Answer:
To answer this question we need the illustration
Explanation:
You have left the question incomplete since we do not have the illustration making this impossible to answer
Answer:
CH₃CH(CH₃)CH(C₃H₇)CH₂CH(CH₃)₂:
4-isopropyl-2-methylpentane.
Explanation:
Step One: Draw the structure formula of this compound. Parentheses in the formula indicate substitute groups that are connected to the carbon atom to the left.
For example, the first (CH₃) indicates that the second carbon atom from the left is connected to:
- the CH₃- on the left-hand side,
- the -CH(C₃H₇)CH₂CH(CH₃)₂ on the right-hand side,
- a hydrogen atom, and
- an additional CH₃- group that replaced one hydrogen atom.
Each carbon atom in this compound is connected to four other atoms. All bonds between carbon atoms are single bonds.
The C₃H₇ in the second pair of parentheses is the condensed form of CH₃CH₂CH₂-. See the first sketch attached. Groups in parentheses are highlighted.
Step Two: Find the carbon backbone. The backbone of a hydrocarbon is the longest chain of carbon atoms that runs through the compound. See the second sketch attached. The backbone of this compound consists of seven carbon atoms and is highlighted in green. The name for this backbone shall be heptane.
Step Three: Identify and name the substitute groups.
The two substitute groups are circled in blue in the second sketch.
- The one on the right -CH₃ is a methyl group.
- The one on the left is branched.
This group can be formed by removing one hydrogen from the central carbon atom in propane. The name for this group is isopropyl.
Step Four: Number the atoms.
Isopropyl shall be placed before methyl. Start from the right end to minimize the index number on all substitute groups. The methyl group is on carbon number two and the isopropyl group on carbon number four. Hence the name:
4-isopropyl-2-methylheptane.
Answer:
Isopropyl propionate
Explanation:
1. Information from formula
The formula is C₆H₁₂O₂. A six-carbon alkane would have the formula C₆H₁₄. The deficiency of two H atoms indicates the presence of either a ring or a double bond.
2. Information from the spectrum
(a) Triplet-quartet
A 3H triplet and a 2H quartet is the classic pattern for a CH₃CH₂- (ethyl) group
(b) Septet-doublet
A 1H septet and a 6H doublet is the classic pattern for a -CH(CH₃)₂ (isopropyl) group
(c) The rest of the molecule
The ethyl and isopropyl groups together add up to C₇H₁₂.
The rest of the molecule must have the formula CO₂ and one unit of unsaturation. That must be a C=O group.
The compound is either
CH₃CH₂-COO-CH(CH₃)₂ or (CH₃)₂CH-COO-CH₂CH₃.
(d) Well, which is it?
The O atom of the ester function should have the greatest effect on the H atom on the adjacent carbon atom.
The CH of an isopropyl is normally at 1.7. The adjacent O atom should pull it down perhaps 3.2 units to 4.9.
The CH₂ of an ethyl group is normally at 1.2. The adjacent O atom should pull it down to about 4.4.
We see a signal at 5.0 but none near 4.4. The compound is isopropyl propionate.
3. Summary
My peak assignments are shown in the diagram below.
Answer:
A
Explanation:
In simple terms: Hammer goes one way, air goes the other way.
Answer: the answer is americium and it has 95 protons
Explanation: