.
Using the given atomic masses: C = 12 u, H = 1 u.
Molar mass of S = $(8 \times 12) + (8 \times 1) = 96 + 8 = 104 \text{ g/mol}$.

Amount of S produced (in grams) = Moles of S $\times$ Molar mass of S
Amount of S = $0.5 \text{ moles} \times 104 \text{ g/mol} = 52 \text{ grams}$.

The final amount of S produced is 52 grams.