Q42 E - Organic Chemistry (Mcmurry)

Question

Assign Cahn–Ingold–Prelog rankings to the following sets of substituents

a.$\mathbf{-}\mathit{C}\mathit{H}\mathbf{=}\mathit{C}{\mathit{H}}_{\mathbf{2}}\mathbf{,}\mathbf{-}\mathit{C}\mathit{H}{\mathbf{\left(}\mathbf{C}{\mathbf{H}}_{\mathbf{3}}\mathbf{\right)}}_{\mathbf{2}}\mathbf{,}\mathbf{-}\mathit{C}{\mathbf{\left(}\mathbf{C}{\mathbf{H}}_{\mathbf{3}}\mathbf{\right)}}_{\mathbf{3}}\mathbf{,}\mathbf{-}\mathit{C}{\mathit{H}}_{\mathbf{2}}\mathit{C}{\mathit{H}}_{\mathbf{3}}$

b.$\mathbf{-}\mathit{C}{\mathit{O}}_{\mathbf{2}}\mathit{C}{\mathit{H}}_{\mathbf{3}}\mathbf{,}\mathbf{-}\mathit{C}\mathit{O}\mathit{C}{\mathit{H}}_{\mathbf{3}}\mathbf{,}\mathbf{-}\mathit{C}{\mathit{H}}_{\mathbf{2}}\mathit{O}\mathit{C}{\mathit{H}}_{\mathbf{3}}\mathbf{,}\mathbf{-}\mathit{C}{\mathit{H}}_{\mathbf{2}}\mathit{C}{\mathit{H}}_{\mathbf{3}}$

c.$\mathbf{-}\mathit{C}{\mathit{O}}_{\mathbf{2}}\mathit{C}{\mathit{H}}_{\mathbf{3}}\mathbf{,}\mathbf{-}\mathit{C}\mathit{O}\mathit{C}{\mathit{H}}_{\mathbf{3}}\mathbf{,}\mathbf{-}\mathit{C}{\mathit{H}}_{\mathbf{2}}\mathit{O}\mathit{C}{\mathit{H}}_{\mathbf{3}}\mathbf{,}\mathbf{-}\mathit{C}{\mathit{H}}_{\mathbf{2}}\mathit{C}{\mathit{H}}_{\mathbf{3}}$

d.$\mathbf{-}\mathit{C}{\mathit{O}}_{\mathbf{2}}\mathit{C}{\mathit{H}}_{\mathbf{3}}\mathbf{,}\mathbf{-}\mathit{C}\mathit{O}\mathit{C}{\mathit{H}}_{\mathbf{3}}\mathbf{,}\mathbf{-}\mathit{C}{\mathit{H}}_{\mathbf{2}}\mathit{O}\mathit{C}{\mathit{H}}_{\mathbf{3}}\mathbf{,}\mathbf{-}\mathit{C}{\mathit{H}}_{\mathbf{2}}\mathit{C}{\mathit{H}}_{\mathbf{3}}$,$\mathbf{-}\mathit{C}{\mathit{O}}_{\mathbf{2}}\mathit{C}{\mathit{H}}_{\mathbf{3}}\mathbf{,}\mathbf{-}\mathit{C}\mathit{O}\mathit{C}{\mathit{H}}_{\mathbf{3}}\mathbf{,}\mathbf{-}\mathit{C}{\mathit{H}}_{\mathbf{2}}\mathit{O}\mathit{C}{\mathit{H}}_{\mathbf{3}}\mathbf{,}\mathbf{-}\mathit{C}{\mathit{H}}_{\mathbf{2}}\mathit{C}{\mathit{H}}_{\mathbf{3}}$

Step-by-Step Solution

Answer

a.Cahn–Ingold–Prelog rankings to the following sets of substituents are:

$-C{\left(C{H}_3\right)}_3$ is first priority group.

$-CH=C{H}_2$ is second priority group.

$-CH{\left(C{H}_3\right)}_2$ is third priority group

$-C{H}_{2}C{H}_3$ is forth priority group.

b.

Cahn–Ingold–Prelog rankings to the following sets of substituents are:

Benzene is first priority group.

