Question

please help with number 11 and 12.

11. what would dept 90 and dept 135

12. what is responsible for the upfield

Question 1 What would DERT and DEPT 135 spectra of the compound shown above look 90 up Odown: 135 Bup, 3 down 090 06 How 135 oua down 90. doi: 135: 8up. Odown 190: & up3 down; 135: B3 down Question 12 12.5 points) What is responsible for the uplicid (-7-8 per center Sydog about the structure below? The intem magnetic field generated by the current low of clectrons of the ring will face in the same direction as the applied magnetic field of the instrument All of the peripheral hydrogens are 4 bonds for more) removed from a Deghboring hydrogen, so no first order splitting will take place. Electronegative atoms 3 bonds away have such electron-withdrawing cffect that this signal may differby - ppm if attached to a saturated hydrocarbon Fradio waves onerated from the instrument are not powertoo Shidrogen's electrons from precession

Answer 1

QUESTION 11

DEPT 90

In the DEPT 90 appears just carbon atoms bound to one hydrogen atom, and all signals have to be up

DEPT 135

In the DEPT 135 appears all carbon atoms bound to hydrogen atoms. Completely unsaturated carbon atoms don't appear. Carbon atoms bonded to two hydrogen atoms must appear down.

As this structure has a symmetry plane, let's count the signals just of one side of the symmetry plane because every specular atom has the same chemical environment, so they will have the same signal. So:

• In DEPT 90 we should expect to see 6 up signals.

• In DEPT 135 we should expect to see 8 signals, not 9, because the two -CH3 in the isopropyl group are chemically identical, they will have the same signal. And all signals must be up because this structure doesn't any carbon atom bound to two hydrogen atoms.

The correct option is the third option.  

QUESTION 12

The correct answer is the first option.

This question is about the hydrogen atoms in bold letters in the figure above. These hydrogen atoms appear in an RMN-1H around 7-8 ppm because they are less protected than hydrogen atoms in sigmaC-H. Circulating π electrons in the conjugate cycle caused a "ring current" which has a magnetic field. In the outer parts of the ring, this magnetic field will reinforce the magnetic field generated by the instrument. So, these hydrogen atoms suffer a higher magnetic field magnitude than other hydrogen atoms in the molecule.