Styrene Reaction with HBr: Products

The reaction of styrene (PhCH = CH$_2$) with HBr is an electrophilic addition reaction to an alkene. This reaction follows Markovnikov's rule, where the hydrogen atom adds to the carbon with more hydrogen atoms, and the bromine atom adds to the more substituted carbon atom.

This leads to the formation of two regioisomers:

• Major product (A): 1-bromo-1-phenylethane (Markovnikov product). The bromine adds to the carbon directly attached to the phenyl group. Structure: Ph-CHBr-CH$_3$.
• Minor product (B): 2-bromo-1-phenylethane (Anti-Markovnikov product). This is formed via a less stable primary carbocation or potentially through a radical mechanism under certain conditions, but in the context of the major/minor designation typical for ionic addition without peroxides, it's the less favored ionic product. Structure: Ph-CH$_2$-CH$_2$Br.

The structures are:

Isomer A (Major):

\( \text{Ph} - \underset{|}{\text{CH}} - \text{CH}_3 \)

\( \phantom{\text{Ph} -} \text{Br} \)

Isomer B (Minor):

\( \text{Ph} - \text{CH}_2 - \text{CH}_2 - \text{Br} \)

NMR Analysis of the Mixture

The ¹H NMR spectrum provides information about the different types of protons in the mixture of isomers A and B. The integral of each signal is proportional to the number of protons giving rise to that signal and the molar amount of the molecule containing those protons.

Let $N_A$ be the number of moles of isomer A and $N_B$ be the number of moles of isomer B in the mixture. The relative integration ratio is given as 2:1:1:6 for signals at δ 5.17, 3.53, 3.15, and 2.00 ppm, respectively.

Assigning NMR Signals

Let's analyze the expected signals for each isomer (ignoring the phenyl protons, which are typically in the 7-8 ppm range and are not among the listed signals):

• Isomer A (Ph-CHBr-CH$_3$):
  • CH proton (adjacent to Ph and Br): Expected chemical shift around 5 ppm. Corresponds to 1 proton.
  • CH$_3$ protons (adjacent to CH): Expected chemical shift around 1.5-2 ppm. Corresponds to 3 protons.
• Isomer B (Ph-CH$_2$-CH$_2$Br):
  • CH$_2$ adjacent to Ph: Expected chemical shift around 2.5-3.5 ppm. Corresponds to 2 protons.
  • CH$_2$ adjacent to Br: Expected chemical shift around 3-4 ppm. Corresponds to 2 protons.

Now we match the given signals and integrations to these expected peaks:

• The signal at δ 5.17 ppm (Integration 2) fits the CH proton in isomer A (1 proton).
• The signal at δ 3.53 ppm (Integration 1) fits one of the CH$_2$ groups in isomer B (2 protons). Likely the one closer to Br.
• The signal at δ 3.15 ppm (Integration 1) fits the other CH$_2$ group in isomer B (2 protons). Likely the one closer to Ph.
• The signal at δ 2.00 ppm (Integration 6) fits the CH$_3$ protons in isomer A (3 protons).

Calculating Molar Ratio

The integration value for a signal is proportional to (number of protons of that type in the molecule) $\times$ (molar amount of the molecule). Let $k$ be the proportionality constant relating integration units to the total molar amount of protons.

From the table, we can set up the following proportions based on the integration values:

• For the signal at 5.17 ppm (CH in A): $1 \times N_A \propto 2 \implies N_A \propto 2$
• For the signal at 2.00 ppm (CH$_3$ in A): $3 \times N_A \propto 6 \implies N_A \propto \frac{6}{3} = 2$

Both signals from isomer A consistently indicate that the molar amount of A is proportional to 2.

• For the signal at 3.53 ppm (CH$_2$ in B): $2 \times N_B \propto 1 \implies N_B \propto \frac{1}{2}$
• For the signal at 3.15 ppm (CH$_2$ in B): $2 \times N_B \propto 1 \implies N_B \propto \frac{1}{2}$

Both signals from isomer B consistently indicate that the molar amount of B is proportional to 0.5.

The molar ratio of A to B is therefore $N_A : N_B \propto 2 : 0.5$.

To express this ratio in whole numbers, we can multiply both sides by 2:

$N_A : N_B = (2 \times 2) : (0.5 \times 2) = 4 : 1$.

Thus, the molar ratio of A and B in the mixture is 4:1.