Which of the following is the correct IUPAC name?
1 answer:
Answer:
(C) 4-ethyl-2,2-dimethylheptane.
Explanation:
Step one: draw the carbon skeletons of each "IUPAC name" as if they were accurate. Note that
- Hexane: a carbon backbone of six carbon atoms, linked in a row with carbon-carbon single bonds.
- Heptane: a carbon backbone of seven carbon atoms, linked in a row with carbon-carbon single bonds.
- A methyl group is a one-carbon substitute -CH₃ for an H atom on the backbone.
- An ethyl group is a two-carbon substitute -CH₂-CH₃ for an H atom on the backbone.
- A propyl group is a three-carbon substitute -CH₂-CH₂-CH₃ for an H atom on the backbone.
Step two: check the structures and the names against the IUPAC rules.
There are many of these rules. Ones that apply here include:
- The backbone shall be named after the longest chain of carbon atoms. There may be right angles but the chain needs to be continuous.
- List the substitutes in alphabetical order (e.g,, ethyl, methyl, and propyl.)
- Number the substitutes from the end that ensures the smallest number for the first substitute group that is encounter. If starting from both ends lead to the same number for the first group, check the second group.
Refer to the sketch attached.
<h3>(A)</h3>Should start numbering from the other end.
<h3>(B) and (D)</h3>The longest backbone contains seven carbon atoms.
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Answer : The correct answer is 96.68 yrs
Radioactivity Decay :
it is a process in which a nucleus of unstable atom emit energy in form of radiations like alpha particle , beta particle etc .
Radioactive decay follows first order kinetics , so its rate , rate constant , amount o isotopes can be calculated using first order equations .
The first order equation for radioactive decay can be expressed as :
----------- equation (1)
Where : N = amount of radioisotope after time "t"
N₀ = Initial amount of radioisotope
k = decay constant and t = time
Following steps can be used to find time :
1) To find deacy constant :
Decay constant can be calculated using half life . Decay constant and half life can be related as :
---------equation (2)
Given : Half life of Strontium -90 = 28.8 years
Plugging value of in above formula (equation 2) :
Multiply both side by k
Dividing both side by 28.8 yrs
(ln 2 = 0.693 )
k = 0.0241 yrs⁻¹
Step 2 : To find time :
Given : N₀ = 10.3 ppm N = 1.0 ppm k = 0.0241 yrs⁻¹
Plugging these value in equation (1) as :
(ln 0.0971 = - 2.33 )
Dividing both side by - 0.0241 yrs⁻¹
t = 96.68 yrs
Hence the concentration of Strontium-90 will drop from 10.3 ppm to 1.0 ppm is 96.68 yrs