Structure Determination

Question	Answer
how do you calculate empirical formula?	Step 1: divide % abundance by atomic weight Step 2: Divide each value from Step 1 by the lowest value
How do you obtain the molecular formula?	Step 1: calculate the molecular weight of the empirical formula Step 2: Obtain the molecular weight from data given/ mass spectrum Step 3: divide molecular weight of sample by molecular weight of empirical formula -the number from step 3 is how many times the empirical formula fits into molecular formula
Alternative method to calculate molecular formula	Step 1: convert % abundance of element in to a decimal Step 2: put it into the following eqn no.of atoms = (decimal*molecular mass)/atomic mass Step 3: repeat for all other elements in the sample
How do you find the number of double bond equivalents?	Step 1: Imagine a linear chain of the C, N and O atoms and work out how many Hs required to give a fully saturated molecule. Step 2: Compare the no. of H atoms needed so that the molecule is saturated. Half of the difference in these numbers is the number of double bond equivalents
1 DBE	1 double bond/1 ring
2 DBE	2 double bonds 1 double bond + 1 ring 2 rings 1 triple bond
4 DBE special case	can be a benzene ring
Beer-Lambert Law	A=εlc
IR 600-1400 cm-1	Fingerprint region
IR 1600- 1850 cm-1	C=C C=O C=N
IR 2100-2300 cm-1	triple bonds CC CN
IR 2700-3000 cm-1	C-H
IR 3000-3200 cm-1	O-H
IR 3200-3450 cm-1	N-H
H NMR δ7-δ8	Aromatic protons
H NMR δ5-δ6	Vinyl Protons C=C-H
H NMR δ2.5	Acetylenic protons R-C-C-H
H NMR δ9-δ10	Aldehyde proton O-C(R)-H
H NMR δ10+	Carboxylic acid proton
H NMR ~δ1-2	Alkane
H NMR ~δ2.5	proton attched to a triple bonded C
C NMR δ165-210	C=O
C NMR δ100-150	C=C
C NMR δ65-90	C-O
C NMR δ0-40	alkanes
C NMR δ75-95	C triple bonded to C
C NMR δ110-150	Aromatic
C NMR δ30-65	C-Cl C-BR
H NMR Withdrawing groups next to the proton	the proton is more deshielded
H NMR Donating groups next to H	H is more shielded
If resonance can occur such as a double bond near a carbonyl it is often the H nearest to the opposite end of the carbonyl that is more deshielded rather than the H nearest the withdrawing O because of the resonance sturucture. The same can be said for the Hs near donating groups. See pics	Image: withdrawing_effect (image/jpg)
Three factors that affect J (coupling constant)	1) Dihedral angle( trans better overlap of orbitals, greater J) 2) Bond Length (shorter bond larger J) 3) Electronegativity (weakens communication)
roofing	when two portons are chemically similar their shift is much more similar and they couple slightly with each other leading to roofing where the direction of the roof points towards the other peak that is being coupled to
Image: bhbhdbs (image/jpg)	X and X' are equivalent They are chemically and spectrscopically identical
Image: sdfsa (image/jpg)	X and X' are enatiotopic. They are only different in a chiral environment
Image: fchdszjv (image/jpg)	X and X' are diastereotopic They are always different chemically and spectroscopically, but not by predictable amounts
1H NMR exchange	Differences in spectroscopic timescales tells us about the rates of proton exchange: Rate of change SLOW = two distinct signals Rate of change FAST = single population average is obsereved
13C NMR Electronegative substituents	Deshield C Low field
Intensity of signals in 13C NMR	In 13C, relaxation effects alter intesities of signals. Relaxation of carbon nuclei occurs via interactions with protons. Hence quaternary carbons that do not have H nearby relax very slowly and usually give weak signale
13C-13C coupling	not usually observed unless 13C enriched molecules is synthesized
13C-1H coupling	Spectra usually recorded decoupled and so only singlets are observed
13C-1H off resonance coupling	Uses weak 1H decoupling and can be used to determine the number of protons attatched to any carbon. Me= quartet, CH2 - triplet ect.

Next up

Structure Determination

Description

Resource summary

Similar

	Created by julia_reast_93 over 10 years ago