Select the single best answer. Compound A exhibits two singlets in its 1H NMR spectrum at 2.64 and 3.69 ppm and the ratio of the
absorbing signals is 2:3. Compound B exhibits two singlets in its 1H NMR spectrum at 2.09 and 4.27 ppm and the ratio of the absorbing signals is 3:2. Which compound corresponds to dimethyl succinate and which compound corresponds to ethylene diacetate? 1455a 1455b dimethyl succinate ethylene diacetate Compound A is dimethyl succinate; Compound B is ethylene diacetate. Compound A is ethylene diacetate; Compound B is dimethyl succinate.
1 answer:
Answer:
Compound A is dimethyl succinate; Compound B is ethylene glycol diacetate.
Explanation:
The structure of dimethyl succinate (DMS) is (CH₂COOCH₃)₂.
The structure of ethylene glycol diacetate (EGD) is (CH₂OCOCH₃)₂.
The most obvious difference I see is that DMS has an -OCH₃ group, while EGD has a CH₃CO- group.
An alkane CH₃ normally appears at 0.9 ppm. The highly electronegative O atom would probably shift the signal downfield to about 3.5 ppm. Add another 0.2 ppm shift for the effect of the C=O group, and the CH₃ group should appear at about 3.7 ppm.
A CH₃CO- group should appear about 1.2 ppm downfield from its normal position at 0.9. That is, it should appear at about 2.1 ppm.
Conclusion:
A is dimethyl succinate; B is ethylene glycol diacetate.
Confirmatory evidence:
A CH₂ group normally appears at 1.3 ppm. A -CH₂CO-would be shifted down about 1.2 ppm to about 2.5 ppm and a -CH₂O- would be at about 4 ppm.
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Answer : The cell emf for this cell is 0.118 V
Solution :
The half-cell reaction is:
In this case, the cathode and anode both are same. So, is equal to zero.
Now we have to calculate the cell emf.
Using Nernest equation :
where,
n = number of electrons in oxidation-reduction reaction = 1
= ?
= 0.0222 M
= 2.22 M
Now put all the given values in the above equation, we get:
Therefore, the cell emf for this cell is 0.118 V