Predict the splitting patterns for the signals given by compounds in Problem 4.

NMR shows a different number of peaks. These are called the splitting patterns in NMR spectroscopy. We know that the simplest signal consists of one line is called a singlet, followed by a doublet, triplet, quartet, and multiplet.

Spitting is caused by the hydrogens on the same or neighboring carbons. Only non-equivalent protons cause splitting.

(a)

As a compound a contains plane of symmetry then spitting in the compound is as shown below;

Spliting in (a)

-CH₃ group shows triplet as it couple with two neighboring hydrogen atoms while the -CH₂ group shows quartet as it couples with three neighboring hydrogen atoms.

(b)

As the compound b contains plane of symmetry then spitting in the compound is as shown below;

Spliting in (b)

Both -CH₂groups are equivalent so that they don’t couple with each other and show singlet.

(c)

Splitting in compound c is showm below:

Spliting in (c)

(d)

As a compound a contains plane of symmetry then spitting in the compound is as shown below;

Spliting in (d)

Hydrogen present on C-2 carbon does not split by hydrogens present on the C-3 carbon as both are equivalent. The hydrogen present on C-5 carbon does not split by hydrogens present on the C-6 carbon as both are equivalent.

C-2,C-3,C-5,C-6 Shows doublet, while C-1 , C-4 shows multiplet.

(e)

Splitting in e is shown below:

Spliting in (e)

As both the hydrogens are equivalent so they are not couple with each other and show singlet.

(f)

-CH₃group next to carbonyl is show singlet as no neighboring hhydrogen is present.

Spliting in (f)

(g)

As a compound a contains plane of symmetry then spitting in the compound is as shown below;

Spliting in (g)

The middle -CH group shows quintet as it couple with four neighboring hydrogens. While -CH₃ the shows triplet as it couples with two hydrogens. -CH₂shows multiplet.

(h)

As a compound a contains plane of symmetry then spitting in the compound is as shown below;

Spliting in (h)

(i)

The group -CH₃ shows doublet while -CH group shows quartet.

Spliting in (i)

(j)

The -OCH₃ group shows singlet. While the -CH₃ group shows singlet.

Spliting in (j)

(k)

As the given molecule contain plane of symmetry so all the benzene hydrogens are identical and show singlet

Spliting in (k)

Also the -CH₃ group shows singlet.

(l)

As the given molecule contain plane of symmetry so hydrogen next to Br shows doublet while the hydogen present in between two group is showing singlet.

Spliting in (l)

(m)

As the given molecule contain plane of symmetry so hydrogen next to -NO₂ shows doublet while the hydrogens on 3 rd carbon shows doublet of doublet.

Spliting in (m)

(n)

Double bonded -CH₃group show doublet while the -CH₃ group next to carbonyl group show singlet. Also the double bonded -CH shows quartet.

Spliting in (n)

(o)

As the both hydrogens are non equivalent so they show doublet.

Spliting in (o)