Question

Below is the H NMR spectra for the molecule styrene.

Consider if a hydroboration reaction was preformed on styrene. How would the H NMR of the expected product change in comparison to styrene? Circle which peaks from the styrene spectra would disappear and not be present in the NMR of the product.

Draw the expected product. Describe the splitting pattern of each new proton signal that would appear in the NMR of the product.

2) Below is the 'H NMR spectra for the molecule styrene - Consider if a hydroboration reaction was performed on styrene. How would the expected product change in comparison to styrene? Circle which peaks from the styrene spectra would disappear and not be present in the NMR of the product. 'H NMR of the Draw the expected product. Describe the splitting pattern (s, d, t, etc.) of each new proton signal that would appear in the NMR of the product. What characteristic signals of styrene would no longer be present in the IR spectra of product? the -What new characteristic signals would be present in the IR spectra of the product? 1) BH3 2) H202 NaOH styrene H NMR of Styrene 5H 1H 1H H 2 PPM 8 5.25 (IH, d,d), 5.76 (1H, d,d), 6.72 (1H, d,d), 7.33 (3H, multiplet), 7.41 (2H, multiplet) ppm;

Answer 1

Styrene (Reactant ) NMR pattern =

Chemical shift (ppm)	number of hydrogen	Multiplicity	Comment
5.25ppm (1H ,dd)	1H	d, d	this proton expected to disappear once product is formed .
5.76 ppm (1H ,dd)	1H	d, d	this proton also expected to disappear from spectrum .
6.72 ppm (1H,dd)	1H	d, d	this peak also disappear from spectrum
7.33ppm (3H ,m)	3H	multiplet	these proton belong to aromatic ring ,so they are remain as such ,only marginal change in chemical shift could observed.
7.41 ppm (2H,m)	2H	multiplet	these proton belong to aromatic ring , so their position could remain as such

Phenylethanol (product)

Chemical Shift (ppm)	Notation	number of hydrogen	Multiplicity	comment
2.01 ppm (1H ,broad singlet)	A	1H	singlet	this proton belong to O-H , so appear as broad singlet , exchangeable .
2.81 ppm (2H, t)	B	2H	triplet , (n=2 ,adjacent proton ), (n+1)= triplets	This peak appears as triplets , due to splitting with neighboring CH2 group .
3.77 ppm (2H , t)	C	2H	triplet	two CH2 group are neighboring partner of one another, so they split as triplet. These protons deshielded more due to oxygen vicinity .
7.05 -7.5 ppm (5H ,m)	D	5H	multiplet	this signal belong to aromatic ring and appears as bunch (multiplet )

Please see assign peak on structure =

(D) B OHA