Question

¹H NMR spectrum A corresponds to a molecule with the formula C₃H₅Cl. The compound shows significant IR bands at 730 (see Problem 53 of Chapter 11), 930, 980, 1630, and 3090 cm^-1.

(a) Deduce the structure of the molecule.

(b) Assign each NMR signal to a hydrogen or group of hydrogens.

(c) The “doublet” at δ = 4.05 ppm has J ≈ 6 Hz. Is this in accord with your assignment in (b)?

(d) This “doublet,” upon five-fold expansion, becomes a doublet of triplets (inset, spectrum A), with J < 1 Hz for the triplet splittings. What is the origin of this triplet splitting? Is it reasonable in light of your assignment in (b)?

Problem 53

From the Hooke’s law equation, would you expect the C – X bonds of common haloalkanes (X = Cl, Br, I) to have IR bands at higher or lower wave-numbers than are typical for bonds between carbon and lighter elements (e.g., oxygen)?

 

 

Answer 1

a.

Compound A

First we need to find degree of unsaturation

For compound A: C3H5Cl

 

Degree of unsaturation = 1/2(2+ 2 x number of C - number of H - number of halide) = 1/2 (2 + 6 -5 -1) = 1

 

There must be a ring or double bond

 

From NMR data it suggests that there is alkene.

 

A doublet appeared at 4.1 ppm is a CH2 attached to the Cl.

A doublet of doublet appeared at 5.2-5.3 ppm for 2 H is terminal CH2 of alkene.

A multiple appeared at 6.0 ppm is of the C-H.

 

 

b & c.

Compound B and C:

Molecular formula is C3H6Cl2O

 

Degree of unsaturation is 0.

 

Addition of Cl2 and H2O takes place across double bond of compound A will produce two isomers compound B and C.

 

Thus there structure is as follows.

 

Signal assignment:

 

d.

Compound D

It\'s a ring formation reaction, KOH abstracts the proton of alcohol, to form oxonium ion which replaces Cl to form epichlorohydrine. As shown below.

 

Protons attached to Cl are enantiotopic thus they appear as two different multiplates at 2.6 and 2.8 ppm

 

A peak at 3.25 is CH attached to the O atom.

A peak at 3.6 is a doublet for CH2

 

Signal assignment

a.

Compound A

First we need to find degree of unsaturation

For compound A: C3H5Cl

 

Degree of unsaturation = 1/2(2+ 2 x number of C - number of H - number of halide) = 1/2 (2 + 6 -5 -1) = 1