Photochemical chlorination of pentane gave a mixture of three isomeric monochlorides. The principal monochloride constituted 46%
of the total, and remaining 54% was approximately a 1:1 mixture of the other two isomers. Write structural formulas for the major product and one of the two minor products of the monochloride isomers.
1 answer:
Answer:
Explanation:
In the chlorination of alkanes, the condition necessary is UV light so free radical substitution can take place. For alkanes like pentane, the primary, secondary and tertiary Hydrogen atoms (Hydrogen atoms bonded to their respective carbon) þare taken into consideration and this is because the tertiary Hydrogen is the most reactive (due to bond dissociation energy) hence the easiest to be substituted. The trend is as follows in the order of their reactivity;
1° < 2° < 3°
So, the products of the chlorination of pentane, the principal monochloride constituted is 3 - chloropentane while the other two monomers are:
2- chloropentane
1- chloropentane
Below is the attachment showing the structural formula of the three monochloride constituted pentane.
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Answer:
Here's what I get.
Explanation:
(b) Wavenumber and wavelength
The wavenumber is the distance over which a cycle repeats, that is, it is the number of waves in a unit distance.
Thus, if λ = 3 µm,
(a) Wavenumber and frequency
Since
λ = c/f and 1/λ = f/c
the relation between wavenumber and frequency is
Thus, if f = 90 THz
(c) Units
(i) Frequency
The units are s⁻¹ or Hz.
(ii) Wavelength
The SI base unit is metres, but infrared wavelengths are usually measured in micrometres (roughly 2.5 µm to 20 µm).
(iii) Wavenumber
The SI base unit is m⁻¹, but infrared wavenumbers are usually measured in cm⁻¹ (roughly 4000 cm⁻¹ to 500 cm⁻¹).
