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UNO [17]
3 years ago
8

Which is the correct statement regarding the relative Rf values of the starting methyl benzoate vs the product, methyl m-nitrobe

nzoate on a silica gel TLC plate. The product has a higher Rf value on a silica gel TLC plate because it is less polar than the starting methyl benzoate. The product has a lower Rf value on a silica gel TLC plate because it is less polar than the starting methyl benzoate. The product has a higher Rf value on a silica gel TLC plate because it is more polar than the starting methyl benzoate. The product has a lower Rf value on a silica gel TLC plate because it is more polar than the starting methyl benzoate. 1.5 points QUESTION 2 Methyl benzoate is more reactive than benzene in Electrophilic Aromatic Substition [EAS] reactions. True False 1 points QUESTION 3 Methyl m-nitrobenzoate is less reactive than methyl benzoate in Electrophilic Aromatic Substition [EAS] reactions. True False
Chemistry
1 answer:
Marianna [84]3 years ago
5 0

Answer:

1. The product has a higher Rf value on a silica gel TLC plate because it is more polar than the starting methyl benzoate.

2. False

3. True

Explanation:

In chromatography, there is a stationary phase and a mobile phase. The ratio of the distance moved by a component and the distance moved by the solvent gives the retention factor (Rf).

Since silica gel is a polar solvent, it will retain the more polar product methyl m-nitrobenzoate compared to the methyl benzoate starting material.

In comparing the electrophillic aromatic substitution of m-nitrobenzoate  and methyl benzoate, we must remember that the presence of electron withdrawing groups (such as -NO2 and -CHO) on the aromatic compound deactivates the compound towards electrophillic aromatic substitution hence, methyl m-nitrobenzoate is less reactive than methyl benzoate in Electrophilic Aromatic Substition and Methyl benzoate is less reactive than benzene in Electrophilic Aromatic Substition

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Answer: Freezing point of a solution will be -1.16^0C

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Depression in freezing point is given by:

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i= vant hoff factor = 1 (for non electrolyte)

K_f = freezing point constant = 5.12^0C/m

m= molality

\Delta T_f=i\times K_f\times \frac{\text{mass of solute}}{\text{molar mass of solute}\times \text{weight of solvent in kg}}

Weight of solvent (benzene)= 1480 g =1.48 kg

Molar mass of solute (octane) = 114.0 g/mol

Mass of solute (octane) = 220 g

(5.50-T_f)^0C=1\times 5.12\times \frac{220g}{114.0 g/mol\times 1.48kg}

(5.50-T_f)^0C=6.68

T_f=-1.16^0C

Thus the freezing point of a solution will be -1.16^0C

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