Acompound with molecular formula c5h10o2 has the following 1h nmr spectrum: 1.15 (t, 3h), 1.25 (t, 3h), 2.33 (q, 2h), 4.13 (q, 2h). draw its structure. hint: how many rings and π bonds does c5h10o2 have

Ethyl propionate

Explanation:

Calculating the double bond equivalence as:

DBE =  C - (H/2) - (X/2) + (N/2) +1

Where  

C is the number of carbon atoms

N is the number of nitrogen atoms

X is the number of halogens

H is the number of hydrogen atoms

So, according to the formula,

DBE =  5 - (10/2) - (0/2) + (0/2) +1 = 1

It means there is one double bond or one ring.

from the NMR signal, it is clear that compound has two triplet and two quardrate group that means two -CH3 and two -CH2 groups are present.

From the splitting information, it is clear that each the -CH2- group is next to a -CH3, and vice-versa. In this case ring structure is not possible.

The chemical shift of two protons 4.13 ppm(q, 2H) is for protons next to an O atom, therefore, compound must have

.

Then the signal at 2.33 is for a -CH2-group next to a C=O. So compound must have a structure CH3-CH2-C=O. Now, there is only one way to complete the structure - by bonding the O to the C=O carbon.

So the compound is ethyl propionate

first, consult an activity series chart to determine if al is more reactive than fe. it is, so it will replace the fe. now, one can continue with the problem.

next, make sure you have a correctly balanced equation.

3feo + 2al ==> 3fe + al2o3

now, determine which reactant (feo or al) is in limiting supply.

moles feo present = 125 g x 1 mole/71.8 g = 1.74 moles

moles al present = 25.0 g x 1 mole/26.98 g = 0.927 moles

since from the balanced equation, it take 3 mol feo for every 2 mol al, al will be limiting.

you can determine this simply by dividing moles of each by their coefficient in the balanced equation. for example: 1.74/3 = 0.58 and 0.927/2 = 0.464. al is less, so it is limiting.

finally, using the stoichiometry in the balanced eq. determine mass of fe that can be produced from 0.927 moles of al.

0.927 moles al x 3 moles fe/2 moles al x 55.8 g/mole = 77.6 g

where's the diagram?