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3 years ago

6

Draw the structure of the compound identified by the following simulated 1H and 13C NMR spectra. The molecular formula of the co

mpound is C10H12O. (Blue numbers next to the lines in the 1H NMR spectra indicate the integration values.)

1 answer:

Luba_88 [7]3 years ago

6 0

Answer: 4-allylanisole

Explanation: The doublets behind the 7 ppm belongs to the

para-substituted benzene ring. The three single-proton multi-plets around 5−6 ppm predicts that there has to be a single subsituted alkene group

A single plus a doublet around 3-4 ppm belongs to CH3 and CH2 Groups as they could be attached to the subsituted alkene group.

Moreover the interpretation of the NMR that there is no peak with a higher intensity for >180 ppm represents an absence of Carbonyl group.

The Predicted Number is attached from a chemical database along with their peaks information

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(b) The conductivity of a 0.01 mol dm–3 solution of a monobasic organic acid in water is 5.07 × 10–2 S m–1. If the molar conduct

Zarrin [17]

Explanation:

The given data is as follows.

$\Lambda^{o}_{m}$ (NaCl) = $1.264 \times 10^{-2}$

$\Lambda^{o}_{m}$ (H-O=C-ONO) = $1.046 \times 10^{-2}$

$\Lambda^{o}_{m}$ (HCl) = $4.261 \times 10^{-2}$

Conductivity of monobasic acid is $5.07 \times 10^{-2} S m^{-1}$

Concentration = 0.01 $mol/dm^{3}$

Therefore, molar conductivity ( $\Lambda_{m}$ ) of monobasic acid is calculated as follows.

$\Lambda_{m} = \frac{conductivity}{concentration}$

= $\frac{5.07 \times 10^{-2} S m^{-1}}{0.01 mol/dm^{3}}$

= $\frac{5.07 \times 10^{-2} S m^{-1}}{0.01 mol \times 10^{3}}$

= $5.07 \times 10^{-3} S m^{2} mol^{-1}$

Also, $\Lambda^{o}_{m}$ = $\Lambda^{o}_{m}_{(HCl)} + \Lambda^{o}_{m}_{(H-O=C-ONO)} - \Lambda^{o}_{m}_{(NaCl)}$

= $4.261 \times 10^{-2} + 1.046 \times 10^{-2} - 1.264 \times 10^{-2}$

= $4.043 \times 10^{-2} S m^{2} mol^{-1}$

Relation between degree of dissociation and molar conductivity is as follows.

$\alpha = \frac{\Lambda_{m}}{\Lambda^{o}_{m}}$

= $\frac{5.07 \times 10^{-2} S m^{-1}}{4.043 \times 10^{-2} S m^{2} mol^{-1}}$

= 0.1254

Whereas relation between acid dissociation constant and degree of dissociation is as follows.

K = $\frac{c \times \alpha^{2}}{1 - \alpha}$

Putting the values into the above formula we get the following.

K = $\frac{c \times \alpha^{2}}{1 - \alpha}$

= $\frac{0.01 \times (0.1254)^{2}}{1 - 0.1254}$

= $0.017973 \times 10^{-2}$

= $1.7973 \times 10^{-4}$

Hence, the acid dissociation constant is $1.7973 \times 10^{-4}$ .

Also, relation between $pK_{a}$ and $K_{a}$ is as follows.

$pK_{a} = -log K_{a}$

= $-log (1.7973 \times 10^{-4})$

= 3.7454

Therefore, value of $pK_{a}$ is 3.7454.

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