$-CCH$ is second priority group. 

$-C{\left(C{H}_3\right)}_3$ is third priority group

$-C{H}_{2}C{H}_3$ is forth priority group

c.

Cahn–Ingold–Prelog rankings to the following sets of substituents are:

$-C{O}_{2}C{H}_3$ is first priority group.

$-COC{H}_3$ is second priority group.

$-C{H}_{2}OC{H}_3$ is third priority group

 $-C{H}_{2}C{H}_3$is forth priority group.

d. Cahn–Ingold–Prelog rankings to the following sets of substituents are:

$-Br$ is first priority group.

$-C{H}_{2}Br$ is second priority group. 

$-CN$ is third priority group

$-C{H}_{2}C{H}_{2}Br$ is forth priority group.

Step 1: Sequence Rules for specifying rankings

• Rule 1

Number all the four atoms directly attached to the chirality center according to their atomic number. The atom with the highest atomic

number has the highest ranking (first), and the atom with the lowest

atomic number has the lowest ranking (fourth). 

.

• Rule 2

If decision can’t be reached by ranking the first atoms in the

Substituent then look at the second, third, or fourth atoms away from the

chirality center until able to find the first difference.

 

• Rule 3

If two atoms are multiple-bonded then it is counted as equivalent to the same number of single-bonded atoms.

 

 

 

Step 2: Given substituents ranking:

a.

It can be seen from question that all have first directly linked atom as carbon. Now it will be decided by 2^nd atoms attached:

• In $-CH=C{H}_2$, C is attached to C,C,H
• In$-CH{\left(C{H}_3\right)}_2$ , C is attached to C,C,H
• In$-C{\left(C{H}_3\right)}_3$, C is attached to C,C,C so, this becomes highest priority group.
• In$-C{H}_{2}C{H}_3$, C is attached to C,H,H so, this becomes lowest priority group.
• To determine between  $-CH=C{H}_2$ and $-CH{\left(C{H}_3\right)}_2$ we will see atoms attached to second atom. The second carbon in $-CH=C{H}_2$ is attached to C,C,H,H and in$-CH{\left(C{H}_3\right)}_2$ it is attached to C,H,H,H. So,  $-CH=C{H}_2$is second priority group and  $-CH{\left(C{H}_3\right)}_2$is third priority group.

b.

 

It can be seen from question that all have first directly linked atom as carbon. Now it will be decided by 2^nd atoms attached:

• In$-CCH$ , C is attached to C,C,C 
• In $-CH=C{H}_2$, C is attached to C,C,H so, this becomes lowest priority group.
• In$-C{\left(C{H}_3\right)}_3$ , C is attached to C,C,C 
• In benzene, C is attached to C,C,C 

Now, look for second carbon:

• In $-CCH$, the second carbon has C,C,C,H and nothing following it so, this becomes second priority group.
• In $-C{\left(C{H}_3\right)}_3$, the second carbon has C,H,H,H so, this becomes third priority group.
• In benzene, the second carbon has C,C,C,H and has carbon following it so, this becomes highest priority group.

c.

 

 

It can be seen from question that all have first directly linked atom as carbon. Now it will be decided by 2^nd atoms attached:

• In $-C{O}_{2}C{H}_3$, C is attached to O,O,O so, this becomes highest priority group.
• In $-COC{H}_3$, C is attached to O,O,C so, this becomes second priority group.
• In$-C{H}_{2}OC{H}_3$ , C is attached to O,H,H so, this becomes third priority group.
• In $-C{H}_{2}C{H}_3$, C is attached to C,H,H so, this becomes lowest priority group.

 

.

d. It can be seen from question that all have first directly linked atom as carbon except $-Br$ it becomes first priority group.

 

 . Now remaining will be decided by 2^nd atoms attached:

• In$-C{H}_{2}C{H}_{2}Br$ , C is attached to C,H,H so, this becomes least priority group.
• In$-C{H}_{2}Br$ , C is attached to Br,H,H so, this becomes second priority group.
• In $-CN$, C is attached to N,N,N so, this becomes third priority group.