What is the IUPAC name of the following compound?
1 answer:
Answer:
2,2,3-trimethylbutane
Step-by-step explanation:
Here are the rules for naming alkanes.
1. Find the longest continuous chain of carbon atoms.
There is a <em>four-carbon chain</em> (blue).
2. Name the main chain.
The chain has four C atoms, so its base name is <em>butane</em>.
3. Identify and name all the substituents.
There are three CH₃ groups (<em>methyl</em><em>,</em> blue bonds).
4. Number the main chain from the end that gives the lowest possible locants to the substituents.
From the left-hand end, the locants are 2,2,3. From the right-hand end, they are 2,3,3. The first point of difference is the second locant. 2 < 3, so we <em>number from the left</em>.
5. Put the names of the substituents with their locants in alphabetical order in front of the base name with no spaces.
Use numbers to connect numbers and hyphens to connect numbers and letters.
The IUPAC name is 2,2,3-trimethylbutane.
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Answer:
The volume of nitrogen oxide formed is 35.6L
Explanation:
The reaction of nitric acid with copper is:
Cu(s) + 4HNO₃ → Cu(NO₃)₂ + 2NO₂(g) + 2H₂O(l)
Moles of copper are:
Moles of nitric acid are:
As 1 mol of Cu reacts with 4 moles of HNO₃:
0.697 mol Cu × (4mol HNO₃ / 1mol Cu) = 2.79 moles of HNO₃ will react. That means Cu is limiting reactant.
Moles of NO₂ produced are:
0.697 mol Cu × (2mol NO₂ / 1mol Cu) = <em>1.394 moles of NO₂</em>
Using PV = nRT
<em>Where P is pressure (735torr / 760 = 0.967atm); n are moles (1.394mol); R is gas constant (0.082atmL/molK); T is temperature (28.2°C + 273.15 = 301.35K). </em>
Thus, volume is:
V = nRT / P
V = 1.394mol×0.082atmL/molK×301.35K / 0.967atm
V = 35.6L
<em>The volume of nitrogen oxide formed is 35.6L</em>