What products are formed from monochlorination of (2r)−2−bromobutane at c1 and c4? draw the products using skeletal structures?
1 answer:
<span>From monochlorination of (2R)−2−bromobutane at C1 is formed </span>(2S)-2-bromo-1-chlorobutane.
From monochlorination of (2R)−2−bromobutane at C4 is formed (2R)-3-bromo-1-chlorobutane.
According to Cahn-Ingold-Prelog rules, the priority of halogen substituents decreases in the following sequence:
<span>-I > -Br > -Cl > -F
</span>
According to the rule, chlorine has priority over bromine, so the configuration changes at C1 substitution.
From monochlorination of (2R)−2−bromobutane at C4 is formed (2R)-3-bromo-1-chlorobutane.
According to Cahn-Ingold-Prelog rules, the priority of halogen substituents decreases in the following sequence:
<span>-I > -Br > -Cl > -F
</span>
According to the rule, chlorine has priority over bromine, so the configuration changes at C1 substitution.
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Answer:
ν = 7.04 × 10¹³ s⁻¹
λ = 426 nm
It falls in the visible range
Explanation:
The relation between the energy of the radiation and its frequency is given by Planck-Einstein equation:
E = h × ν
where,
E is the energy
h is the Planck constant (6.63 × 10⁻³⁴ J.s)
ν is the frequency
Then, we can find frequency,
Frequency and wavelength are related through the following equation:
c = λ × ν
where,
c is the speed of light (3.00 × 10⁸ m/s)
λ is the wavelength
A 426 nm wavelength falls in the visible range (≈380-740 nm)
The concentration of sodium and sulphate ions are [] = 0.4 M, [
] = 0.2 M
Explanation:
The molar concentration is defined as the number of moles of a molecule or an ion in 1 liter of a solution.
In the given solution, the concentration of the salt sodium sulphate is 0.2M. So, 0.2 moles of sodium sulphate is present in 1 liter of solution.
Assuming 100% dissociation,
1 molecule of sodium sulphate gives 2 ions of sodium and 1 ion of sulphate.
So 0.2 moles of sodium sulphate will give 0.4 moles of sodium ions and 0.2 moles of sulphate ions.