Give the major organic products from the oxidation with KMnO4 for the following compounds. Assume an excess of KMnO4.

Chemistry

1 answer:

Firdavs [7]3 years ago

7 0

Answer:

Explanation:

a ) Benzoic acid is formed . In any alkyl benzene derivative , potassium permanganate reacts to form carboxylic acid . It oxidises side chains to carboxylic acid .

C₆H₅CH₃ + 0 = C₆H₅COOH + H₂O

O is provided by KMnO₄

b ) In this reaction isophthalic acid is formed .

C₆H₄(CH₃)₂ +O = C₆H₄(COOH)₂

4-Propyl-3-t-butyltoluene

In this oxidation , three side chains of ring are 1 ) 1-methyl 2 ) 3- butyl 3 ) 4 propyl .

The methyl and 4 - propyl groups are oxidised to di- carboxylic acid and 3 butyl group remains intact ( unoxidised )

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The average atomic mass of Eu is 151.96 amu. There are only two naturally occurring isotopes of europium, Eu with a mass of 151.

earnstyle [38]

Answer:

The percentage abundance of Eu isotopes are 52 % and 48 % .

Explanation:

The formula for the calculation of the average atomic mass is:

$Average\ atomic\ mass=(\frac {\%\ of\ the\ first\ isotope}{100}\times {Mass\ of\ the\ first\ isotope})+(\frac {\%\ of\ the\ second\ isotope}{100}\times {Mass\ of\ the\ second\ isotope})$

Given that:

Since the element has only 2 isotopes, so the let the percentage of first be x and the second is 100 -x.

For first isotope,:

% = x %

Mass = 151.0 amu

For second isotope :

% = 100 - x

Mass = 153.0 amu

Given, Average Mass = 151.96 amu

Thus,

$151.96=\frac{x}{100}\times {151.0}+\frac{100-x}{100}\times {153.0}$

Solving for x, we get that:

x = 52 %

<u>Thus percentage abundance of Eu isotopes are 52 % and 48 % .</u>

7 0

3 years ago

1. The emission spectrum of mercury atoms has a bright green line with wavelength

SashulF [63]

Emission spectrum results from the movement of an electron from a higher to a lower energy level. The frequency of the photon is 5.5 * 10^14 Hz.

From the formula;

E = hc/λ

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