Percent occurrence and isotope masses for oxygen: 99.759 % at 15.99491 amu, 0.037% at 16.99913 amu and 0.204% is at what amu val
1 answer:
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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:
trans-1,3-pentadiene is more stable than 1,4-pentadiene due to presence of a conjugated double bond.
Explanation:
Here,
H(hydrogenated pdt.) is same for both 1,4-pentadiene and 1,3-pentadiene as they both produce pentane after hydrogenation
H(diene) depends on stability of diene.
More stable a diene, lesser will be it's H(diene) value (more neagtive).
trans-1,3-pentadiene is more stable than 1,4-pentadiene due to presence of a conjugated double bond.
Hence, is higher (less negative) for trans-1,3-pentadiene