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jekas [21]
2 years ago
5

Predict the splitting pattern for each of the labeled hydrogens in the following molecules. Assume that all coupling constants a

re equal. a) Ha: Hb: Hc: Hd: b) Ha:

Chemistry
1 answer:
Ghella [55]2 years ago
3 0

Complete Question

The complete question is shown on the first uploaded image

Answer:

a) Splitting pattern for Ha= 2+1 , Triplet

For Proton Hb and Hc both are equivalent to each other. non equivalent protons n= 3

Splitting pattern for Hb and Hc= 3+1 , Quartet

For Proton Hd, number of non equivalent protons n= 0

Splitting pattern for Hd= 0+1 , Singlet

b) Splitting pattern for Ha= 2+1 , Triplet

For Proton Hb and Hc both are equivalent to each other. non equivalent protons n= 3

Splitting pattern for Hb and Hc= 3+1 , Quartet

For Proton Hd, number of non equivalent protons n= 0

Splitting pattern for Hd= 0+1 , Singlet

c) The IUPAC name is Butan-2-ol

Explanation:

Considering the first question the rule used for prediction of splitting pattern is n+1 (Pascal's Triangle rule), where n is number of H atom on the adjacent carbon which are non equivalent.

According to that for molecule 1 as shown on the second uploaded image

For Proton Ha, number of non equivalent protons n= 2

Splitting pattern for Ha= 2+1 , Triplet

For Proton Hb and Hc both are equivalent to each other. non equivalent protons n= 3

Splitting pattern for Hb and Hc= 3+1 , Quartet

For Proton Hd, number of non equivalent protons n= 0

Splitting pattern for Hd= 0+1 , Singlet

Considering the second question for Molecule 2 as shown on the third uploaded image  

For Proton Ha, number of non equivalent protons n= 1

Splitting pattern for Ha= 1+1=2 , Doublet

For Proton Hb, number of non equivalent protons n= 3

Splitting pattern for Hb= 3+1=4 , Quartet

For Proton Hc, number of non equivalent protons n= 3

Splitting pattern for Hc= 3+1=4 , Quartet

For Proton Hd, number of non equivalent protons n= 1

Splitting pattern for Ha= 1+1=2 , Doublet

Considering the third question

The name of the given molecule  is gotten according to longest carbon chain  = 4 (Prefix 'Butan')

Functional group = -OH (Suffix 'ol') at C-2

The IUPAC name is Butan-2-ol

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4 0
3 years ago
Be sure to answer all parts. The equilibrium constant (Kp) for the reaction below is 4.40 at 2000. K. H2(g) + CO2(g) ⇌ H2O(g) +
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<u>Answer:</u>

<u>For 1:</u> The value of \Delta G for the chemical equation is -24.636 kJ/mol

<u>For 2:</u> The value of \Delta G for the chemical equation is -20.925 kJ/mol

<u>Explanation:</u>

For the given chemical equation:

H_2(g)+CO_2(g)\rightleftharpoons H_2O(g)+CO(g)

  • <u>For 1:</u>

To calculate the \Delta G for given value of equilibrium constant, we use the relation:

\Delta G=-RT\ln K_p      .....(1)

where,

\Delta G = ? kJ/mol

R = Gas constant = 8.314J/K mol

T = temperature = 2000 K

K_p = equilibrium constant in terms of partial pressure = 4.40

Putting values in above equation, we get:

\Delta G=-(8.314J/Kmol)\times 2000K\times \ln (4.40)\\\\\Delta G=-24636.12J/mol

Converting this into kilo joules, we use the conversion factor:

1 kJ = 1000 J

So, -24636.12 J/mol = -24.636 kJ/mol

Hence, the value of \Delta G for the chemical equation is -24.636 kJ/mol

  • <u>For 2:</u>

The expression of K_p for the given chemical equation is:

K_p=\frac{p_{CO}p_{H_2O}}{p_{H_2}p_{CO_2}}

We are given:

p_{CO}=1.18atm\\p_{H_2O}=0.66atm\\p_{CO_2}=0.82atm\\p_{H_2}=0.27atm

Putting values in above equation, we get:

K_p=\frac{1.18\times 0.66}{0.27\times 0.82}\\\\K_p=3.52

Now, calculating the value of \Delta G by using equation 1:

R = Gas constant = 8.314J/K mol

T = temperature = 2000 K

K_p = equilibrium constant in terms of partial pressure = 3.52

Putting values in equation 1, we get:

\Delta G=-(8.314J/Kmol)\times 2000K\times \ln (3.52)\\\\\Delta G=-20925.68J/mol

Converting this into kilo joules, we use the conversion factor:

1 kJ = 1000 J

So, -20925.68 J/mol = -20.925 kJ/mol

Hence, the value of \Delta G for the chemical equation is -20.925 kJ/mol

3 0
3 years ago
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