The 1H-NMR of a compound with molecular formula C6H12O2 consists of four signals: 1.1 (triplet, integrating to 3 Hydrogens), 1.2
(doublet, integrating to 6 Hydrogens), 2.3 (quartet, integrating to 2 Hydrogens) and 5.0 (septet, integrating to 1 Hydrogen). Propose a structural formula for this compound consistent with this information.
1 answer:
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.
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Answer: Between 1 and 2.
Explanation:
Half life is the amount of time taken by a radioactive material to decay to half of its original value.
............(1)
where,
a = amount of reactant left after n-half lives = 40
= Initial amount of the reactant = 100
n = number of half lives
Putting in the values we get:
taking log on both sides
Thus half-lives that have elapsed is between 1 and 2
I used the genetic code table. mRNA codon ===> amino acid
1st base 2nd base 3rd base
A U U ===> Isoleucine
A U C ===> Isoleucine
The point mutation of codon AUU to AUC is a neutral mutation because it neither benefits nor deter the ability of the organism to survive and reproduce.
As you can see, Both codons result to the Isoleucine amino acid.
Another codon that will still result to the Isoleucin amino acid is AUA.
1st base 2nd base 3rd base
A U U ===> Isoleucine
A U C ===> Isoleucine
The point mutation of codon AUU to AUC is a neutral mutation because it neither benefits nor deter the ability of the organism to survive and reproduce.
As you can see, Both codons result to the Isoleucine amino acid.
Another codon that will still result to the Isoleucin amino acid is AUA.